HDU6027 Easy Summation 2017-05-07 19:02 23人阅读 评论(0) 收藏

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;
}