poj 1721 CARDS 洗牌机

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本文介绍了一个基于卡片的游戏问题及其解决方案，该问题涉及置换群的快速幂运算。通过分析游戏规则，文章提出了一种高效算法来逆向求解经过多次双倍洗牌后的初始卡片排列。

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CARDS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1684		Accepted: 877

Description

Alice and Bob have a set of N cards labelled with numbers 1 ... N (so that no two cards have the same label) and a shuffle machine. We assume that N is an odd integer.
The shuffle machine accepts the set of cards arranged in an arbitrary order and performs the following operation of double shuffle : for all positions i, 1 <= i <= N, if the card at the position i is j and the card at the position j is k, then after the completion of the operation of double shuffle, position i will hold the card k.

Alice and Bob play a game. 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.

This way, cards are put in some order x1, x2, ..., xN, where xi is the card at the i ^th position.

Now she sequentially performs S double shuffles using the shuffle machine described above. After that, the cards are arranged in some final order p1, p2, ..., pN which Alice reveals to Bob, together with the number S. Bob's task is to guess the order x1, x2, ..., xN in which Alice originally put the cards just before giving them to the shuffle machine.

Input

The first line of the input contains two integers separated by a single blank character : the odd integer N, 1 <= N <= 1000, the number of cards, and the integer S, 1 <= S <= 1000, the number of double shuffle operations.
The following N lines describe the final order of cards after all the double shuffles have been performed such that for each i, 1 <= i <= N, the (i+1) ^st line of the input file contains pi (the card at the position i after all double shuffles).

Output

The output should contain N lines which describe the order of cards just before they were given to the shuffle machine.
For each i, 1 <= i <= N, the ith line of the output file should contain xi (the card at the position i before the double shuffles).

Sample Input

Sample Output

Source

CEOI 1998

翻译在

中有

要注意这个题目的每一次操作都不是一次置换，要按照它规定的方式操作。

反复操作找出周期circle，题目告诉了你终点状态ed,和操作了s次，要你求起始状态st。

首先s=s%circle。

相当于变换了这么多次。

然后从终点再变换 circle-s 次，即可得到起点。

很像链表中的一个操作。

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
#include<cctype>
#include<utility>
#pragma comment(linker, "/STACK:102400000,102400000")
#define PI 3.1415926535897932384626
#define eps 1e-10
#define sqr(x) ((x)*(x))
#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)
#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)
#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)
#define  lson   num<<1,le,mid
#define rson    num<<1|1,mid+1,ri
#define MID   int mid=(le+ri)>>1
#define zero(x)((x>0? x:-x)<1e-15)
#define mk    make_pair
#define _f     first
#define _s     second
using namespace std;
//const int INF=    ;
typedef long long ll;
//const ll inf =1000000000000000;//1e15;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
const int INF =0x3f3f3f3f;
const int maxn= 1000+20    ;
//const int maxm=    ;
//by yskysker123
int n,s,circle;
int A[maxn],B[maxn],C[maxn];
void GetCircle()
{
    circle=0;
    while(1)
    {
        circle++;
        for(int i=1;i<=n;i++)
          B[i]=  C[C[i]];

        int i;
        for(i=1;i<=n;i++)
            if(B[i]!=A[i])  break;

        if(i>n)  break;
        for(int i=1;i<=n;i++)  C[i]=B[i];
    }


}


void Work(int k)
{
    for(int i=1;i<=k;i++)
    {
        for(int j=1;j<=n;j++)  B[j]=A[A[j]];
        for(int j=1;j<=n;j++)  A[j]=B[j];
    }
    for(int i=1;i<=n;i++)
        printf("%d\n",B[i]);
}
int main()
{
   while(~scanf("%d%d",&n,&s))
   {
       for(int i=1;i<=n;i++)  scanf("%d",&A[i]),B[i]=C[i]=A[i];
       GetCircle();
       s=s%circle;
       int k=circle-s;
       Work(k);


   }

    return 0;
}