2015 CCPC Secrete Master Plan-优快云博客

本文链接：https://blog.youkuaiyun.com/wilver/article/details/50018879

本文介绍了一道简单的编程题目，任务是判断两个2x2矩阵是否可以通过旋转达到一致状态。通过C++代码实现了这一逻辑，并提供了样例输入输出。

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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 $2\times 2$ 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.

title

Input
The first line of the input gives the number of test cases, $T$ ( $1\leq T\leq 10^4$ ). $T$ test cases follow. Each test case contains $4$ lines. Each line contains two integers
$a_{i0}$ and $a_{i1}$ ( $1\leq a_{i0}, a_{i1}\leq 100$ ). The first two lines stands for the original plan, the $3_{rd}$ and $4_{th}$ line stands for the plan Fei opened.

Output
For each test case, output one line containing Case #x: y, where $x$ is the test case number (starting from $1$ ) and 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

题意：两个2*2的矩阵，看是否能够通过旋转得到；
简单题。

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int n;
    cin>>n;
    int a[4],b[4];
    for(int k=0; k<n; k++)
    {
        for(int i=0; i<4; i++)
            cin>>a[i];
        for(int i=0; i<4; i++)
            cin>>b[i];
        int flag=0;
        for(int i=0; i<4; i++)
        {
            int tmp;
            tmp=a[0];
            a[0]=a[1];
            a[1]=a[3];
            a[3]=a[2];
            a[2]=tmp;
            if(a[0]==b[0]&&a[1]==b[1]&&a[2]==b[2]&&a[3]==b[3])
            flag=1;
        }
        if(flag==1)
        printf("Case #%d: POSSIBLE\n",k+1);
        if(flag==0)
        printf("Case #%d: IMPOSSIBLE\n",k+1);
    }
    return 0;
}