HDU6027（快速幂）

Easy Summation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 449 Accepted Submission(s): 207

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

解题思路：因为n不是很大，所以直接快速幂就行。

#include<bits/stdc++.h>using namespace std;typedef long long ll;const ll mod = 1e9 + 7;ll n,k;ll quick_pow(ll i,ll x){    ll ans = 1;    while(i)    {        if(i&1) ans = (ans*x)%mod;        i >>= 1;        x = (x*x)%mod;    }    return ans;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%lld%lld",&n,&k);        ll sum = 0;        for(ll i = 1; i <= n; i++)        {            sum += quick_pow(k,i);        }        sum %= mod;        printf("%lld\n",sum);    }    return 0;}