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.
Write to the output file a single word YES if it is possible to dispose of the bricks through the hole or NO otherwise.
//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=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;
}
//**************************************************************************************
double a[3],d,e;
int flag=0;
void solve(int x,int y)
{
for(double i=0;i<=90;i+=0.001)
{
double ttt = i*pi/180;
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=1;
return;
}
}
}
int main()
{
freopen("bricks.in","r",stdin);
freopen("bricks.out","w",stdout);
for(int i=0;i<3;i++)
cin>>a[i];
cin>>d>>e;
for(int i=0;i<3;i++)
{
for(int j=0;j<3;j++)
{
if(i==j)
continue;
solve(i,j);
if(flag)
return 0;
}
}
printf("NO\n");
}
