1 方法
(1)使用一个计数器来判断终止条件
(2)用除以2后的余数来判断奇偶
2 代码
#include<stdio.h>
int main()
{
int count, sumEven, sumOdd;
count = sumEven = sumOdd = 0;
while (++count <= 100)
count % 2 == 0 ? (sumEven += count) : (sumOdd += count);
printf("偶数和: %d\n奇数和: %d\n", sumEven, sumOdd);
getchar();
return 0;
}
3 运行结果
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/a2cc7cd98d1001e9e7fab642be0e7bec55e797bd?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)