质数

试除法判定质数

bool is_prime(int x)
{
    if (x < 2) return false;
    for (int i = 2; i <= x / i; i ++ )
        if (x % i == 0)
            return false;
    return true;
}

试除法分解质因数

质因数：每个合数都可以写成几个质数（也可称为素数）相乘的形式，这几个质数就都叫做这个合数的质因数。
C++

void divide(int x)
{
    for (int i = 2; i <= x / i; i ++ )
        if (x % i == 0)
        {
            int s = 0;
            while (x % i == 0) x /= i, s ++ ;
            cout << i << ' ' << s << endl;
        }
    if (x > 1) cout << x << ' ' << 1 << endl;
    cout << endl;
}

筛质数

朴素筛法（埃式筛）

每次筛掉已确定质数的倍数，这样筛的话会有数被重复筛

int primes[N], cnt;     // primes[]存储所有素数
bool st[N];         // st[x]存储x是否被筛掉

void get_primes(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (st[i]) continue;
        primes[cnt ++ ] = i;
        for (int j = i + i; j <= n; j += i)
            st[j] = true;
    }
}

线性筛（欧拉筛）

核心：n只会被它的最小质因子筛掉

每次筛掉当前质数和已确定质数的乘积，所有的数只会被筛一遍

int primes[N], cnt;     // primes[]存储所有素数
bool st[N];         // st[x]存储x是否被筛掉

void get_primes(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] <= n / i; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;  //这个时候primes[j]一定是i的最小质因子
        }
    }
}

练习

01 试除法判定质数

题目

题解

import java.io.*;
import java.util.*;

public class Main {
    public static final int N = 110;
    public static int[] q = new int[N];
    public static int n, idx;

    public static boolean is_prime(int x) {
        if (x < 2) {
            return false;
        }

        for (int i = 2; i <= x / i; i++) {
            if (x % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static void solve() {
        if (is_prime(q[idx++])) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        n = Integer.parseInt(br.readLine());
        for (int i = 0; i < n; i++) {
            q[i] = Integer.parseInt(br.readLine());
        }

        idx = 0;
        while (n-- > 0) {
            solve();
        }
        br.close();
    }
}

02 分解质因数

题目

题解

import java.io.*;
import java.util.*;

public class Main {
    public static final int N = 110;
    public static int[] q = new int[N];
    public static int n, idx;

    public static void solve(int x) {
        for (int i = 2; i <= x / i; i++) {
            if (x % i == 0) {
                int s = 0;
                while (x % i == 0) {
                    x /= i;
                    s++;
                }
                System.out.println(i + " " + s);
            }
        }
        if (x > 1) {
            System.out.println(x + " " + 1);
        }
        System.out.println();
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        n = Integer.parseInt(br.readLine());
        for (int i = 0; i < n; i++) {
            q[i] = Integer.parseInt(br.readLine());
        }

        idx = 0;
        while (n-- > 0) {
            solve(q[idx++]);
        }
        br.close();
    }
}

03 筛质数

题目

题解1

//埃式筛
import java.io.*;
import java.util.*;

public class Main {
    public static final int N = 1000010;
    //存储已经确定的质数
    public static int[] prime = new int[N];
    //存储st[i]是否被筛掉
    public static boolean[] st = new boolean[N];
    public static int n, idx;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        n = Integer.parseInt(br.readLine());

        idx = 0;
        for (int i = 2; i <= n; i++) {
            if (st[i]) {
                continue;
            }

            prime[idx++] = i;
            for (int j = i + i; j <= n; j += i) {
                st[j] = true;
            }
        }

        pw.println(idx);
        pw.close();
        br.close();
    }
}

题解2

//线性筛
import java.io.*;
import java.util.*;

public class Main {
    public static final int N = 1000010;
    //存储已经确定的质数
    public static int[] prime = new int[N];
    //存储st[i]是否被筛掉
    public static boolean[] st = new boolean[N];
    public static int n, idx;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        n = Integer.parseInt(br.readLine());

        idx = 0;
        for (int i = 2; i <= n; i++) {
            if (!st[i]) {
                prime[idx++] = i;
            }

            for (int j = 0; prime[j] <= n / i; j++) {
                st[prime[j] * i] = true;
                if (i % prime[j] == 0) {
                    break;
                }
            }
        }

        pw.println(idx);
        pw.close();
        br.close();
    }
}