数学--数论--随机算法--Pollard Rho 大数分解算法 （带输出版本）
2024-09-07 11:57:10
RhoPollard Rho是一个著名的大数质因数分解算法，它的实现基于一个神奇的算法：MillerRabinMillerRabin素数测试。
操作流程
首先，我们先用MillerRabinMillerRabin判断当前数xx是否为质数，若是，则可直接统计信息并退出函数
然后是各种证明及优化，我觉得不大实用，这个板子是我改了很多遍了，也过了很多题的板子。用着很舒服，无论卡常，不卡常，速度相差不大，也可以加read.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll pr;
ll pmod(ll a, ll b, ll p) { return (a * b - (ll)((long double)a / p * b) * p + p) % p; } //普通的快速乘会T
ll gmod(ll a, ll b, ll p)
{
ll res = 1;
while (b)
{
if (b & 1) res = pmod(res, a, p);
a = pmod(a, a, p);
b >>= 1;
}
return res;
}
inline ll gcd(ll a, ll b)
{ //听说二进制算法特快
if (!a) return b;
if (!b)return a;
int t = __builtin_ctzll(a | b);
a >>= __builtin_ctzll(a);
do
{
b >>= __builtin_ctzll(b);
if (a > b)
{
ll t = b;
b = a, a = t;
}
b -= a;
} while (b);
return a << t;
}
bool Miller_Rabin(ll n)
{
if (n == 46856248255981ll || n < 2)
return false; //强伪素数
if (n == 2 || n == 3 || n == 7 || n == 61 || n == 24251)
return true;
if (!(n & 1) || !(n % 3) || !(n % 61) || !(n % 24251))
return false;
ll m = n - 1, k = 0;
while (!(m & 1))
k++, m >>= 1;
for (int i = 1; i <= 20; ++i) // 20为Miller-Rabin测试的迭代次数
{
ll a = rand() % (n - 1) + 1, x = gmod(a, m, n), y;
for (int j = 1; j <= k; ++j)
{
y = pmod(x, x, n);
if (y == 1 && x != 1 && x != n - 1)
return 0;
x = y;
}
if (y != 1)
return 0;
}
return 1;
}
ll Pollard_Rho(ll x)
{
ll n = 0, m = 0, t = 1, q = 1, c = rand() % (x - 1) + 1;
for (ll k = 2;; k <<= 1, m = n, q = 1)
{
for (ll i = 1; i <= k; ++i)
{
n = (pmod(n, n, x) + c) % x;
q = pmod(q, abs(m - n), x);
}
t = gcd(x, q);
if (t > 1)
return t;
}
}
void fid(ll n)
{
if (n == 1)
return;
if (Miller_Rabin(n))
{
pr = max(pr, n);
return;
}
ll p = n;
while (p >= n)
p = Pollard_Rho(p);
fid(p);
fid(n / p);
}
int main()
{
int T;
ll n;
scanf("%d", &T);
while (T--)
{
scanf("%lld", &n);
pr = 0;
fid(n);
if (pr == n)
puts("Prime");
else
printf("%lld\n", pr);
}
return 0;
}
带输出的我也写了
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll pr;
ll pmod(ll a, ll b, ll p) { return (a * b - (ll)((long double)a / p * b) * p + p) % p; } //普通的快速乘会T
ll gmod(ll a, ll b, ll p)
{
ll res = 1;
while (b)
{
if (b & 1)
res = pmod(res, a, p);
a = pmod(a, a, p);
b >>= 1;
}
return res;
}
inline ll gcd(ll a, ll b)
{ //听说二进制算法特快
if (!a)
return b;
if (!b)
return a;
int t = __builtin_ctzll(a | b);
a >>= __builtin_ctzll(a);
do
{
b >>= __builtin_ctzll(b);
if (a > b)
{
ll t = b;
b = a, a = t;
}
b -= a;
} while (b);
return a << t;
}
bool Miller_Rabin(ll n)
{
if (n == 46856248255981ll || n < 2)
return false; //强伪素数
if (n == 2 || n == 3 || n == 7 || n == 61 || n == 24251)
return true;
if (!(n & 1) || !(n % 3) || !(n % 61) || !(n % 24251))
return false;
ll m = n - 1, k = 0;
while (!(m & 1))
k++, m >>= 1;
for (int i = 1; i <= 20; ++i) // 20为Miller-Rabin测试的迭代次数
{
ll a = rand() % (n - 1) + 1, x = gmod(a, m, n), y;
for (int j = 1; j <= k; ++j)
{
y = pmod(x, x, n);
if (y == 1 && x != 1 && x != n - 1)
return 0;
x = y;
}
if (y != 1)
return 0;
}
return 1;
}
ll Pollard_Rho(ll x)
{
ll n = 0, m = 0, t = 1, q = 1, c = rand() % (x - 1) + 1;
for (ll k = 2;; k <<= 1, m = n, q = 1)
{
for (ll i = 1; i <= k; ++i)
{
n = (pmod(n, n, x) + c) % x;
q = pmod(q, abs(m - n), x);
}
t = gcd(x, q);
if (t > 1)
return t;
}
}
map<long long, int> m;
void fid(ll n)
{
if (n == 1)
return;
if (Miller_Rabin(n))
{
pr = max(pr, n);
m[n]++;
return;
}
ll p = n;
while (p >= n)
p = Pollard_Rho(p);
fid(p);
fid(n / p);
}
int main()
{
int T;
ll n;
scanf("%d", &T);
while (T--)
{
m.clear();
scanf("%lld", &n);
pr = 0;
fid(n);
if (pr == n)
puts("Prime");
else
{
printf("%lld\n", pr);
for (map<long long, int>::iterator c = m.begin(); c != m.end();)
{
printf("%lld^%d", c->first, c->second);
if ((++c) != m.end())
printf(" * ");
}
printf("\n");
}
}
return 0;
}
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