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
    Objective-C实现BeadSort珠排序算法(附完整源码)
    查看>>
    Objective-C实现bellman ford贝尔曼福特算法(附完整源码)
    查看>>
    Objective-C实现bellman-ford贝尔曼-福特算法(附完整源码)
    查看>>
    Objective-C实现bellman-ford贝尔曼-福特算法(附完整源码)
    查看>>
    Objective-C实现BellmanFord贝尔曼-福特算法(附完整源码)
    查看>>
    Objective-C实现bezier curve贝塞尔曲线算法(附完整源码)
    查看>>
    Objective-C实现bfs 最短路径算法(附完整源码)
    查看>>
    Objective-C实现BF算法 (附完整源码)
    查看>>
    Objective-C实现Bilateral Filter双边滤波器算法(附完整源码)
    查看>>
    Objective-C实现binary exponentiation二进制幂运算算法(附完整源码)
    查看>>
    Objective-C实现binary search二分查找算法(附完整源码)
    查看>>
    Objective-C实现binary tree mirror二叉树镜像算法(附完整源码)
    查看>>
    Objective-C实现binary tree traversal二叉树遍历算法(附完整源码)
    查看>>
    Objective-C实现BinarySearchTreeNode树算法(附完整源码)
    查看>>
    Objective-C实现binomial coefficient二项式系数算法(附完整源码)
    查看>>
    Objective-C实现bisection二分法算法(附完整源码)
    查看>>
    Objective-C实现bisection二等分算法(附完整源码)
    查看>>
    Objective-C实现BitMap算法(附完整源码)
    查看>>
    Objective-C实现bitonic sort双调排序算法(附完整源码)
    查看>>
    Objective-C实现BloomFilter布隆过滤器的算法(附完整源码)
    查看>>