编程求1!+2!+3!+...+n!,用单循环实现
时间: 2024-01-13 20:05:44 浏览: 29
```c
# include <stdio.h>
int main(void) {
int n = 0;
int i = 0;
int m = 1;
int sum=0;
printf("请输入n的值:");
scanf("%d", &n);
for (i=1; i<=n; ++i) {
m = m * i;
sum = sum + m;
}
printf ("sum = %d\n", sum);
return 0;
}
```
相关问题
编程求“1! +2! +·.·+n!”,用单循环实现
可以使用一个循环和一个变量来实现求和,具体实现如下:
```
int n = 5;
int result = 1;
for (int i = 1; i <= n; i++) {
result *= i;
}
cout << result << endl;
```
输出结果为 5! = 120。
.编程求“1! +2! +··…+n!”,用单循环实现和流程图
编程求“1! +2! +··…+n!”,用单循环实现的代码如下(假设n=3):
```
#include <stdio.h>
#include <stdlib.h>
int main() {
int n = 3;
int i, j;
int sum = 0, fac = 1;
for (i = 1; i <= n; i++) {
fac *= i;
sum += fac;
}
printf("%d\n", sum);
system("pause");
return 0;
}
```
流程图如下:
```
start
|
|__> 输入n
|__> i = 1, sum = 0, fac = 1
|__> 判断i <= n
| |__> 是,执行下一步
| |__> 否,跳出循环
|__> fac *= i
|__> sum += fac
|__> i++
|__> 回到第4步
|
|__> 输出sum
|__> 结束
```