HDU - 5540 Secrete Master Plan(水题)

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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 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

题目大意：判断第二个正方形能否由第一个旋转得到。

水题，加一个函数辅助一下。

附上AC代码：

#include<bits/stdc++.h>

using namespace std;
int a,b,c,d;
int a1,b1,c1,d1;
int t;

int judge(int a_,int b_,int c_,int d_)
{
    if(a_==a&&b_==b&&c_==c&&d_==d)
        return 1;
    return 0;
}

int main()
{
    int cases=0;
    cin>>t;
    while(t--)
    {
        cin>>a>>b>>d>>c;
        cin>>a1>>b1>>c1>>d1;
        int flag=0;
        if(judge(a1,b1,d1,c1)) flag=1;
        if(judge(b1,d1,c1,a1)) flag=1;
        if(judge(d1,c1,a1,b1)) flag=1;
        if(judge(c1,a1,b1,d1)) flag=1;
        if(flag)printf("Case #%d: POSSIBLE\n",++cases);
        else printf("Case #%d: IMPOSSIBLE\n",++cases);
    }
    return 0;
}