Easy Summation

                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                             Total Submission(s): 27    Accepted Submission(s): 16


Problem Description
You are encountered with a traditional problem concerning the sums of powers.
Given two integers n and k. Let f(i)=ik, please evaluate the sum f(1)+f(2)+...+f(n). The problem is simple as it looks, apart from the value of n in this question is quite large.
Can you figure the answer out? Since the answer may be too large, please output the answer modulo 109+7.
 

Input
The first line of the input contains an integer T(1≤T≤20), denoting the number of test cases.
Each of the following T lines contains two integers n(1≤n≤10000) and k(0≤k≤5).
 

Output
For each test case, print a single line containing an integer modulo 109+7.
 

Sample Input
3
2 5
4 2
4 1
 

Sample Output
33
30
10
 

Source
2017中国大学生程序设计竞赛 - 女生专场




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题意：给出一个n和k，求从1^k到n^k的和

解题思路：暴力，注意取模即可


#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;
const int mod=1e9+7;

int n,m;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&m,&n);
        LL ans=0;
        for(int i=1; i<=m; i++)
        {
            LL x=1;
            for(int j=1; j<=n; j++)
            {
                x*=i;
                x%=mod;
            }
            ans+=x;
            ans%=mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}