HDU 6027 Easy Summation

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 .

题意

求 Σni=1ik

解题思路

对于 k=0,1,2,3,4,5 ，分别预处理出 Σni=1ik ，对每个询问直接给出答案。

由于 T 组数不多，也可直接针对 n 和 k 计算再给出答案。

代码

#include
using namespace std;
const int mod = 1e9+7;
long long f[6][10010], pre[6][10010];
void init()
{
    for(int k=0;k<=5;k++)
        for(int n=1;n<=10000;n++)
        {
            f[k][n] = k?(f[k-1][n]*n%mod):1;
            pre[k][n] = (pre[k][n-1] + f[k][n]) % mod;
        }
}
int main()
{
    init();
    int T, n, k;
    scanf("%d",&T);
    while(T-- && scanf("%d %d",&n,&k))
        printf("%I64d\n", pre[k][n]);
}