博客
关于我
java 牛客:因子个数
阅读量:749 次
发布时间:2019-03-22

本文共 2746 字,大约阅读时间需要 9 分钟。

To solve this problem, we need to determine the number of factors for each given positive integer. The solution involves understanding the prime factorization of a number and using it to compute the total number of factors.

Approach

The approach can be broken down into the following steps:

  • Prime Factorization: Decompose the given number into its prime factors. For example, the number 36 can be decomposed into (2^2 \times 3^2).

  • Exponent Tracking: For each prime factor, determine its exponent in the factorization. For instance, in the case of 36, the exponent of 2 is 2, and the exponent of 3 is also 2.

  • Calculate Factors: The total number of factors of a number can be found by taking the product of each prime factor's exponent incremented by one. For example, using the prime factors of 36, the total number of factors is ((2+1) \times (2+1) = 9).

  • Efficient Looping: Use efficient looping techniques to iterate through potential factors, and stop early when further division isn't possible. This optimization prevents unnecessary computations.

  • Solution Code

    import java.util.Scanner;public class Main {    public static void main(String[] args) {        Scanner scanner = new Scanner(System.in);        while (scanner.hasNextInt()) {            int n = scanner.nextInt();            System.out.println(countFactors(n));        }    }    private static int countFactors(int n) {        if (n <= 1) {            return 1;        }        int factors = 1;        for (int i = 2; i * i <= n; ) {            if (n % i == 0) {                int exponent = 0;                while (n % i == 0) {                    exponent++;                    n /= i;                }                factors *= (exponent + 1);            } else {                i++;            }        }        if (n > 1) {            factors *= 2;        }        return factors;    }}

    Explanation

  • Reading Input: The code reads each integer from the standard input.
  • Handling Special Cases: If the input number is 1, it directly returns 1 as it is the only factor.
  • Prime Factorization Loop: The loop iterates from 2 up to the square root of the number. For each potential factor, it checks if it divides the number. If it does, it counts how many times it divides (the exponent) and then divides the number by this factor until it no longer can.
  • Updating Factors Count: The number of factors is updated by multiplying the product of each exponent incremented by one.
  • Remaining Prime Check: If after processing all factors up to the square root, the remaining number is greater than 1, it means it is a prime factor itself, contributing one more factor.
  • This approach efficiently computes the number of factors for each positive integer, ensuring correct and optimal results.

    转载地址:http://nvewk.baihongyu.com/

    你可能感兴趣的文章
    MySQL数据库操作
    查看>>
    MySQL数据库故障排错
    查看>>
    MySQL数据库无法远程连接的解决办法
    查看>>
    mysql数据库时间类型datetime、bigint、timestamp的查询效率比较
    查看>>
    MySQL数据库服务器端核心参数详解和推荐配置(一)
    查看>>
    mysql数据库死锁的产生原因及解决办法
    查看>>
    MySQL数据库的事务管理
    查看>>
    mysql数据库的备份与恢复
    查看>>
    Mysql数据库的条件查询语句
    查看>>
    MySQL数据库的高可用
    查看>>
    Mysql数据库相关各种类型的文件
    查看>>
    MYSQL数据库简单的状态检查(show processlist)
    查看>>
    MYSQL数据库简单的状态检查(show status)
    查看>>
    MySQL数据库系列
    查看>>
    MYSQL数据库自动本地/异地双备份/MYSQL增量备份
    查看>>
    mysql数据库表增添字段,删除字段、修改字段的排列等操作,还不快来
    查看>>
    MySQL数据库被黑了
    查看>>
    mysql数据库设计
    查看>>
    MySQL数据库设计与开发规范
    查看>>
    MYSQL数据库进阶操作
    查看>>