A - Secrete Master Plan

最新推荐文章于 2022-08-06 19:26:50 发布

原创最新推荐文章于 2022-08-06 19:26:50 发布 · 406 阅读

0 ·

CC 4.0 BY-SA版权

简单题专栏收录该内容

36 篇文章

订阅专栏

本文介绍了一个简单的算法问题：如何通过旋转和平移操作来判断两个2x2矩阵是否相同。问题来源于一个具体的编程挑战，需要验证两个矩阵是否可以通过某种方式转换为彼此。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 2×22×2 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

The first line of the input gives the number of test cases, T(1≤T≤104)T(1≤T≤104). TT test cases follow. Each test case contains 4 lines. Each line contains two integers ai0ai0and ai1ai1 (1≤ai0,ai1≤1001≤ai0,ai1≤100). The first two lines stands for the original plan, the 3rd3rdand 4th4th 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 yy is either "POSSIBLE" or "IMPOSSIBLE" (quotes for clarity).

Sample Input

Sample Output

Case #1: POSSIBLE
Case #2: POSSIBLE
Case #3: IMPOSSIBLE
Case #4: POSSIBLE

首先这道题的题意就是要我们输入两个2*2的矩阵，通过变换看这两个矩阵是否相等。这个题没啥技巧，
就是四个数字去对比。我当时做的时候还想有什么技巧，想了半天没想出来，最后还是老老实实的用4个if语句。




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

int main()
{
    int T,counts=1;
    scanf("%d",&T);
    int a[4][4],b[4][4];
    while(T--)
    {
      for(int i=1;i<=2;i++)
        for(int j=1;j<=2;j++)
       scanf("%d",&a[i][j]);
      for(int i=1;i<=2;i++)
        for(int j=1;j<=2;j++)
       scanf("%d",&b[i][j]);
       int flag=0;
        if(a[1][1]==b[1][1]&&a[1][2]==b[1][2]&&a[2][1]==b[2][1]&&a[2][2]==b[2][2])
            flag=1;
         else if(a[1][1]==b[2][1]&&a[1][2]==b[1][1]&&a[2][1]==b[2][2]&&a[2][2]==b[1][2])
            flag=1;
         else if(a[1][1]==b[2][2]&&a[1][2]==b[2][1]&&a[2][1]==b[1][2]&&a[2][2]==b[1][1])
            flag=1;
         else if(a[1][1]==b[1][2]&&a[1][2]==b[2][2]&&a[2][1]==b[1][1]&&a[2][2]==b[2][1])
            flag=1;
      if(flag)
      {
           printf("Case #%d: POSSIBLE\n",counts++);
     
      }

      else
      {
           printf("Case #%d: IMPOSSIBLE\n",counts++);
          
      }

    }

    return 0;
}