Codeforces Round #322 (Div. 2) D. Three Logos 暴力
D. Three Logos
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/581/problem/D
Description
Three companies decided to order a billboard with pictures of their logos. A billboard is a big square board. A logo of each company is a rectangle of a non-zero area.
Advertisers will put up the ad only if it is possible to place all three logos on the billboard so that they do not overlap and the billboard has no empty space left. When you put a logo on the billboard, you should rotate it so that the sides were parallel to the sides of the billboard.
Your task is to determine if it is possible to put the logos of all the three companies on some square billboard without breaking any of the described rules.
Input
The first line of the input contains six positive integers x1, y1, x2, y2, x3, y3 (1 ≤ x1, y1, x2, y2, x3, y3 ≤ 100), where xi and yi determine the length and width of the logo of the i-th company respectively
Output
If it is impossible to place all the three logos on a square shield, print a single integer "-1" (without the quotes).
If it is possible, print in the first line the length of a side of square n, where you can place all the three logos. Each of the next n lines should contain n uppercase English letters "A", "B" or "C". The sets of the same letters should form solid rectangles, provided that:
- the sizes of the rectangle composed from letters "A" should be equal to the sizes of the logo of the first company,
- the sizes of the rectangle composed from letters "B" should be equal to the sizes of the logo of the second company,
- the sizes of the rectangle composed from letters "C" should be equal to the sizes of the logo of the third company,
Note that the logos of the companies can be rotated for printing on the billboard. The billboard mustn't have any empty space. If a square billboard can be filled with the logos in multiple ways, you are allowed to print any of them.
See the samples to better understand the statement.
Sample Input
5 1 2 5 5 2
Sample Output
5
AAAAA
BBBBB
BBBBB
CCCCC
CCCCC
HINT
题意
给你三个矩形,问你是否能拼成一个正方形
题解:
啊,能拼的就题目给你的样例的两种方式
那我们就都去尝试咯~
直接暴力枚举就好了,总共就2^3*6*2种搭配,都试试就好了……
@)1%KBO0HM418$J94$1R.jpg)
代码:
//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#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 1205000
#define mod 1000000007
#define eps 1e-9
#define e exp(1.0)
#define PI acos(-1)
#define lowbit(x) (x)&(-x)
const double EP = 1E- ;
int Num;
//const int inf=0x7fffffff;
const ll inf=;
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;
}
//************************************************************************************* int a[][];
int check(int l1,int r1,int l2,int r2,int l3,int r3,int A,int B,int C)
{
if(l1==l2&&l2==l3&&(r1+r2+r3)==l1)
{
cout<<l1<<endl;
for(int i=;i<=l1;i++)
for(int j=;j<=r1;j++)
a[i][j]=A;
for(int i=;i<=l1;i++)
for(int j=r1+;j<=r1+r2;j++)
a[i][j]=B;
for(int i=;i<=l1;i++)
for(int j=r1+r2+;j<=r1+r2+r3;j++)
a[i][j]=C;
for(int i=;i<=l1;i++)
{
for(int j=;j<=l2;j++)
{
if(a[i][j]==)cout<<"A";
else if(a[i][j]==)cout<<"B";
else cout<<"C";
}
cout<<endl;
}
return ;
}
if(l2+l3!=l1)return ;
if(r1+r2!=l1)return ;
if(r2!=r3)return ;
for(int i=;i<=l1;i++)
for(int j=;j<=r1;j++)
a[i][j]=A;
for(int i=;i<=l2;i++)
for(int j=r1+;j<=r1+r2;j++)
a[i][j]=B;
for(int i=;i<=l1;i++)
for(int j=;j<=l1;j++)
if(a[i][j]==)a[i][j]=C;
cout<<l1<<endl;
for(int i=;i<=l1;i++)
{
for(int j=;j<=l1;j++)
{
if(a[i][j]==)cout<<"A";
else if(a[i][j]==)cout<<"B";
else cout<<"C";
}
cout<<endl;
}
return ; }
int main()
{
int l1,r1,l2,r2,l3,r3;
int L1,R1,L2,R2,L3,R3;
L1=read(),R1=read(),L2=read(),R2=read(),L3=read(),R3=read();
//000 001 010 100 110 101 011 111
//123 132 213 231 312 321
//
l1=L1,r1=R1,l2=L2,r2=R2,l3=L3,r3=R3;
if(check(l1,r1,l2,r2,l3,r3,,,))return ;
if(check(l1,r1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,l2,r2,l3,r3,,,))return ;
if(check(r1,l1,r2,l2,l3,r3,,,))return ;
if(check(r1,l1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,r2,l2,r3,l3,,,))return ;
//
l1=L1,r1=R1,l2=L3,r2=R3,l3=L2,r3=R2;
if(check(l1,r1,l2,r2,l3,r3,,,))return ;
if(check(l1,r1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,l2,r2,l3,r3,,,))return ;
if(check(r1,l1,r2,l2,l3,r3,,,))return ;
if(check(r1,l1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,r2,l2,r3,l3,,,))return ;
//
l1=L2,r1=R2,l2=L1,r2=R1,l3=L3,r3=R3;
if(check(l1,r1,l2,r2,l3,r3,,,))return ;
if(check(l1,r1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,l2,r2,l3,r3,,,))return ;
if(check(r1,l1,r2,l2,l3,r3,,,))return ;
if(check(r1,l1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,r2,l2,r3,l3,,,))return ;
//
l1=L2,r1=R2,l2=L3,r2=R3,l3=L1,r3=R1;
if(check(l1,r1,l2,r2,l3,r3,,,))return ;
if(check(l1,r1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,l2,r2,l3,r3,,,))return ;
if(check(r1,l1,r2,l2,l3,r3,,,))return ;
if(check(r1,l1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,r2,l2,r3,l3,,,))return ;
//
l1=L3,r1=R3,l2=L1,r2=R1,l3=L2,r3=R2;
if(check(l1,r1,l2,r2,l3,r3,,,))return ;
if(check(l1,r1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,l2,r2,l3,r3,,,))return ;
if(check(r1,l1,r2,l2,l3,r3,,,))return ;
if(check(r1,l1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,r2,l2,r3,l3,,,))return ; //
l1=L3,r1=R3,l2=L2,r2=R2,l3=L1,r3=R1;
if(check(l1,r1,l2,r2,l3,r3,,,))return ;
if(check(l1,r1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,l2,r2,l3,r3,,,))return ;
if(check(r1,l1,r2,l2,l3,r3,,,))return ;
if(check(r1,l1,l2,r2,r3,l3,,,))return ;
if(check(l1,r1,r2,l2,r3,l3,,,))return ;
if(check(r1,l1,r2,l2,r3,l3,,,))return ;
cout<<"-1"<<endl;
return ;
