Secrete Master Plan HDU - 5540

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本文探讨了一个有趣的问题：诸葛亮给费张的秘密计划可能被错误地解读。通过比较原始计划与费张打开的计划，判断是否一致。文章提供了一个简单的C++实现方案，该方案检查两个2x2矩阵是否相同或者其中一个是否为另一个的转置。

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 ai0ai0 and ai1ai1 (1≤ai0,ai1≤1001≤ai0,ai1≤100). The first two lines stands for the original plan, the 3rd3rd and 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

AC代码（其实就是矩阵转置问题去讨论他会和转置后的哪一个相同）

Select Code

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int a[5], b[5];
bool check()
{
    if(a[0]==b[0]&&a[1]==b[1]&&a[2]==b[2]&&a[3]==b[3]) return 1;
    if(a[0]==b[1]&&a[1]==b[3]&&a[2]==b[0]&&a[3]==b[2]) return 1;
    if(a[0]==b[3]&&a[1]==b[2]&&a[2]==b[1]&&a[3]==b[0]) return 1;
    if(a[0]==b[2]&&a[1]==b[0]&&a[2]==b[3]&&a[3]==b[1]) return 1;
    return 0;
}
int main()
{

    int k, t, i;
    scanf("%d",&t);
    for(k = 1;k<=t;k++)
    {
        for(i = 0;i<4;i++) scanf("%d",&a[i]);
        for(i = 0;i<4;i++) scanf("%d",&b[i]);
        if(check()) printf("Case #%d: POSSIBLE\n",k);
        else printf("Case #%d: IMPOSSIBLE\n",k);
    }
    return 0;
}