zoukankan      html  css  js  c++  java
  • Prime Distance(二次筛素数)

    Description

    The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers. 
    Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).

    Input

    Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000.

    Output

    For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes.

    Sample Input

    2 17
    14 17
    

    Sample Output

    2,3 are closest, 7,11 are most distant.
    There are no adjacent primes.

    解题思路:

    这题做得我都是泪。不断地TLE,好不easy优化好了。又RE。代码也写得非常龊。就是正常的二次筛选素数。

    因为数据非常大。第一次筛出46500以内的素数。再依据此筛选出区间内的素数。

    注意:尽管给的数没有超int范围,但两数相乘是会超int范围的,我也是在这里RE了。

    AC代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    using namespace std;
    const int N = 1000005;
    const int M = 46500;
    const int INF = 999999999;
    bool notprime[N];
    int prime_1[M + 1], prime_2[N];
    int num_1 = 0, num_2;
    void Prime1()  // 第一次筛出46500以内的素数
    {
        memset(notprime, false, sizeof(notprime));
        for(int i = 2; i <= M; i++)
            if(!notprime[i])
            {
                prime_1[num_1++] = i;
                for(int j = 2 * i; j <= M; j += i)
                    notprime[j] = true;
            }
    }
    void Prime2(int l, int u)  // 第二次筛出给定范围内的素数
    {
        memset(notprime, false, sizeof(notprime));
        num_2 = 0;
        if(l < 2)
            l = 2;
        int k = sqrt(u * 1.0);
        for(int i = 0; i < num_1 && prime_1[i] <= k; i++)
        {
            int t = l / prime_1[i];
            if(t * prime_1[i] < l)
                t++;
            if(t <= 1)
                t = 2;
            for(int j = t; (long long)j * prime_1[i] <= u; j++)  // 相乘会超范围,用long long
                notprime[j * prime_1[i] - l] = 1;
        }
        for(int i = 0; i <= u - l; i++)
            if(!notprime[i])
                prime_2[num_2++] = i + l;
    }
    int main()
    {
        int l, u, dis, a_1, b_1, a_2, b_2, minn, maxx;;
        Prime1();
        while(scanf("%d%d", &l, &u) != EOF)
        {
            minn = INF, maxx = -1;
            Prime2(l, u);
            if(num_2  < 2)
            {
                printf("There are no adjacent primes.
    ");
                continue;
            }
            for(int i = 1; i < num_2 && prime_2[i] <= u; i++)
            {
                dis = prime_2[i] - prime_2[i - 1];
                if(dis > maxx)
                {
                    a_1 = prime_2[i - 1];
                    a_2 = prime_2[i];
                    maxx = dis;
                }
                if(dis < minn)
                {
                    b_1 = prime_2[i-1];
                    b_2 = prime_2[i];
                    minn = dis;
                }
            }
            printf("%d,%d are closest, %d,%d are most distant.
    ", b_1, b_2, a_1, a_2);
        }
        return 0;
    }
    



  • 相关阅读:
    容斥原理算法总结(bzoj 2986 2839)
    网络流系列算法总结(bzoj 3438 1061)
    bzoj 2746: [HEOI2012]旅行问题 AC自动机fail树
    bzoj 3283: 运算器 扩展Baby Step Giant Step && 快速阶乘
    计算几何考场绘图技巧
    bzoj 1845: [Cqoi2005] 三角形面积并 扫描线
    bzoj 3784: 树上的路径 堆维护第k大
    BZOJ 1231: [Usaco2008 Nov]mixup2 混乱的奶牛
    BZOJ 1112: [POI2008]砖块Klo
    BZOJ 1003: [ZJOI2006]物流运输trans DP+最短路
  • 原文地址:https://www.cnblogs.com/gccbuaa/p/6843870.html
Copyright © 2011-2022 走看看