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 InnoDB存储引擎中缓冲池Buffer Pool、Redo Log、Bin Log、Undo Log、Channge Buffer
    查看>>
    MySQL InnoDB引擎的锁机制详解
    查看>>
    Mysql INNODB引擎行锁的3种算法 Record Lock Next-Key Lock Grap Lock
    查看>>
    mysql InnoDB数据存储引擎 的B+树索引原理
    查看>>
    mysql innodb通过使用mvcc来实现可重复读
    查看>>
    mysql interval显示条件值_MySQL INTERVAL关键字可以使用哪些不同的单位值?
    查看>>
    Mysql join原理
    查看>>
    mysql order by多个字段排序
    查看>>
    MySQL Order By实现原理分析和Filesort优化
    查看>>
    mysql problems
    查看>>
    mysql replace first,MySQL中处理各种重复的一些方法
    查看>>
    MySQL replace函数替换字符串语句的用法(mysql字符串替换)
    查看>>
    Mysql Row_Format 参数讲解
    查看>>
    mysql select, from ,join ,on ,where groupby,having ,order by limit的执行顺序和书写顺序
    查看>>
    MySQL Server 5.5安装记录
    查看>>
    mysql slave 停了_slave 停止。求解决方法
    查看>>
    MySQL SQL 优化指南:主键、ORDER BY、GROUP BY 和 UPDATE 优化详解
    查看>>
    mysql sum 没返回,如果没有找到任何值,我如何在MySQL中获得SUM函数以返回'0'?
    查看>>
    mysql Timestamp时间隔了8小时
    查看>>
    Mysql tinyint(1)与tinyint(4)的区别
    查看>>