Double Dealing Time Limit: 50000/20000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1924    Accepted Submission(s): 679


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 kth to player k, the k+1st 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




 


Source



 


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求置换群循环节的lcm


注意lcm(x1..xn)=lcm(x1,lcm(x2..xn)!=x1*..*xn/gcd







#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
char s[]="no solution\n";

class Math
{
public:
    ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);}
    ll abs(ll x){if (x>=0) return x;return -x;}
    ll exgcd(ll a,ll b,ll &x, ll &y)
    {
        if (!b) {x=1,y=0;return a;}
        ll g=exgcd(b,a%b,x,y);
        ll t=x;x=y;y=t-a/b*y;
        return g;
    }
    ll pow2(ll a,int b,ll p)
    {
        if (b==0) return 1;
        if (b==1) return a;
        ll c=pow2(a,b/2,p);
        c=c*c%p;
        if (b&1) c=c*a%p;
        return c;
    }
    ll Modp(ll a,ll b,ll p)
    {
        ll x,y;
        ll g=exgcd(a,p,x,y),d;
        if (b%g) {return -1;}
        d=b/g;x*=d,y*=d;
        x=(x+abs(x)/p*p+p)%p;
        return x;
    }
    int h[MAXN];
    ll hnum[MAXN];
    int hash(ll x)
    {
        int i=x%MAXN;
        while (h[i]&&hnum[i]!=x) i=(i+1)%MAXN;
        hnum[i]=x;
        return i;
    }
    ll babystep(ll a,ll b,int p)
    {
        MEM(h) MEM(hnum)
        int m=sqrt(p);while (m*m<p) m++;
        ll res=b,ans=-1;

        ll uni=pow2(a,m,p);
        if (!uni) if (!b) ans=1;else ans=-1; //特判
        else
        {

            Rep(i,m+1)
            {
                int t=hash(res);
                h[t]=i+1;
                res=(res*a)%p;
            }
            res=uni;

            For(i,m+1)
            {
                int t=hash(res);
                if (h[t]) {ans=i*m-(h[t]-1);break;}else hnum[t]=0;
                res=res*uni%p;
            }

        }
        return ans;
    }
}S;

int a[10000+10];
bool b[10000+10];
int p[10000+10];
int main()
{
//    freopen("","r",stdin);
//    freopen(".out","w",stdout);

    int n,k;
    while(cin>>n>>k)
    {
        if (n+k==0) return 0;
        int s=0;
        For(j,k)
            for(int i=n/k*k+j>n?n/k*k+j-k:n/k*k+j;i>=1;i-=k) a[++s]=i;

    //    For(i,n) cout<<a[i]<<' ';


        int tot=0;

        MEM(b)
        For(i,n)
        {
            if (!b[i])
            {
                int t=i; b[i]=1;
                int len=1;
                do {
                    b[t]=1;
                    t=a[t]; ++len;
              //       cout<<t<<endl;

                } while (!b[t]);
                len--;

                p[++tot]=len;
            }
        }

        sort(p+1,p+1+tot);
        tot=unique(p+1,p+1+tot)-(p+1);

//        For(i,tot) cout<<p[i]<<' ';

        ll ans=1;
        For(i,tot) ans=ans/S.gcd(p[i],ans)*p[i];


        cout<<ans<<endl;

    }


    return 0;
}