Secrete Master Plan
Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
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 correctdirection 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 giveFei 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 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
题目大意:给你两个2*2矩阵的矩阵,问第一个矩阵是否通过旋转的到第二个矩阵;
水题,直接贴代码;
# include <stdio.h>
# include <math.h>
#include<string.h>
#include<algorithm>
using namespace std;
void zhuanhuan(int a[2][2])
{
int t=a[0][0];
a[0][0] = a[1][0];
a[1][0] = a[1][1];
a[1][1] = a[0][1];
a[0][1] = t;
}
bool judge(int a[2][2],int mark[2][2])
{
for(int i=0; i<2; i++)
for(int j=0; j<2; j++)
if(a[i][j]!=mark[i][j])
return false;
return true;
}
int main()
{
int t;
scanf("%d",&t);
for(int c=1; c<=t; c++)
{
int mp[2][2],mark[2][2];
for(int i=0; i<2; i++)
for(int j=0; j<2; j++)
scanf("%d",&mp[i][j]);
for(int i=0; i<2; i++)
for(int j=0; j<2; j++)
scanf("%d",&mark[i][j]);
bool in = false;
for(int i=0; i<4; i++)
{
if(judge(mp,mark))
{
printf("Case #%d: POSSIBLE\n",c);
in = true;
break;
}
zhuanhuan(mp);
}
if(in == false) printf("Case #%d: IMPOSSIBLE\n",c);
}
return 0;
}