2015南阳CCPC A

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

The first line of the input gives the number of test cases, T(1≤T≤104). T test cases follow. Each test case contains 4 lines. Each line contains two integers ai0 and ai1 (1≤ai0,ai1≤100). The first two lines stands for the original plan, the 3rd and 4th 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 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[4][4];
int b[4][4];
int c[4][4];
int check()
{
    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])
        return 1;
    if(a[1][2]==b[1][1]&&a[2][2]==b[1][2]&&a[1][1]==b[2][1]&&a[2][1]==b[2][2])
        return 1;
    if(a[2][2]==b[1][1]&&a[2][1]==b[1][2]&&a[1][2]==b[2][1]&&a[1][1]==b[2][2])
        return 1;
    if(a[2][1]==b[1][1]&&a[1][1]==b[1][2]&&a[2][2]==b[2][1]&&a[1][2]==b[2][2])
        return 1;
    return 0;
}
int main()
{
    int t;scanf("%d",&t);
    for(int cas = 1;cas <= t;cas++)
    {
        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]);
        if(check())
            printf("Case #%d: POSSIBLE
",cas);
        else
            printf("Case #%d: IMPOSSIBLE
",cas);
    }
}