Codeforces Gym 100002 B Bricks 枚举角度

Problem B Bricks"

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100002

Description

The prisoner of the "IF" castle has decided to run away by disassembling the brick wall in his prison cell. To hide his work from his jailors he shall get rid of the bricks that he removes from the wall. All bricks have a shape of rectangular parallelepiped with the size of A × B × C inches and are so strong that they are impossible to break. However, there's a small rectangular sewer hole in the cell's floor with the size of D × E inches that goes deep down as a rectangular well of the same size (so deep it is, that its depth could not be measured and can be neglected). The prisoner have precisely (up to a tenth of an inch!) measured all the sizes A, B, C, D, E and wants to know if it is possible to dispose of the castle's bricks through the hole in the floor. Please, answer this question for him.

Input

The input file consists of a single line with 5 numbers A, B, C, D, and E separated by spaces. A, B, C are the lengths of brick's sides, and D, E are the lengths of hole's sides. All lengths are at least 1 and at most 10 inches and have at most 1 digit after decimal point.

Output

Write to the output file a single word YES if it is possible to dispose of the bricks through the hole or NO otherwise.

Sample Input

1.0 2.0 1.5 1.4 1.0

Sample Output

HINT

题意

给你一个无限深矩形的洞和一个立方体，问你这个立方体能否放入这个洞里面

题解：

注意可以斜着放

那就枚举角度往里面塞就行了

代码：

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <bitset>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 110000

#define mod 10007

#define eps 1e-9

#define pi 3.1415926

int Num;

//const int inf=0x7fffffff;   //§ß§é§à§é¨f§³

const ll Inf=0x3f3f3f3f3f3f3f3fll;

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//**************************************************************************************

double a[],d,e;

int flag=;

void solve(int x,int y)

{

    for(double i=;i<=;i+=0.001)

    {

        double ttt = i*pi/;

        double l = a[x]*cos(ttt)+a[y]*sin(ttt);

        double k = a[x]*sin(ttt)+a[y]*cos(ttt);

        if(l<=d&&k<=e)

        {

            cout<<"YES"<<endl;

            flag=;

            return;

        }

    }

}

int main()

{

    freopen("bricks.in","r",stdin);

    freopen("bricks.out","w",stdout);

    for(int i=;i<;i++)

        cin>>a[i];

    cin>>d>>e;

    for(int i=;i<;i++)

    {

        for(int j=;j<;j++)

        {

            if(i==j)

                continue;

            solve(i,j);

            if(flag)

                return ;

        }

    }

    printf("NO\n");

}