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
    mysql5.5最简安装教程
    查看>>
    mysql5.6 TIME,DATETIME,TIMESTAMP
    查看>>
    mysql5.6.21重置数据库的root密码
    查看>>
    Mysql5.6主从复制-基于binlog
    查看>>
    MySQL5.6忘记root密码(win平台)
    查看>>
    MySQL5.6的Linux安装shell脚本之二进制安装(一)
    查看>>
    MySQL5.6的zip包安装教程
    查看>>
    mysql5.7 for windows_MySQL 5.7 for Windows 解压缩版配置安装
    查看>>
    Webpack 基本环境搭建
    查看>>
    mysql5.7 安装版 表不能输入汉字解决方案
    查看>>
    MySQL5.7.18主从复制搭建(一主一从)
    查看>>
    MySQL5.7.19-win64安装启动
    查看>>
    mysql5.7.19安装图解_mysql5.7.19 winx64解压缩版安装配置教程
    查看>>
    MySQL5.7.37windows解压版的安装使用
    查看>>
    mysql5.7免费下载地址
    查看>>
    mysql5.7命令总结
    查看>>
    mysql5.7安装
    查看>>
    mysql5.7性能调优my.ini
    查看>>
    MySQL5.7新增Performance Schema表
    查看>>
    Mysql5.7深入学习 1.MySQL 5.7 中的新增功能
    查看>>