（HDU 6027 女生专场）Easy Summation 水题预处理

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本文介绍了一种解决大数求和问题的有效方法，通过预处理所有可能的值并在需要时提取，解决了当n值较大时传统求和方法的局限性。文章提供了完整的AC代码实现。

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Easy Summation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 675 Accepted Submission(s): 286

Problem Description
You are encountered with a traditional problem concerning the sums of powers.
Given two integers n and k. Let f(i)=i^k, 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

分析：
直接预处理出所有的值，然后进行取值就可以了

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

const int maxn = 10010;
const long long M = 1e9 + 7;

long long a[maxn][8];

void init()
{
    for(int i=1;i<=10000;i++)
    {
        a[i][0] = 1;
        for(int k=1;k<=5;k++)
        {
            a[i][k] = (a[i][k-1] * i) % M;
        }
    }
}

int main()
{
    init();
    int t,n,k;
    scanf("%d",&t);
    while(t--)
    {
        long long ans = 0;
        scanf("%d%d",&n,&k);
        for(int i=1;i<=n;i++)
        {
            ans = (ans + a[i][k]) % M;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}