hdoj Secrete Master Plan 5540 （矩阵翻转）

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Secrete Master Plan

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 121 Accepted Submission(s): 88

Problem 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

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

The first line of the input gives the number of test cases,

T(1≤T≤10

)

test cases follow. Each test case contains 4 lines. Each line contains two integers

and

(

1≤a

≤100

). The first two lines stands for the original plan, the

and

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

41 23 41 23 41 23 43 14 21 23 43 24 11 23 44 32 1

Sample Output

Case #1: POSSIBLECase #2: POSSIBLECase #3: IMPOSSIBLECase #4: POSSIBLE
因为是二阶的矩阵，所以它的所有情况可以枚举出来，例如
1   2       3    1      4   3      2   4
3   4、、、、4    2、、、 2   1、、  1   3
#include<stdio.h>#include<string.h>int a[4][4];int b[4][4];int main(){int t;int T=1;scanf("%d",&t);while(t--){scanf("%d%d%d%d",&a[1][1],&a[1][2],&a[2][1],&a[2][2]);scanf("%d%d%d%d",&b[1][1],&b[1][2],&b[2][1],&b[2][2]);int flag=0;printf("Case #%d: ",T++);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("POSSIBLE\n");elseprintf("IMPOSSIBLE\n");}return 0;}

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