POJ-1721 CARDS 置换群分数幂运算

　　题目链接：http://poj.org/problem?id=1721

　　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 }