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  • 素数判断的多种方法

    包括线性筛,朴素判断方法,以及miller-rabin(非加强版)

    #include <iostream>
    #include <cstdlib>
    #include <vector>
    #include <map>
    using namespace std;
    
    inline bool sqrt_judge(int x)  //复杂度O(sqrt(n))
    {
        if(x == 2) return 1;
        if(x == 1) return 0;
        for(int i = 2; i*i <= x; i++) if(x % i == 0) return 0;
        return 1;
    }
    
    inline int Pow(int x, int P)
    {
        int ans = 1;
        for(int i = 1; i <= P; i++) ans = ans*x%(P+1);
        return ans;
    }
    
    inline bool Miller_Rabin(int x)  //不稳定随机判断,复杂度O(1)
    {
        if(x == 2) return 1;
        if(x == 1) return 0;
        int p[7] = {2, 3, 5, 7, 11, 13, 17};
        for(int i = 0; i < 7 && p[i] != x; i++) if(Pow(x, p[i]-1) != 1) return 0;
        return 1;
    }
    vector <int> prime;
    map <int, int> check;
    void Select()  //素数筛,筛选复杂度O(n),判断复杂度O(1)
    {
        prime.clear(); check.clear();
        for(int i = 2; i <= 100000; i++)
        {
            if(! check[i]) prime.push_back(i);
            for(auto x : prime)
            {
                if(i * x > 100000) break;
                check[i * x] = 1;
                if(i % x == 0) break;
            }
        }
    }
    
    int main()
    {
        Select();
        int n;
        while(cin>>n)
        {
            int ans = 0, i;
            switch(rand()%3)
            {
                case 0: for(i = 1; i < n; i++) if(sqrt_judge(i)) ans += i; break;
                case 1: for(i = 1; i < n; i++) if(Miller_Rabin(i)) ans += i; break;
                case 2: for(auto x : prime) if(x < n) ans += x; else break; break;
            }
            cout<<ans<<endl;
        }
    }
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  • 原文地址:https://www.cnblogs.com/Saurus/p/6017219.html
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