A - Secrete Master Plan【ccpc】

A - Secrete Master Plan

Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%lld & %llu

Submit Status

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

判断一个矩阵旋转后能否与另一个矩阵重合

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>int map1[5][5];int map2[5][5];using namespace std;int main(){int t,i,j,k=1;scanf("%d",&t);while(t--){scanf("%d%d%d%d%d%d%d%d",&map1[1][1],&map1[1][2],&map1[2][1],&map1[2][2],&map2[1][1],&map2[1][2],&map2[2][1],&map2[2][2]);int flag=0;if(map1[1][1]==map2[1][1]&&map1[1][2]==map2[1][2]&&map1[2][1]==map2[2][1]&&map1[2][2]==map2[2][2])flag=1;else if(map1[1][1]==map2[1][2]&&map1[1][2]==map2[2][2]&&map1[2][1]==map2[1][1]&&map1[2][2]==map2[2][1])flag=1;else if(map1[1][1]==map2[2][2]&&map1[1][2]==map2[2][1]&&map1[2][1]==map2[1][2]&&map1[2][2]==map2[1][1])flag=1;else if(map1[1][1]==map2[2][1]&&map1[1][2]==map2[1][1]&&map1[2][1]==map2[2][2]&&map1[2][2]==map2[1][2])flag=1;if(flag==1){printf("Case #%d: POSSIBLE\n",k++);}else{printf("Case #%d: IMPOSSIBLE\n",k++);}}return 0;}