}
Codeforces Round #322 (Div. 2) D. Three Logos 暴力的更多相关文章
- Codeforces Round #322 (Div. 2) D. Three Logos 模拟
D. Three Logos Three companies decided to order a ...
- Codeforces Round #322 (Div. 2) C. Developing Skills 优先队列
C. Developing Skills Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/p ...
- Codeforces Round #322 (Div. 2) B. Luxurious Houses 水题
B. Luxurious Houses Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/pr ...
- Codeforces Round #322 (Div. 2) A. Vasya the Hipster 水题
A. Vasya the Hipster Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/p ...
- Codeforces Round #322 (Div. 2)
水 A - Vasya the Hipster /************************************************ * Author :Running_Time * C ...
- Codeforces Round #322 (Div. 2) —— F. Zublicanes and Mumocrates
It's election time in Berland. The favorites are of course parties of zublicanes and mumocrates. The ...
- Codeforces Round #322 (Div. 2) E F
E. Kojiro and Furrari 题意说的是 在一条直线上 有n个加油站, 每加一单位体积的汽油 可以走1km 然后每个加油站只有一种类型的汽油,汽油的种类有3种 求从起点出发到达终点要求使 ...
- 树形dp - Codeforces Round #322 (Div. 2) F Zublicanes and Mumocrates
Zublicanes and Mumocrates Problem's Link Mean: 给定一个无向图,需要把这个图分成两部分,使得两部分中边数为1的结点数量相等,最少需要去掉多少条边. ana ...
- Round #322 (Div. 2) 581D Three Logos (模拟)
先枚举两个矩形,每个矩形横着放或竖着放,把一边拼起来, 如果不是拼起来有缺口就尝试用第三个矩形去补. 如果没有缺口就横着竖着枚举一下第三个矩形和合并的矩形x或y拼接. #include<bits ...
随机推荐
- CodeWars可以学习的
http://www.codewars.com/kata/54ff3102c1bad923760001f3/solutions/csharp 判断给定的字符串有多少个a e i o u using S ...
- java之内部类与匿名内部类
Java 内部类 分四种:成员内部类.局部内部类.静态内部类和匿名内部类. 1.成员内部类: 即作为外部类的一个成员存在,与外部类的属性.方法并列. 注意:成员内部类中不能定义静态变量,但可以访问外部 ...
- sharedevelop iis express
sharedevelop 的IIS express的配置文件在 %userprofile%\documents\IISExpress\config\applicationhost.config 自动会 ...
- [swustoj 404] 最小代价树
最小代价树(0404) 问题描述 以下方法称为最小代价的字母树:给定一正整数序列,例如:4,1,2,3,在不改变数的位置的条件下把它们相加,并且用括号来标记每一次加法所得到的和. 例如:((4+1)+ ...
- 25、BroadCastRecevier
BroadCastRecevier 有两种注册方式 1. 清单文件里注册: 一旦应用程序被部署到手机, 广播接受者就会生效 2. 代码里面注册: 一旦代码所在的进程被杀死了, 广播接受者就失效了. 广 ...
- 卸载sql server 2012
好不容易装上了sql server 2012数据库,可是却不能连接本地的数据库,后来发现缺少一些服务,于是决定重新安装,但是卸载却很麻烦,如果卸载不干净的话,重新安装会出问题,所以下面就总结一些方法: ...
- Velocity - 单例还是非单例
在Velocity1.2版本以后,开发者现在又两种选择来使用Velocity引擎,单例模型(singleton model)和单独实例模型(separate instance model).这是相同的 ...
- Channel 详解
java.nio.channels.FileChannel封装了一个文件通道和一个FileChannel对象,这个FileChannel对象提供了读写文件的连接. 1.接口 2.通道操作 a.所有通道 ...
- C 语言控制台实现五子棋项目
花了一天时间实现了控制台五子棋项目,把项目贴上来.也算是告一段落了. 为了进一步了解C语言编程,熟悉优秀的编码风格,提升编码能力,丰富项目经验.所以在编程初期选择了控制台小游戏<单机五子棋> ...
- Linux与Windows的8个不同
(整理来自网络) 对刚刚接触Linux的人来说,很容易从windows的观念去理解Linux系统.今天扒一扒Win和Linux之间常见的8个区别. 一.Linux终端输入密码不回显字符 用户的密码在L ...