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
    NB-IOT使用LWM2M移动onenet基础通信套件对接之APN设置
    查看>>
    NBear简介与使用图解
    查看>>
    Vue过滤器_使用过滤器进行数据格式化操作---vue工作笔记0015
    查看>>
    Ncast盈可视 高清智能录播系统 IPSetup.php信息泄露+RCE漏洞复现(CVE-2024-0305)
    查看>>
    NCNN中的模型量化解决方案:源码阅读和原理解析
    查看>>
    NCNN源码学习(1):Mat详解
    查看>>
    nc命令详解
    查看>>
    NC综合漏洞利用工具
    查看>>
    ndarray 比 recarray 访问快吗?
    查看>>
    ndk-cmake
    查看>>
    NdkBootPicker 使用与安装指南
    查看>>
    ndk特定版本下载
    查看>>
    NDK编译错误expected specifier-qualifier-list before...
    查看>>
    Neat Stuff to Do in List Controls Using Custom Draw
    查看>>
    Necurs僵尸网络攻击美国金融机构 利用Trickbot银行木马窃取账户信息和欺诈
    查看>>
    Needle in a haystack: efficient storage of billions of photos 【转】
    查看>>
    NeHe OpenGL教程 07 纹理过滤、应用光照
    查看>>
    NeHe OpenGL教程 第四十四课:3D光晕
    查看>>
    Neighbor2Neighbor 开源项目教程
    查看>>
    neo4j图形数据库Java应用
    查看>>