C语言用for语句求求1 + 1/2! +....+ 1/n!
时间: 2023-11-26 09:04:20 浏览: 36
以下是C语言用for语句求1 + 1/2! +....+ 1/n!的程序:
```c
#include <stdio.h>
int main() {
int n;
double sum = 0, factorial = 1;
printf("请输入一个整数n:\n");
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
factorial *= i; // 计算i的阶乘
sum += 1.0 / factorial; // 累加1/i!到sum中
}
printf("1 + 1/2! +....+ 1/n!的和为%.6lf", sum);
return 0;
}
```
相关问题
c语言求用循环语句计算e=1+1/1!+1/2!+1/3!+...1/n!
以下是两种C语言求解e的例子:
1.使用for循环计算e
```c
#include <stdio.h>
int main(void){
int i, n;
double sum = 1, t = 1;
printf("输入n:\n");
scanf("%d", &n);
for(i = 1; i <= n; i++){
t = t * i;
sum = sum + 1 / t;
}
printf("e=%lf", sum);
return 0;
}
```
2.使用while循环计算e
```c
#include <stdio.h>
int main(void){
int i = 1, n;
double sum = 1, t = 1;
printf("输入n:\n");
scanf("%d", &n);
while(i <= n){
t = t * i;
sum = sum + 1 / t;
i++;
}
printf("e=%lf", sum);
return 0;
}
```
C语言求1+1/2+1/3+...+1/n
这个问题可以用循环语句来实现,以下是C语言代码:
#include <stdio.h>
int main() {
int n;
double sum = .;
printf("请输入n的值:");
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
sum += 1. / i;
}
printf("1 + 1/2 + 1/3 + ... + 1/n = %lf", sum);
return ;
}