C语言用循环结构实现：输入正整数n，计算s=1/2! +1/4! +1/6! +…+1/n！

Question

如题所述

Answer 1

#include <stdio.h>

int main()
{
int n,i=1,j=1;
double s=0;
//printf("Enter A Integer:\n");
scanf("%d",&n);
/*s=1/2! +1/4! +1/6! ++1/n！*/
for(i;i<=n/2;i++)
{
int f=1;
// printf("i=%d\n",i);
for(j;j<=i*2;j++)
{
// printf("j=%d\n",j);
f*=j;
// printf("f1=%d\n",f);
}
// printf("f2=%d\n f3=%f\n",f,1.0/f);
s+=1.0/f;
j=1;
printf("s1=%f\n",s);
}
printf("%f\n",s);
}本回答被提问者采纳

Answer 2

Answer 3

int func1(int n)
{
if(n == 0) return 1;
return n*func(n-1);

}

int num = 0;
double sum = 0;
scanf("%d",&num);
for(int i = 2;i<=num;i++)
{
sum+= 1.0/func1(i);

}
或者：
int num = 0;
double sum = 0;
int n_ = 2;
scanf("%d",&num);
for(int i = 2;i<=num;i++)
{
sum+= 1.0/n_ ;

n_ *= i+1;

}

Answer 4

int sum=0；
int shu=1；
for(int i=1；i＜=5；i++)：
{
shu=shu×1/i
sum=sum+shu
}

Answer 5

double sum =0;
if(n%2 != 0 || n<2)
{pf(重新输入)；}
else{
for(int i=2;i<=n;i=i+2)
{long int l=1;
for(int j=1;j<=i;j++)
l = j*l;
}
sum + = 1/l;
}
pf(sum);