uva725（除法）

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

Output

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所有形如abcde/fghij=n的表达式，其中a~j恰好为数字0~9的一个排列（可以有0前导）

思路：

枚举fghij就可以算出abcde,然后判断符不符合条件。

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("\n");
        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\n",b,i,j,k,l,m,N);
          flag=1;
        }
        else
            continue;

       }

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

    }
    return 0;

}