编程语言 | C | 代码整理 | 4月

八月拍了拍你，并对你说：“好运就要开始了”！

编程语言 | C | 代码整理 | 4月

2019/4/1

1.递归和非递归分别实现求第n个斐波那契数。
法一
#include <stdio.h>
int Fib(int n){if (n <= 2){return 1;}return Fib(n - 1) + Fib(n - 2);
}
int main(){int ret = 0;printf("ret = %d\n", Fib(10));system("pause");return 0;
}
//法二
#include <stdio.h>
int Fib(int n){//if(n == 1 || n == 2)if (n <= 2){return 1;}int last2 = 1;  //第i-2项int last1 = 1;  //第i-1项int cur = 0;    //第i项for (int i = 3; i <= n; ++i){cur = last1 + last2;last2 = last1;last1 = cur;}return cur;
}
int main(){printf("cur = %d\n", Fib(6));system("pause");return 0;
}
//法三 0
#include <stdio.h>
int Fib(int n,int last2,int last1){//if (n == 1 || n == 2)if (n <= 2){return 1;}return Fib(n - 1,last1,last1+last2);
}
int main(){int ret = 0;printf("ret = %d\n", Fib(20,1,1));system("pause");return 0;
}
2.编写一个函数实现n^k，使用递归实现 0
#include <stdio.h>
#include <stdlib.h>
int myPow(int n, int k)
{if (k == 1){return k;}return n * myPow(n, k - 1);
}
int main(){int ret = 0;printf("ret=%d\n", myPow(6, 2));system("pause");return 0;
}
3. 写一个递归函数DigitSum(n)，输入一个非负整数，返回组成它的数字之和，
例如，调用DigitSum(1729)，则应该返回1 + 7 + 2 + 9，它的和是19   0000
#include <stdio.h>
#include <stdlib.h>
int DigitNum(int n){if (n == 0){return 0;}return n % 10 + DigitNum(n / 10);
}
int main(){printDigitNum(1729);printf("\n%d\n", DigitNum(1729));system("pause");return 0;
}
4. 编写一个函数 reverse_string(char * string)（递归实现）
实现：将参数字符串中的字符反向排列。
//要求：不能使用C函数库中的字符串操作函数。
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
void reverse_string(char* string) {char p;                                                          //定义一个字符来存储第一个地址自增后第一个字符p = *string;if (*(string++) == '\0')return;                                                       //如果是结束符号,那么返回elsereverse_string(string++);printf("%c", p);                                               //顺序执行输出p存储的字符,再返回上一层
}
int main() {char string[20];gets(string);reverse_string(string);printf("\n");system("pause");return 0;
}5.递归和非递归分别实现strlen
void reverse_string(char * str)
{int len = strlen(str) - 1;char tmp;if (str[0] == '\0'){return;}tmp = str[0];str[0] = str[len];str[len] = '\0'; //保证下一次能找到尾部reverse_string(str + 1);str[len] = tmp;
}int main()
{printDigitNum(1729, 16);printf("\n%d\n", DigitNum(1729));printf("%d\n", myStrlenN("bitekeji"));printf("%d\n", myStrlen("bitekeji"));char test[] = "bitekeji";reverse_string(test);puts(test);return 0;
}6.递归和非递归分别实现求n的阶乘
#include <stdio.h>
int Factor(int n){if (n == 1){return 1;}return n*Factor(n - 1);
}
int main(){int ret = 0;printf("ret=%d\n",Factor(5));system("pause");return 0;
}
7.递归方式实现打印一个整数的每一位
#include <stdio.h> 
int  fun(int number) {                                                         if(number>9)                                                   {                       fun(number/10);                                     }printf("%d ",number%10);                       
}
int main() {int number;printf("输入一个数字:\n");scanf("%d",&number);fun(number);printf("\n");return 0;
}

2019/4/2

//创建一个字符型二维数组(3*3)表示棋盘(x表
//x示玩家落子,o表示电脑落子,' '表示未落子)
//1.游戏开始是, 进行初始化棋盘, 把所有元素都设为''
//2.提示玩家落子(输入一个坐标)
//3.判定胜负
//4.电脑落子(基于随机数的方式生成一个坐标)
//5.判定胜负
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
int Menu(){printf("==================\n");printf("1.开始游戏\n");printf("0.结束游戏\n");printf("==================\n");printf("请输入您的选择: ");int choice = 0;scanf("%d",&choice);return choice;
}
#define MAX_ROW 3
#define MAX_COL 3
char chess_board[MAX_ROW][MAX_COL];void Init(){for (int row = 0; row < MAX_ROW; ++row){for (int col = 0; col < MAX_COL; ++col){chess_board[row][col] = ' ';}}//设置随机种子srand((unsigned int)time(0));
}
void print(){for (int row = 0; row < MAX_ROW; ++row){printf("| %c | %c | %c |\n", chess_board[row][0],chess_board[row][1], chess_board[row][2]);if (row != MAX_ROW - 1){printf("|---|---|---|\n");}}//为了调试程序临时加的代码,最好给他加上一个统一风格的注释//TODO//system("pause");
}
void PlayerMove(){printf("玩家落子!\n");while (1){printf("请输入落子位置的坐标(row col): ");int row = 0;int col = 0;scanf("%d %d", &row, &col);//检查用户输入的坐标是否合法if (row < 0 || row >= MAX_ROW ||col < 0 || col >= MAX_COL){printf("您输入的坐标非法!请重新输入!\n");continue;}if (chess_board[row][col] != ' '){printf("您要落子的位置已经被占了!\n");continue;}chess_board[row][col] = 'x';break;}printf("玩家落子完毕!\n");
}
void ComputerMove(){printf("电脑落子\n");while (1){int row = rand() % 3;int col = rand() % 3;if (chess_board[row][col] != ' '){continue;}chess_board[row][col] = 'o';break;}printf("电脑落子完毕\n");
}
//返回值表示胜利者是谁
//x表示玩家胜
//o表示电脑胜利
//' '表示未分出胜负
//如果棋盘满了返回1;否则返回0;
int IsFull(){for (int row = 0; row < MAX_ROW; ++row){for (int col = 0; col < MAX_COL; ++col){//未满return 0;}}return 1;
}
//返回值表示胜利者是谁
//x表示玩家胜
//q表示和棋
//' '表示胜负未分
char CheckWinner(){//检查所有行是否连成一条线for (int row = 0; row < MAX_ROW; ++row){if (chess_board[row][0] == chess_board[row][1]&& chess_board[row][0] == chess_board[row][2]){return chess_board[row][0];}}//检查所有列是否连成一条线for (int col = 0; col < MAX_COL; ++col){if (chess_board[0][col] == chess_board[1][col]&& chess_board[0][col] == chess_board[0][2]){return chess_board[0][col];}}//检查所有对角线是否连成一条线if (chess_board[0][0] == chess_board[1][1]&& chess_board[0][0] == chess_board[2][2]){return chess_board[0][0];}if (chess_board[0][2] == chess_board[1][1]&& chess_board[0][0] == chess_board[2][0]){return chess_board[0][2];}//棋盘满并且未分出胜负if (IsFull()){return 'q';}return ' ';
}
//自顶向下式的程序开发方法
void Game(){//1.初始化棋盘Init();char winner = ' ';while (1){//2.打印棋盘print();//3.玩家落子PlayerMove();//4.检测胜负winner = CheckWinner();if (winner != ' '){//胜负已分break;}//5.电脑落子ComputerMove();//6.检测胜负winner = CheckWinner();if (winner != ' '){break;}}print();if (winner == 'x'){9+-printf("您赢了!\n");}else if (winner == 'o'){printf("您真菜!\n");}else if (winner == 'q'){printf("您和电脑五五开!\n");}else{printf("代码好像出bug了!\n");}
}
int main(){while (1){int choice = Menu();if (choice == 1){Game();}else if (choice == 0){printf("goodbye!\n");break;}else{printf("您的输入有误!\n");}}system("pause");return 0;
}

