poj 2429 GCD & LCM Inverse（大数质因数分解+DFS）

GCD & LCM Inverse

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 16973 Accepted: 3137

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

题意：

给你两个数n和m，让你求出一组a和b满足

①a<=b；②Gcd(a, b) = n；③Lcm(a, b) = m；④a+b的和尽可能的小

思路：

m/n = Lcm(a, b)/Gcd(a, b) = a/Gcd(a, b)*b/Gcd(a, b)

也就是要找到一对a, b满足

①a和b互质

②a+b尽可能小

③a*b = n

很显然a和b一定都是m/n的若干个质因子的乘积

用Pollard_rho分解m/n的质因数，然后爆搜就行了，注意剪枝

#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;#define LL long longint t, cnt;LL fat[101], val, ans, c, d;LL Multi(LL a, LL b, LL mod){LL ans = 0;a %= mod;while(b){if(b%2==1)  ans = (ans+a)%mod, b--;else  a = (a+a)%mod, b /= 2;}return ans;}LL Pow(LL a, LL b, LL mod){LL ans = 1;a %= mod;while(b){if(b&1)  ans = Multi(ans, a, mod), b--;else  a = Multi(a, a, mod), b /= 2;}return ans;}LL Gcd(LL a, LL b){if(b==0)return a;return Gcd(b, a%b);}int Miller_Rabin(LL n){int i, j, k;LL a, x, y, mod;if(n==2)  return 1;if(n<2 || n%2==0)  return 0;k = 0, mod = n-1;while(mod%2==0){k++;mod /= 2;}for(i=1;i<=10;i++){a = rand()%(n-1)+1;x = Pow(a, mod, n);y = 0;for(j=1;j<=k;j++){y = Multi(x, x, n);if(y==1 && x!=1 && x!=n-1)return 0;x = y;}if(y!=1)return 0;}return 1;}LL Divi(LL n){LL i, k, x, y, p, c;if(n==1)return 1;k = 2, p = 1;y = x = rand()%n, c = rand()%(n-1)+1;for(i=1;p==1;i++){x = (Multi(x, x, n)+c)%n;p = x-y;if(p<0)p = -p;p = Gcd(n, p);if(i==k)y = x, k *= 2;}return p;}void Pollard_rho(LL n){LL p;if(n==1)return;if(Miller_Rabin(n))fat[++cnt] = n;else{p = Divi(n);Pollard_rho(n/p);Pollard_rho(p);}}void Sech(int x, LL now){if(x>=cnt+1 || x>sqrt(val)+1)return;if(now+val/now<ans){c = now;d = val/now;ans = now+val/now;}Sech(x+1, now*fat[x]);Sech(x+1, now);}int main(void){int i, k;LL n, m, now;while(scanf("%lld%lld", &n, &m)!=EOF){k = cnt = 0;val = m/n;Pollard_rho(val);sort(fat+1, fat+cnt+1);for(i=1;i<=cnt;i++){if(fat[i]!=fat[i-1]){if(i!=1)fat[++k] = now;now = fat[i];}elsenow *= fat[i];}fat[++k] = now;cnt = k;ans = 1e18;Sech(1, 1);if(c>d)swap(c, d);printf("%lld %lld\n", c*n, d*n);}return 0;}