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Double Dealing

1 3
10 3
52 4
0 0

1
4
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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("C.in","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;
}

HDU 4259(Double Dealing-lcm(x1..xn)=lcm(x1,lcm(x2..xn))

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原文地址：http://blog.csdn.net/nike0good/article/details/47448319