LightOJ 1213 Fantasy of a Summation(快速幂)

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题目链接：
LightOJ 1213 Fantasy of a Summation
题意：
给定n,k,m求解：

int res = 0;for( i1 = 0; i1 < n; i1++ ) {    for( i2 = 0; i2 < n; i2++ ) {        for( i3 = 0; i3 < n; i3++ ) {            ...            for( iK = 0; iK < n; iK++ ) {                res = ( res + A[i1] + A[i2] + ... + A[iK] ) % MOD;            }            ...        }    }}

分析：
其实就是k层循环，每层遍历n个数字，把遍历到的数字相加和取模。模拟一下可以得到：ans = sum * n^(k -1) * k。sum是n个数字之和。然后就是快速幂计算下就好了。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <climits>#include <cmath>#include <ctime>#include <cassert>#include <set>#define IOS ios_base::sync_with_stdio(0); cin.tie(0);using namespace std;typedef long long ll;const int MAX_N = 1010;int T, n, total, cases = 0, K;ll sum, mod, data[MAX_N];ll quick_pow(ll a, ll b){    ll res = 1, tmp = a;    while(b){        if(b & 1) res = res * tmp % mod;        tmp = tmp * tmp % mod;        b >>= 1;    }    return res;}int main(){    scanf("%d", &T);    while(T--){        sum = 0;        scanf("%d%d%lld", &n, &K, &mod);        for(int i = 0; i < n; i++){            scanf("%lld", &data[i]);            sum += data[i];            sum %= mod;        }        ll ans = quick_pow(n, K - 1) * K % mod * sum % mod;        printf("Case %d: %lld\n", ++cases, ans);    }    return 0;}

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