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<bits/stdc++.h>
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]
= k?(f[k-1]
*n%mod):1;
pre[k]
= (pre[k][n-1] + f[k]
) % mod;
}
}
int main()
{
init();
int T, n, k;
scanf("%d",&T);
while(T-- && scanf("%d %d",&n,&k))
printf("%I64d\n", pre[k]
);
}