除法(Division)

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Division

Time Limit:3000MS Memory Limit:Unknown 64bit IO Format:%lld & %llu

Submit Status

Description

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits0 through 9 once each, such that the first number divided by the second is equal to an integerN, where $2\le N \le 79$ . 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

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

Sample Input

Sample Output

There are no solutions for 61.79546 / 01283 = 6294736 / 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;elseSystem.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;}}

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