1.
Maximum Multiple
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3985 Accepted Submission(s): 926
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
3Sample Output
-1
-1
1题目大意：
已知n，n=x+y+z,且n可以整除x,y,z,问符合条件的xyz的最大乘积
想法：
”|“原来是整除符号的意思。数学推理啊，没细想，就交给队友了。听了直播题解，是只要推式子就好了。签到题。
思路：
1、n=x+y+z 1= x/n+y/n+z/n 1=1/(n/x)+1/(n/y)+1/(n/z) 因为n整除x,y,z，所以
1=1/a+1/b+1/c 且a,b,c为整数。问题就成了abc的最小乘积
2、从小到大枚举一下，有3种: 2 4 4 , 3 3 3, 2 3 6（什么？你说枚举的不够多？你再举个试试？）
3、244：32，333：27，236：36。对应的xyz就是，n*n*n/32,/27,/36，所以n整除4取第一种，整除3取第三种，整除6。。。呃，就抛弃了。
标准题解
.
AC代码：
// A
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
int n,_;
int main() {
for (scanf("%d",&_);_;_--) {
scanf("%d",&n);
if (n%3==0) printf("%lld\n",1ll*n*n*n/27); //1ll是把long long 的1,防计算溢出
else if (n%4==0) printf("%lld\n",1ll*n*n*n/32);
else puts("-1");
}
}
参考：