pollard_rho讲解

大神orz

poj 1811Prime Test http://poj.org/problem?id=1811

坑点:re了好多发,一直以为dfs次数太多,检查了很多次都不觉得会爆,数组也开得足够大,最后一种可能就是除0了.。。。。pollard_rho中有一步(y-x)的操作可能出现<=0的情况。。。。这是个pollard_rho的基本题,可以用来做模板

/**************************************************************
    Problem:poj 1811
    User: youmi
    Language: C++
    Result: Accepted
    Time:704MS
    Memory:708K
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <cmath>
#include <ctime>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#define zeros(a) memset(a,0,sizeof(a))
#define ones(a) memset(a,-1,sizeof(a))
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define sclld(a) scanf("%I64d",&a)
#define pt(a) printf("%d\n",a)
#define ptlld(a) printf("%I64d\n",a)
#define rep0(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define rep_1(i,n) for(int i=n;i>=1;i--)
#define rep_0(i,n) for(int i=n-1;i>=0;i--)
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define lson (step<<1)
#define rson (lson+1)
#define esp 1e-6
#define oo 0xffffffffffffff
#define TEST cout<<"*************************"<<endl

using namespace std;
typedef long long ll;
ll ans;
const int maxn=100000;
ll fact[maxn];
const int Times=10;
const int C=201;
int cnt=0;
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
ll q_mul(ll a,ll b,ll mod)
{
    ll res=0;
    a%=mod;
    while(b)
    {
        if(b&1)
            res=(res+a)%mod;
        b>>=1;
        a=(a<<1)%mod;
    }
    return res;
}
ll q_pow(ll a,ll b,ll mod)
{
    ll res=1;
    a%=mod;
    while(b)
    {
        if(b&1)
            res=q_mul(res,a,mod);
        b>>=1;
        a=q_mul(a,a,mod);
    }
    return res;
}
bool miller_rabin(ll n)
{
    if(n<2)
        return false;
    if(n==2)
        return true;
    if(!(n&1))
        return false;
    ll u=n-1;
    int tot=0;
    while(!(u&1))
    {
        u>>=1;tot++;
    }
    rep1(i,Times)
    {
        ll x=rand()%(n-2)+2;
        if(x==n)
            continue;
        x=q_pow(x,u,n);
        ll pre=x;
        rep0(j,tot)
        {
            x=q_mul(x,x,n);
            if(x==1&&pre!=1&&pre!=n-1)
                return false;
            pre=x;
        }
        if(x!=1)
            return false;
    }
    return true;
}
ll pollard_rho(ll n,ll c)
{
    ll x,y,d,i=1,k=2;
    x=rand()%(n-1)+1;
    y=x;
    while(1)
    {
        i++;
        x=(q_mul(x,x,n)+c)%n;
        d=gcd(n,(y-x+n)%n);//注意!!!!!!!!!
        if(d>1&&d<n)
            return d;
        if(y==x)
            return n;
        if(i==k)
        {
            y=x;
            k<<=1;
        }
    }
}
void Find(ll n,int k)
{
    if(n==1)
        return ;
    if(miller_rabin(n))
    {
        fact[++cnt]=n;
        if(n<ans)
            ans=n;
        return;
    }
    ll p=n;
    while(p>=n)
        p=pollard_rho(p,k--);
    Find(p,k);
    Find(n/p,k);
}
int main()
{
    //freopen("in.txt","r",stdin);
    int T_T;
    scanf("%d",&T_T);
    ll n;
    for(int kase=1;kase<=T_T;kase++)
    {
        scanf("%lld",&n);
        if(miller_rabin(n))
            printf("Prime\n");
        else
        {
            cnt=0;
            ans=n;
            Find(n,C);
            printf("%lld\n",ans);
        }
    }
    return 0;
}

 

poj 2429 GCD & LCM Inverse  http://poj.org/problem?id=2429 

思路:now=lcm/gcd 的因子是a,b所不同的,所以只需对now进行整数分解(因为Now比较大,所以用pollard_rho),分解出来的因子,相同因子一定分在一起,要不然gcd应该更大(至少应该乘这个相同因子)所以,又因为11个素数相乘的大小就差不多能达到10^18,所以只需对用dfs深搜一下(2^11层),或者把2^tot-1个状态一一列举就行了。。。dfs的速度更快

