2015南阳CCPC A - Secrete Master Plan 水题
D. Duff in Beach
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
无
Description
Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners of the city wall. Unfortunately, when Fei opened the pocket he found there are only four numbers written in dots on a piece of sheet. The numbers form a 2×2 matrix, but Fei didn't know the correct direction to hold the sheet. What a pity!
Given two secrete master plans. The first one is the master's original plan. The second one is the plan opened by Fei. As KongMing had many pockets to hand out, he might give Fei the wrong pocket. Determine if Fei receives the right pocket.
Input
Output
Case #x: y
, where x is the test case number (starting from 1) and y is either "POSSIBLE
" or "IMPOSSIBLE
" (quotes for clarity).Sample Input
4
1 2
3 4
1 2
3 4 1 2
3 4
3 1
4 2 1 2
3 4
3 2
4 1 1 2
3 4
4 3
2 1
Sample Output
Case #1: POSSIBLE Case #2: POSSIBLE Case #3: IMPOSSIBLE Case #4: POSSIBLE
HINT
题意
给你俩2*2的矩阵,问你能不能通过旋转从第一个得到第二个
题解:
暴力转圈圈就好了
代码:
#include<iostream>
#include<stdio.h>
using namespace std; int a[][];
int b[][];
int c[][];
int check()
{
if(a[][]==b[][]&&a[][]==b[][]&&a[][]==b[][]&&a[][]==b[][])
return ;
if(a[][]==b[][]&&a[][]==b[][]&&a[][]==b[][]&&a[][]==b[][])
return ;
if(a[][]==b[][]&&a[][]==b[][]&&a[][]==b[][]&&a[][]==b[][])
return ;
if(a[][]==b[][]&&a[][]==b[][]&&a[][]==b[][]&&a[][]==b[][])
return ;
return ;
}
int main()
{
int t;scanf("%d",&t);
for(int cas = ;cas <= t;cas++)
{
for(int i=;i<=;i++)
for(int j=;j<=;j++)
scanf("%d",&a[i][j]);
for(int i=;i<=;i++)
for(int j=;j<=;j++)
scanf("%d",&b[i][j]);
if(check())
printf("Case #%d: POSSIBLE\n",cas);
else
printf("Case #%d: IMPOSSIBLE\n",cas);
}
}
2015南阳CCPC A - Secrete Master Plan 水题的更多相关文章
- 2015南阳CCPC A - Secrete Master Plan A.
D. Duff in Beach Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a ...
- 2015南阳CCPC L - Huatuo's Medicine 水题
L - Huatuo's Medicine Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Huatuo was a famous ...
- The 2015 China Collegiate Programming Contest A. Secrete Master Plan hdu5540
Secrete Master Plan Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Othe ...
- hdu 5540 Secrete Master Plan(水)
Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed × matrix, but ...
- ACM Secrete Master Plan
Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. T ...
- HDU-5540 Secrete Master Plan
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission( ...
- 2015南阳CCPC E - Ba Gua Zhen 高斯消元 xor最大
Ba Gua Zhen Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description During the Three-Kingdom perio ...
- 2015南阳CCPC F - The Battle of Guandu 多源多汇最短路
The Battle of Guandu Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description In the year of 200, t ...
- 2015南阳CCPC H - Sudoku 暴力
H - Sudoku Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Yi Sima was one of the best cou ...
随机推荐
- 【转】./a.out 2>&1 > outfile
原文网址:http://www.cnblogs.com/zhaoyl/archive/2012/10/22/2733418.html APUE 3.5关于重定向有个容易迷惑人的问题: ./a.out ...
- 如何使用 orachk 工具
Oracle RAC 安装完毕后的健壮性是一个令人头疼的问题.之前Oracle为之专门推出了raccheck工具,确实方便了我们这些个苦逼的DBA.现在Oracle在raccheck的基础之上又推出了 ...
- 动态 SQL
MyBatis 的强大特性之一便是它的动态 SQL.如果你有使用 JDBC 或其他类似框架的经验,你就能体会到根据不同条件拼接 SQL 语句有多么痛苦.拼接的时候要确保不能忘了必要的空格,还要注意省掉 ...
- window下版本控制工具Git 客户端安装
安装使用 1.下载msysgit http://code.google.com/p/msysgit/ 2.下载tortoisegit客户端安装 http://code.google.com/p/tor ...
- How to enable DateTimePicker to use both date and time z
Recently in one of my project I needed to have an option to display the DateTimePicker allowing user ...
- 《Python 学习手册4th》 第十章 Python语句简介
''' 时间: 9月5日 - 9月30日 要求: 1. 书本内容总结归纳,整理在博客园笔记上传 2. 完成所有课后习题 注:“#” 后加的是备注内容 (每天看42页内容,可以保证月底看完此书) “重点 ...
- Ye.云狐J2刷机笔记 | 完美切换内部存储卡和SD卡的改法.vold.fstab
================================================================================Ye.完美切换内部存储卡和SD卡成功.v ...
- 直线相交 POJ 1269
// 直线相交 POJ 1269 // #include <bits/stdc++.h> #include <iostream> #include <cstdio> ...
- The solution to Force.Com IDE 29.0 PassWord Problem
我最近使用Force.com IDE 时,经常提示密码错误.从Google 中终于发现一个解决方法,分享给大家,以供大家参考. 在Force.com IDE 29.0中,IDE 存储我们开发Org ...
- J2EE到底是什么
目前所有的B/S系统应用可以分为:有状态(statefull)和无状态(stateless)两大类别. 有状态是指在整个系统的处理过程中要保留记住一些信息,而无状态则相反,每次request都是独立的 ...