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

32 54 24 1

 

Sample Output

333010
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;}