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
思路：这个题我刚开始想麻烦了，本来想通过公式变换凑出来找答案，其实是想麻烦了，就是分情况讨论被3和被4整除的情况分类讨论一下。
首先这个数
1,n/3,n/3,n/3可以
2,n/2,n/4,n/4可以
3,n/2,n/3,n/6不可以，可被1代替变得更优
....
讨论到最后可以发现如果能满足3及后面的，那么前两条必定更优。
int main()
{
ll int n, t;
while(scanf("%lld",&t)!=EOF)
{
while(t--)
{
scanf("%lld",&n);
if(n%3==0)
{
n = n/3;
printf("%lld\n",n*n*n);
}
else if(n%4==0) printf("%lld\n",n*n*n/32);
else printf("-1\n");
}
}
return 0;
}