UVA_725: Division

最新推荐文章于 2020-12-28 21:54:42 发布

原创最新推荐文章于 2020-12-28 21:54:42 发布 · 606 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #stl #uva

算法同时被 2 个专栏收录

141 篇文章

订阅专栏

cpp

42 篇文章

订阅专栏

本文介绍了一个编程挑战，任务是找出所有符合条件的五位数对，其中一个数除以另一个数等于给定整数N，且两个数合起来恰好使用0到9的所有数字一次。文章提供了完整的C++代码实现，并解释了其基本思路。

Description

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 = Nwhere 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 = Nxxxxx / xxxxx = N..
In case there are no pairs of numerals satisfying the condition, you mustwrite ``There are no solutions forN.". 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

分析：大水题啊，暴力破解，b从1234到50000，a=b×n，然后检查a和b的合法性（wa了7发，简直不敢想，就是那个点，点啊！）

#include <cstdio>

using namespace std;

int valid(int a, int b)
{
	int va[10] = {0};
	if(a<10000)va[0]=1;
	while(a){va[a%10]=1;a/=10;}
	while(b){va[b%10]=1;b/=10;}
	for(int i=0; i<10; i++)if(!va[i])return 0;
	return 1;
}

int main()
{
	int n;
	int first = 1;
	while(~scanf("%d",&n)&&n)
	{
		int solu = 0;
		if(first)first=0;
		else putchar('\n');
		for(int i=1234; i<50000; i++)
		{
			int fuck = i*n;
			if(fuck>98765)break;
			if(valid(i,fuck))
			{
				printf("%05d / %05d = %d\n",fuck,i,n);
				solu = 1;
			}
		}
		if(!solu)printf("There are no solutions for %d.\n",n);//这里有个点啊！
	}
    return 0;
}