HDU6298 Maximum Multiple （多校第一场1001）

Maximum Multiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3241    Accepted Submission(s): 1344

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

题目大意：n = x+ y + z, 需要满足  x|n,  y|n,   z|n (整除关系)，输出xyz的最大值，若不存在，输出-1

设 n/x = r     n/y = s      n/z = t  ，    r  <=  s  <=  t

所以1/r + 1/s + 1/t = 1

所以r <= 3

当r=3    s,t = (3,3)

当r=2    s,t = (4, 4) , (3, 6) 

所以三种情况

n/3   n/3   n/3

n/2   n/4   n/4

n/2   n/3   n/6  (没有第一种大，舍去)

所以就判是否能 整除3 或 4

 #include <iostream>
 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <algorithm>
 #include <sstream>
 #include <stack>
 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())
 #define FO freopen("in.txt", "r", stdin)
 #define lowbit(x) (x&-x)
 #define mem(a,b) memset(a, b, sizeof(a))
 typedef vector<int> VI;
 typedef long long ll;
 typedef pair<int,int> PII;
 const ll mod=;
 const int inf = 0x3f3f3f3f;
 ll powmod(ll a,ll b) {ll res=;a%=mod;for(;b;b>>=){if(b&)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% == ) printf("%lld\n", 1ll * n * n * n / );
         else if(n% == ) printf("%lld\n", 1ll * n * n * n / );
         else puts("-1");
     }
 }