（HDU 6027 女生专场）Easy Summation 水题 预处理

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;}