PoRho模板

以O(n^(1/4))的复杂度对一个数进行质因子分解

无敌了已经

ll ksc(int a,int b,int p){
    ll z = (long double)a/p*b;
    ll res =(ull)a*b -(ull)(z*p);
    return (res+p)%p;
}
int ksm(int a,int b,int p){
    int res=1;
    while(b){
        if(b&1)res=ksc(res,a,p);
        a=ksc(a,a,p);
        b>>=1;
    }
    return res%p;
}
int mr(int a,int p){
    if(ksm(a,p-1,p)!=1)return 0;
    int x= p-1,y=0;
    while(x%2==0){
        x>>=1;
        y=ksm(a,x,p);
        if(y!=1&&y!=p-1)return 0;
        if(y==p-1)return 1;
    }
    return 1;
}
int isprime(int x){
    if(x<2)return 0;
    if(x==2||x==3||x==5||x==7||x==43)return 1;
    return mr(2,x)&&mr(3,x)&&mr(5,x)&&mr(7,x)&&mr(43,x);
}
int rho(int n){
    ll x,y,z,c,g;
    ll i,j;
    //f(x) =x*x + c
    //gcd(abs(y-x),n)>1 => 找到非平凡因子
    while(1){
        x=y=rand()%n;
        z=1;c=rand()%n;
        i=0;j=1;
        while(++i){
            x=(ksc(x,x,n)+c)%n;
            z=ksc(z,abs(y-x),n);
            if(x==y||!z)break;
            if(!(i%127)||i==j){
                g=__gcd(z,n);
                if(g>1)return g;
                if(i==j)y=x,j<<=1;
            }
        }
    }
}
vector<int>v;
void work(int n){
    if(n==1)return;
    if(isprime(n)){
        v.pb(n);return;
    }
    int pi = rho(n);
    while(n%pi==0)n/=pi;
    work(pi);
    work(n);
}