hdu acm 4259 Double Dealing (置换群)

Double Dealing

Time Limit: 50000/20000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1708    Accepted Submission(s): 585

Problem Description

Take a deck of n unique cards. Deal the entire deck out to k players in the usual way: the top card to player 1, the next to player 2, the k^th to player k, the k+1^st to player 1, and so on. Then pick up the cards – place player 1′s cards on top, then player 2, and so on, so that player k’s cards are on the bottom. Each player’s cards are in reverse order – the last card that they were dealt is on the top, and the first on the bottom.
How many times, including the first, must this process be repeated before the deck is back in its original order?

 

Input

There will be multiple test cases in the input. Each case will consist of a single line with two integers, n and k (1≤n≤800, 1≤k≤800). The input will end with a line with two 0s.

 

Output

For each test case in the input, print a single integer, indicating the number of deals required to return the deck to its original order. Output each integer on its own line, with no extra spaces, and no blank lines between answers. All possible inputs yield answers which will fit in a signed 64-bit integer.

 

Sample Input

1 3 10 3 52 4 0 0

 

Sample Output

1 4 13

 

置换群水题，主要是置换映射不好表示
如10 3
  1 2 3 4 5 6 7 8 9 10->
10 7 4 1 8 5 2 9 6 3  .

for (int i = 1;i <= K;i++)
{
 for (int j = (N- i)/K *K + i;j > 0;j -= K)
{ a[num++] = j; }
}

只要知道对于第i列，它拥有的元素个数是(N-i)/K +1，就很好理解了。

下面是完整代码

/*
杭电 acm 4259 Double Dealing 
                                    简单置换群
*/
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>

using namespace std;
#define maxn 880
int a[880],low[880];//a数组判断是否找过   low存置换后的数
int gcd(long long a,long long b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
int main()
{
    int n,k;
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        if(n==0||k==0)
            break;
        long long ans=1;
        int i,M=0;
        memset(a,1,sizeof(a));
        for(i = 0; i < k && i < n; i++)
            for(int j=(n-i-1)/k*k+i; j>=0; j-=k)
                low[M++] = j;
        for( i=0; i<n; i++)
        {
            if(a[i]==0)
                continue;
            int t=0,x=i;
            while(a[x])
            {
                a[x]=0;
                x=low[x];
                t++;
            }
            if(t)
                ans=(ans/gcd(ans,t))*t;
        }
        printf("%I64d
",ans);
    }
    return 0;
}