2019/4/2

//1.先搞两个二维数组地图//(a)show_map玩家看到的地图,已经翻开和未翻开两种状态//(b)mine_map地雷布局图,每个位置标记是否是地雷(0/1)
//2.初始化,初始化刚才两个地图//(a)show_map初始化就是把每个元素都设为*//(b)mine_map初始化随机生成10个地雷
//3.打印地图
//4.提示用户输入一个坐标,表示要翻开某个位置(进行必要的合法性判定)
//5.该位置是否是地雷.如果是地雷,游戏结束!
//6.如果不是地雷,先判定是否是游戏胜利(把所有不是地雷的格子翻开)
//7.给当前位置的格子生成一个数字表示周围有几个雷(先不实现翻开一片的功能)
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int Menu(){printf("================\n");printf("   1.开始游戏\n");printf("   0.退出游戏\n");printf("================\n");printf("请输入您的选择: \n");int choice = 0;scanf("%d",&choice);return choice;
}
#define Max_Row 9
#define Max_Col 9
#define Mine_Count 10
void Init(char show_map[Max_Row][Max_Col],char mine_map[Max_Row][Max_Col]){//1.对于show_map,都需要设为*for (int row = 0; row < Max_Row; ++row){for (int col = 0; col < Max_Col; ++col){show_map[row][col] = '*';}}//2.对于mine_map,需要随机生成若干个地雷//使用0表示不是地雷,使用1表示是地雷for (int row = 0; row < Max_Row; ++row){for (int col = 0; col < Max_Col; ++col){mine_map[row][col] = '0';}}int n = Mine_Count;while (n>0){//生辰一组随机坐标int row = rand() % Max_Row;int col = rand() % Max_Col;if (mine_map[row][col] == '1'){//该位置已经设置是地雷了,需要重新生成continue;}mine_map[row][col] = '1';--n;}
}
void PrintMap(char show_map[Max_Row][Max_Col]){//不光能打印出地图.还能带坐标//先打印第一行printf("    ");for (int col = 0; col < Max_Col; ++col){printf("%d ",col);}printf("\n");//打印一个分割线for (int col = 0; col < Max_Col-2; ++col){printf("---");}printf("\n");//再打印其他行for (int row = 0; row < Max_Row; ++row){printf("%d| ",row);//打印本行每一列for (int col = 0; col < Max_Col; ++col){printf("%c ",show_map[row][col]);}printf("\n");}
}
void UpdateShowMap(int row, int col, char show_map[Max_Row][Max_Col],char mine_map[Max_Row][Max_Col]){//根据当前位置来判定这个位置周围8个格子有几个地雷,//并且将这个数字更新到Show_Map中int count = 0;if (row-1 >= 0 && col-1 >= 0 &&row-1 < Max_Row && col-1 < Max_Col &&mine_map[row - 1][col - 1] == '1'){++count;}if (row - 1 >= 0 && col  >= 0 &&row - 1 < Max_Row && col  < Max_Col &&mine_map[row - 1][col] == '1'){++count;}if (row - 1 >= 0 && col + 1 >= 0 &&row - 1 < Max_Row && col + 1 < Max_Col &&mine_map[row - 1][col + 1] == '1'){++count;}if (row  >= 0 && col - 1 >= 0 &&row  < Max_Row && col - 1 < Max_Col &&mine_map[row][col - 1] == '1'){++count;}if (row  >= 0 && col + 1 >= 0 &&row  < Max_Row && col + 1 < Max_Col &&mine_map[row][col + 1] == '1'){++count;}if (row + 1 >= 0 && col - 1 >= 0 &&row + 1 < Max_Row && col - 1 < Max_Col &&mine_map[row + 1][col - 1] == '1'){++count;}if (row +1 >= 0 && col >= 0 &&row + 1 < Max_Row && col < Max_Col &&mine_map[row + 1][col] == '1'){++count;}if (row + 1 >= 0 && col + 1 >= 0 &&row +1 < Max_Row && col + 1 < Max_Col &&mine_map[row + 1][col + 1] == '1'){++count;}//得到了周围8个人格子中的地雷个数//假设count为2,实际想看到的内容是字符2 '2',也就是ascii中的50//show_map[row][col] ='0'+ count;
}
void Game(){
//1.先创建地图,并初始化char show_map[Max_Row][Max_Col];char mine_map[Max_Row][Max_Col];//已经翻开的空格的个数(非地雷)int blank_count_already_show = 0;Init(show_map, mine_map);while (1){//2.打印地图PrintMap(show_map);//TODO 这个打印是为了调临时加的PrintMap(mine_map);//3.让用户输入坐标并且进行合法性检测printf("请输入一组坐标(row col):");int row = 0;int col = 0;scanf("%d %d", &row, &col);//清掉之前打印的内容system("cls");if (row < 0 || row >= Max_Row || col < 0 ||col >= Max_Col){printf("您的输入非法!请重新输入!\n");continue;}if (show_map[row][col] != '*'){printf("您输入的位置已经翻开了!\n");continue;}//4.判定是否是地雷if (mine_map[row][col] == '1'){printf("游戏结束!\n");PrintMap(mine_map);break;}//5.判定游戏是否胜利//判定所有的非地雷位置都被翻开了++blank_count_already_show;if (blank_count_already_show == Max_Row*Max_Col- Mine_Count){printf("游戏胜利!\n");PrintMap(mine_map);break;}//6.统计当前位置周围的累的个数UpdateShowMap(row,col,show_map,mine_map);}
}
int main(){while (1){int choice = Menu();if (choice == 1){Game();}else if (choice == 0){printf("goodbye!\n");break;}else{printf("您的输入有误!\n");}}system("pause");return 0;
}

2019/4/3

1. 5位运动员参加了10米台跳水比赛，有人让他们预测比赛结果
A选手说：B第二，我第三；
B选手说：我第二，E第四；
C选手说：我第一，D第二；
D选手说：C最后，我第三；
E选手说：我第四，A第一；
比赛结束后，每位选手都说对了一半，请编程确定比赛的名次。
#include <stdio.h>
int checkRank(int * player, int n){int i, res = 0;for (i = 0; i < n; i++){res |= 1 << player[i];}return res == 0x3e;
}
int main(){int player[5] = { 0 };for (player[0] = 1; player[0] <= 5; player[0]++){for (player[1] = 1; player[1] <= 5; player[1]++){for (player[2] = 1; player[2] <= 5; player[2]++){for (player[3] = 1; player[3] <= 5; player[3]++){for (player[4] = 1; player[4] <= 5; player[4]++){A选手说：B第二，我第三；B选手说：我第二，E第四；C选手说：我第一，D第二；D选手说：C最后，我第三；E选手说：我第四，A第一；if ((player[0] == 3) + (player[1] == 2) == 1 &&(player[1] == 2) + (player[4] == 4) == 1 &&(player[2] == 1) + (player[3] == 2) == 1 &&(player[2] == 5) + (player[3] == 3) == 1 &&(player[4] == 4) + (player[0] == 1) == 1 &&checkRank(player, 5)){printf("a是第%d\n" \"b是第%d\n" \"c是第%d\n" \"d是第%d\n" \"e是第%d\n",player[0], player[1], player[2], player[3], player[4]);}}}}}}system("pause");return 0;
}
2.日本某地发生了一件谋杀案，警察通过排查确定杀人凶手必为4个
嫌疑犯的一个。以下为4个嫌疑犯的供词。
A说：不是我。
B说：是C。
C说：是D。
D说：C在胡说
已知3个人说了真话，1个人说的是假话。
现在请根据这些信息，写一个程序来确定到底谁是凶手。
#include <stdio.h>
#include <stdlib.h>
int main(){int murder = 0;for (murder = 'A'; murder < 'E'; ++murder){A说：不是我。B说：是C。C说：是D。D说：C在胡说if ((murder != 'A') + (murder == 'C') + (murder == 'D') + (murder != 'D') == 3){printf("%c是凶手\n", murder);}}system("pause");return 0;
}3.在屏幕上打印杨辉三角。
1
1 1
1 2 1
1 3 3 1
#include <stdio.h>
void printArray(int * arr, int n){int i;for (i = 0; i < n; i++){printf("%d ", arr[i]);}putchar('\n');
}
int main(){int n = 10;int i, j;
#if 0int data[20][20] = { 0 };printf("%d\n", data[0][0] = 1);for (i = 1; i < n; i++){printf("%d ", data[i][0] = 1);for (j = 1; j < i; j++){printf("%d ", data[i][j] = data[i - 1][j - 1] + data[i - 1][j]);}printf("%d\n", data[i][j] = 1);}
#elseint data[20] = { 1, 1 };puts("1");puts("1 1");for (i = 2; i < n; i++){data[i] = 1;for (j = i - 1; j > 0; j--){data[j] += data[j - 1];}printArray(data, i + 1);}
#endifsystem("pause");return 0;
}

2019/4/4

1.编写函数：unsigned int reverse_bit(unsigned int value);
这个函数的返回值value的二进制位模式从左到右翻转后的值。
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
unsigned int reverse(unsigned int data){int tmp1, tmp2;int i, j;for (i = 0, j = 31; i < j; i++, j--){tmp1 = !!(data & 1 << i);tmp2 = !!(data & 1 << j);data &= ~(1 << i);data &= ~(1 << j);data |= tmp1 << j;data |= tmp2 << i;}return data;
}
int main(){unsigned int n, sn = 2;scanf("%u", &n);int i;unsigned int tmp, sum = 0;printf("%u\n", reverse(n));return 0;for (i = 0; i < 32; n /= sn, i++){tmp = n % sn;sum = sum * sn + tmp;}printf("%u\n", sum);system("pause");return 0;
}2.编程实现：一组数据中只有一个数字出现了一次。其他所有数字都是成对
出现的。请找出这个数字。（使用位运算）
#include <stdio.h>
int main(){int arr[9] = { 1, 3, 5, 2, 1, 2, 4, 3, 4 };int res = 0;int i;for (i = 0; i < 9; i++){res ^= arr[i];}printf("%d\n", res);system("pause");return 0;
}3.有一个字符数组的内容为:"student a am i",请你将数组的内容改为"i am a student".要求：不能使用库函数。只能开辟有限个空间（空间个数和字符串的长度无关）。student a am ii ma a tnedutsi am a student
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>
void reversestr(char * str, int start, int end){char tmp;int i, j;for (i = start, j = end - 1; i < j; i++, j--){tmp = str[i];str[i] = str[j];str[j] = tmp;}
}
char * reverseWordN(char * str){int i;int start = 0;for (i = 0; str[i]; i++){if (str[i] == ' '){reversestr(str, start, i);start = i + 1;}}reversestr(str, start, i);reversestr(str, 0, i);return str;
}
char * reverseWordY(char * str){char * ptmp;char tmp[100] = { 0 };while (ptmp = strrchr(str, ' ')){strcat(tmp, ptmp + 1);strcat(tmp, " ");*ptmp = '\0';}strcat(tmp, str);strcpy(str, tmp);return str;
}
int main(){char str[] = "I don't like study";puts(reverseWordY(str));system("pause");return 0;
}4.实现一组数字,第二个数是第一个的二倍,第三个数是第一个数的三倍
#include <stdio.h>
int checkNum(int a, int b, int c){int res = 0;res |= 1 << a % 10;res |= 1 << a / 10 % 10;res |= 1 << a / 100;res |= 1 << b % 10;res |= 1 << b / 10 % 10;res |= 1 << b / 100;res |= 1 << c % 10;res |= 1 << c / 10 % 10;res |= 1 << c / 100;return res == 0x3fe;
}
int main(){int a, b, c;for (a = 123; a <= 329; a++){for (b = 246; b <= 658; b++){for (c = 369; c <= 987; c++)			{if (b == a * 2 && c == a * 3 && checkNum(a, b, c)){printf("%d %d %d\n", a, b, c);}}}}system("pause");return 0;
}

