Codeforces 51A Cheaterius's Problem

最新推荐文章于 2024-09-24 23:09:19 发布

GAO1O24

最新推荐文章于 2024-09-24 23:09:19 发布

阅读量871

点赞数 1

CC 4.0 BY-SA版权

分类专栏：水题

本文链接：https://blog.youkuaiyun.com/cainiaoxiaojunjun/article/details/22687889

水题专栏收录该内容

18 篇文章

订阅专栏

本文解析了CodeForces平台上的一道A级题目——Cheaterius的问题，该题涉及通过旋转来判断2x2矩阵（代表魔法护身符）的相似性，并统计不同堆的数量。文章详细介绍了算法实现思路，包括如何通过旋转比较矩阵以及使用哈希表记录和统计相似矩阵。

Cheaterius's Problem

http://codeforces.com/problemset/problem/51/A

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest

inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself.

The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino

pairs with super glue to get squares 2 × 2which are the Cheaterius' magic amulets!

$\text{[math]}$ That's what one of Cheaterius's amulets looks like

After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6

dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be

rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters

should be the same. It is forbidden to turn over the amulets.

Write a program that by the given amulets will find the number of piles on Cheaterius' desk.

Input

The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are

contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of

amulets the line "**" is located.

Output

Print the required number of piles.

Sample Input

Input

Output

Input

Output

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
bool t[70000000];
int s[4];
int main()
{
    int n,u,i,x,y,tem,ans;
    char r[5];
    while(scanf("%d",&n)!=EOF)
    {
        for(i=11111111; i<=66666666; i++)
            t[i]=0;
        for(u=1; u<=n; u++)
        {
            scanf("%d%d",&x,&y);
            if(u!=n) scanf("%s",r);
            s[0]=x;
            s[1]=x%10*10+y%10;
            s[2]=y%10*10+y/10;
            s[3]=y/10*10+x/10;
            sort(s,s+4);
            tem=s[0]*1000000+s[1]*10000+s[2]*100+s[3];
            t[tem]=1;
        }
        ans=0;
        for(i=11111111; i<=66666666; i++)
            if(t[i]) ans++;
        printf("%d\n",ans);
    }
    return 0;
}