HDU 多校对抗赛 A Maximum Multiple

Maximum Multiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 494    Accepted Submission(s): 215

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

 

题意：

找到三个正整数x，y和z，使得：n = x + y + z，x /n，y / n，z / n和xyz是最大的。

 题解：

x+y+z=n

设r=n/x,s=n/y,t=n/z

所以 1/r+1/s+1/t=1;

设 r<=s<=t;

所以r<=3

r=2     1/s+1/t=1/2;   s=t=4  | s=3 ,t=6

r=3      s=t=3; 

如果3/n的话 就分成 n/3，n/3，n/3；

n/2，n/4，n/4；

n/2，n/3，n/6； //这种的乘积小于其余的，舍弃

n可以被3整除的话就分成n/3，n/3，n/3的情况

n可以被4整除的情况就分成n/2,n/4,n/4的情况

 

代码如下

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
int main(){
  int T;
  scanf("%d",&T);
  while(T--){
     ll n;
     scanf("%lld",&n);
     if(n%3==0){
       ll x=n/3;
       ll ans=x*x*x;
       printf("%lld
",ans);
     }
     else if (n%4==0){
       ll x=n/2;
       ll ans=x*n/4*n/4;
       printf("%lld
",ans);
     }else{
       printf("-1
");
     }
  }
  return 0;
}