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=1000000007;
const int inf = 0x3f3f3f3f;
ll powmod(ll a,ll b) {ll res=1;a%=mod;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);
        else if(n%4 == 0) printf("%lld\n", 1ll * n * n * n / 32);
        else puts("-1");
    }
}