cf466B Wonder Room

B. Wonder Room

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The start of the new academic year brought about the problem of accommodation students into dormitories. One of such dormitories has aa × b square meter wonder room. The caretaker wants to accommodate exactly n students there. But the law says that there must be at least 6 square meters per student in a room (that is, the room for n students must have the area of at least 6n square meters). The caretaker can enlarge any (possibly both) side of the room by an arbitrary positive integer of meters. Help him change the room so as all nstudents could live in it and the total area of the room was as small as possible.

Input

The first line contains three space-separated integers n, a and b (1 ≤ n, a, b ≤ 109) — the number of students and the sizes of the room.

Output

Print three integers s, a1 and b1 (a ≤ a1; b ≤ b1) — the final area of the room and its sizes. If there are multiple optimal solutions, print any of them.

Sample test(s)

input

3 3 5

output

18
3 6

input

2 4 4

output

16
4 4

第二题题意是有n*m的方格，可以增加n、m的值，但不能减少。要求使得(n')*(m')>=6k，求n' * m' 的最小值，及此时的n' * m'

以为是很厉害的数论……当时比赛中没想出来……后来写个爆搜竟然A了……后悔莫及

就是从6k开始判断可行性，不行就一直+1+1……

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<ctime>
#define LL long long
using namespace std;
LL n,m,k;
bool rev;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int main()
{
	k=read();n=read();m=read();
	if (n>m)swap(n,m),rev=1;
	if (n*m>=6*k)
	{
		printf("%lld\n%lld %lld",n*m,n,m);
		return 0;
	}
	k*=6;
	while (1)
	{
		bool mrk=0;LL a=0;
		for (int i=n;i<=sqrt(k);i++)
		  if (k%i==0&&k/i>=m)
		  {
		  	mrk=1;
		  	a=i;
		  	break;
		  }
		if (mrk)
		{
			if (!rev)printf("%lld\n%lld %lld",k,a,k/a);
			else printf("%lld\n%lld %lld",k,k/a,a);
			return 0;
		}else k++;
	}
}