HDOJ Secrete Master Plan (求矩阵翻转是否和给定矩阵相同)

Secrete Master Plan

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 28    Accepted Submission(s): 18

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 integersai0 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

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题意：两个2*2的矩阵，判断是否前者可以通过翻转和后者相同思路：因为是2阶矩阵，那直接写就行了ac代码：
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 100100#define MOD 1000000007#define LL long long#define INF 0xfffffffusing namespace std;int map[5][5];int num[5][5];int main(){    int t,n;    int cas=0;    scanf("%d",&t);    while(t--)    {    int bz=0;    scanf("%d%d%d%d",&map[1][1],&map[1][2],&map[2][1],&map[2][2]);    scanf("%d%d%d%d",&num[1][1],&num[1][2],&num[2][1],&num[2][2]);if(num[1][1]==map[1][1]&&num[1][2]==map[1][2]&&num[2][1]==map[2][1]&&num[2][2]==map[2][2])        bz=1;        else if(num[1][1]==map[2][1]&&num[1][2]==map[1][1]&&num[2][1]==map[2][2]&&num[2][2]==map[1][2])        bz=1;        else if(num[1][1]==map[2][2]&&num[1][2]==map[2][1]&&num[2][1]==map[1][2]&&num[2][2]==map[1][1])        bz=1;        else if(num[1][1]==map[1][2]&&num[1][2]==map[2][2]&&num[2][1]==map[1][1]&&num[2][2]==map[2][1])        bz=1;        if(bz)        printf("Case #%d: POSSIBLE\n",++cas);        else        printf("Case #%d: IMPOSSIBLE\n",++cas);}    return 0;}