2019/4/5

//1.用函数求两个数中的最大值
#include <stdio.h>
#include <stdlib.h>
int Max(int x, int y){if (x > y){return x;}return y;
}
int main(){int a = 10;int b = 20;//函数的调用,实际参数int ret = Max(a, b);printf("%d\n",ret);system("pause");return 0;
}//2.两个数相除
#include <stdio.h>
#include <math.h>
int Divide(int x, int y, int* ok){if (y==0){*ok = 0;}*ok = 1;return x / y;
}
int main(){int ok = 0;int ret = Divide(10, 3, &ok);printf("%d\n",ret);system("pause");return 0;
}//3.两个数的交换
#include <stdio.h>
void Swap(int* x,int* y){int tmp = *x;*x = *y;*y = tmp;
}
int main(){int a = 10;int b = 20;Swap(&a, &b);    //  等价   int* x=&a;printf("%d %d\n",a, b); //  int* y=&b;system("pause");        //  int tmp=*x;return 0;               //  *x = *y;
}                           //  *y = tmp;  *///4.判断一个数是否为素数
#include <stdio.h>
#include <math.h>
//如果是素数返回1,否则返回0
int IsPrime(int x){if (x <= 0){return 0;}if (x == 1){return 0;}for (int i = 2; i < x; ++i){if (x%i == 0){return 0;}}return 1;
}
int main(){printf("%d\n",IsPrime(108));system("pause");return 0;
}//5.编写一个函数,实现一个整形有序数组的二分查找
int Binarysearch(int* arr[],int size,int to_find){
//[left,right]int left = 0;int right = size - 1;while (left<right){int mid = (left + right) / 2;if (to_find < arr[mid]){right = mid - 1;}else if (to_find > arr[mid]){left = mid + 1;}else{return 1;}}
}
int main(){int arr[4] = { 1,2,3,4 };int size = sizeof(arr) / sizeof(arr[0]);int ret = Binarysearch(arr, size, 2);printf("%d\n",ret);system("pause");return 0;
}
//6.编写一个函数,判断是不是闰年
#include <stdio.h>
#include <math.h>
int IsLeapYear(int year){if (year % 100 == 0){//判断是不是世纪闰年if (year % 400 == 0){return 1;}else{return 0;}}else{//判断是不是普通闰年if (year % 4 == 0){return 1;}else{return 0;}}
}
int main(){printf("%d\n", IsLeapYear(2008));system("pause");return 0;
}//7.编写一个函数,每调用一次函数,就会将num的值增加1
//方法一
void Func(int* x){*x += 1;
}
int main(){int x = 0;Func(&x);printf("%d\n",x);system("pause");return 0;
}//方法二
int Strlen(char str[]){if (str[0]=='\0 '){return 0;}return 1 + Strlen(str + 1);
}
int main(){char str[] = "abcd";int ret = Strlen(str);printf("ret = %d\n",ret);system("pause");return 0;
}

2019/4/6

//实现一个通讯录,完成联系人信息的存储.
//1.新增  
//2.删除
//3.修改
//4.查找记录
//5.打印全部记录
//6.排序记录
//7.清空全部记录//管理
//1.把基本信息抽象并描述出来(结构体)
//2.需要管理很多数据,就需要组织起来(数据结构)#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <windows.h>//通讯录每个条目的抽象
typedef struct PersonInfo{char name[1024];char phone[1024];       //电话号码用字符串表示
}PersonInfo;//通讯录的最大容量
#define MAX_PERSON_INFO_SIZE 1000//通信录本体的抽象
typedef struct AddressBook{PersonInfo persons[MAX_PERSON_INFO_SIZE];//[0,size)int size;    //控制通讯录的有效和无效
}AddressBook;AddressBook g_address_book;       //全局变量void Init(){g_address_book.size = 0;for (int i = 0; i < MAX_PERSON_INFO_SIZE; ++i){g_address_book.persons[i].name[0] = '\0';g_address_book.persons[i].phone[0] = '\0';}
}int Menu(){printf("===================\n");printf(" 1.新增联系人\n");printf(" 2.删除联系人\n");printf(" 3.查找联系人\n");printf(" 4.修改联系人\n");printf(" 5.打印全部联系人\n");printf(" 6.排序联系人\n");printf(" 7.清空联系人\n");printf(" 0.退出\n");printf("===================\n");printf(" 请输入您的选择: ");int choice = 0;scanf("%d",&choice);return choice;
}void Empty(){
//这个函数就是用来凑数
}void AddPersonInfo(){printf("新增联系人\n");if (g_address_book.size >= MAX_PERSON_INFO_SIZE){printf("您的通讯录已满,请重试!\n");return;}PersonInfo* person_info = &g_address_book.persons[g_address_book.size];printf("请输入联系人姓名: ");//必须获取到一个指针修改的内容是一个预期的内容scanf("%s",person_info -> name);printf("请输入联系人电话: ");scanf("%s",person_info -> phone);++g_address_book.size;printf("新增联系人成功!\n");system("pause");system("cls");
}void DelPersonInfo(){printf("删除联系人\n");if (g_address_book.size <= 0){printf("删除失败!通讯录为空!\n");}printf("请输入要删除的序号: ");int id = 0;scanf("%d",&id);if (id < 0 || id >= g_address_book.size){printf("删除失败!输入的序号有误!\n");return;}g_address_book.persons[id] = g_address_book.persons[g_address_book.size - 1];--g_address_book.size;printf("删除联系人成功!\n");system("pause");system("cls");
}void FindPersonInfo(){printf("查找联系人\n");if (g_address_book.size <= 0){printf("查找失败,通信录为空!\n");return;}//根据姓名查找电话printf("请输入要查找的姓名: ");char name[1024] = { 0 };scanf("%s",name);int size = 0;for (int i = 0; i < g_address_book.size;++size){//元素地址取出来PersonInfo* info = &g_address_book.persons[i];  if (strcmp(info ->name,name) == 0){//序号iprintf("[%d] %s\t%s\n", i, info->name, info->phone);}}printf("查找联系人成功!\n");system("pause");system("cls");
}void UpdatePersonInfo(){printf("更新联系人\n");if (g_address_book.size<=0){printf("修改失败,通信录为空!\n");return;}printf("请输入要修改的序号: ");int id = 0;scanf("%d",&id);if (id < 0||id>=g_address_book.size){printf("修改失败,输入的序号有误!\n");return;}PersonInfo* info = &g_address_book.persons[id]; printf("请输入新的姓名: (%s)\n",info -> name);char name[1024] = { 0 };scanf("%s", name);if (strcmp(name, "*") != 0){strcmp(info->name, name);}char phone[1024] = { 0 };printf("请输入新的电话: (%s)\n", info-> phone);scanf("%s", phone);if (strcmp(phone, "*") != 0){strcmp(info->phone,phone);}printf("更新联系人成功!\n");system("pause");system("cls");
}void PrintAllPersonInfo(){printf("打印全部联系人\n");for (int i = 0; i < g_address_book.size; ++i){PersonInfo* info = &g_address_book.persons[i];printf("[%d] %s\t%s\n", i, info->name, info->phone);}printf("共打印了 %d 条数据!\n", g_address_book.size);printf("打印全部联系人成功!\n");system("pause");system("cls");
}void SortPersonInfo(){//按照字节序排序,取结构体中的姓名字段char name[1024] = { '\0' };char phone[1024] = { '\0' };int bound;int i;int size;for (i = 0; i < g_address_book.size; i++) {for (bound = 0; bound < g_address_book.size; bound++) {for (size = bound; size<g_address_book.size - 1; size++) {if (strcmp(g_address_book.persons[size].name,g_address_book.persons[size + 1].name)>0) {strcpy(name, g_address_book.persons[size].name);strcpy(g_address_book.persons[size].name,g_address_book.persons[size + 1].name);strcpy(g_address_book.persons[size + 1].name,name);//copy 电话号码strcpy(phone, g_address_book.persons[size].phone);strcpy(g_address_book.persons[size].phone,g_address_book.persons[size + 1].phone);strcpy(g_address_book.persons[size + 1].phone, phone);}}}}printf("排序所有联系人:\n");printf("排序成功!\n");system("pause");system("cls");
}void ClearAllPersonInfo(){printf("清空全部数据\n");printf("您真的要清空全部数据吗?Y/N\n");char choice[1024] = { 0 };scanf("%s",choice);if (strcmp(choice,"Y")==0){g_address_book.size = 0;printf("清空全部数据成功!\n");}else{printf("清空操作取消!\n");}
}typedef void(*Func)();    //函数指针的类型
// Vim操作风格
int main(){Func arr[] = {Empty,AddPersonInfo,DelPersonInfo,FindPersonInfo,UpdatePersonInfo,PrintAllPersonInfo,SortPersonInfo,ClearAllPersonInfo};Init();while (1){int choice = Menu();if (choice < 0 || choice >= sizeof(arr) / sizeof(arr[0])){printf("您的输入有误!\n");system("pause");system("cls");continue;}else if (choice==0){printf("goodbye!\n");break;}else {Sleep(200);system("cls");arr[choice]();}}system("pause");return 0;
}

2019/4/7

//1.打印数组中的元素以及各自的地址
#include <stdio.h>
#include <stdlib.h>
int main(){int arr[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };for (int i = 0; i < 10;++i){printf("%d\n",arr[i]);printf("%p\n",&arr[i]);  //%p打印指针变量的值,打印地址}system("pause");return 0;
}//2.打印出二维数组中的元素以及各自的地址
#include <stdio.h>
#include <stdlib.h>
int main(){int arr[3][4] = { (1,2),(3,4),5 };int row;int col;for (row  = 0; row < 3; ++row){for (col = 0; col < 4; ++col){printf("%d ", arr[row][col]);printf("%p\n",&arr[row][col]);}printf("\n");  }system("pause");return 0;
}#include <stdio.h>
#include <stdlib.h>
int main(){int arr[3][4] = { 0 };int row;int col;for (row = 0; row < 3; ++row){for (col = 0; col < 4; ++col){printf("&arr[%d][%d]=%p\n",row,col, &arr[row][col]);}printf("\n");}system("pause");return 0;
}//3.操作符
#include <stdio.h>
#include <stdlib.h>
int main(){//按位与&两个数据都为1,结果为1,否则都为0;//按位或|两个数据都为0,结果为0,否则都为1;//按位异或^,相同位0,相异为1;//按位取反~,0变1,1变0;//16进制的数字就是4个2进制位//%x打印一个无符号的十六进制的整数int x = 0x1;int y = 0x2;int n = 10;n = n | (1 << 4);printf("%x\n", x & y);printf("%x\n", x | y);printf("%x\n", x ^ y);printf("%x\n", ~x);printf("%x\n", ~ y);printf("%x\n",n);system("pause");return 0;
}//4.编写代码实现,求一个整数存储中的二进制中1的个数
//解法一 0
#include <stdio.h>
#include <stdlib.h>
int main(){int num = -1;int i = 0;int count = 0;while (num){++count;num = num &(num - 1);}printf("%d\n", count);system("pause");return 0;
}//解法二
#include <stdio.h>
#include <stdlib.h>
//打印1的个数
int CountBit(int num){int count = 0;for (int i = 0; i < 32;++i){if (num&(1<<i)){++count;}}return count;
}int main(){int x = -10;// +10  1010// 1000 0000 0000 0000 0000 0000 0000 1010  原码(-10)// 1111 1111 1111 1111 1111 1111 1111 0101  反码// 1111 1111 1111 1111 1111 1111 1111 0110  补码(反码+1)printf("%d\n",CountBit(x));printf("%p",&x);system("pause");return 0;
}//5.不创建临时变量,实现两个数的交换
#include <stdio.h>
#include <stdlib.h>
int main(){int x = 10;int y = 20;x = x^y;y = x^y;x = x^y;printf("%d %d",x,y);system("pause");return 0;
}//6.360笔试题
#include <stdio.h>
#include <stdlib.h>
int main(){int i = 0, a = 0, b = 2, c = 3, d = 4;//对于逻辑与&&操作符来说,如果左侧的表达式值为0,那么右侧不再计算//对于逻辑或||操作符来说,如果左侧的表达式值为1,那么右侧不再计算//i = a++ && ++b && d++ && c++; i = a++ || ++b || c++ || d++;//printf(" a=%d b=%d c=%d d=%d",a, b, c,d);   //1 2 3 4printf("\n");printf(" a=%d b=%d c=%d d=%d", a, b, c, d);   //1 3 3 4system("pause");return 0;
}
//7.
#include <stdio.h>
int main(){char a = 0xff;a = (a << 1) >> 1;printf("%x",a);        //ffffffffsystem("pause");return 0;
}//8.操作符的属性
#include <stdio.h>
int main(){int a = 10;int b = 20;int c = 30;int ret = ++a + ++b + ++c;printf("%d",ret);           //63 system("pause");return 0;
}

