题解
找质因子用Miller-Rabin和Pollard-Rho实现，然后求解的时候考虑对勾函数，当$a$和$b$越接近$\sqrt{ab}$时$a+b$越小。就做完了。1
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using namespace std;
typedef long long ll;
ll g, l, prime[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
ll dd[55], ans = -1;
int e1[55] = {0}, e2[55] = {0}, cnt = 0;
double lev, dis;
ll gcd(ll a, ll b){
return (!b) ? a : gcd(b, a % b);
}
ll mul(ll a, ll b, ll M){
ll res = 0;
while(b){
if(b & 1ll) {
res += a;
if(res >= M) res -= M;
}
a <<= 1, b >>= 1;
if(a >= M) a -= M;
}
return res;
}
ll modpow(ll a, ll b, ll M){
a %= M;
ll res = 1;
while(b){
if(b & 1ll) res = mul(res, a, M);
a = mul(a, a, M), b >>= 1;
}
return res;
}
bool witness(ll a, ll n, ll t, ll u){
ll x = modpow(a, u, n);
for(ll i = 1; i <= t; ++i){
ll xx = mul(x, x, n);
if(xx == 1 && x != 1 && x != n - 1)
return true;
x = xx;
}
if(x != 1) return true;
return false;
}
bool miller_rabin(ll n){
for(int i = 0; i < 10; ++i){
if(n == prime[i]) return true;
else if(n % prime[i] == 0) return false;
}
ll t, u;
for(t = 1; ; ++t)
if((n - 1) % (1 << t) == 0) break;
u = (n - 1) / (1 << t);
for(int i = 0; i < 10; ++i)
if(witness(rand() % (n - 1) + 1, n, t, u))
return false;
return true;
}
ll rho(ll n, ll c){
ll x = modpow(rand(), rand(), n), y = x, d = 1;
int k = 2;
for(int i = 1; d == 1; ++i){
x = mul(x, x, n) + c;
if(x >= n) x -= n;
if(x > y) d = gcd(x - y, n);
else d = gcd(y - x, n);
if(i == k) y = x, k <<= 1;
}
return d;
}
ll Pollard(ll n){
ll d = n;
while(d == n)
d = rho(n, rand() % (n - 1) + 1);
return d;
}
void addFac(ll &x, ll d, int *e){
int cur;
for(cur = 0; cur < cnt; ++cur)
if(dd[cur] == d) break;
dd[cur] = d;
do{
x /= d, e[cur]++;
}while(x % d == 0);
if(cur == cnt) cnt++;
}
void getFac(ll x, int *e){
if(!miller_rabin(x)){
ll d = Pollard(x);
getFac(d, e), getFac(x / d, e);
}else{
addFac(x, x, e);
}
}
void dfs(int ind, ll x){
if(ind == cnt){
if(1.0 * x > lev && ans > x)
ans = x;
return ;
}
ll res = 1;
for(int i = 0; i < e2[ind]; ++i) res *= dd[ind];
dfs(ind + 1, x * res);
for(int i = e2[ind]; i < e1[ind]; ++i) res *= dd[ind];
dfs(ind + 1, x * res);
}
void init(){
srand(122144);
}
void solve(){
cnt = 0;
memset(dd, 0, sizeof(dd));
memset(e1, 0, sizeof(e1));
memset(e2, 0, sizeof(e2));
if(g == l) cout << g << " " << l << endl;
else{
getFac(l, e1);
if(g != 1) getFac(g, e2);
lev = sqrt(1.0 * l * g);
ans = l, dfs(0, 1);
cout << (l / ans) * g << " " << ans << endl;
}
}
int main(){
while(scanf("%lld %lld", &g, &l) == 2){
init();
solve();
}
return 0;
}