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不会性能调优?看看这份清华架构师编写的MySQL性能优化手册吧
    查看>>
    MySQL不同字符集及排序规则详解:业务场景下的最佳选
    查看>>
    Mysql不同官方版本对比
    查看>>
    MySQL与Informix数据库中的同义表创建:深入解析与比较
    查看>>
    mysql与mem_细说 MySQL 之 MEM_ROOT
    查看>>
    MySQL与Oracle的数据迁移注意事项,另附转换工具链接
    查看>>
    mysql丢失更新问题
    查看>>
    MySQL两千万数据优化&迁移
    查看>>
    MySql中 delimiter 详解
    查看>>
    MYSQL中 find_in_set() 函数用法详解
    查看>>
    MySQL中auto_increment有什么作用?(IT枫斗者)
    查看>>
    MySQL中B+Tree索引原理
    查看>>
    mysql中cast() 和convert()的用法讲解
    查看>>
    mysql中datetime与timestamp类型有什么区别
    查看>>
    MySQL中DQL语言的执行顺序
    查看>>
    mysql中floor函数的作用是什么?
    查看>>
    MySQL中group by 与 order by 一起使用排序问题
    查看>>
    mysql中having的用法
    查看>>
    MySQL中interactive_timeout和wait_timeout的区别
    查看>>
    mysql中int、bigint、smallint 和 tinyint的区别、char和varchar的区别详细介绍
    查看>>