分数拆分(Fractions Again?!)

​​Fractions Again?!​​

Time Limit:3000MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

​​Submit  ​​​​Status​​

Description

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It is easy to see that for every fraction in the form

(k > 0), we can always find two positive integers x and y, x ≥ y, such that:

.

Now our question is: can you write a program that counts how many such pairs of x and y there are for any given k?

​​Input​​

Input contains no more than 100 lines, each giving a value of k (0 < k ≤ 10000).

​​Output​​

For each k, output the number of corresponding (x, y) pairs, followed by a sorted list of the values of x and y, as shown in the sample output.

​​Sample Input​​

212

​​Sample Output​​

21/2 = 1/6 + 1/3
1/2 = 1/4 + 1/4
8
1/12 = 1/156 + 1/13
1/12 = 1/84 + 1/14
1/12 = 1/60 + 1/15
1/12 = 1/48 + 1/16
1/12 = 1/36 + 1/18
1/12 = 1/30 + 1/20
1/12 = 1/28 + 1/21
1/12 = 1/24 + 1/24

【分析】

       当 x 和 y 满足题意时，由于 x >= y，则 1/x <= 1/y，又因为1/k = 1/x + 1/y，所以 1/x = 1/k - 1/y <= 1/y 即 y <= 2k。通过枚举 y，算出 x 的值，那么x 和 y 便满足题意要求。 x = ( k * y) / (y - k)，因为 x 为正整数，所以 y - k > 0 即 y > k。综上，y 枚举的范围为 [k + 1, 2k]。

用java语言编写程序，代码如下：

import java.io.BufferedInputStream;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner input = new Scanner(new BufferedInputStream(System.in));
    while(input.hasNext()) {
      int k = input.nextInt();
      
      ArrayList<FA> list = new ArrayList<FA>();
      for(int y = k + 1; y <= 2 * k; y++) {
        if((k * y) % (y - k) == 0) {
          int x = (k * y) / (y - k);
          list.add(new FA(x, y));
        }
      }
      int n = list.size();
      System.out.println(n);
      for(int i = 0; i < n; i++)
        System.out.println("1/" + k + " = " + "1/" + 
            list.get(i).x + " + " + "1/" + list.get(i).y);
    }
  }
  
  static class FA {
    int x, y;
    public FA(int x, int y) {
      this.x = x;
      this.y = y;
    }
  }
}