GCD & LCM Inverse
题意：
给你两个数的gcd和lcm，[1, 2^63)。求a,b。使得a+b最小。
思路：
lcm = a * b / gcd 将lcm/gcd之后进行大数分解，形成a^x1 * b^x2 * c^x3…… 的形式，其中a,b,c为互不相同的质数。然后暴力枚举即可。
代码：
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, n) for (int i = 0; i < n; i++)
#define debug puts("===============")
typedef long long ll;
using namespace std;
const int S = 20;
//计算 (a*b)%c. a,b都是long long的数，直接相乘可能溢出的
// a,b,c <2^63
ll mult_mod(ll a, ll b, ll c) {
a %= c;
b %= c;
ll ret = 0;
while(b) {
if(b & 1) {
ret += a;
ret %= c;
}
a <<= 1;
if(a >= c) a %= c;
b >>= 1;
}
return ret;
}
//计算 x^n %c
ll pow_mod(ll x, ll n, ll mod) { //x^n%c
if(n == 1)return x % mod;
x %= mod;
ll tmp = x;
ll ret = 1;
while(n) {
if(n & 1) ret = mult_mod(ret, tmp, mod);
tmp = mult_mod(tmp, tmp, mod);
n >>= 1;
}
return ret;
}
//以a为基,n-1=x*2^t a^(n-1)=1(mod n) 验证n是不是合数
//一定是合数返回true,不一定返回false
bool check(ll a, ll n, ll x, ll t) {
ll ret = pow_mod(a, x, n);
ll last = ret;
for(int i = 1; i <= t; i++) {
ret = mult_mod(ret, ret, n);
if(ret == 1 && last != 1 && last != n - 1) return true; //合数
last = ret;
}
if(ret != 1) return true;
return false;
}
// Miller_Rabin()算法素数判定
//是素数返回true.(可能是伪素数，但概率极小)
//合数返回false;
bool Miller_Rabbin(ll n) {
if(n < 2)return false;
if(n == 2)return true;
if((n & 1) == 0) return false;
ll x = n - 1;
ll t = 0;
while((x & 1) == 0) {
x >>= 1;
t++;
}
for(int i = 0; i < S; i++) {
ll a = rand() % (n - 1) + 1;
if(check(a, n, x, t))
return false;//合数
}
return true;
}
//************************************************
//pollard_rho 算法进行质因数分解
//************************************************
ll factor[100];//质因数分解结果（刚返回时是无序的）
int tot;//质因数的个数。数组小标从0开始
ll gcd(ll a, ll b) {
if(a == 0) return 1;
if(a < 0) return gcd(-a, b);
while(b) {
ll t = a % b;
a = b;
b = t;
}
return a;
}
ll Pollard_rho(ll x, ll c) {
ll i = 1, k = 2;
ll x0 = rand() % x;
ll y = x0;
while(1) {
i++;
x0 = (mult_mod(x0, x0, x) + c) % x;
ll d = gcd(y - x0, x);
if(d != 1 && d != x) return d;
if(y == x0) return x;
if(i == k) {
y = x0;
k += k;
}
}
}
//对n进行素因子分解
void findfac(ll n) {
if(Miller_Rabbin(n)) { //素数
factor[tot++] = n;
return;
}
ll p = n;
while(p >= n) p = Pollard_rho(p, rand() % (n - 1) + 1);
findfac(p);
findfac(n / p);
}
int vis[111] = {0};
ll g, lcm;
ll ans = (1LL << 61) , s;
int main () {
while(~scanf("%lld%lld", &g, &lcm)) {
lcm /= g;
if (lcm == 1) {
printf("%lld %lld\n", g, g);
continue;
}
tot = 0;
findfac(lcm);
sort(factor, factor + tot);
int cnt = 0;
ll a[111];
ll res = 1, now = factor[0];
for (int i = 0; i < tot; i++) {
if (factor[i] == now) {
res = res * now;
} else {
a[cnt++] = res;
now = factor[i];
res = now;
}
}
a[cnt++] = res;
int t = (1 << cnt);
ll A = 1, B = lcm;
for (int i = 0; i < t; i++) {
ll TA = 1, TB = 1;
for (int j = 0; j < cnt; j++)
if (i & (1 << j)) TA *= a[j];
else TB *= a[j];
if (TA + TB < A + B) {
A = TA;
B = TB;
}
}
if (A > B) swap(A, B);
printf("%lld %lld\n", A * g, B * g);
}
return 0;
}