请问C语言S=1^k-2^k+3^k-...+(-1)^(k+1)*n^k?

S=1^k-2^k+3^k-...+(-1)^(k+1)*n^k?

应为 S=1^k-2^k+3^k-...+(-1)^(n+1)*n^k?

程序如下

double s(int n, int k)
{

double sum=0;
int flag=-1;
for(int i=1;i<=n;i++)
{
flag*=-1;
temp=1;
for(int t=1;t<=k;t++)
temp*=i;

sum+=temp*flag;
}
}

#include<stdio.h>

main()
{
int i,k,n,S1,S2,S;
printf("Input n,k:");
scanf("%d,%d",&n,&k);
for(i=1;i<=n;i++)
{
S1=pow(-1,n+1);
S2=pow(n,k);
S=S+S1S2;
}
printf("%d",S);
}

S=1^k-2^k+3^k-...+(-1)^(k+1)*n^k?
S=1^k-2^k+3^k-...+(-1)^(n+1)*n^k吗?

double pownk(double n,double k)
{
double r;
for(;n;n--)
r+=((int)n%2?1:-1)*pow(n,k);
return r;
}

递归
double pownk(double n,double k)
{
if(n==1)