hdu6027Easy Summation(快速幂取模)

Easy Summation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4112    Accepted Submission(s): 1676

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

题意：给出n,k，定义f(i)=i^k,求出[f(1)+f(2)+……+f(n)]mod 1e9+7

题解：直接套快速幂，注意用long long

 #include<bits/stdc++.h>
 using namespace std;
 const long long mod=1e9+;
 long long PowerMod(long long a, long long b,long long c) {
     long long ans = ;
     a = a % c;
     while(b>) {
 
         if(b %  == )
             ans = (ans * a) % c;
         b = b/;
         a = (a * a) % c;
     }
     return ans;
 }
 int main()
 {
     int t;
     while(~scanf("%d",&t))
     {
         while(t--)
         {
             long long n,k,sum=;
             scanf("%lld %lld",&n,&k);
             for(int i=;i<=n;i++)
             {
                 sum+=PowerMod(i,k,mod);
                 sum=sum%mod;
             }
             printf("%lld\n",sum);
         }
     }
     return ;
 }