Codeforce 287A - IQ Test (模拟)
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4 × 4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2 × 2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2 × 2 square, consisting of cells of the same color.
Four lines contain four characters each: the j-th character of the i-th line equals "." if the cell in the i-th row and the j-th column of the square is painted white, and "#", if the cell is black.
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
####
.#..
####
....
YES
####
....
####
....
NO
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.
题解:模拟,枚举看是否有一个2*2矩形有3个字符相等
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll gcd(ll a,ll b){
return b?gcd(b,a%b):a;
}
const int N=;
const int mod=1e9+;
char a[][];
bool flag=;
int main()
{
std::ios::sync_with_stdio(false);
for (int i=;i<=;i++) scanf("%s",a[i]+);
for (int i=;i<;i++){
for (int j=;j<;j++){
if(a[i][j]==a[i][j+]&&a[i][j]==a[i+][j]) flag=;
if(a[i][j]==a[i][j+]&&a[i][j]==a[i+][j+]) flag=;
if(a[i][j]==a[i+][j]&&a[i][j]==a[i+][j+]) flag=;
if(a[i+][j]==a[i][j+]&&a[i+][j]==a[i+][j+]) flag=;
}
}
if (flag) printf("YES\n");
else printf("NO\n");
return ;
}
Codeforce 287A - IQ Test (模拟)的更多相关文章
- Codeforce#354_B_Pyramid of Glasses(模拟)
题目连接:http://codeforces.com/contest/676/problem/B 题意:给你一个N层的杯子堆成的金字塔,倒k个杯子的酒,问倒完后有多少个杯子的酒是满的 题解:由于数据不 ...
- 用 IQ分布模拟图来测试浏览器的性能
今天天气太凉快,跟这个日历上属于夏天的那一页显得格格不入!就连我我床下那台废弃的ThinkPad,居然也十分透凉气,那外壳连我的体温高都没有,于是,我就开始想一个方法,让我那个废弃的电脑发热,顺便用它 ...
- Codeforce 25A - IQ test (唯一奇偶)
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of th ...
- Codeforces Round #176 (Div. 1 + Div. 2)
A. IQ Test 模拟. B. Pipeline 贪心. C. Lucky Permutation 每4个数构成一个循环. 当n为偶数时,n=4k有解:当n为奇数时,n=4k+1有解. D. Sh ...
- Codeforce 294A - Shaass and Oskols (模拟)
Shaass has decided to hunt some birds. There are n horizontal electricity wires aligned parallel to ...
- CodeForce 7 B - Memory Manager(模拟)
题目大意:给你一段内存,要你进行如下的三个操作. 1.分配内存 alloc X ,分配连续一段长度为X的内存. 如果内存不够应该输出NULL,如果内存够就给这段内存标记一个编号. 2.擦除编号为 ...
- CodeForce 439C Devu and Partitioning of the Array(模拟)
Devu and Partitioning of the Array time limit per test 1 second memory limit per test 256 megabytes ...
- Kilani and the Game-吉拉尼的游戏 CodeForce#1105d 模拟 搜索
题目链接:Kilani and the Game 题目原文 Kilani is playing a game with his friends. This game can be represente ...
- 格子游戏Grid game CodeForce#1104C 模拟
题目链接:Grid game 题目原文 You are given a 4x4 grid. You play a game — there is a sequence of tiles, each o ...
随机推荐
- mysql show prifile基本详解
show profile默认情况下,参数处于关闭状态,并保存最近15次的运行结果查看profile是否开启 show variables like '%profi%';开启profile记录功能 se ...
- what's the python之函数及装饰器
what's the 函数? 函数的定义:(return是返回值,可以没有,不过没有的话就返回了None) def wrapper(参数1,参数2,*args,默认参数,**kwargs): '''注 ...
- Jmeter原理
Jmeter结构体系及运行原理 Jmeter结构体系 把Jmeter的结构体系拆分为三维空间,如图: X1~X5:是负载模拟的一个过程,使用这些组件来完成负载的模拟: X1:选择协议,模拟用户请求 ...
- Dockerfile语法解析
Dockfile介绍 从上到下依次执行 每次执行一条指令就创建一个镜像层 第一条指令必须是FROM 表示需要构建的镜像是由哪个镜像为基础镜像 后续的指令运行于此基准镜像所提供的运行环境 可以 ...
- UDP网络通信
网络概念 一.目的 二.IP地址 三.端口 一.目的 目的 : 主要用于让两个用户端的服务器或者客户端,可以实现资源共享和信息传递 二.IP地址 1.作用 : 计算机网络中一台计算机的标识 2.种类 ...
- studio-3t-x64 下载地址
https://download.studio3t.com/studio-3t/windows/2018.4.6/studio-3t-x64.zip Studio 3T 破解教程1.创建文件studi ...
- [转]如何快速转载CSDN中的博客
原文:https://blog.csdn.net/bolu1234/article/details/51867099 前言 对于喜欢逛CSDN的人来说,看别人的博客确实能够对自己有不小的提高,有时 ...
- 【UML】-NO.40.UML.1.UML.1.001-【UML】- uml
1.0.0 Summary Tittle:[UML]-NO.40.UML.1.UML.1.001-[UML]- uml Style:DesignPattern Series:DesignPattern ...
- 阿里云RDS备份在本地mysql快速还原
本地准备: ##安装和RDS相同的mysql版本,拿mysql5.6为例 http://www.cnblogs.com/37yan/p/7513605.html ##安装Xtrabackup 包 cd ...
- linux curl命令如何上传本地文件夹和下载文件
本地有一个文件夹为my_dir,里面有四个文件,分别是test1.txt,user_account,tools_user,plans 要把这个my_dir文件夹传到ftp 192.168.8.251 ...