2019/4/8

//冒泡排序基本原理：
//1.比较相邻的元素,如果第一个比第二个大，就交换他们两个;
//2.对每一对相邻元素做同样的工作，从开始第一对到结尾的最后一对;
//3.在这一点，最后的元素应该会是最大的数。
//4.针对所有的元素重复以上的步骤，除了最后一个。
//5.持续每次对越来越少的元素重复上面的步骤，直到没有任何一对数字需要比较。//普通解法
#include <stdio.h>
#include <stdlib.h>void Swap(int* x,int* y){int tmp = *x;*x = *y;*y = tmp;
}//数组作为函数参数,就会隐式转换成指针
void BubbleSort(int arr[],int size){printf("size=%d\n",size);//每次找最小值,[0,bound)已排序区间,[bound,size)未排序//初始情况下,所有元素都是待排序的,已排序区间为空,//待排序区间为整个数组//每次找出一个最小值,放在数组开头位置//当已排序区间扩充和整个数组一样,数组就排好序了//一种找size次就完成整个排序,每次找到一个元素,已排序区间就扩充//一个元素,待排序区间就减少一个元素int bound = 0;for (bound = 0; bound < size; ++bound){//具体完成找最小元素和交换的过程for (int cur = size - 1; cur < size;--cur){if (arr[cur-1]>arr[cur]){//如果两个元素不满足升序要求就交换Swap(&arr[cur - 1], &arr[cur]);}}}
}int main(){int arr[4] = { 9, 5, 2, 7 };int size = sizeof(arr) / sizeof(arr[0]);BubbleSort(arr, size);for (int i = 0; i < 4;++i){printf("%d ",arr[i]);}system("pause");return 0;
}//优化  函数调用
#include <stdio.h>
#include <stdlib.h>void Swap(int* x, int* y){int tmp = *x;*x = *y;*y = tmp;
}typedef int(*Cmp)(int x, int y);
//数组作为函数参数,就会隐式转换成指针
void BubbleSort(int*arr[], int size,Cmp cmp){printf("size=%d\n", size);//每次找最小值,[0,bound)已排序区间,[bound,size)未排序//初始情况下,所有元素都是待排序的,已排序区间为空,//待排序区间为整个数组//每次找出一个最小值,放在数组开头位置//当已排序区间扩充和整个数组一样,数组就排好序了//一种找size次就完成整个排序,每次找到一个元素,已排序区间就扩充//一个元素,待排序区间就减少一个元素int bound = 0;for (bound = 0; bound < size; ++bound){//具体完成找最小元素和交换的过程for (int cur = size - 1; cur > bound; --cur){if (!cmp(arr[cur - 1],arr[cur])){//如果两个元素不满足升序要求就交换Swap(&arr[cur - 1], &arr[cur]);}}}
}int Less(int x, int y){if (x < y){return 1;}return 0;
}int Greater(int x, int y){if (x > y){return 1;}return 0;
}int main(){int arr[4] = { 9, 5, 2, 7 };int size = sizeof(arr) / sizeof(arr[0]);BubbleSort(arr, size,Greater);//BubbleSort(arr, size, Less);for (int i = 0; i < 4; ++i){printf("%d ", arr[i]);}system("pause");return 0;
}

2019/4/9

//1.指针
//定义:是编程语言中的一个对象,利用地址,它的值直接指向存在电脑存储器中// 另一个地方的值.由于通过地址能找到所需的变量单元,可以说,地址指向该//变量单元, 因此, 将地址形象化成为"指针",通过找到地址的内存单元.
/*#include <stdio.h>
int main(){int x = 10;int* p = &x;double y = 10.0;double* p2 = &y;printf("%d\n",sizeof(p));printf("%d\n",*p);printf("%p\n",p);printf("%p\n",p);printf("%f\n",*p2);printf("%d\n",sizeof(p2));system("pause");return 0;
}*//*#include <stdio.h>
int main(){char*p1 = 0x100;char*p2 = 0x200;printf("%x\n",p2-p1);//指针相减,通常情况下是没有意义的,变量在哪个地址取决于操作系统,//再去做差,可能得到的值就会发生变化//指针相减其实就是求两个指针之间隔了多少个元素short*p3 = 0x100;short*p4 = 0x200;printf("%x\n", p4 - p3);system("pause");return 0;
}*///指针比较(指针中存的地址是否相同,或者是否指向同一块内存空间)
/*int main(){int n1 = 10;int*p1 = &n1;int n2 = 10;int*p2 = &n2;if (p1==p2){printf("呵呵\n");}else{printf("哈哈\n");}system("pause");return 0;
}*///2.指针和数组
/*int main(){int arr[] = { 0 };printf("%d\n",sizeof(arr));printf("%d\n", sizeof(arr+0));printf("%d\n", sizeof(arr[0]));system("pause");return 0;
}*///3.二级指针
/*#include <stdio.h>
int main(){int arr[4] = { 1, 2, 3, 4 };//arr是一个指向首元素的指针//&arr是一个指向整个数组的指针(数组指针)printf("%p\n", &arr[0]);   //打印://0073FA7Cprintf("%p\n", arr);          //0073FA7Cprintf("%p\n", arr + 1);      //0073FA80printf("%p\n", &arr);	      //0073FA7Cprintf("%p\n", &arr + 1);     //0073FA8Csystem("pause");return 0;
}*///4.结构体
#include <stdio.h>
#include <stdlib.h>
int main(){struct Student{long int num;char name[20];   //member/成员/属性/字段char sex;char addr[20];}a = { 10101, "Li Lin",'M',"123 Beijing Road" };printf("NO.:%ld\nname:%s\nsex:%c\naddress:%s\n", a.num, a.name,a.sex, a.addr);system("pause");return 0;
}

2019/4/10

//1.设计一个程序来判断当前机器的字节序.
//解法一
/*#include <stdio.h>
//如果是小端就返回1,否则返回0
int IsLittleEnd(){int num = 10;char*p = (char*)&num;if (*p==0){return 0;}return 1;
}
int main(){printf("%d\n",IsLittleEnd());    // 1int num = 10;system("pause");return 0;
}*///解法二
/*#include <stdio.h>
int check_sys(){int i = 10;return(*(char*)&i);
}
int main(){int ret = check_sys();if (ret==1){printf("小端\n");}else{printf("大端\n");}system("pause");return 0;
}*///2.
/*#include <stdio.h>
int main(){char a = -1;signed char b = -1;unsigned char c = -1;printf("a=%d,b=%d,c=%d",a,b,c);   // a=-1 b=-1 c=255system("pause");return 0;
}*///3.
//%u无符号整形
//解题思路  char->int->unsigned int
/*#include <stdio.h>
int main(){char a = -128;printf("%u\n", a);      //4294967168(溢出)system("pause");return 0;
}*///4.
/*#include <stdio.h>
int main(){char a = 128;printf("%u\n", a);      //4294967168(溢出)system("pause");return 0;
}*///5.
//解题思路 i+j-->unsigned-->int
/*#include <stdio.h>
int main(){int i = -20;unsigned int j = 10;printf("%d\n", i+j);      //-10system("pause");return 0;
}*///6.
//解题思路:-1-i如果当成int来理解,不可能是0,如果当成char来理解,就有机会;
//转成char的过程中高位就截断,高位是什么都不重要,重要的是得让低位8位为0
//即可,构造一个Ascii值为0的字符,0-254这样的下标,对应的元素就是字符串的
//有效数字
/*#include <stdio.h>
#include <string.h>
int main(){char a[1000];int i;for (i = 0; i < 1000;++i){a[i] = -1 - i;}printf("%d",strlen(a));  //255system("pause");return 0;
}*///7.溢出,死循环
/*#include <stdio.h>
int main(){unsigned char i = 0;for (i = 0; i <= 255;++i){             //255  无符号printf("hello word\n");            //1111 1111}                                      //0000 0001 system("pause");                      //10000 0000return 0;
}*///8.浮点型在内存中的存储
/*int main(){int n = 9;float*pf = (float*)&n;printf("n的值为:%d\n",n);     printf("*pf的值为:%f\n",*pf);*pf = 9.0;printf("num的值为:%d\n",n);printf("*pf的值为:%f\n",*pf);system("pause");return 0;
}*/
//程序运行结果
//n的值为:9
//* pf的值为 : 0.000000
//num的值为 : 1091567616
//* pf的值为 : 9.000000//9.误差区间
/*#include <stdio.h>
int main(){float i = 19.0;float j = i / 7.0;if (j * 7.0 - i < 0.0001 && j * 7.0 - i > -0.0001){printf("呵呵\n");}else{printf("哈哈\n");}system("pause");return 0;
}*/

