Question

编程计算1!+3!+5!+…+(2N-1)!的值,其中N值由键盘输入

Answer 1

int jiecheng(int n);int main()
{
int i,n;
int sum=0;
printf("please input n:");
scanf("%d",&n);
for (i=1;i<=n;i++)
{
sum+=jiecheng(2*i-1);
}
printf("\n1!+3!+……+(2n-1)!=%d\n",sum);
}int jiecheng(int n)
{
if (n==1)
return 1;
else
return n*jiecheng(n-1);
}

Answer 2

int fun(int n)
{
int mult = 1;
while(n >= 1)
{
mult = mult * n;
n--;
}
return mult;
}
void main()
{
int sum = 0;
int i;
int n;
scanf("%d",&n);
for(i = 1;i <= 2n-1; i = i + 2)
{
sum = sum + fun(n);
}
printf("%d",sum);
}

Answer 3

#include<stdio.h>
int N(int n)
{
int i,sum=1;
for(i=1;i<=n;i++)
sum*=i;
return sum;
}
int main()
{
int i,n,sum=0;
scanf("%d",&n);

for(i=1;i<=2*n-1;i=i+2)
sum+=N(i);
printf("%d\n",sum);
return 0;
}
希望采纳

Answer 4

#include <stdio.h>
void main()
{
int i,j,k,n;
long int sum=0;
printf("please input N:");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
k=1;
for(j=1;j<=(2*i-1);j++)//计算（2*n-1的阶乘）
{
k=k*j;
}
sum+=k;
}
printf("%d\n",sum);
}
不懂的追问