Maximum Multiple

Desprition

 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≤10^6), indicating the number of test cases. For each test case: 
 The first line contains an integer n (1≤n≤10^6). 

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

解析

 首先先理解题意：

1. n=x+y+z

2.x,y,z都是n的因子

3.输出最大的x*y*z的值

 思路

找出符合以上三个条件的数。我们假设n=1；

1=1/2+1/3+1/6=1/3+1/3+1/3=1/2+1/3+1/6。我们发现当n能被3整除，xyz的最大值是pow(n/3,3).当n能被4整除时，xyz的最大值是n/2*n/4*n/4，当n能被6整除，它的最大值其实也是pow(n/3,3)。其余的数都不符合条件，输出-1

代码

#include<stdio.h>
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        long long n;
        scanf("%lld",&n);
        if(n%3==0)
        {
            long long temp=n/3;
            printf("%lld\n",temp*temp*temp);
        }
        else if(n%4==0)
        {
            long long temp=n/4;
            printf("%lld\n",temp*temp*2*temp);
        }
        else
            printf("-1\n");
    }
    return 0;
}