代码语言：javascript复制
Given an integer nn, Chiaki would like to find three positive integers xx, yy and zzsuch that: n=x y zn=x y z, x∣nx∣n, y∣ny∣n, z∣nz∣n and xyzxyz is maximum. Input
代码语言：javascript复制There are multiple test cases. The first line of input contains an integer TT (1≤T≤1061≤T≤106), indicating the number of test cases. For each test case:
The first line contains an integer nn (1≤n≤1061≤n≤106). Output
代码语言：javascript复制For each test case, output an integer denoting the maximum xyzxyz. If there no such integers, output −1−1 instead. Sample Input
代码语言：javascript复制3
1
2
3Sample Output
代码语言：javascript复制-1
-1
1 只有因子中有4或者有3才能被拆成 X Y Z=N,然后打了表验证。 最后wa了好几次，是因为int和int计算之后还是int就算赋值给long long . 打表代码
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
int n;
for (int n = 1; n <= 100; n )
{
int maxt = -1;
int a, b, c;
for (int x = 1; x <= n; x )
{
for (int y = 1; y <= n - x; y )
{
int z = n - x - y;
if (z && n % x == 0 && n % y == 0 && n % z == 0)
{
if (maxt < x * y * z)
{
a = x;
b = y;
c = z;
}
maxt = max(maxt, x * y * z);
}
}
}
printf("%d:] %d %d %dn", n, maxt, a, b, c);
if (n % 12 == 0)
printf("n");
}
}
return 0;
}AC
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
int main()
{
int T;
long long n,x,y,z;
long long sum;
scanf("%d", &T);
while (T--)
{
scanf("%lld", &n);
if ((n % 3) == 0)
{
x = y = z = n / 3;
sum = x * y * z;
if (x y z == n)
printf("%lldn", sum);
else
puts("-1");
}
else if ((n % 4) == 0)
{
x = y = n / 4, z = n / 2;
sum = x * y * z;
if (x y z == n)
printf("%lldn", sum);
else
puts("-1");
}
else
puts("-1");
}
}
