码迷,mamicode.com
首页 > 其他好文 > 详细

POJ 2429 long long 质因数分解

时间:2017-07-24 09:58:56      阅读:127      评论:0      收藏:0      [点我收藏+]

标签:int   dfs   which   算法   ret   mis   mon   ios   bin   

GCD & LCM Inverse
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 16206   Accepted: 3008

Description

Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a and b. But what about the inverse? That is: given GCD and LCM, finding a and b.

Input

The input contains multiple test cases, each of which contains two positive integers, the GCD and the LCM. You can assume that these two numbers are both less than 2^63.

Output

For each test case, output a and b in ascending order. If there are multiple solutions, output the pair with smallest a + b.

Sample Input

3 60

Sample Output

12 15

Source

题意:
给出gcd和lcm求和最小的两个数a,b
代码:
//a*b=gcd*lcm,x=a/gcd,y=b/gcd,s=lcm/gcd => x*y=s;
//想要a+b小 => x*gcd+y*gcd小 => x*y小,x和y越接近sqrt(S)时x+y越小(对勾函数)
//所以求出s的素因子然后搞一搞就行了
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<algorithm>
using namespace std;
typedef long long ll;
const int S=20;//随机算法判定次数,S越大判错概率越小
//计算(a*b)%c;a,b,c<2^63
ll mult_mod(ll a,ll b,ll c){
    a%=c;
    b%=c;
    ll ret=0;
    while(b){
        if(b&1){
            ret+=a;
            ret%=c;
        }
        a<<=1;
        if(a>=c) a%=c;
        b>>=1;
    }
    return ret;
}
//计算(x^n)%c
ll pow_mod(ll x,ll n,ll c){
    if(n==1) return x%c;
    x%=c;
    ll tmp=x;
    ll ret=1;
    while(n){
        if(n&1) ret=mult_mod(ret,tmp,c);
        tmp=mult_mod(tmp,tmp,c);
        n>>=1;
    }
    return ret;
}
//以a为基,n-1=x*2^t      a^(n-1)=1(mod n)  验证n是不是合数
//一定是合数返回true,不一定返回false
bool check(ll a,ll n,ll x,ll t){
    ll ret=pow_mod(a,x,n);
    ll last=ret;
    for(int i=1;i<=t;i++){
        ret=mult_mod(ret,ret,n);
        if(ret==1&&last!=1&&last!=n-1) return true;
        last=ret;
    }
    if(ret!=1) return true;
    return false;
}
// Miller_Rabin()算法素数判定
//是素数返回true.(可能是伪素数,但概率极小)
//合数返回false;
bool Miller_Rabin(ll n){
    if(n<2) return false;
    if(n==2) return true;
    if((n&1)==0) return false;
    ll x=n-1;
    ll t=0;
    while((x&1)==0){
        x>>=1;
        t++;
    }
    for(int i=0;i<S;i++){
        ll a=rand()%(n-1)+1;
        if(check(a,n,x,t)) return false;
    }
    return true;
}
ll gcd(ll a,ll b){
    if(a==0) return 1;
    if(a<0) return gcd(-a,b);
    while(b){
        ll t=a%b;
        a=b;
        b=t;
    }
    return a;
}
//Pollard_rho质因数分解算法
ll factor[1000];//质因数分解结果(无序的)
int tot;//质因数的个数,数组下标从0开始
ll Pollard_rho(ll x,ll c){
    ll i=1,k=2;
    ll x0=rand()%x;
    ll y=x0;
    while(1){
        i++;
        x0=(mult_mod(x0,x0,x)+c)%x;
        ll d=gcd(y-x0,x);
        if(d!=1&&d!=x) return d;
        if(y==x0) return x;
        if(i==k){
            y=x0;
            k+=k;
        }
    }
}
//对n进行素因子分解
void findfac(ll n){
    if(Miller_Rabin(n)){//素数
        factor[tot++]=n;
        return;
    }
    ll p=n;
    while(p>=n)
        p=Pollard_rho(p,rand()%(n-1)+1);
    findfac(p);
    findfac(n/p);
}
ll ans,fa[100];
int nu;
void dfs(int st,ll x,ll maxx)
{
    if(st>nu){
        if(x<=maxx&&x>ans)
            ans=x;
        return;
    }
    dfs(st+1,x,maxx);
    dfs(st+1,x*fa[st],maxx);
}
int main()
{
    //srand(time(NULL));
    ll a,b;
    while(scanf("%lld%lld",&a,&b)==2){
        if(a==b){
            printf("%lld %lld\n",a,b);
            continue;
        }
        tot=nu=0;
        ll s=b/a;
        findfac(s);
        sort(factor,factor+tot);
        fa[0]=factor[0];
        for(int i=1;i<tot;i++){//合并相同的素因子
            if(factor[i]==factor[i-1]) fa[nu]*=factor[i];
            else
                fa[++nu]=factor[i];
        }
        ll x=(ll)sqrt(s*1.0);
        ans=1;
        dfs(0,1,x);
        printf("%lld %lld\n",ans*a,s/ans*a);
    }
    return 0;
}

 

POJ 2429 long long 质因数分解

标签:int   dfs   which   算法   ret   mis   mon   ios   bin   

原文地址:http://www.cnblogs.com/--ZHIYUAN/p/7227129.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!