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
- php类 静态变量赋值 static $name="abc"
<?php class test { static $name="abc"; } echo test::$name; 1.巴斯科范式 class_statement: var ...
- Nhibernate与Dapper对比,及Nhibernate增删改和9种查询语法
1,Sql语法. NH:HQL Dapper:原生Sql. 点评:原生Sql可以直接放在数据库里执行,Hql不行,且Hql增加学习负担.(Hn也可以原生Sql,但好像用的不多呀) 2,开发速度. NH ...
- POJ 3683 Priest John's Busiest Day (2-SAT,常规)
题意: 一些人要在同一天进行婚礼,但是牧师只有1个,每一对夫妻都有一个时间范围[s , e]可供牧师选择,且起码要m分钟才主持完毕,但是要么就在 s 就开始,要么就主持到刚好 e 结束.因为人数太多了 ...
- MATLAB函数表(转自:http://bbs.06climate.com/forum.php?mod=viewthread&tid=16041&extra=page%3D4)
MATLAB函数表 4.1.1特殊变量与常数 ans 计算结果的变量名 computer 确定运行的计算机 eps 浮点相对精度 Inf 无穷大 I 虚数单位 inputname 输入参数名 NaN ...
- 多线程程序设计学习(8)Thread-Per-Message
Thread-Per-Message[这个工作交给你模式]一:Thread-Per-Message的参与者--->Client(委托人)--->host(中介开线程)--->hepl ...
- WCF配置文件详解(一)
<?xml version="1.0" encoding="utf-8" ?><configuration> <!-- &l ...
- u-boot向linux内核传递启动参数(详细)
U-BOOT 在启动内核时,会向内核传递一些参数.BootLoader 可以通过两种方法传递参数给内核,一种是旧的参数结构方式(parameter_struct),主要是 2.6 之前的内核使用的方式 ...
- mac 修改xcode的版本
http://blog.csdn.net/yangzhenping/article/details/50266245
- mssql server 2005还原数据库bak文件与“备份集中的数据库备份与现有的xx数据库不同”解决方法
mssql server 2005还原数据库bak文件,网站使用虚拟主机建站会经常遇到,一般情况下,主机商有在线的管理程序,但有时候没有的话,就需要本地还原备份sql数据库了.这种情况mssql se ...
- activemq 的小实验
package ch02.chat; import javax.jms.Connection; import javax.jms.ConnectionFactory; import javax.jms ...