2019/4/11

//指针的进阶
//1.
#include <stdio.h>
#include <stdlib.h>
int main(){char w = 'w';char* p = &w;char* p2 = "hehehehe";char str[] = { 'a', 'b', 'c' };char *p3 = str;printf("%c\n",*p2);printf("%s\n", p2);printf("%s\n", p3);printf("%s\n", p2+1);system("pause");return 0;
}
//程序运行结果:
h
hehehehe
abc烫烫烫烫汤X ?
ehehehe//2.
//解析:虽然指针和数组在C语言中经常混用,但是两者之间也是有本质差别
//数组是自带内存空间的,往往是把原来的数据又复制一份放到当前新的内
//存空间中;指针是不自带内存空间的,自身内存空间就是4个字节,不管指向
//的是什么内容,这4个字节只是保存对应内存空间的地址,不涉及数据拷贝的
//过程.
int main(){char str1[] = "hehe";char str2[] = "hehe";if (str1==str2){printf("hehe\n");}char* str3 = "hehe";char* str4 = "hehe";if (str3==str4){printf("haha");}system("pause");return 0;
}//3.
int main(){int arr[4] = { 0 };int* p = arr;printf("%p\n",&arr[0]);//普通指针printf("%p\n",arr);printf("%p\n", arr+1);//数组指针printf("%p\n",&arr);printf("%p\n", &arr+1);//二级指针printf("%p\n",&p);system("pause");return 0;
}
//程序运行结果:
00B7FB30
00B7FB30
00B7FB34
00B7FB30
00B7FB40
00B7FB24//4.一级指针传参
#include <stdio.h>
void Func(){printf("hehe\n");
}
int main(){printf("%p\n",Func);printf("%p\n", &Func);system("pause");return 0;
}//5.函数指针的用途:转移表
//exp:计算器
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
int Add(int x, int y){return x + y;
}
int Sub(int x, int y){return x - y;
}
int Mul(int x, int y){return x * y;
}
int Div(int x, int y){return x / y;
}
typedef int(*Func)(int x, int y);
int main(){Func arr[] = { Add, Sub, Mul, Div };while (1){printf("1.相加\n");printf("2.相减\n");printf("3.相乘\n");printf("4.相除\n");printf("请输入您要进行的运算: ");int choice = 0;scanf("%d",&choice);//转移表，表驱动的方式arr[choice - 1](10, 20);}system("pause");return 0;
}//6.//对一个数组进行逆序
void Reverse(char arr[],int size){int left = 0;int right = size - 1;while (left<=right){swap(&arr[left],&arr[right]);++left;--right;}
}void Reverse(char arr[],int size){std::reverse(arr,arr+size)
}

2019/4/12

//指针和数组笔试题
//一维数组
//1.
#include <stdio.h>
#include <stdlib.h>
int main(){int a[] = { 1, 2, 3, 4 };printf("%d\n", sizeof(a));         //16  整个数组占有字节的内存printf("%d\n", sizeof(a + 0));     //4   隐式转换为指针printf("%d\n", sizeof(*a));        //4   指向首元素    printf("%d\n", sizeof(a + 1));     //4   指针printf("%d\n", sizeof(a[1]));      //4printf("%d\n", sizeof(&a));        //4    数组指针printf("%d\n", sizeof(*&a));       //16   整个数组printf("%d\n", sizeof(&a + 1));    //4    数组指针printf("%d\n", sizeof(&a[0]));     //4    首元素指针printf("%d\n", sizeof(&a[0] + 1)); //4    第二个元素指针printf("%d\n", sizeof(&*a));       //4    首元素地址system("pause");return 0;
}
总结:
1.只要是指针,sizeof()就是4个字节
2.&a数组指针,数组指针解引用到了整个数组//字符数组
//2.
#include <stdio.h>
#include <stdlib.h>
int main(){char a[] = { 'a', 'b', 'c', 'd', 'e', 'f' };   printf("%d\n", sizeof(a));         //6   不包括'\0'printf("%d\n", sizeof(a + 0));     //4   指向首元素地址printf("%d\n", sizeof(*a));        //1   指向首元素'a'    printf("%d\n", sizeof(a + 1));     //4   printf("%d\n", sizeof(a[1]));      //1    指向元素'b' printf("%d\n", sizeof(&a));        //4    数组指针printf("%d\n", sizeof(*&a));       //6    printf("%d\n", sizeof(&a + 1));    //4    数组指针printf("%d\n", sizeof(&a[0]));     //4    首元素指针printf("%d\n", sizeof(&a[0] + 1)); //4    第二个元素指针printf("%d\n", sizeof(&*a));       //4    首元素地址//未定义行为printf("%d\n",strlen(a));printf("%d\n", strlen(a+0));printf("%d\n", strlen(*a));printf("%d\n", strlen(a[1]));printf("%d\n", strlen(&a));printf("%d\n", strlen(&a + 1));printf("%d\n", strlen(&a[0]+1));system("pause");return 0;
}//3.
#include <stdio.h>
#include <stdlib.h>
int main(){char a[] = "abcdef";printf("%d\n", sizeof(a));           //7    包括'\0'printf("%d\n", sizeof(a + 0));       //4    指向首元素的指针printf("%d\n", sizeof(*a));          //1    sizeof("a")    printf("%d\n", sizeof(a + 1));       //4   printf("%d\n", sizeof(a[1]));        //1    sizeof("b")   printf("%d\n", sizeof(&a));          //4    数组指针printf("%d\n", sizeof(*&a));         //6   printf("%d\n", sizeof(&*a));         //4    首元素地址printf("%d\n", sizeof(&a + 1));      //4    数组指针printf("%d\n", sizeof(&a[0]));       //4    首元素指针printf("%d\n", sizeof(&a[0] + 1));   //4    第二个元素地址printf("%d\n", strlen(a));           //6printf("%d\n", strlen(a + 0));       //6printf("%d\n", strlen(*a));          //未定义行为 首元素'a'=97printf("%d\n", strlen(a[1]));        //未定义行为printf("%d\n", strlen(&a));          //6printf("%d\n", strlen(&a + 1));      //未定义行为 越界printf("%d\n", strlen(&a[0] + 1));   //5system("pause");return 0;
}//4.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){char *p = "abcdef";printf("%d\n", sizeof(p));           //4    指针printf("%d\n", sizeof(p + 0));       //4    指向首元素的指针printf("%d\n", sizeof(*p));          //1    sizeof("a")    printf("%d\n", sizeof(p + 1));       //4   printf("%d\n", sizeof(p[1]));        //1    sizeof("b")   printf("%d\n", sizeof(&p));          //4    二级指针printf("%d\n", sizeof(*&p));         //4   printf("%d\n", sizeof(&*p));         //4    printf("%d\n", sizeof(&p + 1));      //4    printf("%d\n", sizeof(&p[0]));       //4    printf("%d\n", sizeof(&p[0] + 1));   //4    第二个元素指针printf("%d\n", strlen(p));           //6printf("%d\n", strlen(p + 0));       //5    从b开始,遇到'\0'结束printf("%d\n", strlen(*p));          //未定义行为 首元素'a'=97printf("%d\n", strlen(p[0]));        //未定义行为  a//未定义行为  &p二级指针,对应的内存中计提有多少字节还不确定printf("%d\n", strlen(&p));         printf("%d\n", strlen(&p + 1));      //未定义行为 越界printf("%d\n", strlen(&p[0] + 1));   //5    从下一个元素地址开始system("pause");return 0;
}//二维数组
#include <stdio.h>
#include <stdlib.h>
int main(){int a[3][4] = { 0 };printf("%d\n", sizeof(a));           //48    4*12printf("%d\n", sizeof(a[0][0]));     //4     printf("%d\n", sizeof(a[0]));        //16    4*4 a[0]=>int[4]   printf("%d\n", sizeof(a[0]+ 1));     //4     指向第二个元素的指针printf("%d\n", sizeof(*a[0]+1));     //4      =>sizeof(a[0][1])printf("%d\n", sizeof(a+1));         //4      数组指针printf("%d\n", sizeof(*(a + 1)));    //16     *(a + 1)=>a[1]printf("%d\n", sizeof(&a[0]+1));     //4      数组指针printf("%d\n", sizeof(&*a));         //4    printf("%d\n", sizeof(&a + 1));      //4 //a[0]长度为4个元素的数组,&a[0]得到了一个数组指针,&a[0]+1还是//数组指针,*(&a[0]+1)便得到了一各数组printf("%d\n", sizeof(*(&a[0]+1)));  //16     printf("%d\n", sizeof(*a));          //16   printf("%d\n", sizeof(a[3]));        //16    system("pause");return 0;
}

2019/4/13

1..调整数组使奇数全部都位于偶数前面。 
//例如：
//输入一个整数数组，实现一个函数，来调整该数组中数字的顺序使得数组中
//所有的奇数位于数组的前半部分，所有偶数位于数组的后半部分。
//解法一:
#include <stdio.h>
#define SIZE(a) (sizeof(a) / sizeof(a[0]))
void divide(int * a, int n){int s[128] = { 0 };int d[128] = { 0 };int i, counts = 0, countd = 0;for (i = 0; i < n; i++){if (a[i] % 2){s[counts++] = a[i];}else{d[countd++] = a[i];}}for (i = 0; i < counts; i++){a[i] = s[i];}for (; i < n; i++){a[i] = d[i - counts];}
}
void printArray(int * a, int n){int i;for (i = 0; i < n; i++){printf("%d ", a[i]);}putchar('\n');
}
int main(){int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };divide(a, SIZE(a));printArray(a, SIZE(a));system("pause");return 0;
}//解法二:
#include <stdio.h>
#define SIZE(a) (sizeof(a) / sizeof(a[0]))
void divide(int * a, int n){int start = 0;int end = n - 1;int tmp;while (a[start++] % 2);while (a[end--] % 2 == 0);while (start < end){tmp = a[end + 1];a[end + 1] = a[start - 1];a[start - 1] = tmp;while (a[start++] % 2);while (a[end--] % 2 == 0);}
}
void printArray(int * a, int n){int i;for (i = 0; i < n; i++){printf("%d ", a[i]);}putchar('\n');
}
int main(){int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };divide(a, SIZE(a));printArray(a, SIZE(a));system("pause");return 0;
}
2.2. 
//杨氏矩阵 
//有一个二维数组.数组的每行从左到右是递增的，每列从上到下是递增的.
//在这样的数组中查找一个数字是否存在。时间复杂度小于O(N);
//数组：1 2 3
//      4 5 6
//      7 8 9
#include <stdio.h>
int findnum(int a[][3], int x, int y, int f){int i = 0, j = x - 1;while (j >= 0 && i < y){if (a[i][j] < f){++i;}else if (a[i][j] > f){--j;}else{return 1;}}return 0;
}
int main(){int a[][3] = { { 1, 3, 5 },{ 3, 6, 7 },{ 7, 8, 9 } };if (findnum(a, 3, 3, 2)){printf("It has been found!\n");}else{printf("It hasn't been found!\n");}system("pause");return 0;
}

