sum of power

Think:

PS : 题目出处“浪潮杯”山东省第八届ACM大学生程序设计竞赛

题目要求 就是 计算 从1到n的m次幂运算的和， 同时注意mod取余运算；

Problem Description

Calculate mod (1000000000+7) for given n，m.
Input

Input contains two integers n,m(1≤n≤1000,0≤m≤10).
Output

Output the answer in a single line.
Example Input

10 0

Example Output

10

Hint
Author
“浪潮杯”山东省第八届ACM大学生程序设计竞赛（感谢青岛科技大学）

#include<bits/stdc++.h>using namespace std;#define mod 1000000007;long long int ans;long long int sum;long long int p(int a, int n)  {     ans = 1;     for (int i = 0;i < n;i ++)        {          ans = ans * a;          ans = ans % mod;        }        return ans;  }int main() {    int n, m;    scanf("%d %d", &n, &m);         sum = 0;          for (int i = 1;i <= n;i ++)            {               sum = sum + p(i, m);               sum = sum % mod;            }         printf("%lld\n", sum); }