POJ1721 CARDS 置换群分数幂运算

　　置换群的分数幂运算，分数幂运算考虑的是置换的合并，但这道题简化了很多，首先注意到“Alice first writes down all the numbers from 1 to N in some random order: a1, a2, ..., aN. Then she arranges the cards so that the position ai holds the card numbered a_i+1, for every 1 <= i <= N-1, while the position aN holds the card numbered a1.”，即置换只有一个循环节且长度 l 为奇数，其次开放数为2。则有gcd(l,2)=1，即永远都只有一个循环节，直接模拟就可以了。

 1 //STATUS:C++_AC_63MS_192KB
 2 #include<stdio.h>
 3 #include<stdlib.h>
 4 #include<string.h>
 5 #include<math.h>
 6 #include<iostream>
 7 #include<string>
 8 #include<algorithm>
 9 #include<vector>
10 #include<queue>
11 #include<stack>
12 #include<map>
13 using namespace std;
14 #define LL __int64
15 #define pii pair<int,int>
16 #define Max(a,b) ((a)>(b)?(a):(b))
17 #define Min(a,b) ((a)<(b)?(a):(b))
18 #define mem(a,b) memset(a,b,sizeof(a))
19 #define lson l,mid,rt<<1
20 #define rson mid+1,r,rt<<1|1
21 const int N=1010,INF=0x3f3f3f3f,MOD=10000,STA=8000010;
22 const LL LNF=0x3f3f3f3f3f3f3f3f;
23 const double DNF=1e13;
24 
25 int T[2][N],vis[N],C[N];
26 int n,s;
27 
28 int main()
29 {
30  //   freopen("in.txt","r",stdin);
31     int i,j,k,p,u,d;
32     while(~scanf("%d%d",&n,&s))
33     {
34         for(i=1;i<=n;i++)
35             scanf("%d",&T[0][i]);
36 
37         p=1;
38         while(s--){
39             mem(vis,0);
40             for(i=1;i<=n;i++){
41                 if(!vis[i]){
42                     mem(vis,0);
43                     u=i;k=0;
44                     while(!vis[u]){
45                         vis[u]=1;
46                         C[k++]=u;
47                         u=T[!p][u];
48                     }
49                     d=(k+1)/2;
50                     for(j=0;j<k;j++){
51                         T[p][C[j]]=C[(j+d)%k];
52                     }
53                 }
54             }
55             p=!p;
56         }
57 
58         p=!p;
59         for(i=1;i<=n;i++)
60             printf("%d\n",T[p][i]);
61     }
62     return 0;
63 }