第一题
示例代码：
#include<stdio.h>int main()
{printf("Enter a two-digit number: ");int number,ten_n,g_n;scanf_s("%d", &number);ten_n = number / 10;g_n = number % 10;printf("The reversal is %d%d", g_n, ten_n);return 0;
}输出：
第二题
解题思路：这道题的解题思路有很多，一种思路就是分别求出每个位上的数字，然后排序输出，不难给出代码。
示例代码：
#include<stdio.h>int main()
{printf("Enter a two-digit number: ");int number, hund_n, ten_n, g_n;scanf_s("%d", &number);hund_n = number / 100;ten_n = (number-hund_n*100) / 10;g_n = (number-hund_n*100) % 10;//g_n = number-hund_n*100-ten_n*10;printf("The reversal is %d%d%d", g_n, ten_n, hund_n);return 0;
}输出
第三题
解题思路：%1d代表一位整数，所以输入的数字是按照个数组合起来的，而不是眼睛看起来通常意义上的几百，如此可以解答。
示例代码：
#include<stdio.h>int main()
{printf("Enter a two-digit number: ");int hund_n, ten_n, g_n;scanf_s("%1d", &hund_n);scanf_s("%1d", &ten_n);scanf_s("%1d", &g_n);//g_n = number-hund_n*100-ten_n*10;printf("The reversal is %d%d%d", g_n, ten_n, hund_n);return 0;
}输出
第四题
解题思路：根据题中提示，需要用累除连续除4次8，然后将余数排列出来即可。
示例代码
#include<stdio.h>int main()
{printf("Enter a number betweeen 0 and 32767: ");int number;scanf_s("%d", &number);int a, b, c, d, e;int temp_y, temp_r;temp_y = number % 8;a = temp_y;temp_r = number / 8;b = temp_r % 8;temp_r = temp_r / 8;c = temp_r % 8;temp_r = temp_r / 8;d = temp_r % 8;temp_r = temp_r / 8;e = temp_r % 8;printf("%d%d%d%d%d", e, d, c, b, a);return 0;
}输出
第五题
题目分析：这个题不难，关键在于审题，计算规则要看清楚，别搞错了。
示例代码
#include<stdio.h>
int main()
{int b, a1, a2, a3, a4, a5, b1, b2, b3, b4, b5;printf("Enter the first 11 digits of UPC: ");scanf_s("%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d",&b, &a1, &a2, &a3, &a4, &a5, &b1, &b2, &b3, &b4, &b5);int result;int first_sum,second_sum, total;first_sum = b + a2 + a4 + b1 + b3 + b5;second_sum = a1 + a3 + a5 + b2 + b4;total = 3 * first_sum + second_sum;result = 9 - (total - 1) % 10;printf("Check digit: %d", result);return 0;
}输出
第六题
题目分析：这种长臭的题，耐心看，提取出有用信息转换为代码语言。
示例代码
#include<stdio.h>
int main()
{int a1, a2, a3, a4, a5,a6, b1, b2, b3, b4, b5, b6;printf("Enter the first 11 digits of UPC: ");scanf_s("%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%d", &a1, &a2, &a3, &a4, &a5,&a6, &b1, &b2, &b3, &b4, &b5,&b6);int result;int first_sum, second_sum, total;first_sum = a6 + a2 + a4 + b2 + b4 + b6;second_sum = a1 + a3 + a5 + b1 + b3+b5;total = 3 * first_sum + second_sum;result = 9 - (total - 1) % 10;printf("Check digit: %d", result);return 0;
}输出
