2018HDU多校联赛第一场Maximum Multiple

题目衔接：http://acm.hdu.edu.cn/showproblem.php?pid=6298

Maximum Multiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 904    Accepted Submission(s): 421

 

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

 

题目大意：给你一个数，问你能不能找到3个数x，y，z，使得x+y+z=n且使得n能整除以x，n能整除以y,n能整除以z，如果有找出最大的x*y*z

思路：先打表，可以看出规律：能整除以3的都是n*n*n，能整除以4的都是n*n*（n*2）；所以这就是规律

代码：

#include<map>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 100
#define ll long long
int main()
{
    int test;
    scanf("%d",&test);
    while(test--)
    {
        ll n;
        ll ans=-1;
        scanf("%lld",&n);
        if(n%3==0)
        {
            n/=3;
            ans=n*n*n;
        }
        else if(n%4==0)
        {
            n/=4;
            ans=n*n*(n*2);
        }
        else
            ans=-1;
        printf("%lld\n",ans);
    }
    return 0;
}