c语言s=1+2+3+…+n,当s刚超过某个数m时,停止运算。求n和s的值
时间: 2023-12-13 18:03:42 浏览: 248
同样可以使用循环来实现,具体实现如下:
```c
#include <stdio.h>
int main() {
int m = 100; // 假设 m = 100
int n = 0;
int s = 0;
while (s < m) {
n++;
s = n * (n + 1) / 2;
}
printf("n = %d\n", n);
printf("s = %d\n", s);
return 0;
}
```
运行结果为:
```
n = 14
s = 105
```
因此,当s刚好超过100时,n的值为14,s的值为105。
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