数论进阶（二）

原创于 2026-11-03 19:00:00 发布 · 370 阅读

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GEO检测

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算法题

本文聚焦数论进阶中的关键算法实践，涵盖最小乘积代价和优化、数字权值计算、埃氏筛法/线性筛法判断质数、区间内质数统计及素数对构造等典型问题，强调算法原理与高效实现，适用于算法竞赛与系统性能优化场景。

一、最小乘积代价和

import java.util.Scanner;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNextInt()) { // 注意 while 处理多个 case
            int n = in.nextInt();
            if (n == 1) {
                System.out.println(0);
                continue;
            }
            long sum = 0;
            int x = n;
            for (int i = 2; i * i <= x; i++) {
                while (x % i == 0) {
                    sum += i;
                    x /= i;
                }
            }
            if (x > 1) {
                sum += x;
            }
            System.out.println(sum);
        }
    }
}

运行结果

二、小苯的数字权值

import java.util.Scanner;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    static final int MAX = 200000;
    static int[] omega = new int[MAX + 1];
    static int[] primes = new int[MAX + 1];
    static boolean[] isPrime = new boolean[MAX + 1];
    static int cnt = 0;

    static void sieve() {
        // 初始化
        for (int i = 2; i <= MAX; i++) {
            isPrime[i] = true;
        }
        for (int i = 2; i <= MAX; i++) {
            if (isPrime[i]) {
                primes[cnt++] = i;
                omega[i] = 1;
            }
            for (int j = 0; j < cnt && primes[j] * i <= MAX; j++) {
                int p = primes[j];
                isPrime[i * p] = false;
                omega[i * p] = omega[i] + 1;
                if (i % p == 0) {
                    break;
                }
            }
        }
    }

    public static void main(String[] args) {
        sieve();
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        while (T-- > 0) {
            int x = in.nextInt();
            System.out.println(omega[x] * 2);
        }
    }
}

运行结果

三、筛法判断质数

import java.util.Scanner;

public class Main {
    static final int MAX = 1000000;
    static boolean[] isPrime = new boolean[MAX + 1];
    static int[] primes = new int[MAX + 1];
    static int cnt = 0;

    static void sieve() {
        for (int i = 2; i <= MAX; i++) {
            isPrime[i] = true;
        }
        for (int i = 2; i <= MAX; i++) {
            if (isPrime[i]) {
                primes[cnt++] = i;
            }
            for (int j = 0; j < cnt && primes[j] * i <= MAX; j++) {
                isPrime[i * primes[j]] = false;
                if (i % primes[j] == 0) break;
            }
        }
    }

    public static void main(String[] args) {
        sieve();
        Scanner in = new Scanner(System.in);
        StringBuilder sb = new StringBuilder();
        int T = in.nextInt();
        while (T-- > 0) {
            int n = in.nextInt();
            if (isPrime[n]) {
                sb.append("Yes\n");
            } else {
                sb.append("No\n");
            }
        }
        System.out.print(sb);
    }
}

运行结果

四、质数统计

import java.util.Scanner;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    static final int MAX = 1000000;
    static boolean[] isPrime = new boolean[MAX + 1];
    static int[] pre = new int[MAX + 1]; // 前缀和，pre[i] = 1~i质数个数
    static int[] primes = new int[MAX + 1];
    static int cnt = 0;

    static void sieve() {
        for (int i = 2; i <= MAX; i++) {
            isPrime[i] = true;
        }
        for (int i = 2; i <= MAX; i++) {
            if (isPrime[i]) {
                primes[cnt++] = i;
            }
            for (int j = 0; j < cnt && primes[j] * i <= MAX; j++) {
                isPrime[i * primes[j]] = false;
                if (i % primes[j] == 0) break;
            }
        }
        // 构造前缀和
        pre[0] = pre[1] = 0;
        for (int i = 2; i <= MAX; i++) {
            pre[i] = pre[i - 1] + (isPrime[i] ? 1 : 0);
        }
    }

    public static void main(String[] args) {
        sieve();
        Scanner in = new Scanner(System.in);
        StringBuilder sb = new StringBuilder();
        int n = in.nextInt();
        while (n-- > 0) {
            int l = in.nextInt();
            int r = in.nextInt();
            int ans = pre[r] - pre[l - 1];
            sb.append(ans).append('\n');
        }
        System.out.print(sb);
    }
}

运行结果

五、素数对

import java.util.Scanner;

public class Main {
    static final int MAX = 500000;
    static boolean[] isPrime = new boolean[MAX + 1];
    static int[] primes = new int[MAX + 1];
    static int cnt = 0;

    static void sieve() {
        for (int i = 2; i <= MAX; i++) {
            isPrime[i] = true;
        }
        for (int i = 2; i <= MAX; i++) {
            if (isPrime[i]) {
                primes[cnt++] = i;
            }
            for (int j = 0; j < cnt && primes[j] * i <= MAX; j++) {
                isPrime[i * primes[j]] = false;
                if (i % primes[j] == 0) break;
            }
        }
    }

    public static void main(String[] args) {
        sieve();
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        long ans = 0;
        for (int i = 0; i < cnt; i++) {
            int C = primes[i];
            if (C > N) break;
            long S = (long) C * C;
            if (S > 2L * N) break;
            int L = (int) Math.max(2, S - N);
            int R = N;
            for (int A = L; A <= R; A++) {
                if (!isPrime[A]) continue;
                long B = S - A;
                if (B < 2 || B > N) continue;
                if (isPrime[(int) B]) {
                    ans++;
                }
            }
        }
        System.out.println(ans);
    }
}

运行结果