UVa 725(简单枚举)

最新推荐文章于 2020-05-20 21:00:11 发布

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最新推荐文章于 2020-05-20 21:00:11 发布

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分类专栏： ACM-暴力枚举

本文链接：https://blog.youkuaiyun.com/Cui_csdn/article/details/73650217

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本文介绍了一种算法，用于寻找所有符合条件的五位数对，这些数对使用0至9的数字各一次，并且第一个数除以第二个数的结果为整数N。程序通过枚举并检查每对数来解决问题。

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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 $2\le N \le 79$ . 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 for N.". Separate the output for two different 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

//思路:从b等于1234开始枚举生成a，然后判断a和b是否满足要求(巧用了sprintf先保存到buf，再sort) 还有%05d解决了前导0的问题

AC源码:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
	char buf[15];
	int kase=0,n;
	while(cin>>n&&n)
	{
		int cnt=0;
		if(kase++)
			printf("\n");
		for(int b=1234;;++b)
		{
			int a=b*n;
			sprintf(buf,"%05d%05d",a,b);
			if(strlen(buf)>10)
				break;
			sort(buf,buf+10);
			bool ok=true;
			for(int i=0;i<10;++i)
				if(buf[i]!='0'+i)
					ok=false;
			if(ok)
			{
				++cnt;
				printf("%05d / %05d = %d\n",a,b,n);
			}	
		}
		if(!cnt)
			printf("There are no solutions for %d.\n",n);

	}
	return 0;
}