【模板】SPOJ FACT0 大数分解 miller-rabin & pollard-rho

http://www.spoj.com/problems/FACT0/en/
给一个小于1e15的数,将他分解。

miller-rabin & pollard-rho模板

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int xjb=10;
ll mmul(ll a, ll b, ll m){
	ll d=((long double)a/m*b+1e-8);
	ll r=a*b-d*m;
	return r<0?r+m:r;
}
ll mpow(ll a, ll b, ll m){ll r=1;for(;b;b>>=1,a=mmul(a,a,m))if(b&1)r=mmul(r,a,m);return r;}
ll gcd(ll a, ll b){return a?gcd(b%a,a):b;}
int prime(ll n){
	if(n==1) return 0;
	if(n==2||n==3||n==5) return 1;
	if(!(n&1)||(n%3==0)||(n%5==0)) return 0;
	ll m=n-1; int k=0;
	while(!(m&1)) m>>=1, k++;
	for(int tt=0; tt<xjb; ++tt){
		ll x=mpow(rand()%(n-2)+2,m,n), y=x;
		for(int i=0; i<k; ++i){
			x=mmul(x,x,n);
			if(x==1&&y!=1&&y!=n-1) return 0;
			y=x;
		}
		if(x!=1) return 0;
	}
	return 1;
}
ll f[105]; int M;
ll rho(ll n, ll c){
	ll x=rand()%n, y=x, t=1;
	for(int i=1, k=2; t==1; ++i){
		x=(mmul(x,x,n)+c)%n;
		t=gcd(x>y?x-y:y-x, n);
		if(i==k) y=x, k<<=1;
	}
	return t;
}
void work(ll n){
	if(n==1) return;
	if(prime(n)){f[M++]=n; return;}
	ll t=n;
	while(t==n) t=rho(n, rand()%5+1);
	work(t); work(n/t);
}
int main(){
	srand(19260817);
	ll n; 
	while(scanf("%lld", &n), n){
		if(n==1){puts(""); continue;}
		M=0;
		work(n);
		sort(f, f+M);
		for(int i=0, c=1; i<M; ++i){
			if(f[i]!=f[i+1]) printf("%lld^%d ", f[i], c), c=1;
			else c++;
		}
		puts("");
	}
	return 0;
}
posted @ 2017-03-20 15:09  will7101  阅读(350)  评论(0编辑  收藏  举报