uva725（除法）

Description

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 . 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

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

题意：

输入正整数n,按从小到大的顺序输出全部形如abcde/fghij=n的表达式，当中a~j恰好为数字0~9的一个排列（能够有0前导）

思路：

枚举fghij就能够算出abcde,然后推断符不符合条件。

代码：

#include<cstdio>
using namespace std;
int main()
{
    int i,j,k,l,m,a,b;
    int b1,b2,b3,b4,b5;
    int flag;
    int N;
    int casex=0;
    while(scanf("%d",&N)&&N)
    {
        if(casex++)   printf("
");
        flag=0;
      for( i=0;i<=9;i++)
          for(j=0;j<=9;j++)
            for(k=0;k<=9;k++)
                for(l=0;l<=9;l++)
                    for(m=0;m<=9;m++)
    {
        if(i!=j&&i!=k&&i!=l&&i!=m&&j!=k&&j!=l&&j!=m&&k!=l&&k!=m&&l!=m)
            a=i*10000+j*1000+k*100+l*10+m;
            else continue;
         b=N*a;
         if(b>99999)
            continue;
        b1=b/10000;
        b2=b%10000/1000;
        b3=b%10000%1000/100;
        b4=b%10000%1000%100/10;
        b5=b%10;
        if(b1!=b2&&b1!=b3&&b1!=b4&&b1!=b5&&b2!=b3&&b2!=b4&&b2!=b5&&b3!=b4&&b3!=b5&&b4!=b5&&b1!=i&&b1!=j&&b1!=k&&b1!=l&&b1!=m&&b2!=i&&b2!=j&&b2!=k&&b2!=l&&b2!=m&&b3!=i&&b3!=j&&b3!=k&&b3!=l&&b3!=m&&b4!=i&&b4!=j&&b4!=k&&b4!=l&&b4!=m&&b5!=i&&b5!=j&&b5!=k&&b5!=l&&b5!=m)
        {
          printf("%d / %d%d%d%d%d = %d
",b,i,j,k,l,m,N);
          flag=1;
        }
        else
            continue;

       }

    if(!flag)  printf("There are no solutions for %d.
",N);

    }
    return 0;

}

---

Miguel Revilla
2000-08-31