POJ 2429 GCD & LCM Inverse 因式分解

最新推荐文章于 2020-04-09 17:47:27 发布

原创最新推荐文章于 2020-04-09 17:47:27 发布 · 2.0k 阅读

0 ·

CC 4.0 BY-SA版权

数学专栏收录该内容

153 篇文章

订阅专栏

本文介绍了一种解决GCD（最大公约数）和LCM（最小公倍数）逆问题的方法，即已知GCD和LCM求最小的a和b。通过分析给出的数学关系，采用整数分解和深度优先搜索策略找到最优解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

点击打开链接

GCD & LCM Inverse

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8706		Accepted: 1649

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

POJ Achilles

给你gcd(a,b)和lcm(a,b)，让你找最小的a和b。

因为(a*b)/gcd=lcm.那么(a/gcd*b/gcd)*gcd=lcm因此(a/gcd*b/gcd)=lcm/gcd。题目即可转换为把lcm/gcd分解成两个互质的数使这两个数和最小，只需要将key=lcm/gcd整数分解，然后dfs一下即可得到结果。

//408K	500MS	
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define Times 11
#define inf ((long long)1<<61)
#define C 201
using namespace std;
long long jl[501],numfactor[501],mini,mina,minb;//jl里面存的是大数的所有质因子，mini为最小的质因数
int ct,num[65];
long long key,gc;
int len;
long long random(long long n)//生成随机数
{
    return (long long)((double)rand()/RAND_MAX*n+0.5);
}
long long gcd(long long a,long long b)//最大公约数
{
    return b==0?a:gcd(b,a%b);
}
long long multi(long long a,long long b,long long m)//a*b%m
{
    long long ret=0;
    while(b>0)
    {
        if(b&1)ret=(ret+a)%m;
        b>>=1;
        a=(a<<1)%m;
    }
    return ret;
}
long long quick_mod(long long a,long long b,long long m)//a^b%m
{
    long long ans=1;
    a%=m;
    while(b)
    {
        if(b&1)
        {
            ans=multi(ans,a,m);
            b--;
        }
        b/=2;
        a=multi(a,a,m);
    }
    return ans;
}
bool Witness(long long a,long long n)
{
    long long m=n-1;
    int j=0;
    while(!(m&1))
    {
        j++;
        m>>=1;
    }
    long long x=quick_mod(a,m,n);
    if(x==1||x==n-1)return false;
    while(j--)
    {
        x=x*x%n;
        if(x==n-1)return false;
    }
    return true;
}
bool miller_rabin(long long n)//素数测试
{
    if(n<2)return false;
    if(n==2)return true;
    if(!(n&1))return false;
    for(int i=1; i<=Times; i++)
    {
        long long a=random(n-2)+1;
        if(Witness(a,n))return false;
    }
    return true;
}
long long pollard_rho(long long n,int c)//整数n分解,c一般为201
{
    long long x,y,d,i=1,k=2;
    x=random(n-1)+1;
    y=x;
    while(1)
    {
        i++;
        x=(multi(x,x,n)+c)%n;
        d=gcd(y-x,n);
        if(1<d&&d<n)return d;
        if(y==x)return n;
        if(i==k)
        {
            y=x;
            k<<=1;
        }
    }
}
void find(long long n,int k)
{
    if(n==1)return;
    if(miller_rabin(n))
    {
        jl[++ct]=n;
        return;
    }
    long long p=n;
    while(p>=n)p=pollard_rho(p,k--);
    find(p,k);
    find(n/p,k);
}
void dfs(int cur,long long value)
{
    long long s=1,a,b;
    if(cur==len+1)
    {
        a=value;
        b=key/value;
        if(gcd(a,b)==1)
        {
            a*=gc;
            b*=gc;
            if(a+b<mini)
            {
                mini=a+b;
                mina=a<b?a:b;
                minb=a<b?b:a;
            }
        }
        return;
    }
    for(int i=0;i<=num[cur];i++)
    {
        if(value*s>=mini) return;
        dfs(cur+1,value*s);
        s*=numfactor[cur];
    }
}
void solve(long long n)
{
    ct=0;
    find(n,C);
    sort(jl+1,jl+ct+1);
    memset(num,0,sizeof(num));
    len=0;
    num[0]=1;
    numfactor[0]=jl[1];
    for(int i=2;i<=ct;i++)
    {
        if(numfactor[len]!=jl[i])numfactor[++len]=jl[i];
        num[len]++;
    }
    dfs(0,1);
    printf("%lld %lld\n",mina,minb);
}
int main()
{
    long long  lc;
    while(scanf("%lld%lld",&gc,&lc)!=EOF)
    {
        if(gc==lc){printf("%lld %lld\n",gc,lc);continue;}
        key=lc/gc;
        mini=inf;
        solve(key);
    }
    return 0;
}