UVA - 725 ACM/ICPC Division

最新推荐文章于 2018-08-15 18:18:55 发布

LLer_Dagu

最新推荐文章于 2018-08-15 18:18:55 发布

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本文链接：https://blog.youkuaiyun.com/LLer_Dagu/article/details/81159250

本文介绍了一款程序设计挑战，该程序旨在找出所有符合条件的五位数对，这些数对使用0到9的所有数字恰好一次，并且第一个数除以第二个数的结果为指定整数N，其中2≤N≤79。程序输出所有符合条件的数对，并按分子递增顺序排列。

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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 ≤ N ≤ 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

Sample Output

There are no solutions for 61.

79546 / 01283 = 62

#include <stdio.h>
int num(int shu);
void wei(int shu,int a[],int n);

int main()
{
	int n;
	int nn=0;
	while(scanf("%d",&n)!=EOF&&n)
	{
		int f=0;
		if(nn++>0)
		printf("\n");
		int i;
		for(i=1;i<=99999/n;i++)
		{
			int a,b;
			a=i*n;b=i;
			int numa,numb;
			numa=num(a);
			numb=num(b);
			int zi[10]={0};
			int mu[10]={0};
			wei(a,zi,numa);
			wei(b,mu,numb);
			int shu[10]={0};
			int j;
			for(j=0;j<5;j++)
				{
					shu[zi[j]]++;
					shu[mu[j]]++;
				}
			int flag=0;
			for(j=0;j<=9;j++)
				if(shu[j]>=2)
					flag=1;
			if(flag==0)
				{
					printf("%05d / %05d = %d\n",a,b,n);
					f=1;
					nn++;
				}
		}
		if(f==0)
			printf("There are no solutions for %d.\n",n);
	}
	return 0;
}

int num(int shu)
{
	int n=0;
	while(shu>0)
	{
		n++;
		shu=shu/10;
	}
	return n;
}

void wei(int shu,int a[],int n)
{
	int i;
	for(i=0;i<n;i++)
	{
		a[i]=shu%10;
		shu=shu/10;
	}
}

94736 / 01528 = 62