E - Fantasy of a Summation LightOJ1213

E - Fantasy of a Summation

Time Limit: 2000/1000 MS (Java/Others)      Memory Limit: 128000/64000 KB (Java/Others)

Submit Status

Problem 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 ... An-1, you have to write K nested loops and calculate the summation of all Ai 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 ... An-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

#include <stdio.h>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
typedef long long LL;
const int N=1e5+10;
const int INF=0x3f3f3f3f;
int cas=1,T;
int n,k,mod;
int quick_mod(int t,int n)
{
    int ans=1;
    while(n)
    {
        if(n&1) ans=ans*t%mod;
        t=t*t%mod;
        n>>=1;
    }
    return ans;
}
int main()
{
    //freopen("1.in","w",stdout);
//    freopen("1.in","r",stdin);
//    freopen("1.out","w",stdout);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&k,&mod);
        int ans=0;
        for(int i=0,x;i<n;i++)
        {
            scanf("%d",&x);
            ans+=x%mod;
        }
        ans%=mod;
        ans=k%mod * quick_mod(n,k-1) % mod *ans %mod;
        printf("Case %d: %d\n",cas++,ans);
    }
    //printf("time=%.3lf\n",(double)clock()/CLOCKS_PER_SEC);
    return 0;
}

 

题解：

仔细研究一下代码会发现只是一条公式：k*n^(k-1)*(a1+a2+...+an)
然后快速幂即可，注意不要被循环绕晕