/**************************************************************
    Problem:poj 2429
    User: youmi
    Language: C++
    Result: Accepted
    Time:79MS
    Memory:148K
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <ctime>
#include <sstream>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#define zeros(a) memset(a,0,sizeof(a))
#define ones(a) memset(a,-1,sizeof(a))
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define sclld(a) scanf("%I64d",&a)
#define pt(a) printf("%d\n",a)
#define ptlld(a) printf("%I64d\n",a)
#define rep0(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define rep_1(i,n) for(int i=n;i>=1;i--)
#define rep_0(i,n) for(int i=n-1;i>=0;i--)
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define lson (step<<1)
#define rson (lson+1)
#define esp 1e-6
#define oo 0x3fffffff
#define TEST cout<<"*************************"<<endl

using namespace std;
typedef unsigned long long ll;
const int maxn=100000;
ll fact[maxn];
ll a[maxn];
const int C=201;
const int Times=5;
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
ll q_mul(ll a,ll b,ll mod)
{
    ll res=0;
    while(b)
    {
        if(b&1)
            res=(res+a)%mod;
        b>>=1;
        a=(a<<1)%mod;
    }
    return res;
}
ll q_pow(ll a,ll b,ll mod)
{
    ll res=1;
    while(b)
    {
        if(b&1)
            res=q_mul(res,a,mod);
        b>>=1;
        a=q_mul(a,a,mod);
    }
    return res;
}
bool miller_rabbin(ll n)
{
    if(n<2)
        return false;
    if(n==2)
        return true;
    if(!(n&1))
        return false;
    ll u=n-1;
    int tot=0;
    while(!(u&1))
    {
        u>>=1;tot++;
    }
    rep0(i,Times)
    {
            ll x=rand()%(n-2)+2;
            x=q_pow(x,u,n);
            ll pre=x;
            for(int j=0;j<tot;j++)
            {
                x=q_mul(x,x,n);
                if(x==1&&pre!=1&&pre!=n-1)
                    return false;
                pre=x;
            }
            if(x!=1)
                return false;
    }
    return true;
}
ll pollard_rho(ll n,ll c)
{
    ll x,y,d,i=1,k=2;
    x=rand()%(n-1)+1;
    y=x;
    while(1)
    {
        i++;
        x=(q_mul(x,x,n)+c)%n;
        d=gcd(n,(y-x+n)%n);
        if(d>1&&d<n)
            return d;
        if(x==y)
            return n;
        if(i==k)
        {
            y=x;
            k<<=1;
        }
    }
}
int cnt;
void Find(ll n,int k)
{
    if(n==1)
        return ;
    if(miller_rabbin(n))
    {
        fact[++cnt]=n;
        return ;
    }
    ll p=n;
    while(p>=n)
        p=pollard_rho(n,k--);
    Find(p,k);
    Find(n/p,k);
}
ll ans,n,m,maxa;
int all;
/**< 148K    94MS 这是另一种思路
void solve()
{
    ll x,y;
    pair<ll,ll>ans=make_pair(1,n/m);
    ll mini=ans.first+ans.second;
    ll bit=(1ll<<all)-1;
    for(int i=1;i<=bit;i++)
    {
        x=1;
        for(int j=1;j<=all;j++)
        {
            if(i&(1ll<<(j-1)))
                x*=a[j];
        }
        y=maxa/x;
        if(x<y&&x+y<mini)
        {
            mini=x+y;
            ans.first=x;
            ans.second=y;
            //printf("x->%llu y->%llu mini->%llu\n",x,y,mini);
        }
    }
    printf("%llu %llu\n",ans.first*m,m*ans.second);
}
 */
void dfs(ll cur,int temp)
{
    if(temp>all)
        return;
    if(ans+maxa/ans>cur*a[temp]+maxa/(cur*a[temp]))
    {
        ans=cur*a[temp];
    }
    dfs(cur*a[temp],temp+1);
    dfs(cur,temp+1);
}
bool cmp(ll c,ll b)
{
    return c<b;
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%llu%llu",&m,&n))
    {
        if(n==m)
        {
            printf("%llu %llu\n",n,m);
            continue;
        }
        maxa=n/m;
        cnt=0;
        Find(maxa,C);
        sort(fact+1,fact+cnt,cmp);
        all=1;
        a[all]=fact[1];
        for(int i=2;i<=cnt;i++)
        {
            if(fact[i]!=fact[i-1])
            {
                a[++all]=fact[i];
            }
            else
                a[all]*=fact[i];
        }
        ans=a[1];
        dfs(a[1],2);
        ll ans2=maxa/ans;
        ll temp;
        if(ans>ans2)
        {
            temp=ans;
            ans=ans2;
            ans2=temp;
        }
        printf("%llu %llu\n",ans*m,m*ans2);
    }
    return 0;
}
(*  ̄3)(ε ̄ *)

 

posted on 2015-10-29 21:32  中子星  阅读(208)  评论(0编辑  收藏  举报