UVa 725 Division（枚举）

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本文提供了一个程序解决方案来寻找并显示所有在指定范围内使用0到9的唯一数字组成的五位数对，其中一个五位数除以另一个等于整数N。程序通过输入整数N并输出符合条件的数对，按升序排列，每一对数字作为分子和分母。若无解，则输出相应的提示。

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UVA 725 Division

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits0 through9 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 ofN by a blank line.

Sample Input

61
62
0

Sample Output

There are no solutions for 61.

79546 / 01283 = 62
94736 / 01528 = 62

#include <stdio.h>
#include <string.h>
using namespace std;

int n, a[10];

int judge(int x, int y) {
	int tx = x, ty = y;
	memset(a, 0, sizeof(a));
	while (tx) {
		if (a[tx % 10])
			return 0;
		else
			a[tx % 10] = 1;
		tx /= 10;
	}
	while (ty) {
		if (a[ty % 10])
			return 0;
		else
			a[ty % 10] = 1;
		ty /= 10;
	}
	for (int i = 1; i <= 9; i++)
		if (!a[i])
			return 0;
	printf("%05d / %05d = %d\n", x, y, n);
	return 1;
}

int main() {
	int index = 0, flag = 0;
	while (scanf("%d", &n) == 1) {
		if (n == 0)
			break;
		if (index > 1)
			printf("\n");
		for (int i = 12345; i <= 98765; i++) {
			if (i % n == 0) 
				if (judge(i, i / n)) 
					flag = 1;
		}
		if (!flag)
			printf("There are no solution for %d.\n", n);
		flag = 0;
		if (!index)
			printf("\n");
		index++;
	}
	return 0;
}