UESTC 1215 (思维题 旋转)

Secrete Master Plan

Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)

Submit Status

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 2×22×2 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, TT(1≤T≤1041≤T≤104). TT test cases follow. Each test case contains 44 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 xx is the test case number (starting from 11) and yy is either "

POSSIBLE

" or "

IMPOSSIBLE

" (quotes for clarity).

Sample input and output

Sample Input	Sample Output
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	Case #1: POSSIBLE Case #2: POSSIBLE Case #3: IMPOSSIBLE Case #4: POSSIBLE

Source

The 2015 China Collegiate Programming Contest

#include <iostream>
using namespace std;
int main()
{
int t,a,b,c,d,x,y,z,w,tmp,k=1;
cin>>t;
while(t--)
{
int flag=0;
cin>>a>>b>>d>>c;
cin>>x>>y>>w>>z;
while (a!=x||b!=y||d!=w||c!=z)
{
tmp=a;
a=b;
b=c;
c=d;
d=tmp;
flag++;
if (flag==4)
break;
}
if (flag<4)
{
cout<<"Case #"<<k++<<": POSSIBLE"<<endl;
}
else
cout<<"Case #"<<k++<<": IMPOSSIBLE"<<endl;

}
return 0;
}