bnuoj 51277(魔方复原-手推置换群)

来源：互联网发布：知床热气球编辑：程序博客网时间：2024/05/01 15:56

题意:给一个魔方，定义一堆操作，现给出操作序列，问这个操作序列重复多少次之后魔方复原？

用了大半天时间设计程序计算6个置换……结果没做出

其实手推最省事。

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;}ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);}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};#define MAXN (6*9+1)const int n = 54;const char c[7]="FLRUDB";int h[1000];void work(vector<int> &p,int type) { // 一个循环节     Rep(k,5) {        swap(p[d[type][k][2]],p[d[type][k][3]]);        swap(p[d[type][k][1]],p[d[type][k][2]]);        swap(p[d[type][k][0]],p[d[type][k][1]]);    }}int t[MAXN],t2[MAXN];void multi(vector<int> &p,vector<int> q) {    For(i,n) t[i]=p[q[i]];    For(i,n) p[i]=t[i];} char s[100000+10];int x;vector<int> solve() {    vector<int> p(n+1);    For(i,n) p[i]=i;    while (s[x] && s[x]!=')') {          if (isdigit(s[x])) { //数字()             ll r=0;             while (isdigit(s[x])) {                r=r*10+s[x++]-'0';            }            x++; //去掉左括号            vector<int> q=solve();             while(r) {                if (r&1) multi(p,q);                multi(q,q);                r>>=1;            }               } else { //字母            work(p,h[s[x++]]);        }    }    x++; //去掉右括号    return p;}bool vis[MAXN];ll get_ans(vector<int> p) {    MEM(vis)    ll res=1;    For(i,n) if (!vis[i]) {        int u=i,s=0;        while (!vis[u]) {            vis[u]=1;            u=p[u];            s++;        }        res=res/gcd(res,s)*s;    }    return res;}int main(){//  freopen("B.in","r",stdin);//  freopen(".out","w",stdout);    Rep(i,6) h[c[i]]=i;    int T=read();    while(T--) {        scanf("%s",s);        x=0;        cout<<get_ans(solve())<<endl;           }    return 0;}

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