LightOj 1213

题目链接：http://lightoj.com/volume_showproblem.php?problem=1213

#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
", ++caseno, res);
    }
    return 0;
}

　　

题意：告诉你这段代码，然后优化，求res； n (1 ≤ n ≤ 1000), K (1 ≤ K < 2³¹), MOD (1 ≤ MOD ≤ 35000)

我们很容易就知道最内成的加法式子执行了n^K次，每次加了K个数，所以一共加了K*n^K个数，一共有n个数，每个数加的次数一定是相同的，所以每个数都加了K*n^(K-1)次，所以结果就是Sum*K*n^(K-1)%mod; 快速幂求一下即可；

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
using namespace std;
typedef long long LL;
const int oo = 0xfffffff;
const int N = 1e3+5;

LL K, mod, a[N];

int n;

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

int main()
{
    int T, t = 1;

    scanf("%d", &T);

    while(T--)
    {

        scanf("%d %lld %lld", &n, &K, &mod);

        LL sum = 0;

        for(int i=1; i<=n; i++)
        {
            scanf("%lld", &a[i]);

            sum = (sum + a[i])%mod;
        }

        sum = (sum * K)%mod * Pow((LL)n, K-1)%mod;

        printf("Case %d: %lld
", t++, sum);
    }
    return 0;
}