Equipment Box

题目链接:

http://acm.hust.edu.cn/vjudge/contest/130510#problem/B

Description


There is a large room in the Pyramid called Room-of-No-Return. Its floor is covered by rectangular tiles of equal size. The name of the room was chosen because of the very high number of traps and mechanisms in it. The ACM group has spent several years studying the secret plan of this room. It has made a clever plan to avoid all the traps. A specially trained mechanic was sent to deactivate the most feared trap called Shattered Bones. After deactivating the trap the mechanic had to escape from the room. It is very important to step on the center of the tiles only; he must not touch the edges. One wrong step and a large rock falls from the ceiling squashing the mechanic like a pancake. After deactivating the trap, he realized a horrible thing: the ACM plan did not take his equipment box into consideration. The box must be laid onto the ground because the mechanic must have both hands free to prevent contact with other traps. But when the box is laid on the ground, it could touch the line separating the tiles. And this is the main problem you are to solve.

Input


The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case consists of a single line. The line contains exactly four integer numbers separated by spaces: A, B, X and Y. A and Bindicate the dimensions of the tiles, X and Y are the dimensions of the equipment box (1

Output


Your task is to determine whether it is possible to put the box on a single tile -- that is, if the whole box fits on a single tile without touching its border. If so, you are to print one line with the sentence "Escape is possible.". Otherwise print the sentence "Box cannot be dropped.".

Sample Input


```
2
10 10 8 8
8 8 10 10
```

Sample Output


```
Escape is possible.
Box cannot be dropped.
```

Source


2016-HUST-线下组队赛-5


##题意:

判断一个小矩形能否放入一个大矩形.


##题解:

横着和竖着的情况很容易判断.
关键在于斜着放的情况:可以直接暴力枚举小矩形的旋转角.
数据量比较大,一开始用0.01的差值T掉了.


##代码:
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define maxn 101000
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
#define mod 1000000007
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

int a,b,A,B;

bool solve() {

for(double arg=0; arg<=90.0; arg+=0.1) {

double p = argpi / 180.0;

double t1 = b
sin(p) + acos(p);

double t2 = a
sin(p) + b*cos(p);

if(t1 < A && t2 < B) return 1;

}

return 0;

}

int main()

{

//IN;

int T;
cin >> T;
while (T--){
scanf("%d %d %d %d",&A,&B,&a,&b);
if(A > B) swap(A,B);
if(a > b) swap(a,b); bool flag = 0;
if(a < A && b < B) flag = 1; if(solve()) flag = 1; if(flag) printf("Escape is possible.\n");
else printf("Box cannot be dropped.\n");
} return 0;

}

POJ 1380 Equipment Box (暴力枚举)的更多相关文章

  1. poj 1753 Flip Game(暴力枚举)

    Flip Game   Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 52279   Accepted: 22018 Des ...

  2. POJ 3522 Slim Span 暴力枚举 + 并查集

    http://poj.org/problem?id=3522 一开始做这个题的时候,以为复杂度最多是O(m)左右,然后一直不会.最后居然用了一个近似O(m^2)的62ms过了. 一开始想到排序,然后扫 ...

  3. poj 2585 Window Pains 暴力枚举排列

    题意: 在4*4的格子中有9个窗体,窗体会覆盖它之下的窗体,问是否存在一个窗体放置的顺序使得最后的结果与输入同样. 分析: 在数据规模较小且不须要剪枝的情况下能够暴力(思路清晰代码简单),暴力一般分为 ...

  4. 字符串截取模板 && POJ 3450、3080 ( 暴力枚举子串 && KMP匹配 )

    //截取字符串 ch 的 st~en 这一段子串返回子串的首地址 //注意用完需要根据需要最后free()掉 char* substring(char* ch,int st,int en) { ; c ...

  5. poj 3080 Blue Jeans (暴力枚举子串+kmp)

    Description The Genographic Project is a research partnership between IBM and The National Geographi ...

  6. POJ - 1426 暴力枚举+同余模定理 [kuangbin带你飞]专题一

    完全想不到啊,同余模定理没学过啊,想起上学期期末考试我问好多同学'≡'这个符号什么意思,都说不知道,你们不是上了离散可的吗?不过看了别人的解法我现在会了,同余模定理介绍及运用点这里点击打开链接 简单说 ...

  7. POJ 3080 Blue Jeans (字符串处理暴力枚举)

    Blue Jeans  Time Limit: 1000MS        Memory Limit: 65536K Total Submissions: 21078        Accepted: ...

  8. POJ 2739 Sum of Consecutive Prime Numbers( *【素数存表】+暴力枚举 )

    Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19895 ...

  9. POJ 3174 暴力枚举

    思路: 暴力枚举三个点 判一判 搞定 (x1*y1=x2*y2) x1.y1.x2.y2为他们两两的差 //By SiriusRen #include <cstdio> using nam ...

随机推荐

  1. XSS注入常用语句(整理)

    <script>alert('hello,gaga!');</script> //经典语句,哈哈! >"'><img src="javas ...

  2. python UnicodeEncodeError: 'gbk' codec can't encode character ...

    使用Python写文件的时候,或者将网络数据流写入到本地文件的时候,大部分情况下会遇到:UnicodeEncodeError: 'gbk' codec can't encode character ' ...

  3. windows和linux安装rabbitmq

    一.windows安装rabbitmq 1.安装erlang 点击进入官网下载:http://erlang.org/download/ 2.安装rabbitmq 点击进入官网下载:http://www ...

  4. mac chromedriver error

    问题一:MAC 使用splinter error Traceback (most recent call last): from splinter.browser import Browser b = ...

  5. Laravel 学习笔记之文件上传

    自定义添加磁盘——upload 位置:config/filesystems.php 'disks' => [ 'local' => [ 'driver' => 'local', 'r ...

  6. python-xss攻击和单例模式

    1.python中单例模式 class Foo: instance = None def __new__(cls, *args, **kwargs): if not Foo.instance: Foo ...

  7. Django模板层2

    一.单表操作 1.1 开启test from django.test import TestCase import os # Create your tests here. if __name__ = ...

  8. oracle给用户赋dblink权限

    create database link 别名(可任意起) connect to 需要连接库的用户名identified by 需要连接库的用户名 using '(DESCRIPTION =(ADDR ...

  9. 十一、Boostrap-X-editable

    一.官网 http://vitalets.github.io/x-editable/index.html 二.实践 在jQuery中ajax配置项中的使用type与method的区别: type 和m ...

  10. 计蒜客 蓝桥模拟 G. 数列求值

    递归式移项得Ai+1 = 2Ai + 2Ci - Ai-1; 1.A2 = 2A1 + 2C1 - A0; 2.A3 = 2A2 + 2C2 - A1; . . . n.An+1 = 2An + 2C ...