Problem B. Beer Refrigerator

http://codeforces.com/gym/241680/problem/B
比赛的时候考虑的是，它们3个尽可能接近，然后好麻烦，不如暴力枚举，这里不需要质因数分解，而是两重循环枚举所有因数，第3个因数也就随之确定

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define ls rt<<1
#define rs rt<<1|1
#define lson ls,nl,mid,l,r
#define rson rs,mid+1,nr,l,r
#define N 100010
#define For(i,a,b) for(int i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()

using namespace std;
int x;
int a[1010];
int cnt;
int k;
int ans=inf;
int ANS[10];
void in(int &x){
    int y=1;
    char c=g();x=0;
    while(c<'0'||c>'9'){
        if(c=='-')y=-1;
        c=g();
    }
    while(c<='9'&&c>='0'){
        x=(x<<1)+(x<<3)+c-'0';c=g();
    }
    x*=y;
}
void o(int x){
    if(x<0){
        p('-');
        x=-x;
    }
    if(x>9)o(x/10);
    p(x%10+'0');
}
int main(){
    freopen("beer.in","r",stdin);
    freopen("beer.out","w",stdout);
    in(x);
    For(i,1,x)
        if(x%i==0)
            a[++cnt]=i;
    For(i,1,cnt)
        For(j,1,cnt){
            k=x/a[i]/a[j];
            if(x!=a[i]*a[j]*k)
                continue;
            if(ans>a[i]*a[j]+a[i]*k+a[j]*k){
                ans=a[i]*a[j]+a[i]*k+a[j]*k;
                ANS[1]=a[i];
                ANS[2]=a[j];
                ANS[3]=k;
            }
        }
    o(ANS[1]);p(' ');
    o(ANS[2]);p(' ');
    o(ANS[3]);
    return 0;
}