hdu 6298

Maximum Multiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1500    Accepted Submission(s): 650

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

 

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 /*
  经过数学变形知：
  x=n/k1,y=n/k2,z=n/k3
  1/k1+1/k2+1/k3=1
 3 3 3
 2 4 4
 2 3 6
只有这三组解，可令k1<=k2<=k3来推、
最后的结果即x*y*z=n*n*n/(k1*k2*k3)

 1 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <utility>
#include <queue>
using namespace std;
typedef long long ll;
int t;
int main()
{    scanf("%d",&t);
     ll n;
    while(t--){
    scanf("%lld",&n);
    if(n%3==0){
        printf("%lld
",n*n*n/27);
        continue;
    }
    if(n%4==0){
        printf("%lld
",n*n*n/32);
    }
    else{
        printf("-1
");//n%6=0，那么n%3也一定=0
    }
             }
return 0;
}