CCPC 2015 A题

最新推荐文章于 2023-11-12 12:05:35 发布

aiterator

最新推荐文章于 2023-11-12 12:05:35 发布

阅读量426

点赞数

CC 4.0 BY-SA版权

分类专栏：题解文章标签： CCPC 竞赛

本文链接：https://blog.youkuaiyun.com/TXT003/article/details/49701197

题解专栏收录该内容

25 篇文章

订阅专栏

本文介绍了一道关于判断两个2x2矩阵是否可以通过旋转达到相同状态的问题，并提供了一个简单的解决方案。通过四次旋转尝试来验证原始矩阵与目标矩阵是否一致。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

title

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

The 2015 China Collegiate Programming Contest

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;
}