编写一C程序，求数列1,-3!,5!,-7!,…,(-1)n-1(2n-1)!前n项的和。

Question

程序模板：
#include<stdio.h>
main()
{
int n,i,j,sign=1;
float fac,sum;
printf("please input value of n:");
scanf("%d",&n);
sum=0.0;
for(i=1;i<=2*n-1;i=i+2)
{
fac=1.0;
/*计算第n项*/
/*求前n项的和*/
sign=-sign;
}
printf("sum=%.0f\n",sum);
}

Answer 1

#include<stdio.h>
main()
{
int n,i,j,sign=1;
float fac,sum;
printf("please input value of n:");
scanf("%d",&n);
sum=0.0;
for(i=1;i<=2*n-1;i=i+2)
{
fac=1.0;
for(j=1;j<=i;j++)
fac*=j; /*计算第n项*/

sum+=sign*fac; /*求前n项的和*/
sign=-sign;
}
printf("sum=%.0f\n",sum);
}本回答被提问者采纳

Answer 2

#include<stdio.h>
main()
{
int
n,i,j,sign=1;
float
fac,sum;
printf("please
input
value
of
n:");
scanf("%d",&n);
sum=0.0;
for(i=1;i<=2*n-1;i=i+2)
{
fac=1.0;
for(j=1;j<=i;j++)
fac*=j;
/*计算第n项*/
sum+=sign*fac;
/*求前n项的和*/
sign=-sign;
}
printf("sum=%.0f\n",sum);
}

Answer 3

#include<stdio.h>
main()
{
int n,i,j,sign=1;
float fac,sum;
printf("please input value of n:");
scanf("%d",&n);
sum=0.0;
for(i=1;i<=2*n-1;i=i+2)
{
fac=1.0;
/*计算第n项*/
/*求前n项的和*/
sign=-sign;
}
printf("sum=%.0f\n",sum);
}