uva 725Division

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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 $2\le N \le 79$ . 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 forN.". 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 = 6294736 / 01528 = 62

根据除数暴力，第一次写的很乱，看了别人的题解感觉很简洁。

#include <iostream>#include<cstdio>#include<cstring>using namespace std;int a[10];int judge(int x){    int y;    while(x)    {        y=x%10;        x=x/10;        a[y]++;    }    return 0;}int main(){    int n,m;    int num=0;    while(scanf("%d",&n)!=EOF && n)    {        if(num) printf("\n");        num++;        m=0;        int flag=0;        for(int i=0;i<10;i++)            for(int j=0;j<10;j++)                if(i!=j)                   for(int k=0;k<10;k++)                   if(i!=k && j!=k)                     for(int x=0;x<10;x++)                        if(i!=x && j!=x && k!=x)                          for(int y=0;y<10;y++)                             if(i!=y && j!=y && k!=y && x!=y)                             {                                 int cnt=0;                                 memset(a,0,sizeof(a));                                 a[i]=1;                                 a[j]=1;                                 a[k]=1;                                 a[x]=1;                                 a[y]=1;                                 m=y+x*10+k*100+j*1000+i*10000;                                 int s=m*n;                                 judge(s);                                 for(int i=0;i<10;i++)                                    if(a[i]==1)                                    cnt++;                                 if(cnt==10)                                 {                                     flag=1;                                     printf("%05d / %05d = %d\n",s,m,n);                                 }                             }        if(!flag) printf("There are no solutions for %d.\n",n);    }    return 0;}

#include<cstdio>#include<iostream>#include<cstring>using namespace std;int a[10];int judge(int x,int y){    if(y>98765) return 0;    memset(a,0,sizeof(a));    if(x<10000) a[0]=1;    while(x)    {        a[x%10]++;        x=x/10;    }    while(y)    {        a[y%10]++;        y=y/10;    }    int cnt=0;    for(int i=0;i<10;i++)        if(a[i]==1)        cnt++;    if(cnt == 10) return 1;    else return 0;}int main(){    int n;    int num=0;    while(scanf("%d",&n)!=EOF && n)    {        if(num) printf("\n");        num++;        int flag=0;        for(int i=1234;i<=98765;i++)        {            if(judge(i,i*n))            {                flag=1;                printf("%05d / %05d = %d\n",i*n,i,n);            }        }        if(!flag) printf("There are no solutions for %d.\n",n);    }}

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