UVA - 725 Division （暴力枚举）

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本文介绍了一个编程挑战，任务是找出所有由0至9这十个数字组成的五位数对abcde/fghij，使得这两个数之间的除法等于给定的整数N，并确保每个数字恰好使用一次。文章提供了一种解决方案，通过遍历可能的数对来检查条件是否满足。

Division

Write a program that finds and displays all pairs of 5-digit numbers thatbetween them use the digits 0 through 9 once each, such that thefirst number divided by the second is equal to an integer N, where $2\le N \le 79$ .That is,

abcde / fghij = N

where each letter represents a different digit. The first digit of one ofthe 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 terminatethe program.

Output

Your program have to display ALL qualifying pairs of numerals, sorted byincreasing 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 mustwrite ``There are no solutions for N.". Separate the output for twodifferent values of N by a blank line.

Sample Input

61
62
0

Sample Output

There are no solutions for 61.

79546 / 01283 = 62
94736 / 01528 = 62

题目大意：

输入正整数n，按照从小到大的顺序输出所有形如abcde/fghij = n 的表达式，其中a～j恰好为数组0～9的一个排列（可以有前导0），2<= n <= 79

解析：

不需要枚举0～9所有的排列。只要枚举fghij就可以算出abcde，然后判断所有的数字都不相同即可。

#include <stdio.h>
#include <string.h>
bool judge(int ans1,int ans2) {
	char str[11];	
	bool vis[10];
	int num;
	memset(vis,false,sizeof(vis));
	sprintf(str,"%05d%05d",ans1,ans2);
	for(int i = 0; i < 10; i++) {
		num = str[i] - '0';
		if(vis[num]) {
			return false;
		}
		vis[num] = true;
	}
	return true;
}
int main() {
	int n;
	int ans1,ans2;
	int cas = 0;
	while(scanf("%d",&n) != EOF && n) {
		if(cas++) {
			printf("\n");
		}
		bool flag = false;
		for(int i = 1234; i <= 100000; i++) {
			if(i % n == 0) {
				ans2 = i / n;
				if( judge(i,ans2) ) {
					flag = true;
					printf("%05d / %05d = %d\n",i,ans2,n);	
				}
			}
		}
	
		if(!flag) {
			printf("There are no solutions for %d.\n",n);
		}
	}
	return 0;
}