Division

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0through 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

暴力求解，枚举第二个数

#include<iostream>
#include <stdio.h>
#include<algorithm>
#include<cstring>
#include<map>
#include<set>
using namespace std;
int a[11],n;
int is(int i)
{
    if(a[i]==1)return 1;
    return 0;
}
int judge(int i,int j,int k,int o,int p)
{
    int x,s,s1,s2;
    memset(a,0,sizeof(a));
    a[i]++;a[j]++;a[k]++;a[o]++;a[p]++;
    s1=10000*i+1000*j+100*k+10*o+p;
    s=s1*n;
    s2=s;
    while(s){
        a[s%10]++;
        s/=10;
    }
    if(is(1)&&is(2)&&is(3)&&is(4)&&is(5)&&is(6)&&is(7)&&is(8)&&is(9)&&is(0))
        {printf("%d / %05d = %d\n",s2,s1,n);return 1;}
    else return 0;
}
int main()
{
    int i,j,k,o,p,t=0;
    while(scanf("%d",&n),n){
       t++;
       if(t!=1)puts("");
       int e=0;
       for(i=0; i<=9; i++)
       for(j=0; j<=9; j++)
       for(k=0; k<=9; k++)
       for(o=0; o<=9; o++)
       for(p=0; p<=9; p++)
            if(judge(i,j,k,o,p))
                e=1;
       if(!e)printf("There are no solutions for %d.\n",n);
    }
    return 0;
}

uva 725 - Division

原文：http://blog.csdn.net/u014705854/article/details/43057249