HDU6027-Easy Summation

最新推荐文章于 2019-05-20 20:14:34 发布

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本文介绍了一种解决幂次求和问题的算法实现，通过暴力枚举的方式计算从1^k到n^k的和，并对结果进行10^9+7取模处理。适用于n和k值较大情况下的快速求解。

Easy Summation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

Sample Output

33
30
10

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;
}