POJ1721_CARDS【置換群】

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CARDSTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 1452Accepted: 776Description


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 ai+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 ith 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


7 4
6
3
1
2
4
7
5
Sample Output


4
7
5
6
1
2
3
Source

CEOI 1998

題目大意：

第i個位置的牌是a[i],一次交換後第i個位置的牌變成a[a[i]]。序列所有位置經過一次交換為一次交換，

已知交換m次之後的序列，求原先序列

思路：想了一天，一直沒明白什麼意思，晚上才想明白。題目大意如上，不過需要注意的是 只能在

原有序列的基礎上得到新的序列，而不能直接將原序列進行交換。

比如：              4 7 5 6 1 2 3

進行一次變換    6 3 1 2 4 7 5

而不能直接將原來第i個位置的牌改變，不能變成 6 3 1 2 6 3 5

意思就是說新的序列式在原有序列的基礎上，在交換的過程中沒有改變原有序列的值，從而得到一個

新的序列。

現在我們考慮變換次數和原先序列的關係

考慮                 4 7 5 6 1 2 3

進行一次變換    6 3 1 2 4 7 5

進行兩次變換    7 1 6 3 2 5 4

進行三次變換    4 7 5 6 1 2 3

原序列在第三次變換的時候出現了迴圈

考慮一般情況。因為原序列為1~N個不同的數，則序列能構成至少一個置換群，則經過多次變換後肯定

會出現迴圈，現在我們求出置換的迴圈節，已知新的序列，和進行交換的次數，則再進行res - s % res

次交換即可得到原序列。//res為迴圈節，s為給出的變換次數

#include <stdio.h>#include <string.h>int Array[1010],ArrayA[1010],ArrayB[1010];void GetNext(int n){    for(int i = 1; i <= n; i++)        ArrayB[i] = ArrayA[ArrayA[i]];    for(int i = 1; i <= n; i++)        ArrayA[i] = ArrayB[i];}bool IsSame(int n){    for(int i = 1; i <= n; i++)        if(ArrayB[i]!=Array[i])            return false;    return true;}int main(){    int n,s;    while(~scanf("%d%d",&n,&s))    {        for(int i = 1; i <= n; i++)        {            scanf("%d",&Array[i]);            ArrayA[i] = Array[i];        }        int res;        for(res = 1;;res++)        {            GetNext(n);            if(IsSame(n))                break;        }        int m = res - s % res;        int num = 0;        while(num < m)        {            GetNext(n);            num++;        }        for(int i = 1; i <= n; i++)            printf("%d\n",ArrayB[i]);    }    return 0;}

POJ1721_CARDS【置換群】