HDU 6027

题意

给出n, k, 已知: f(i)=i^k , 求 sum = f(1)+f(2)+…+f(n) , 并将结果模10e9+7

思路

其实这个题直接暴力就能暴过去
用快速幂的时候忘了中间的a也要用LL结果WA了两发

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快速幂取模算法详解
快速幂
快速幂板子

ll quickmi(ll a, ll b){
    ll ans = 1;
    a %= mod;
    while(b){
        if( b & 1 )  ans = (ans*a) % mod;
        b >>= 1;
        a = (a*a) % mod;
    }
    return ans;
}

AC代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
const int maxn = 1e4+5;

ll quickmi(ll a, ll b){
    ll ans = 1;
    a %= mod;
    while(b){
        if( b & 1 )  ans = (ans*a) % mod;
        b >>= 1;
        a = (a*a) % mod;
    }
    return ans;
}

int main()
{
    int T;
    ll n, k;
    scanf("%d",&T);
    while(T--){
        scanf("%lld%lld", &n, &k);
        ll sum = 0;
        for( ll i = 1; i <= n; i++ )
            sum = (sum + quickmi(i, k)) % mod;
        printf("%lld
", sum % mod);
    }
    return 0;
}