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
- vim常用指令整理小结
启动Vim后,默认是在 Normal 模式下,但是我们有时不知道是在编辑模式还是normal模式,按ESC键就可以返回normal模式.因为所有的命令都需要在Normal模式下使用,所以建议多按几下E ...
- 怎么修改TOMCAT的默认主页为你自己项目的主页
如果webapp下有一个abc的文件来下有一个index.html,想设置为首页怎么操作 方法: 修改tomcat/conf/web.xml文件.在web.xml文件中,有一段如下:<welco ...
- 报错解决——DateTimeField *** received a naive datetime (***) while time zone support is active
这是一个跟时区有关的问题,报错中说到datetime字段得到一个naive datetime,而不是支持time zone的active datetime由于Django的设置中米哦人USE_TZ设置 ...
- 组合覆盖与PICT的使用
组合覆盖法是一种有效减少测试用例个数的测试用例设计方法.根据覆盖程度的不同,可以分为单因素覆盖.成对组合覆盖.三三组合覆盖等.其中又以成对组合覆盖最常用. 关于组合覆盖的更多内容,参考:http:// ...
- 下载ez_setup
1.下载ez_setup链接:https://pypi.org/project/ez_setup/#files
- 并发编程---开启进程方式---查看进程pid
1.开启进程的两种方式 方式一: from multiprocessing import Process import time def task(name): print('%s is runnin ...
- c# ThreadPool 判断子线程全部执行完毕的四种方法
1.先来看看这个 多线程编程 多线程用于数据采集时,速度明显很快,下面是基本方法,把那个auto写成采集数据方法即可. using System; using System.Collections.G ...
- 外部盒模型大小固定 内部有边框div设置浮动时 缩放窗口内部div溢出的解决办法
原因分析: chorme和firefox浏览器下当缩放窗口大小时,边框的计算宽度变大造成内部div宽度的计算宽度变大,外部div放不下内部div而溢出. 解决办法: 给内部div设置 box-sizi ...
- Http post/get
什么是HTTP? 超文本传输协议(HyperText Transfer Protocol -- HTTP)是一个设计来使客户端和服务器顺利进行通讯的协议. HTTP在客户端和服务器之间以request ...
- 带上RESTful的金手铐,你累吗?
1. 首先RESTful是一套规范,不是框架,它是来约束你的.也不关心生产效率的提高.就好像使用汇编开发应用,性能是快了,但是生产效率很低.RESTful它需要你在路由上定义很多规则来解释的URL,假 ...