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