LightOj 1220 Fantasy of a Summation（快速幂）

Fantasy of a Summation

Description

If you think codes, eat codes then sometimes you may get stressed. In your dreams you may see huge codes, as I have seen once. Here is the code I saw in my dream.

#include <stdio.h>

int cases, caseno;
int n, K, MOD;
int A[1001];

int main() {
    scanf("%d", &cases);
    while( cases-- ) {
        scanf("%d %d %d", &n, &K, &MOD);

        int i, i1, i2, i3, ... , iK;

        for( i = 0; i < n; i++ ) scanf("%d", &A[i]);

        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;
                    }
                    ...
                }
            }
        }
        printf("Case %d: %d\n", ++caseno, res);
    }
    return 0;
}

Actually the code was about: 'You are given three integers n, K, MOD and n integers: A₀, A₁, A₂ ... A_n-1, you have to write K nested loops and calculate the summation of all A_i where i is the value of any nested loop variable.'

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with three integers: n (1 ≤ n ≤ 1000), K (1 ≤ K < 2³¹), MOD (1 ≤ MOD ≤ 35000). The next line contains n non-negative integers denoting A₀, A₁, A₂ ... A_n-1. Each of these integers will be fit into a 32 bit signed integer.

Output

For each case, print the case number and result of the code.

Sample Input

2

3 1 35000

1 2 3

2 3 35000

1 2

Sample Output

Case 1: 6

Case 2: 36

解题思路：

代码中有k层循环，每层循环n次，即进行了n^k次加法。每一次加法从n个数中可重复的选取k个数，每个数被选择的概率是k/n，所

有每个数对答案的贡献就是k*n^(k-1).

AC代码：

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