E.Multiply Pollard_rho质因数分解

本文主要是介绍E.Multiply Pollard_rho质因数分解，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

2019 icpc xuzhou
思路很简单，但是这个Pollard_rho的模板要选好，不然不是wa 就是 tle ，我太难了

#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <algorithm>
#include <iostream>
const int S=20;
using namespace std;typedef long long ll;
#define maxn 1000000ll factor[maxn];
int tot;ll muti_mod(ll a,ll b,ll c){    //返回(a*b) mod c,a,b,c<2^63a%=c;b%=c;ll ret=0;while (b){if (b&1){ret+=a;if (ret>=c) ret-=c;}a<<=1;if (a>=c) a-=c;b>>=1;}return ret;
}ll pow_mod(ll x,ll n,ll mod){  //返回x^n mod c ,非递归版if (n==1) return x%mod;int bit[64],k=0;while (n){bit[k++]=n&1;n>>=1;}ll ret=1;for (k=k-1;k>=0;k--){ret=muti_mod(ret,ret,mod);if (bit[k]==1) ret=muti_mod(ret,x,mod);}return ret;
}bool check(ll a,ll n,ll x,ll t){   //以a为基，n-1=x*2^t，检验n是不是合数ll ret=pow_mod(a,x,n),last=ret;for (int i=1;i<=t;i++){ret=muti_mod(ret,ret,n);if (ret==1 && last!=1 && last!=n-1) return 1;last=ret;}if (ret!=1) return 1;return 0;
}bool Miller_Rabin(ll n){ll x=n-1,t=0;while ((x&1)==0) x>>=1,t++;bool flag=1;if (t>=1 && (x&1)==1){for (int k=0;k<S;k++){ll a=rand()%(n-1)+1;if (check(a,n,x,t)) {flag=1;break;}flag=0;}}if (!flag || n==2) return 0;return 1;
}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,x0=rand()%x,y=x0,k=2;while (1){i++;x0=(muti_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;}}
}void findfac(ll n){           //递归进行质因数分解Nif (!Miller_Rabin(n)){factor[tot++] = n;return;}ll p=n;while (p>=n) p=Pollard_rho(p,rand() % (n-1) +1);findfac(p);findfac(n/p);
}
int idx = 0 ;
struct node
{ll val , num ;node() {}node(ll val , ll num) : val(val) , num(num) {} bool operator<(const node &a) const {val < a.val ;}
}fac[maxn];
ll b[maxn] ;
bool solve(ll n)
{if (!Miller_Rabin(n)) {fac[++ idx] = {n , 1} ;}else{tot = 0;findfac(n);sort(factor , factor + tot) ;idx = 0 ;for(int i = 0 ; i < tot ;i ++){if(factor[i] != fac[idx].val) fac[++ idx] = {factor[i] , 1};else fac[idx].num ++ ;}}
}
ll get(ll n , ll p)
{ll res = 0 ;while(n){res += n / p ;n /= p ;}return res ;
}
int main(){srand(time(NULL));int t;scanf("%d",&t);while (t--){ll n , x , y ;scanf("%lld%lld%lld",&n , &x , &y);idx = 0 ;solve(x) ;for(int j = 1; j <= idx ;j ++) b[j] = 0 ;for(int i = 1; i <= n ;i ++){ll p ;scanf("%lld" , &p) ;for(int j = 1 ;j <= idx ;j ++)b[j] += get(p , fac[j].val)  ;}ll minx = 0 ;for(int j = 1; j <= idx ;j ++)b[j] = get(y , fac[j].val) - b[j] , minx = max(minx , b[j]);for(int j = 1; j <= idx ;j ++)if(b[j] > 0)minx = min(minx , b[j] / fac[j].num) ; printf("%lld\n" , minx) ;}
}