zoukankan      html  css  js  c++  java
  • POJ 2429(GCD&LCM Inverse)

    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

    思路:

    其实思路不复杂,题目也短小精炼,但是这个涉及到 Miller_Rabin 和 Pollard_rho 算法

     

    直接上代码:

    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<queue>
    #include<stack>
    #include<cmath>
    #include<set>
    #include<algorithm>
    #include<vector>
    
    using namespace std;
    #define clc(a,b) memset(a,b,sizeof(a))
    #define inf  (LL)1<<61
    #define LL long long
    
    const int Times = 10;
    const int N = 5500;
    
    LL n, m, ct, cnt;
    LL minn, mina, minb, ans;
    LL fac[N], num[N];
    
    LL gcd(LL a, LL b)
    {
        return b? gcd(b, a % b) : a;
    }
    
    LL multi(LL a, LL b, LL m)
    {
        LL ans = 0;
        a %= m;
        while(b)
        {
            if(b & 1)
            {
                ans = (ans + a) % m;
                b--;
            }
            b >>= 1;
            a = (a + a) % m;
        }
        return ans;
    }
    
    LL quick_mod(LL a, LL b, LL m)
    {
        LL ans = 1;
        a %= m;
        while(b)
        {
            if(b & 1)
            {
                ans = multi(ans, a, m);
                b--;
            }
            b >>= 1;
            a = multi(a, a, m);
        }
        return ans;
    }
    
    bool Miller_Rabin(LL n)
    {
        if(n == 2) return true;
        if(n < 2 || !(n & 1)) return false;
        LL m = n - 1;
        int k = 0;
        while((m & 1) == 0)
        {
            k++;
            m >>= 1;
        }
        for(int i=0; i<Times; i++)
        {
            LL a = rand() % (n - 1) + 1;
            LL x = quick_mod(a, m, n);
            LL y = 0;
            for(int j=0; j<k; j++)
            {
                y = multi(x, x, n);
                if(y == 1 && x != 1 && x != n - 1) return false;
                x = y;
            }
            if(y != 1) return false;
        }
        return true;
    }
    
    LL pollard_rho(LL n, LL c)
    {
        LL i = 1, k = 2;
        LL x = rand() % (n - 1) + 1;
        LL y = x;
        while(true)
        {
            i++;
            x = (multi(x, x, n) + c) % n;
            LL d = gcd((y - x + n) % n, n);
            if(1 < d && d < n) return d;
            if(y == x) return n;
            if(i == k)
            {
                y = x;
                k <<= 1;
            }
        }
    }
    
    void find(LL n, int c)
    {
        if(n == 1) return;
        if(Miller_Rabin(n))
        {
            fac[ct++] = n;
            return ;
        }
        LL p = n;
        LL k = c;
        while(p >= n) p = pollard_rho(p, c--);
        find(p, k);
        find(n / p, k);
    }
    
    void dfs(LL dept, LL tem=1)
    {
        if(dept == cnt)
        {
            LL a = tem;
            LL b = ans / a;
            if(gcd(a, b) == 1)
            {
                a *= n;
                b *= n;
                if(a + b < minn)
                {
                    minn = a + b;
                    mina = a;
                    minb = b;
                }
            }
            return ;
        }
        for(int i=0; i<=num[dept]; i++)
        {
            if(tem > minn) return;
            dfs(dept + 1, tem);
            tem *= fac[dept];
        }
    }
    
    int main()
    {
        while(~scanf("%llu %llu", &n, &m))
        {
            if(n == m)
            {
                printf("%llu %llu
    ",n,m);
                continue;
            }
            minn = inf;
            ct = cnt = 0;
            ans = m / n;
            find(ans, 120);
            sort(fac, fac + ct);
            num[0] = 1;
            int k = 1;
            for(int i=1; i<ct; i++)
            {
                if(fac[i] == fac[i-1])
                    ++num[k-1];
                else
                {
                    num[k] = 1;
                    fac[k++] = fac[i];
                }
            }
            cnt = k;
            dfs(0, 1);
            if(mina > minb) swap(mina, minb);
            printf("%llu %llu
    ",mina, minb);
        }
        return 0;
    }
    

      

    补一个 扩展欧几里得算法

     分析:

    ax1+by1=gcd(a,b)
    =gcd(b,a%b)=bx2+a%by2
    =bx2+(a-(a/b)*b)y2
    =bx2+ay2-(a/b)*by2
    =b(x2-(a/b)*y2)+ay2

      

    待定系数:
    x1=y2
    y1=x2-(a/b)*y2
    // 说明:a/b是取整除法

    代码:

    int ext_gcd(int a,int b,int& x,int& y){
    
      int t,ret;
    
      if (!b){
    
        x=1,y=0;
    
        return a;
    
      }
    
      ret=ext_gcd(b,a%b,x,y);
    
      t=x,x=y,y=t-a/b*y;
    
      return ret;
    
    }
    

      

  • 相关阅读:
    1211.分割平衡字符串
    1282.用户分组
    分模块配置
    Spring Bean相关配置
    Spring IOC是什么
    Spring简介
    小黄衫感想
    团队展示
    原型设计
    结对作业
  • 原文地址:https://www.cnblogs.com/jaszzz/p/12693721.html
Copyright © 2011-2022 走看看