函式_例題：編一函式求1^k+2^k+...n^k的和
阿新 • • 發佈：2019-01-31
/*
程式功能：求1^k+2^k+...n^k的和
*/
# include <stdio.h>
double power(int m , int k) /*求m的n次方並返回*/
{
int i;
double pw = 1;
for(i=1; i<=k; ++i) //k代表k次方，比如k=4,i的k次方=i*i*i*i;
pw*=m;
return (pw);
}
double sum_power(int k , int n) /*求1到n的k次方的和並返回*/
{
int i;
double sum = 0;
for(i=1; i<=n; ++i)
sum+=power(i, k);
return (sum);
}
int main(void)
{
double sum = 0;
int k, n;
printf("input k:");
scanf("%d", &k);
printf("\ninput n:");
scanf("%d", &n);
sum = sum_power(k, n);
printf("\n1^%d+2^%d+...%d^%d = %.0lf\n", k,k,n,k,sum);
}
/*
-----------------------------------
input k:4
input n:6
1^4+2^4+...6^4 = 2275
-----------------------------------
*/