1001. Maximum Multiple

1001. Maximum Multiple

• Time limit: 2 seconds
• Memory limit: 32 megabytes
  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≤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

题意：把给的一个n分解为三个数，这三个数必须分别被n整除，输出这三个书的最大乘积，若无法找出，则输出-1
思路：1 = （1/2+1/3+1/6 =） 1/4+1/4+1/2 = 1/3+1/3+1/3
也就是说，这个数一定是3或者4的倍数,需要注意的是，题目中给的范围是10^6三次方后会超出范围，所以用lld

so：

#include <cstdio>

int main() {
  int T;
  scanf("%d", &T);
  for (int cas = 1; cas <= T; ++cas) {
    int n;
    scanf("%d", &n);
    if (n % 3 == 0) printf("%lld\n", 1ll * n * n * n / 27);
    else if (n % 4 == 0) printf("%lld\n", 1ll * n * n * n / 32);
    else puts("-1");
  }
  return 0;
}