编程思路:
(1)使用一个计数器count来判断终止条件;
(2)用除以2后的余数来判断是不是偶数。
程序如下:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/314e251f95cad1c829c5e88b753e6709c93d511b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
编译连接:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/2e2eb9389b504fc28f69a4a9efdde71190ef6d63?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
运行结果:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/6c224f4a20a44623197fb1ce9222720e0df3d7e0?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
由于偶数相差为2,所以可以优化代码:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/f703738da977391267b3956af7198618367ae220?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
源代码:
#include<stdio.h>
int main() /*浙江宁海陆继信*/
{
/* 浙江宁海陆继信程序优化 */
int sumEven = 0;
int i=0;
while(i <= 100) /* 循环直到100 */
{
sumEven=sumEven+i; //偶数求和;
i=i+2; //偶数相差为2,不必判断;
}
printf("sumEven: %d\n" , sumEven); //打印结果
return 0;
}