2019/4/14

//1.实现一个函数，可以左旋字符串中的k个字符。
//ABCD左旋一个字符得到BCDA;          ABCD左旋两个字符得到CDAB
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int round(char s[], int k){int len = strlen(s);char * tmp = (char *)calloc(len + 1, sizeof(char));if (tmp == NULL){return 1;}int i;k %= len;for (i = 0; i < len - k; i++){tmp[i] = s[i + k];}for (; i < len; i++){tmp[i] = s[i + k - len];}for (i = 0; i < len; i++){s[i] = tmp[i];}free(tmp);return 0;
}
int main(){char s[] = "ABCDE";round(s, 2);puts(s);system("pause");return 0;
}//2.2.判断一个字符串是否为另外一个字符串旋转之后的字符串。 
//例如：给定s1 =AABCD和s2 = BCDAA，返回1，
//给定s1 = abcd和s2 = ACBD，返回0.
//AABCD左旋一个字符得到ABCDA
//AABCD左旋两个字符得到BCDAA
//AABCD右旋一个字符得到DAABC
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int find(char * a, char * b){char * tmp = (char *)calloc(strlen(a) * 2 + 1, sizeof(char));if (tmp == NULL){return 0;}strcpy(tmp, a);strcat(tmp, a);if (strstr(tmp, b)){free(tmp);return 1;}free(tmp);return 0;
}
int main(){char a[] = "AABCD";char b[] = "BCDAA";if (find(a, b)){printf("找到了\n");}else{printf("没找到\n");}system("pause");return 0;
}

2019/4/15

//1.一个数组中只有两个数字是出现一次，
//其他所有数字都出现了两次,找出这两个数字，编程实现。
#include <stdio.h>
#define SIZE(a) (sizeof(a) / sizeof(a[0]))
int main(){int a[] = { 8, 10, 8, 2, 11, 9, 12, 11, 9, 2 };int i, sum = 0;int pos;int num1 = 0, num2 = 0;for (i = 0; i < SIZE(a); i++){sum ^= a[i];}for (i = 0; i < 32; i++){if (sum & 1 << i){pos = i;break;}}for (i = 0; i < SIZE(a); i++){if (a[i] & 1 << pos){num1 ^= a[i];}else{num2 ^= a[i];}}printf("%d %d\n", num1, num2);system("pause");return 0;
}//2.喝汽水，1瓶汽水1元，2个空瓶可以换一瓶汽水， 给20元，可以多少汽水。
#include <stdio.h>
int main(){int i;int sum = 0;for (i = 20; i > 1; i /= 2){sum += i - i % 2;i += i % 2;}printf("%d", sum + 1);system("pause");return 0;
}//3.模拟实现strcpy
//char * strcpy(char * destination, const char * source);
//Return Value : destination is returned.
//destination=dest;source=src;#include<stdio.h>
#include <assert.h>
char* Strcpy(char* dest, const char* src) {int i;assert(src != NULL);assert(dest != NULL);for (i = 0; src[i] != '\0'; ++i) {dest[i] = src[i];}dest[i] = '\0';return dest;
}
int main() {char* p = "abcdef";char string[1024] = {0};Strcpy(string, p);printf("%s\n",string);system("pause");return 0;
}
//运行结果
//abcdef//4.模拟实现strcat 
//char * strcat(char * destination, const char * source);
//Return Value : destination is returned.#include<stdio.h>
#include <assert.h>
char* Strcat(char* dest, const char* src) {int i;int j;assert(src != NULL);assert(dest != NULL);for(i = 0;dest[i]!='\0';++i);for (j = 0; src[j] != '\0'; ++i, ++j) {dest[i] = src[j];}dest[i] = '\0';return dest;
}
int main() {char string1[1024]="aaa";char string2[1024]="bbb";Strcat(string2, string1);printf("%s\n",string1);printf("%s\n", string2);system("pause");return 0;
}

2019/4/16

//1.模拟实现strcpy
//char * strcpy(char * destination, const char * source);
//Return Value : destination is returned.
//destination=dest;source=src;#include<stdio.h>
#include <assert.h>
char* Strcpy(char* dest, const char* src) {int i;assert(src != NULL);assert(dest != NULL);for (i = 0; src[i] != '\0'; ++i) {dest[i] = src[i];}dest[i] = '\0';return dest;
}
int main() {char* p = "abcdef";char string[1024] = { 0 };Strcpy(string, p);printf("%s\n", string);system("pause");return 0;
}
//运行结果
//abcdef//2.模拟实现strcat 
//char * strcat(char * destination, const char * source);
//Return Value : destination is returned.#include <stdio.h>
#include <assert.h>
char* Strcat(char* dest, const char* src) {int i;int j;assert(src != NULL);assert(dest != NULL);for (i = 0; dest[i] != '\0'; ++i);for (j = 0; src[j] != '\0'; ++i, ++j) {dest[i] = dest[j];}dest[i] = '\0';return dest;
}
int main() {char string1[1024] = "aaa";char string2[1024] = "bbb";Strcat(string2, string1);printf("%s\n", string1);printf("%s\n", string2);system("pause");return 0;
}//3.实现strstr
//char * strstr(const char *, const char *);
//Return Value:
//A pointer to the first occurrence in str1 of the entire sequence of
//characters specified in str2, or a null pointer if the sequence is 
//not present in str1.#define _CRT_SECURE_NO_WARNINGS
#include <string.h>
#include<stdio.h>
#include <assert.h>
char* Strstr(const char * str1,const char * str2){assert(str1 != NULL);assert(str2 != NULL);if (*str2 == '\0'){return NULL;}//黑指针功能是记录从哪个位置找字符串const char* black_ptr = str1;while (*black_ptr !='0'){//红指针帮我们完成具体的字符串比较const char* red_ptr = black_ptr;const char* sub_ptr = str2;while (*red_ptr == *sub_ptr && *red_ptr != '\0'&& *sub_ptr != '\0') {++red_ptr;++sub_ptr;}if (*sub_ptr == '\0') {return black_ptr;}++black_ptr;}return NULL;
}//4.实现strchr
char * strchr(const char *, int);
Returns a pointer to the first occurrence of character in the 
C string str.
The terminating null - character is considered part of the C string.
Therefore, it can also be located in order to retrieve a pointer to 
the end of a string.#include <string.h>
#include<stdio.h>
#include <assert.h>const char* Strchr(const char* str,int character) {assert(str != NULL);while (*str++) {if (character == *str) {return str;}}return NULL;
}
int main () {const char* search_char;char str[1024];char character;gets(str);character = getchar();search_char = Strchr(str, character);puts(search_char);return 0;
}//5.实现strcmp
int strcmp(const char * str1, const char * str2);#include <string.h>
#include<stdio.h>
#include <assert.h>int Strcmp(const char* str1, const char* str2) {assert(str1 != NULL);assert(str2 != NULL);int i;for (i = 0; str1[i] != '\0'&&str2[i] != '\0'; ++i){if (str1[i] < str2[i]){return -1;}else{//什么都不做，直接比较下一个字符}}if (str1[i] < str2[i]){return -1;}else if (str1[i] > str2[i]){return 1;}else{return 0;}
}int main(){char str1[1024] = "aaa";char str2[1024] = "bbb";//strcmp  o(n)=nint ret = Strcmp(str1,str2);if (ret>0){printf("str1 > str2\n");}else if (ret<0){printf("str1 < str2\n");}else{ printf("str1==str2\n");}system("pause");return 0;
}//6.实现memcpy
//void * memcpy(void * destination, const void * source, size_t num);
//Return Value:
//destination is returned#include <string.h>
#include<stdio.h>
void* Memcpy(void* dest, const void* src, size_t num) {assert(dest != NULL);assert(src != NULL);const char* psrc = (const char*)src;const char* pdest = (const char*)dest;for (size_t i = 0; i < num; ++i){pdest[i] = psrc[i];}return dest;
}
int main() {int arr1[4] = { 0 };int arr2[4] = { 1, 2, 3, 4 };Memcpy(arr1,arr2,16);for (int i = 0; i < 4;++i){printf("%d\n",arr1[i]);}system("pause");return 0;
}//7.实现memmove
//void * memmove(void * destination, const void * source, size_t num);
#include <string.h>
#include<stdio.h>
#include <assert.h>void *Memmove(void* dest, const void* src, size_t num) {char* pdest = (char*)dest;const char* psrc = (char*)src;if (dest == NULL || src == NULL) {return NULL;}//地址不重叠时候,从前面开始复制if (pdest <= psrc || pdest >= psrc + num) {while (num--) {*pdest++ = *psrc++;}}//地址重叠时候,从尾部开始复制else {//减1 就是获取下标,到了尾指针pdest = pdest + num - 1;psrc = psrc + num - 1;while (num--) {*pdest-- = *psrc--;}}return dest;
}
int main() {const char string1[1024] = "we are the best!";char string2[1024];Memmove(string1 + 3, string1, strlen(string1) + 1);printf("%s\n", string1 + 3);system("pause");return 0;
}

2019/4/20

//num1
#include <stdio.h>
int main(){
int a[5] = {1,2,3,4,5};
int* ptr = (int*)(&a+1);
printf("%d,%d",*(a+1),*(ptr-1));
system("pause");
return 0;
}
运行结果:
2,5//num2
#include <stdio.h>
//结构体大小为20个字节
struct test{int num;char* pcname;char cha[2];short sba[4];
}*p;                           //结构体指针(未初始化,其实默认为0)
int main(){//%p十六进制打印printf("%d\n", p + 0x1);printf("%d\n", (unsigned long)p + 0x1);printf("%d\n", (unsigned int*)p + 0x1);system("pause");return 0;
}
运行结果:
0x14   0x1  0x4//num3
int main(){int a[4] = {1,2,3,4};          总结://1.(int)a+1,其实地址+1int* ptr1 = (int*)(&a+1);           //2.对int*解引用的含义int* ptr2 = (int*)((int)a+1);       //3.整数在内存中如何存储printf("%x,%x",ptr1[-1],*ptr2);system("pause");return 0;
}
运行结果:
4, 2000000//num4
int main(int argc,char* argv[]){int a[3][2] = {(0,1),(2,3),(4,5)};int*p;p = a[0];printf("%d\n", p[0]);              //a[0][0]=1system("pause");return 0;
}//num5
int main(){//指针相减=>相间隔元素int a[5][5];int(*p)[4];p = a;printf("%p,%d\n",&p[4][2]-&a[4][2],&p[4][2]-&a[4][2]);system("pause");return 0;
}
运行结果:
FFFFFFFC, -4//num6.
int main(){int a[2][5] = {1, 2, 3, 4, 5,6, 7, 8, .9, 10};int *ptr1 = (int*)(&a + 1);          //&a+1会跳出整个二维数组int *ptr2 = (int *)(*(a + 1));       //*(a+1)=>a[1]printf("%d,%d",*(ptr1-1),*(ptr2-1)); system("pause");return 0;
}
运行结果:
10,5//num7
int main(){char* a[] = {"work","at","alibaba"};char** pa = a;                         //pa二级指针pa++;printf("%s\n",*pa);system("pause");return 0;
}
运行结果:
at//num8.
int main(){char* c[] = {"ENTER","NEW","POINT","FIRST"};char** cp[] = {c+3,c+2,c+1};char*** cpp = cp;printf("%s\n",**++cpp);printf("%s\n",*--*++cpp+3);printf("%s\n", *cpp[-2]+3);printf("%s\n", cpp[-1][-1]+1);system("pause");return 0;
}
运行结果:
POINT
ER
ST
EW

