poj1811(pollard_rho模板)

题目链接: http://poj.org/problem?id=1811

题意: 判断一个数 n (2 <= n < 2^54)是否为质数, 是的话输出 "Prime", 否则输出其第一个质因子.

思路: 大数质因子分解, 直接用 pollard_rho (详情参见: http://blog.csdn.net/maxichu/article/details/45459533) 模板即可.

代码:

 #include <iostream>
 #include <algorithm>
 #define ll long long
 using namespace std;

 const int MAXN = 1e2;
 const int repeat = ;//repeat为检测次数,判断错误率为4^-repeat,一般8~10就够了

 ll get_mult(ll a, ll b, ll c){//返回a*b%c
     a %= c;
     b %= c;
     ll ret = ;
     while(b){
         if(b & ){
             ret += a;
             if(ret >= c) ret -= c;
         }
         b >>= ;
         a <<= ;
         if(a >= c) a -= c;
     }
     return ret;
 }

 ll get_pow(ll x, ll n, ll mod){//返回x^n%mod
     ll ret = ;
     x %= mod;
     while(n){
         if(n & ) ret = get_mult(ret, x, mod);
         x = get_mult(x, x, mod);
         n >>= ;
     }
     return ret;
 }

 //通过 a^(n-1) = 1(mod n) 来判断n是否为素数
 //n-1 = x*2^t 中间使用二次探测定理
 //是合数返回true,不一定是合数返回false
 bool cherk(ll a, ll n, ll x, ll t){
     ll ret = get_pow(a, x, n);
     ll last = ret;
     for(int i = ; i <= t; i++){
         ret = get_mult(ret, ret, n);
         if(ret ==  && last !=  && last != n - ) return true;
         last = ret;
     }
     if(ret != ) return true;
     return false;
 }

 bool Miller_Rabin(ll n){
     if(n < ) return false;
     if(n == ) return true;
     if(!(n & )) return false;//偶数
     ll x = n - , t = ;
     while(!(x & )){
         x >>= ;
         t++;
     }
     // srand(time(NULL));
     for(int i = ; i < repeat; i++){
         ll a = rand() % (n - ) + ;
         if(cherk(a, n, x, t)) return false;
     }
     return true;
 }

 ll factor[MAXN];
 int tot;//n的质因子个数

 ll get_gcd(ll a, ll b){
     ll tmp;
     while(b){
         tmp = a;
         a = b;
         b = tmp % b;
     }
     return a >=  ? a : -a;
 }

 ll pollard_rho(ll x, ll c){//返回x的一个因子
     ll i = , k = ;
     // srand(time(NULL));
     ll x0 = rand() % (x - ) + ;
     ll y = x0;
     while(){
         i++;
         x0 = (get_mult(x0, x0, x) + c) % x;
         ll d = get_gcd(y - x0, x);
         if(d !=  && d != x) return d;
         if(y == x0) return x;
         if(i == k){
             y = x0;
             k += k;
         }
     }
 }

 void findfac(ll n, int k){//对n质因分解并将结果保存到factor中
     if(n == ) return;
     if(Miller_Rabin(n)){
         factor[tot++] = n;
         return;
     }
     ll p = n;
     int c = k;//c防止死循环
     while(p >= n){
         p = pollard_rho(p, c--);
     }
     findfac(p, k);
     findfac(n / p, k);
 }

 int main(void){
     ll n;
     int t;
     cin >> t;
     while(t--){
         cin >> n;
         if(Miller_Rabin(n)) cout << "Prime" << endl;
         else{
             tot = ;
             findfac(n, );
             ll sol = factor[];
             for(int i = ; i < tot; i++){
                 sol = min(sol, factor[i]);
             }
             cout << sol << endl;
         }
     }
     return ;
 }