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 编译安装 window篇
    查看>>
    mysql 网络目录_联机目录数据库
    查看>>
    MySQL 聚簇索引&&二级索引&&辅助索引
    查看>>
    Mysql 脏页 脏读 脏数据
    查看>>
    mysql 自增id和UUID做主键性能分析,及最优方案
    查看>>
    Mysql 自定义函数
    查看>>
    mysql 行转列 列转行
    查看>>
    Mysql 表分区
    查看>>
    mysql 表的操作
    查看>>
    mysql 视图,视图更新删除
    查看>>
    MySQL 触发器
    查看>>
    mysql 让所有IP访问数据库
    查看>>
    mysql 记录的增删改查
    查看>>
    MySQL 设置数据库的隔离级别
    查看>>
    MySQL 证明为什么用limit时,offset很大会影响性能
    查看>>
    Mysql 语句操作索引SQL语句
    查看>>
    MySQL 误操作后数据恢复(update,delete忘加where条件)
    查看>>
    MySQL 调优/优化的 101 个建议!
    查看>>
    mysql 转义字符用法_MySql 转义字符的使用说明
    查看>>
    mysql 输入密码秒退
    查看>>