Easy Summation

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 $10^9+7$.

 

Input

The first line of the input contains an integer $T(1leq Tleq20)$, denoting the number of test cases. 
Each of the following $T$ lines contains two integers $n(1leq nleq 10000)$ and $k(0leq kleq 5)$. 

 

Output

For each test case, print a single line containing an integer modulo $10^9+7$.

 

Sample Input

3

2 5

4 2

4 1

 

Sample Output

33

30

10

 

 

题目意思：给定两个数n,k。求1的k次方 + 2的k次方 + ....+ n的k次方。结果会很大，所以要对10的9次方+7取余。

 

解题思路：把1-n的每个数的k次方加起来，求幂的时候使用快速幂，值得一提的是数非常大，因此在快速幂求幂的过程中， 每一步也要对10的9次方+7取余。

 

#include<stdio.h>
#define MAX 1000000007
#define ll long long int
ll test(int n,int k)
{
    ll i,t;
    t=1;
    for(i=1; i<=k; i++)
    {
        t=(t*n)%MAX;
    }
    return t;
}
int main()
{
    ll n,k,sum,i,t;
    while(scanf("%lld",&t)!=EOF)
    {
        while(t--)
        {
            sum=0;
            scanf("%lld%lld",&n,&k);
            for(i=1; i<=n; i++)
            {
                sum=(sum+test(i,k))%MAX;
            }
            printf("%lld
",sum);
        }
    }
    return 0;
}