CodeForces 466B Wonder Room

最新推荐文章于 2019-03-10 15:14:00 发布

Clearle

最新推荐文章于 2019-03-10 15:14:00 发布

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分类专栏：杂题

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本文介绍了一个关于学生宿舍分配的问题，需要在确保每名学生至少有6平方米空间的前提下，尽可能缩小房间面积。通过调整房间尺寸来优化空间使用，使得n名学生能够居住在限定面积内。

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The start of the new academic year brought about the problem of accommodation students into dormitories. One of such dormitories has a a × 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 alln students 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
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <bitset>

using namespace std;

#define PB push_back
#define MP make_pair
#define REP(i,n) for(int i=0;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define DWN(i,h,l) for(int i=(h);i>=(l);--i)
#define CLR(vis,pos) memset(vis,pos,sizeof(vis))
#define PI acos(-1.0)
#define INF 0x7FFFFFFF
#define LINF 1000000000000000000LL
#define eps 1e-8

typedef long long ll;

int main()
{
    ll n,a,b;
    cin>>n>>a>>b;

    n*=6;
    int flag=0;
    if(a*b>=n)
    {
        cout<<a*b<<endl;
        cout<<a<<" "<<b<<endl;
        return 0;
    }

    if(a>b)
        swap(a,b),flag=1;
    ll x;
    ll ansa=INF,ansb=INF;
    x=n/b;
    if(x*b<n) x++;
    ll ta,tb;
    FOR(i,a,x)
    {
        ta=i;
        tb=n/ta;
        if(ta*tb<n) tb++;

        if(ta>tb) break;
        //cout<<ansa*ansb<<" "<<ta*tb<<endl;
        ll sum1=ansa*ansb;
        ll sum2=ta*tb;
        if(ta>=a && tb>=b && ta*tb>=n && sum1>sum2)
        {
            ansa=ta,ansb=tb;
            //cout<<ta<<" "<<tb<<endl;
        }
    }

    if(flag && ansa<ansb)
        swap(ansa,ansb);

    cout<<ansa*ansb<<endl;

    cout<<ansa<<" "<<ansb<<endl;




    return 0;
}