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 $T(1\leq T\leq20)$, denoting the number of test cases. 
Each of the following $T$ lines contains two integers $n(1\leq n\leq 10000)$ and $k(0\leq k\leq 5)$. 

 

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。

 

 1 #include<stdio.h>
 2 #define MAX 1000000007
 3 #define ll long long int
 4 ll test(int n,int k)
 5 {
 6     ll i,t;
 7     t=1;
 8     for(i=1; i<=k; i++)
 9     {
10         t=(t*n)%MAX;
11     }
12     return t;
13 }
14 int main()
15 {
16     ll n,k,sum,i,t;
17     while(scanf("%lld",&t)!=EOF)
18     {
19         while(t--)
20         {
21             sum=0;
22             scanf("%lld%lld",&n,&k);
23             for(i=1; i<=n; i++)
24             {
25                 sum=(sum+test(i,k))%MAX;
26             }
27             printf("%lld\n",sum);
28         }
29     }
30     return 0;
31 }