HDU5540-Secrete Master Plan

Secrete Master Plan

                                                                    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
                                                                                                Total Submission(s): 1139    Accepted Submission(s): 680

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≤104). T test cases follow. Each test case contains 4 lines. Each line contains two integers ai0and 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

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

 

Source

The 2015 China Collegiate Programming Contest

 

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wange2014

题意：给你两个2 * 2的矩形，每个格子代表一个数。问你这两个矩形是否可以通过一个做旋转得到另外一个

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f3f;int a[10], b[10];bool slove(){for (int i = 0; i<4; i++){bool flag = 1;for (int j = 0; j<4; j++)if (a[(i + j) % 4] != b[j]) flag = 0;if (flag) return 1; }return 0;}int main(){int t, cas = 1;scanf("%d", &t);while (t--){scanf("%d%d%d%d", &a[0], &a[1],& a[3], &a[2]);scanf("%d%d%d%d", &b[0],& b[1],&b[3],&b[2]);printf("Case #%d: %s\n", cas++, slove() ? "POSSIBLE" : "IMPOSSIBLE");}return 0;}