2015南阳CCPC A - Secrete Master Plan A.

D. Duff in Beach

 

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

 

题意

      给你两个2*2的矩阵，判断一个旋转是否能得到另外一个

题解：

    模拟不多说

///1085422276
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a))
#define TS printf("111111\n")
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define inf 100000
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//****************************************

int main()
{
    int oo=1;
    int a[6],b[5];
     int T=read();
     while(T--){
        for(int i=1;i<=4;i++){

            scanf("%d",&b[i]);

        }
        for(int i=1;i<=4;i++){

            scanf("%d",&a[i]);

        }
        bool flag=0;
        if(b[1]==a[1]&&b[2]==a[2]&&b[3]==a[3]&&b[4]==a[4]){
           flag=1;
        }
        if(a[3]==b[1]&&a[1]==b[2]&&a[2]==b[4]&&a[4]==b[3]){
            flag=1;
        }
         if(a[4]==b[1]&&a[3]==b[2]&&a[1]==b[4]&&a[2]==b[3]){flag=1;}
         if(a[2]==b[1]&&a[4]==b[2]&&a[3]==b[4]&&a[1]==b[3]){flag=1;}
        printf("Case #%d: ",oo++);
        if(flag){
            cout<<"POSSIBLE"<<endl;
        }
        else {
            cout<<"IMPOSSIBLE"<<endl;
        }
     }
    return 0;
}