例题:P4718
题意:
T组数据,输入T个数,对于每个数 n 判断是不是素数,如果是素数输出:"Prim",否则输出他的最大质因子。(1 < T < 350, 1 <= n <= 1e18)。
思路:
很典型的一道模板题,Pollard_Rho算法能够在$ O(n ^ {1/4})$的时间内找到 n 的一个质因子 p ,然后再递归计算 \(n' = n / p\), 直到 n 为素数为止,通过这样的方法将 n 进行质数分解。
模板代码:
// -----------------
#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(0);
#define endl '\n'
typedef long long ll;
using namespace std;
const int N = 1e5 + 10;
ll x, y, a[N], max_factor;
ll n;
struct BigIntegerFactor
{
const static int N = 1e6 + 7;
ll prime[N], p[N], fac[N], sz, cnt = 0;
inline ll mul(ll a, ll b, ll mod)
{
if(mod <= 1e9) return a * b % mod;
return (a * b - (ll)((long double)a / mod * b + 1e-8) * mod + mod ) % mod;
}
void init(int maxn)
{
int tot = 0;
sz = maxn - 1;
for(int i = 1; i <= sz; i ++) p[i] = i;
for(int i = 2; i <= sz; i ++)
{
if(p[i] == i) prime[tot ++] = i;
for(int j = 0; j < tot && 1ll * i * prime[j] <= sz; j ++)
{
p[i * prime[j]] = prime[j];
if(i % prime[j] == 0) break;
}
}
}
ll qpow(ll a, ll x, ll mod)
{
ll res = 1ll;
while(x)
{
if(x & 1) res = mul(res, a, mod);
a = mul(a, a, mod);
x >>= 1;
}
return res;
}
bool check(ll a, ll n) // 二次探测原理检验n
{
ll t = 0, u = n - 1;
while(!(u & 1)) t ++, u >>= 1;
ll x = qpow(a, u, n), xx = 0;
while(t --)
{
xx = mul(x, x, n);
if(xx == 1 && x != 1 && x != n - 1) return false;
x = xx;
}
return xx == 1;
}
bool miller(ll n, int k)
{
if(n == 2) return true;
if(n < 2 || !(n & 1)) return false;
if(n <= sz) return p[n] == n;
for(int i = 0; i <= k; i ++) // 测试k次
{
if( !check(rand() % (n - 1) + 1, n) ) return false;
}
return true;
}
inline ll gcd(ll a, ll b)
{
return b == 0 ? a : gcd(b, a % b);
}
inline ll Abs(ll x)
{
return x < 0 ? -x : x;
}
ll Pollard_rho(ll n) // 基于路径倍增的Pollard_Rho算法
{
ll s = 0, t = 0, c = rand() % (n - 1) + 1, v = 1, ed = 1;
while(1)
{
for(int i = 1; i <= ed; i ++)
{
t = (mul(t, t, n) + c) % n;
v = mul(v, Abs(t - s), n);
if(i % 127 == 0)
{
ll d = gcd(v, n);
if(d > 1) return d;
}
}
ll d = gcd(v, n);
if(d > 1) return d;
s = t;
v = 1;
ed <<= 1;
}
}
void getfactor(ll n)
{
if(n <= sz)
{
while(n != 1) fac[cnt ++] = p[n], n /= p[n];
max_factor = max_factor > p[n] ? max_factor : p[n];
return;
}
if(miller(n, 6))
{
fac[cnt ++] = n;
max_factor = max_factor > n ? max_factor : n;
}
else
{
ll d = n;
while(d >= n) d = Pollard_rho(n);
getfactor(d);
getfactor(n / d);
}
return;
}
} Q;
void solve()
{
max_factor = -1;
cin >> n;
Q.getfactor(n);
if(max_factor == n) cout << "Prime" << endl;
else cout << max_factor << endl;
}
signed main()
{
IOS
//Q.init(N-1); // 仅需要分解大数的质因数且代码超时
int T = 1;
cin >> T;
while(T --) {solve(); }
return 0;
}