徐州网站制作流程_广东seo推广哪里好_seo优化技术_百度手机版下载

循环控制

---

判断素数的程序

①以无法证伪来证明是真的

#include <stdio.h>int main()
{int x;scanf("%d", &x);int i;int isPrime = 1;for ( i=2; i<x; i++ ) {if ( x%i == 0 ) {isPrime = 0;}}if ( isPrime == 1) {printf ("是素数\n");} else {printf("不是素数\n");}return 0;
}

但是循环结束后，还会再判断i<x，才结束循环

有没有方法能直接跳出循环？

#include <stdio.h>int main()
{int x;scanf("%d", &x);int i;int isPrime = 1;for ( i=2; i<x; i++ ) {if ( x%i == 0 ) {isPrime = 0;break;}}if ( isPrime == 1) {printf ("是素数\n");} else {printf("不是素数\n");}return 0;
}

break vs continue

• break：跳出循环
• continue：跳过循环这一轮剩下的语句，进入下一轮

②

#include <stdio.h>int main()
{int x;scanf("%d", &x);int i;//int isPrime = 1;for ( i=2; i<x; i++ ) {if ( x%i == 0 ) {isPrime = 0;}}//if ( isPrime == 1) {if ( i < x) {printf ("不是素数\n");} else {printf("是素数\n");}return 0;
}

嵌套循环

---

输出100以内的素数

#include <stdio.h>int main()
{int x;x = 6;for (x=2; x<100; x++);{int i;int isPrime = 1;for ( i=2; i<x; i++ ) {if ( x%i == 0 ) {isPrime = 0;break;}}if ( isPrime == 1) {printf ("%d", x);} }printf("\n");return 0;
}

从嵌套循环中跳出

---

凑硬币：如何用一角、两角和五角的硬币凑出10元以下的金额

#include <stdio.h>int main()
{int x;int one, two, five;scanf("%", &x);for (one=1; one<x*10; one++ ) {for ( two=1; two<x*10/2; two++ ) {for ( five=1; five<x*10/5; five++) {if (one + two*2 + five*5 == x*10 ) {printf("可以用%d个一角加%d个两角加%d个五角得到%d元\n",one, two, five, x);}}}}return 0;
}

如何让程序找到一种组合方法就停止？

如果在printf之后加一个break/continue：之离开了最内层的循环（也就是第三个for循环），其他循环依旧在继续。break和continue只能对他所在的那层循环做跳出。

法1：接力break（教科书）

#include <stdio.h>int main()
{int x;int one, two, five;scanf("%", &x);for (one=1; one<x*10; one++ ) {for ( two=1; two<x*10/2; two++ ) {for ( five=1; five<x*10/5; five++) {if (one + two*2 + five*5 == x*10 ) {printf("可以用%d个一角加%d个两角加%d个五角得到%d元\n",one, two, five, x);exit = 1;break;}}if (exit) break;}if (exit) break;}return 0;
}

法2：goto（传送门）

#include <stdio.h>int main()
{int x;int one, two, five;scanf("%", &x);for (one=1; one<x*10; one++ ) {for ( two=1; two<x*10/2; two++ ) {for ( five=1; five<x*10/5; five++) {if (one + two*2 + five*5 == x*10 ) {printf("可以用%d个一角加%d个两角加%d个五角得到%d元\n",one, two, five, x);goto out;}}}}
out;return 0;
}