POJ 2429 GCD & LCM Inverse

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题目链接：GCD & LCM Inverse

解题思路：这个给了a和b的GCD和LCM，让求出a和b，这里a和b有很多解，让输出a + b最小的一组。

gcd(a, b) = k

lcm(a, b) = m

a = pk, b = qk

a + b = (p + q)k

p * q * k = m

可以通过以上得出

p * q = m / k，gcd(p, q) = 1，所以我们要求出q + p最小的。

我们可以知道当d是m / k的约数的时候并且gcd(d, m / k / d) = 1，那么我们这里可以看到，当p q的差距越来越小的时候p + q就越来越小，但是我们对于longlong没有sqrt，所以我们要另一种思路。

我们将m / k分解质因数，我们将相同的质因数相乘合成n个约数，这里为什么要合成，因为同一个约数既在p也在q的时候gcd(p, q) ！= 1，所以我们这样合成的都是两两互质的，所以搜索一遍就得出结果了。

这里涉及大整数分解，那就是Pollard rho算法，详细请见

#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#include<ctime>#include<algorithm>#include<cstdlib>#include<vector>#define ll __int64#define MAX 5500const int S = 20;//素数测试误判率=1 / 2^S   S越大误差率越小using namespace std;ll gcd(ll a, ll b){    return b == 0 ? a : gcd(b, a % b);}ll multi_mod(ll a, ll b, ll p){    ll ret = 0, q = a % p;    while(b){        if(b & 1){            ret = (ret + q) % p;        }        q = (q + q) % p;        b >>= 1;    }    return ret % p;}ll pow_mod(ll a, ll b, ll p){    ll ret = 1, q = a % p;    while(b){        if(b & 1){            ret = multi_mod(ret, q, p);        }        q = multi_mod(q, q, p);        b >>= 1;    }    return ret % p;}bool Miller_Rabin(ll p){    int i, j, k;    ll u, t, x, y;    if(p == 2)  return true;    if(p % 2 == 0 || p == 1) return false;    u = p - 1, t = 0;    while(u % 2 == 0){        t++;        u >>= 1;    }    for(int i = 0; i < S; i++){        x = rand() % (p - 1) + 1;        x = pow_mod(x, u, p);        ll y = 0;        for(j = 0; j < t; j++){            y = multi_mod(x, x, p);            if(y == 1 && x != 1 && x != p - 1){                return false;            }            x = y;        }        if(y != 1)  return false;    }    return true;}ll fact[MAX];ll tot = 0;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_mod(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, ll c){    if(n == 1){        return;    }    if(Miller_Rabin(n)){        fact[tot++] = n;        return;    }    ll d = n;    ll k = c;    while(d >= n){        d = pollard_rho(n, c--);    }    find(d, k);    find(n / d, k);}vector<ll> num;int main(){    int i, j, k;    ll a, b, p;    ll n;    //srand(time(NULL));    while(cin >> a >> b){        ll gcd = a;        n = b / a;        tot = 0;        num.clear();        find(n, 120);        sort(fact, fact + tot);        for(i = 0; i < tot; i++){            if(i == 0 || fact[i] != fact[i - 1]){                num.push_back(fact[i]);            }            else{                num[num.size() - 1] *= fact[i];            }        }        p = n + 1;        a = 1, b = n;        ll tem, am;        for(i = 0; i < (1 << num.size()); i++){            int j = i;            tem = 1, am = 0;            while(j){                if(j & 1){                    tem *= num[am];                }                j >>= 1;                am++;            }            ll tt = n / tem;            //cout << tt << " " << tem << endl;            if(tt + tem < p){                p = tt + tem;                a = min(tt, tem);                b = max(tt, tem);            }        }        cout << gcd * a << " " << gcd * b << endl;    }    return 0;}

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