除法(Division)

​​Division​​

​​Submit  ​​​​Status​​

Description

​​

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, where

. That is,

abcde / fghij =N

where each letter represents a different digit. The first digit of one of the numerals is allowed to be zero.

​​Input​​ 

Each line of the input file consists of a valid integer

N. An input of zero is to terminate the program.

​​Output​​ 

Your program have to display ALL qualifying pairs of numerals, sorted by increasing numerator (and, of course, denominator).

Your output should be in the following general form:

xxxxx / xxxxx =N

xxxxx / xxxxx =N

.

.

In case there are no pairs of numerals satisfying the condition, you must write ``There are no solutions for N.". Separate the output for two different values of N by a blank line.

​​Sample Input​​ 

6162 0

​​Sample Output​​ 

There are no solutions for 61.79546 / 01283 = 62 94736 / 01528 = 62

【分析】

       枚举abcde，然后通过N算出fghij，最后判断是否所有数字不相同（统计0到9的个数，如果个数均为1，说明所有数字均不相同）。

用java语言编写程序，代码如下：

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

public class Main {
  public static void main(String[] args) {
    Scanner input = new Scanner(new BufferedInputStream(System.in));
    boolean first = true;
    while(input.hasNext()) {
      int n = input.nextInt();
      if(n == 0)
        break;
      
      if(first)
        first = !first;
      else
        System.out.println();
      
      boolean hasResult = false;
      for(int i = 123; i <= 98765; i++) {
        if(i % n == 0) {
          int m = i / n;
          if(count(i, m)) {
            hasResult = true;
            
            System.out.printf("%05d / %05d = " + n + "\n", i, m);
          }
        }
      }
      
      if(!hasResult)
        System.out.println("There are no solutions for " + n + ".");
    }
  }
  
  public static boolean count(int n, int m) {
    int[] count = new int[10];
    for(int i = 0; i < 5; i++) {
      count[n % 10]++;
      count[m % 10]++;
      
      if(count[n % 10] > 1 || count[m % 10] > 1)
        return false;
      
      n = n / 10;
      m = m / 10;
    }
    
    for(int i = 0; i < 10; i++) {
      if(count[i] != 1)
        return false;
    }
    return true;
  }
}