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 常见错误
    查看>>
    mysql 常见问题
    查看>>
    MYSQL 幻读(Phantom Problem)不可重复读
    查看>>
    mysql 往字段后面加字符串
    查看>>
    mysql 快照读 幻读_innodb当前读 与 快照读 and rr级别是否真正避免了幻读
    查看>>
    MySQL 快速创建千万级测试数据
    查看>>
    mysql 快速自增假数据, 新增假数据,mysql自增假数据
    查看>>
    MySql 手动执行主从备份
    查看>>
    Mysql 批量修改四种方式效率对比(一)
    查看>>
    Mysql 报错 Field 'id' doesn't have a default value
    查看>>
    MySQL 报错:Duplicate entry 'xxx' for key 'UNIQ_XXXX'
    查看>>
    Mysql 拼接多个字段作为查询条件查询方法
    查看>>
    mysql 排序id_mysql如何按特定id排序
    查看>>
    Mysql 提示:Communication link failure
    查看>>
    mysql 插入是否成功_PDO mysql:如何知道插入是否成功
    查看>>
    Mysql 数据库InnoDB存储引擎中主要组件的刷新清理条件:脏页、RedoLog重做日志、Insert Buffer或ChangeBuffer、Undo Log
    查看>>
    mysql 数据库中 count(*),count(1),count(列名)区别和效率问题
    查看>>
    mysql 数据库备份及ibdata1的瘦身
    查看>>
    MySQL 数据库备份种类以及常用备份工具汇总
    查看>>
    mysql 数据库存储引擎怎么选择?快来看看性能测试吧
    查看>>