HDU 5250 三阶魔方（模拟、置换）

题意:

给定N≤100的三阶魔方操作序列
魔方操作如下图：

求问：对一个初状态（六个面都是拼好的）的魔方进行多少次连续的序列操作后，魔方会恢复到初状态

分析:

对魔方的每个方块标号之后，每个操作的置换可以手工推出
由于多次操作的复合仍然是一个置换，因此魔方总是能复原，并且将置换分解成环之后，ans=这些环长度的lcm
小细节:X与X′是互补的,即X′=3X,显然X2=2X

代码:

//
// Created by TaoSama on 2016-02-05
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int L = 54;

const char op[7]="FLRUDB";
const int d[6][5][4]=
{
    {{1,3,9,7},{2,6,8,4},{43,19,48,18},{44,22,47,15},{45,25,46,12}},//F
    {{10,12,18,16},{11,15,17,13},{37,1,46,36},{40,4,49,33},{43,7,52,30}},//L
    {{19,21,27,25},{20,24,26,22},{54,9,45,28},{51,6,42,31},{48,3,39,34}},//R
    {{37,39,45,43},{38,42,44,40},{28,19,1,10},{29,20,2,11},{30,21,3,12}},//U
    {{16,7,25,34},{17,8,26,35},{46,48,54,52},{47,51,53,49},{18,9,27,36}},//D
    {{28,30,36,34},{29,33,35,31},{27,39,10,52},{24,38,13,53},{21,37,16,54}}//B
};

void trans(vector<int>& p, int o){
    for(int i = 0; i < 5; ++i){
        int t = p[d[o][i][3]];
        for(int j = 3; j; --j)
            p[d[o][i][j]] = p[d[o][i][j - 1]];
        p[d[o][i][0]] = t;
    }
}

char s[N];

vector<int> solve(){
    vector<int> p(L + 1);
    for(int i = 1; i <= L; ++i) p[i] = i;
    for(int i = 0; s[i]; ++i){
        int o = strchr(op, s[i]) - op;
        if(isalpha(s[i + 1]) || !s[i + 1]) trans(p, o);
        else {
            int cnt = s[i + 1] == '\'' ? 3 : 2;
            while(cnt--) trans(p, o);
            ++i;
        }
    }
    return p;
}

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--){
        scanf("%s", s);
        vector<int> p = solve();
        int ans = 1;  bool vis[L + 1] = {};
        for(int i = 1; i <= L; ++i){
            if(vis[i]) continue;
            int cycle = 0, tmp = i;
            do{
                ++cycle;
                vis[tmp] = true;
                tmp = p[tmp];
            }while(tmp != i);
            ans = ans / __gcd(ans, cycle) * cycle;
        }
        static int kase = 0;
        printf("Case #%d:\n%d\n", ++kase, ans);
    }
    return 0;
}