2019/4/21

//实现一个通讯录,完成联系人信息的存储.
//1.新增  
//2.删除
//3.修改
//4.查找记录
//5.打印全部记录
//6.排序记录
//7.清空全部记录#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <windows.h>
//通讯录每个条目的抽象
typedef struct PersonInfo{char name[1024];char phone[1024];       //电话号码用字符串表示
}PersonInfo;
typedef struct AddressBook{PersonInfo* persons;//[0,size)int size;    //控制通讯录的有效和无效int capacity;
}AddressBook;AddressBook g_address_book;       //全局变量void Init(){g_address_book.size = 0;//申请100个内存空间g_address_book.capacity = 100;g_address_book.persons = (PersonInfo*)malloc(100*sizeof(PersonInfo));g_address_book.capacity * sizeof(PersonInfo);for (int i = 0; i < g_address_book.capacity; ++i){g_address_book.persons[i].name[0] = '\0';g_address_book.persons[i].phone[0] = '\0';}
}int Menu(){printf("===================\n");printf(" 1.新增联系人\n");printf(" 2.删除联系人\n");printf(" 3.查找联系人\n");printf(" 4.修改联系人\n");printf(" 5.打印全部联系人\n");printf(" 6.排序联系人\n");printf(" 7.清空联系人\n");printf(" 0.退出\n");printf("===================\n");printf(" 请输入您的选择: ");int choice = 0;scanf("%d", &choice);return choice;
}void Empty(){//这个函数就是用来凑数
}void AddPersonInfo(){printf("新增联系人\n");if (g_address_book.size >= g_address_book.capacity){printf("当前空间不足,进行扩容!\n");g_address_book.capacity += 100;/*g_address_book.persons = (PersonInfo*)realloc(g_address_book.persons, g_address_book.capacity*sizeof(PersonInfo));*/PersonInfo* p = (PersonInfo*)malloc(g_address_book.capacity*sizeof(PersonInfo));for (int i = 0; i < g_address_book.size;++i){p[i] = g_address_book.persons[i];}free(g_address_book.persons);g_address_book.persons = p;return;}PersonInfo* person_info = &g_address_book.persons[g_address_book.size];printf("请输入联系人姓名: ");//必须获取到一个指针修改的内容是一个预期的内容scanf("%s", person_info->name);printf("请输入联系人电话: ");scanf("%s", person_info->phone);++g_address_book.size;printf("新增联系人成功!\n");system("pause");system("cls");
}void DelPersonInfo(){printf("删除联系人\n");if (g_address_book.size <= 0){printf("删除失败!通讯录为空!\n");}printf("请输入要删除的序号: ");int id = 0;scanf("%d", &id);if (id < 0 || id >= g_address_book.size){printf("删除失败!输入的序号有误!\n");return;}g_address_book.persons[id] = g_address_book.persons[g_address_book.size - 1];--g_address_book.size;printf("删除联系人成功!\n");system("pause");system("cls");
}void FindPersonInfo(){printf("查找联系人\n");if (g_address_book.size <= 0){printf("查找失败,通信录为空!\n");return;}//根据姓名查找电话printf("请输入要查找的姓名: ");char name[1024] = { 0 };scanf("%s", name);int size = 0;for (int i = 0; i < g_address_book.size; ++size){//元素地址取出来PersonInfo* info = &g_address_book.persons[i];if (strcmp(info->name, name) == 0){//序号iprintf("[%d] %s\t%s\n", i, info->name, info->phone);}}printf("查找联系人成功!\n");system("pause");system("cls");
}void UpdatePersonInfo(){printf("更新联系人\n");if (g_address_book.size <= 0){printf("修改失败,通信录为空!\n");return;}printf("请输入要修改的序号: ");int id = 0;scanf("%d", &id);if (id < 0 || id >= g_address_book.size){printf("修改失败,输入的序号有误!\n");return;}PersonInfo* info = &g_address_book.persons[id];printf("请输入新的姓名: (%s)\n", info->name);char name[1024] = { 0 };scanf("%s", name);if (strcmp(name, "*") != 0){strcmp(info->name, name);}char phone[1024] = { 0 };printf("请输入新的电话: (%s)\n", info->phone);scanf("%s", phone);if (strcmp(phone, "*") != 0){strcmp(info->phone, phone);}printf("更新联系人成功!\n");system("pause");system("cls");
}void PrintAllPersonInfo(){printf("打印全部联系人\n");for (int i = 0; i < g_address_book.size; ++i){PersonInfo* info = &g_address_book.persons[i];printf("[%d] %s\t%s\n", i, info->name, info->phone);}printf("共打印了 %d 条数据!\n", g_address_book.size);printf("打印全部联系人成功!\n");system("pause");system("cls");
}void SortPersonInfo(){//按照字节序排序,取结构体中的姓名字段char name[1024] = { '\0' };char phone[1024] = { '\0' };int bound;int i;int size;for (i = 0; i < g_address_book.size; i++) {for (bound = 0; bound < g_address_book.size; bound++) {for (size = bound; size<g_address_book.size - 1; size++) {if (strcmp(g_address_book.persons[size].name,g_address_book.persons[size + 1].name)>0) {strcpy(name, g_address_book.persons[size].name);strcpy(g_address_book.persons[size].name,g_address_book.persons[size + 1].name);strcpy(g_address_book.persons[size + 1].name, name);//copy 电话号码strcpy(phone, g_address_book.persons[size].phone);strcpy(g_address_book.persons[size].phone,g_address_book.persons[size + 1].phone);strcpy(g_address_book.persons[size + 1].phone, phone);}}}}printf("排序所有联系人:\n");printf("排序成功!\n");system("pause");system("cls");
}void ClearAllPersonInfo(){printf("清空全部数据\n");printf("您真的要清空全部数据吗?Y/N\n");char choice[1024] = { 0 };scanf("%s", choice);if (strcmp(choice, "Y") == 0){g_address_book.size = 0;printf("清空全部数据成功!\n");}else{printf("清空操作取消!\n");}
}typedef void(*Func)();    //函数指针的类型
// Vim操作风格
int main(){Func arr[] = {Empty,AddPersonInfo,DelPersonInfo,FindPersonInfo,UpdatePersonInfo,PrintAllPersonInfo,SortPersonInfo,ClearAllPersonInfo};Init();while (1){int choice = Menu();if (choice < 0 || choice >= sizeof(arr) / sizeof(arr[0])){printf("您的输入有误!\n");system("pause");system("cls");continue;}else if (choice == 0){printf("goodbye!\n");break;}else {Sleep(200);system("cls");arr[choice]();}}system("pause");return 0;
}

2019/4/23

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#pragma once//1.将温度计测量的度数华式法(64F)转换为摄氏度(17.8度)
int main(){float f, c;f = 64.0;c = (5.0 / 9)*(f - 32);printf("f=%f,c=%f\n",f,c);system("pause");return 0;
}//2.计算存款利息.有1000元,想存一年,有三种方法可选:(1)活期,年利率为r1.
//(2)一年期为定期,,年利率为r2.(3)存两次半年期,年利率为r3,,分别计算一
//年后按三种方法的本息和.
//解题思路:
//一年活期存款 : 一年后本息和为p1 = p0*(1 + r1);
//一年定期存款:一年后本息和为p2 = p0*(1 + r2);
//两次半年定期存款, 一年后本息和为p3 = p0*(1 + r3 / 2)*(1 + r3 / 2);int main(){//定义变量float p0 = 1000, r1 = 0.0036, r2 = 0.0225, r3 = 0.0198, p1, p2, p3;p1 = p0*(1 + r1);p2= p0*(1 + r2);p3 = p0*(1 + r3 / 2)*(1 + r3 / 2);printf("p1=%f\np2=%f\np3=%f\n",p1,p2,p3);system("pause");return 0;
}
运行结果:
p1 = 1003.599976
p2 = 1022.500061
p3 = 1019.897949
请按任意键继续. . .//3.给定一个大写字母,要求用小写字母输出.
int main(){char c1, c2;c1 = 'A';            //将字符'A'的ASCII代码放到c1变量中c2 = c1 + 32;        //得到字符'a'的ASCII代码放在c2变量中printf("%c\n",c2);   //输出c2的值,是一个字符printf("%d\n",c2);   //输出c2的值,是字符'a'的ASCII代码system("pause");return 0;
}
运行结果:
a
97//4.先后输出BOY三个字符.
法一:
int main(){char a = 'B', b = 'O', c = 'Y';  //定义三个字符变量并且初始化putchar(a);                      //像显示器输出字符Bputchar(b);                      //像显示器输出字符Oputchar(c);                      //像显示器输出字符Yputchar('\n');                   //像显示器输出一个换行符system("pause");return 0;
}法二:
int main(){int a = 66, b = 79, c = 89;putchar(a);putchar(b);putchar(c);putchar('\n');system("pause");return 0;
}法三:
int main(){putchar(getchar());              //将接受的字符输出Bputchar(getchar());              //像显示器输出字符Oputchar(getchar());              //像显示器输出字符Yputchar('\n');system("pause");return 0;
}
//5.求解ax^2+bx+c=0方程的解.
解题思路:
(1)a = 0, 不是二次方程.
(2)b ^ 2 - 4ac = 0, 方程有两个相等实根.
(3)b ^ 2 - 4ac > 0, 方程有两个不等实根.
(4)b ^ 2 - 4ac < 0, 方程有两个共轭复根, 应当以p + qi和p - qi的形式输出复根,
其中p = -b / 2a, q = sqrt(b ^ 2 - 4ac) / 2a.int main(){double a, b, c, disc, x1, x2, realpart, imagpart;scanf("%lf,%lf,%lf",&a,&b,&c);printf("The equation");if (fabs(a)<=1e-6){printf("is not a quadratic\n");}else{disc = b*b - 4 * a * c;if (fabs(disc) <= 1e-6){printf(" has two equal roots:%8.4f\n",-b/(2 * a));}else{if (disc>1e-6){x1 = (-b + sqrt(disc) / (2 * a));x2 = (-b - sqrt(disc) / (2 * a));printf(" has distinct real roots:%8.4f and %8.4f\n",x1,x2);}else{realpart = -b / (2 * a);              //复根的实部imagpart = sqrt(-disc) / (2 * a);     //复根的虚部printf(" has complex roots:\n");printf("%8.4f + %8.4fi\n", realpart,imagpart);printf("%8.4f - %8.4fi\n", realpart, imagpart);}}}system("pause");return 0;
}
程序分析:
(1)程序中用disc代表b^2-4ac,先计算disc的值.
(2)由于disc(即b ^ 2 - 4ac)是实数,而实数在计算和存储时会有一些小的误差,
因此采取判别disc的绝对值(fabs(disc))是否小于一个很小的数(例如10^-6).
(3)realpart代表实部,imagpart代表虚部.

