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  • Pollard-rho算法:模板

    #include<algorithm>
    #include<cstdio>
    #include<cstdlib>
    #define N 5500
    using namespace std;
    typedef long long ll;
    ll ct,cnt;  
    ll fac[N],num[N];
    const int BASE[]={2,3,5,7,11,13,17,19,23};
    ll Quick_Mul(ll a,ll p,ll MOD)
    {
        if(!p){
            return 0;
        }
        ll ans=Quick_Mul(a,p>>1,MOD);
        ans=(ans+ans)%MOD;
        if((p&1ll)==1ll){
            ans=ans+a%MOD%MOD;
        }
        return ans;
    }
    ll Quick_Pow(ll a,ll p,ll MOD)
    {
        if(!p){
            return 1;
        }
        ll ans=Quick_Pow(a,p>>1,MOD);
        ans=Quick_Mul(ans,ans,MOD);
        if((p&1ll)==1ll){
            ans=a%MOD*ans%MOD;
        }
        return ans;
    }
    bool test(ll n,ll a,ll d){
        if(n==2){
            return 1;
        }
        if(n==a){
            return 0;
        }
        if(!(n&1)){
            return 0;
        }
        while(!(d&1ll)){
            d>>=1;
        }
        ll t=Quick_Pow(a,d,n);
        if(t==1){
            return 1;
        }
        while(d!=n-1ll && t!=n-1ll && t!=1ll){
            t=Quick_Mul(t,t,n);
            d<<=1;
        }
        return t==n-1ll;
    }
    bool Miller_Rabin(ll n){
        if(n==1 || n==3825123056546413051ll){
            return 0;
        }
        for(int i=0;i<9;++i){
            if(n==(ll)BASE[i]){
                return 1;
            }
            if(!test(n,(ll)BASE[i],n-1ll)){
                return 0;
            }
        }
        return 1;
    }
    ll pollard_rho(ll n,ll c){
        ll i=1,k=2;
        ll x=rand()%(n-1)+1;
        ll y=x;
        while(1){
            i++;
            x=(Quick_Mul(x,x,n)+c)%n;
            ll d=__gcd((y-x+n)%n,n);
            if(1ll<d &&d<n){
            	return d;
            }
            if(y==x){
            	return n;
            }
            if(i==k){
                y=x;
                k<<=1;
            }
        }
    }
    void find(ll n,int c){
        if(n==1){
        	return;
        }
        if(Miller_Rabin(n)){
            fac[ct++]=n;
            return;
        }
        ll p=n;
        ll k=c;
        while(p>=n){
        	p=pollard_rho(p,c--);
        }
        find(p,k);
        find(n/p,k);
    }
    ll n;
    int main(){
    	srand(233);
        while(scanf("%I64d",&n)!=EOF){
            ct=0;
            find(n,120);
            sort(fac,fac+ct);
            num[0]=1;
            int k=1;
            for(int i=1;i<ct;++i){
                if(fac[i]==fac[i-1]){
                	++num[k-1];
                }
                else{
                    num[k]=1;
                    fac[k++]=fac[i];
                }
            }
            cnt=k;
            for(int i=0;i<cnt;++i){
            	printf("%I64d^%I64d
    ",fac[i],num[i]);
            }
            puts("");
        }  
        return 0;  
    }  
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  • 原文地址:https://www.cnblogs.com/autsky-jadek/p/7750708.html
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