女生赛hdu6027

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

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

快速幂的裸

#include
#include
#include
#include
using namespace std;
#define ll long long 
ll mod = 1e9+7;
ll a[10005][10];
ll mul(ll a,ll b){
	ll res=1;
	while(b){
		if(b%2){
			res=(res*a)%mod;
		}
		a=(a*a)%mod;
		b/=2;
	}
	return res;
}
int main(){	
	int t;
	cin>>t;
	for(int i=0;i<=5;i++){
		for(int j=1;j<=10001;j++){
			a[j][i]=(a[j-1][i]+mul((ll)j,(ll)i))%mod;
		}
	}
	while(t--){
		int n,m;
		cin>>n>>m;
		cout<