HDU 6298(找规律)

Maximum Multiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 541    Accepted Submission(s): 243

 

Problem Description

Given an integer n, Chiaki would like to find three positive integers x, y and z such that: n=x+y+z, x∣n, y∣n, z∣n and xyz is maximum.

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤106), indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤106).

 

Output

For each test case, output an integer denoting the maximum xyz. If there no such integers, output −1 instead.

 

Sample Input

3 1 2 3

 

Sample Output

-1 -1 1

 

Source

2018 Multi-University Training Contest 1

签到题

打了个表，发现n%3==0的时候是均匀三份，n%4==0的时候是1：1：2

#include <bits/stdc++.h>
#define fir first
#define se second
#define pb push_back
#define ll long long
#define mp make_pair
using namespace std;
const int maxn=1e5+10;
const int maxm=1e6+10;
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
const double eps=1e-7;

int check(int x){
    int Max=-1;
    for (int i=1;i<=x;i++){
        if (x%i) continue;
        for (int j=1;j<=x;j++){
            if (x%j) continue;
            for (int k=1;k<=x;k++){
                if (x%k) continue;
                if (i+j+k!=x) continue;
                Max=max(Max,i*j*k);
            }
        }
    }
    return Max;
}

int main(){

   /* for (int i=1;i<=40;i++){
        cout<<i<<" "<<check(i)<<endl;
    }*/
    int t;
    scanf("%d",&t);
    while (t--){
        int n;
        scanf("%d",&n);
        if (n%3==0){
            ll temp=n/3;
            printf("%lld\n",temp*temp*temp);
        }
        else if (n%4==0){
            ll temp=n/4;
            printf("%lld\n",temp*temp*temp*2);
        }
        else printf("-1\n");
    }
}