阶乘的倒数之和啊
double fun(int n)
{
double sum = 0;
int i=0,j=0,jc=1;
for(i=1;i<=n;i++)
{
jc=1;
for(j=1;j<=i;j++)
jc*=j;
sum += 1.0/jc;
}
return sum;
}
代码:
#include "stdio.h"
double fun(int n)
{
double sum = 0.0;
int i =1;
int j = 1;
double step = 1.0;
for(i = 1;i<=n;i++)
{
step = 1;
for(j=1;j<=i;j++)
step = step * j;
sum = sum + 1.0 / step;
}
return sum;
}
void main()
{
printf("S=:%lf",fun(50));
}
截图
代码只需要填写这一段放在 begin — end中
double sum = 0.0;
int i =1;
int j = 1;
double step = 1.0;
for(i = 1;i<=n;i++)
{
step = 1;
for(j=1;j<=i;j++)
step = step * j;
sum = sum + 1.0 / step;
}
return sum;
如果回答帮您解决了问题,希望您给我点个采纳。
另外询问一下题主,是否是金陵科技学院学生。
供参考:
#include<stdio.h>
double fun(int n)
{
int i=1;
double t=1,s=0;
while(i<=n){
t = t/i;
s += t;
i++;
}
return s;
}
void main()
{
int n;
printf("输入数n:");
scanf("%d",&n);
printf("s=%lf\n",fun(n));
}
double fun(int n)
{
double y = 0;
double temp = 1;
int i,j;
for (i = 1; i < n + 1; i++)
{
for (j = 1; j <= i; j++)
{
temp /= j;
}
y += temp;
}
return y;
}
代码如下
#include <stdio.h>
double fun(int n)
{
int i,j;
int jc = 1;
double sum = 0.0;
for (i=1;i<=n;i++)
{
//计算阶乘
jc =1;
for(j = 1;j<=n;j++)
jc *= i;
//求和
sum += 1.0/jc;
}
return sum;
}