Codeforces round #266 (Div. 2) B (brute force enumeration)

Simple brute-force enumeration, but it took me so long, it can be seen that my basic skills are not solid enough.

If the multiplication of two numbers equals to a number of 6 * n, I can enumerate one of them, and enumerate to SQRT (6 * n, this is a very common nature of brute force solutions.

Find the smallest one in A and B, and start enumeration until the ascending order of SQRT (6 * n. In this way, there may be answers. The other items are repeated or certainly not the smallest area.

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<map>#include<set>#include<vector>#include<algorithm>#include<stack>#include<queue>#include<cctype>#include<sstream>using namespace std;#define INF 1000000000#define eps 1e-8#define pii pair<int,int>#define LL long long intLL a,b,n,s,a1,b1,t;int main(){    //freopen("in8.txt","r",stdin);    //freopen("out.txt","w",stdout);    scanf("%I64d%I64d%I64d",&n,&a,&b);    s=10123456789;    t=ceil(sqrt(6*n));    LL sum=6*n;    if(n*6<=a*b)        cout<<a*b<<endl<<a<<‘ ‘<<b<<endl;    else    {        bool c=0;        if(a>b)        {            c=1;            swap(a,b);        }        for(LL i=a;i<=t;i++)        {            LL t=max(b,sum/i+(sum%i!=0));            if(i*t<s)            {                a1=i,b1=t,s=i*t;            }        }        if(c)            cout<<s<<endl<<b1<<‘ ‘<<a1<<endl;        else            cout<<s<<endl<<a1<<‘ ‘<<b1<<endl;    }    //fclose(stdin);    //fclose(stdout);    return 0;}

 

Codeforces round #266 (Div. 2) B (brute force enumeration)