LightOJ 1213(快速幂)

Fantasy of a Summation

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

 

分析：代码中k个循环执行加法的所有次数为k*n^k次，由k个循环的对称性可知，

数组中每个数被加的次数为k*n^(k-1)次，因此答案就是∑(a[i]*k*n^(k-1))，0<i<n，

即sum*k*n^(k-1)，然后用快速幂做。

#include<cstdio>
long long N,K,mod;
long long pow(long long x,long long k)
{
    long long res=1;
    while(k>0)
    {
        if(k&1) res=res*x%mod;
        k>>=1;
        x=x*x%mod;
    }
    return res;
}
int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        long long sum=0,x;
        scanf("%lld%lld%lld",&N,&K,&mod);
        for(int i=0;i<N;i++)
        {
            scanf("%lld",&x);
            sum+=x;
        }
        sum%=mod;
        long long ans=pow(N,K-1);
        printf("Case %d: %lld\n",++cas,ans%mod*K%mod*sum%mod);
    }
    return 0;
}