POJ 2429 GCD & LCM Inverse Pollard_rho大数因子分解

这题的话。假设原来的数是a和b，

给出GCD和LCM以后

我们只需要求一下

LCM/GCD 的因子 就可以求出 a/gcd 和 b/gcd

然后求出的这两个东西之间不能有公约数

所以一个数取因子的时候要把某个因子全拿走

用的是Pollard_rho进行分解这种比较大的数。

复杂度的话，假设分解出一个因子是p，用的时间是O(sqrt(p))

总体来说比你直接sqrt(n)分解快多了

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <map>
#define MAXN 10
#define INF 1LL<<61
#define C 16381
using namespace std;
typedef long long LL;
LL gcd(LL a, LL b)
{
return b ? gcd(b, a % b) : a;
}

LL multi(LL a, LL b, LL m)
{
LL ans = 0;
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 a, m = n - 1, x, y;
int k = 0;
while((m & 1) == 0)
{
k++;
m >>= 1;
}
for(int i = 0; i < MAXN; i++)
{
a = rand() % (n - 1) + 1;
x = quick_mod(a, m, n);
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 x, y, d, i = 1, k = 2;
y = x = rand() % (n - 1) + 1;
while(true)
{
i++;
x = (multi(x, x, n) + c) % n;
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;
}
}
}
LL fac[555];
int cnt;
map<LL, int>mp;
void find(LL n, int c)
{
if(n == 1) return;
if(Miller_Rabin(n))
{
mp
++;
return ;
}
LL p = n;
LL k = c;
while(p >= n) p = Pollard_rho(p, c--);
find(p, k);
find(n / p,k);
}

int main()
{
LL a, b;
while(scanf("%I64d%I64d", &a, &b) != EOF)
{
cnt = 0;
LL tmp = b / a;
if(a == b)
{
printf("%I64d %I64d\n", a, b);
continue;
}
mp.clear();
find(tmp, C);
cnt = 0;
for(map<LL, int>:: iterator it = mp.begin(); it != mp.end(); it++)
{
LL t = 1;
for(int j = 0; j < it -> second; j++)
t = t * (it -> first);
fac[cnt++] = t;
}
//for(int i = 0; i < cnt; i++)
// printf("%I64d\n", fac[i]);
LL ansa = INF, ansb = INF;
for(int i = 0; i < (1 << cnt); i++)
{
LL ta = 1;
for(int j = 0; j < cnt; j++)
if(i & (1 << j))
ta *= fac[j];
if(ansa + ansb > ta + b / a / ta)
ansa = ta, ansb = b / a / ta;
}
if(ansa > ansb) swap(ansa, ansb);
printf("%I64d %I64d\n", ansa * a, ansb * a);
}
return 0;
}