2017女生赛 1005 Easy Summation【】

Easy Summation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0


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当时一激动没考虑到直接pow超出了longlong范围wa了两发

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b)  memset(a,b,sizeof(a))
const int M=1e5+10;
const int inf=0x3f3f3f3f;
int i,j,k,n,m;
const ll mod=1e9+7;

int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&k);
ll ans=0;
for(int i=1;i<=n;i++){
ll sum=1;
for(int j=1;j<=k;j++)
sum*=i,sum%=mod;
ans+=sum;
ans%=mod;
}
printf("%lld\n",ans);
}
return 0;
}