Codeforces round #266 (div.2) B wonder room -- Enumeration

A rectangle with the length of A and B must be added to a and B to make a * B> = 6 * n and a * B to the minimum.

Solution: If a and B are set to A1 and B1 and a <B, then a <= A1 <= Ceil (SQRT (6 * n )), then we can enumerate A1 and calculate B1. If B1 <B, then b1 = B and then calculate the area. The minimum value of all the areas is not enough. Then we can enumerate B1 again, the processing is the same as that.

Complexity O (SQRT (n ))

Code:

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#define lll __int64using namespace std;int main(){    lll n,a,b;    lll mini,a1,b1;    while(scanf("%I64d%I64d%I64d",&n,&a,&b)!=EOF)    {        mini = (1LL<<62);        lll sum = (lll)6*n;        for(lll i=a;i<=a+50000LL;i++)        {            lll tb = sum/i + (sum%i!=0);            tb = max(b,tb);            lll tarea = i*tb;            if(tarea < mini)            {                mini = tarea;                a1 = i;                b1 = tb;            }        }        for(lll i=b;i<=b+50000LL;i++)        {            lll ta = sum/i + (sum%i!=0);            ta = max(a,ta);            lll tarea = i*ta;            if(tarea < mini)            {                mini = tarea;                a1 = ta;                b1 = i;            }        }        cout<<mini<<endl<<a1<<" "<<b1<<endl;    }    return 0;}

View code

 

Codeforces round #266 (div.2) B wonder room -- Enumeration