2019/4/25

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#pragma once//1.求1+2+3+4+...+100的值
//解法一:
int main(){int i = 1, sum = 0;while (i<=100){sum += i;++i;}printf("sum = %d\n", sum);system("pause");return 0;
}//解法二:
int main(){int i = 1, sum = 0;do{sum += i;++i;} while (i <= 100);printf("sum = %d\n", sum);system("pause");return 0;
}//2.在全系1000名学生中举行慈善募捐,当总数达到10万元时就结束,
//统计此时捐款的人数以及平均捐款的数目.
解析:
(1)用循环来处理,实际循环的次数不知道,可以设为最大值,即最多会有1000人捐款
(2)用if语句检查是否达到10万元,如果达到就不再执行.
(3)定义变量存放捐款数(amount), 总捐款数(total), 人均存款数(aver).
#define  SUM 100000                        //指定符号常量SUM代表100000
int main(){float amount, aver, total;int i;for (i = 1, total = 0; i < 1000;++i){printf("please enter amount:");scanf("%f",&amount);total += amount;if (total>SUM) break;}aver = total / i;printf("num=%d\naver=%10.2f\n",i,aver);system("pause");return 0;
}
运行结果:
please enter amount : 10000
please enter amount : 20000
please enter amount : 30000
please enter amount : 25000
please enter amount : 15000
please enter amount : 20000
num = 6
aver = 20000.00//3.输出以下4 * 5的矩阵.
1  2   3   4   5
2  4   6   8  10
3  6   9  12  15
4  8  12  16  20
int main(){int i, j, n = 0;for (i = 1; i <= 4;++i){for (j = 1; j <= 5;++j,++n){        //n用来累计输出数据的个数if (n % 5 == 0){                //控制输出在5个数据后换行printf("\n");} printf("%d\t",i*j);}}printf("\n");system("pause");return 0;
}int main(){int i, j, n = 0;for (i = 1; i <= 4; ++i){for (j = 1; j <= 5; ++j, ++n){if (n % 5 == 0){printf("\n");}if (i==3&&j==1){break;}printf("%d\t", i*j);}}printf("\n");system("pause");return 0;
}
运行结果:
1  2   3   4   5
2  4   6   8  104  8  12  16  20int main(){int i, j, n = 0;for (i = 1; i <= 4; ++i){for (j = 1; j <= 5; ++j, ++n){if (n % 5 == 0){printf("\n");}if (i == 3 && j == 1){continue;}printf("%d\t", i*j);}}printf("\n");system("pause");return 0;
}
运行结果:
1  2   3   4   5
2  4   6   8  10
6   9  12  15
4  8  12  16  20//4.用公式π/4=1-(1/3)+(1/5)-(1/7)+...求pi的近似值,直到大西安某一项的
//绝对值小于10^(-6).
//解析:
//(1)每项的分子都是1
//(2)后一项的分母是前一项的分母加2
//(3)第一项的符号为正,从第二项起,每一项的符号与前一项的符号相反.
int main(){int flag = 1;                         //用来表示数值的符号//pi代表多项式的值,最后代表π,n代表分母double pi = 0.0, n = 1.0, term = 1.0; while (fabs(term)>1e-6){         //检查绝对值pi += term;n += 2;flag *=-1 ;                  //下一项与上一项符号相反term = flag / n;             //求出下一项的值term}pi = pi * 4;printf("pi = %10.8f\n",pi);      //输出π的近似值system("pause");return 0;
}
运行结果:
pi = 3.14159065

2019/4/27

#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#pragma once//1.对10个数组元素依次赋值为0,1,2,,3,4,5,6,7,8,9,要求按照逆序输出.
int main(){int i, a[10];for (i = 0; i <= 9;++i){         //对数组元素a[0]-a[9]赋值        a[i] = i;}for (i = 9; i >= 0;--i){         //输出a[9]-a[0]这10个数组元素printf("%d ",a[i]);}system("pause");return 0;
}
运行结果:
9 8 7 6 5 4 3 2 1 0//2.将一个二维数组行和列的元素互换,存到另外一个二维数组中.
int main(){int a[2][3] = { { 1, 2, 3 },{4, 5, 6 } };int b[3][2], i, j;printf("array a:\n");for (i = 0; i < 2;++i){         //处理a数组中的每一行中各元素for (j = 0; j < 3;++j){     //处理a数组中的每一列中各元素printf("%5d",a[i][j]);  //输出a数组的一个元素b[j][i] = a[i][j];      //将a数组中元素赋给b数组中相应的元素}printf("\n");}printf("array b:\n");for (i = 0; i < 3; ++i){for (j = 0; j < 2; ++j){printf("%5d", b[i][j]); //输出b数组的一个元素}printf("\n");}system("pause");return 0;
}
运行结果:
Array A :
1    2    3
4    5    6
Array B :
1    4
2    5
3    6//3.有一个3*4的矩阵,要求编程序求出其中值最大的那个元素的值,以及其所在
//的行号和列号.(打擂台算法)
int main(){int i, j, row = 0, col = 0, colum = 0, max;int a[3][4] = { { 1, 2, 3, 4 },{ 9, 8, 7, 6 },{ -10, 10, -5, 2 } };max = a[0][0];                          //先默认a[0][0]最大for (i = 0; i < 3;++i){for (j = 0; j < 4;++j){if (a[i][j]>max){              //如果元素大于max,就取代max = a[i][j];row = i;                   //记下元素的行号col = j;                   //记下元素的列号}}}printf("max=%2d\nrow=%2d\ncolum=%2d\n", max, row, colum);system("pause");return 0;
}
运行结果:
max = 10
row = 2
colum = 0//4.输入一行字符,统计其中有多少个单词,单词之间用空格分隔开.
int main(){char string[1024];int i, num = 0, word = 0;char c;                                    gets(string);                                //输入一个字符串//只要字符不是'\0'就继续执行循环for (i = 0; (c = string[i]) != '\0';++i){if (c == ' '){                  //如果是空格字符,使word置为0word = 0;}else if (word == 0){       //如果不是空格字符且word原值为0word = 1;              //使word加1++num;                 //num累加1,表示增加一个单词}}printf("There are %d words in this line.\n", num);system("pause");return 0;
}
运行结果:
someone like you!
There are 3 words in this line.//5.有三个字符串,要求找出其中"最大者".
int main(){char str[3][20];  //定义二维数组char string[20];  //定义一维数组,作为交换字符串时的临时字符数组int i;for (i = 0; i < 3;++i){//读入三个字符串,分别给str[0]str[1],str[2]gets(str[i]);            }//若str[0]>str[1],把str[0]的字符串赋值给stringif (strcmp(str[0],str[1])>0){strcpy(string,str[0]);}//若str[0]<str[1],把str[1]的字符串赋值给stringelse{strcpy(string, str[1]);}//若str[2]<string,把str[2]的字符串赋值给stringif (strcmp(str[2], string) > 0){strcpy(string, str[2]);}printf("\nthe largest string is:\n%s\n",string);   //输出stringsystem("pause");return 0;
}
运行结果:
China
America
Brazilthe largest string is :
China

2019/4/30

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#pragma once//1.将若干字符串按字母顺序(由小到大)输出.
//解题思路:
//(1)定义一个指针数组name, 用各字符串对它进行初始化, 即把各字符串中第1个
//字符的地址赋给指针数组的各元素;
//(2)用选择法排序, 但不是移动字符串, 而是改变指针数组的各元素的指向.
void sort(char* name[],int n){                 //定义sort函数char* temp;int i, j, k;for (i = 0; i < n - 1;i++){                //使用选择法排序k = i;for (j = i + 1; j < n;j++){if (strcmp(name[k],name[j])>0){k = j;}}if (k != i){temp = name[i];name[i] = name[k];name[k] = temp;}}
}void print(char* name[],int n){                //定义print函数int i;for (i = 0; i < n;i++){//按指针元素的顺序输出它们所指向的字符串printf("%s\n",name[i]);}
}int main(){void sort(char* name[], int n);          //函数声明void print(char* name[], int n);char* name[] = { "Follow me", "BASIC", "Great Wall","FORTRAN", "Computer design" };      //定义指针数组int n = 5;sort(name, n);                           //调用sort函数,对字符串排序print(name,n);                           //调用print函数,输出字符串system("pause");return 0;
}//2.使用指着数据的指针变量.int main(){char* name[] = { "Follow me", "BASIC", "Great Wall","FORTRAN", "Computer design" };char** p;int i;for (i = 0; i < 5;i++){p = name + i;printf("%s\n",*p);}system("pause");return 0;
}
运行结果:
Follow me
BASIC
Great Wall
FORTRAN
Computer design//3.有一个指针数组,其元素分别指向一个整形数组的元素,用指向指针数据的
//指针变量,输出整形数组各元素的值.int main(){int a[5] = { 1, 3, 5, 7, 9 };int* num[5] = { &a[0], &a[1], &a[2], &a[3], &a[4] };int** p, i;                   //p是指向指针型数据的指针变量p = num;                      //使p指向num[0]for (i = 0; i < 5;i++){printf("%d ",**p);p++;}printf("\n");system("pause");return 0;
}
运行结果:
1 3 5 7 9//4.建立动态数组,输入5个学生的成绩,另外用一个函数检查其中有无低于60分的,
//输出不合格的成绩.
//解题思路:
//(1)使用malloc函数开辟一个动态自由区域, 用来存储5个学生的成绩, 得到这个
//动态域第1个字节的地址, 基类型为void型;
//(2)使用一个基类型为int的指针变量p来指向动态数组的各元素, 并且输出它们的值;
//(3)首先必须将malloc函数返回的void指针转换为整数指针,然后赋给p1.
void check(int* p){                   //定义check函数int i;printf("They are fail:");for (i = 0; i < 5;i++){if (p[i]<60){printf("%d ",p[i]);       //输出不合格的成绩}}printf("\n");
}int main(){void check(int*);                   int* p1, i;//开辟动态内存区,将地址转换成int*型,然后放在p1中p1 = (int*)malloc(5 * sizeof(int)); for (i = 0; i < 5; i++){scanf("%d",p1+i);                //输入5个学生的成绩}check(p1);                           //调用check函数system("pause");return 0;
}
运行结果:
59 98 67 57 78
They are fail : 59 57