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

---

Miguel Revilla 
2000-08-31

题目大意：

如样例所示。

解题思路：

暴力枚举。从分母下手找分子更快，还有就是用到sprintf函数更简单的判断是否重。

代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>

using namespace std;

int n,num,den;
char buffer[20];
bool visited[10];

bool haveCommonElement(int a,int b){
    int temp;
    sprintf(buffer,"%05d%05d",a,b);//这里的buffer要为char,不能为string.
    memset(visited,false,sizeof(visited));
    for(int i=0;i<strlen(buffer);i++){
        temp=buffer[i]-'0';
        if(visited[temp])
            return true;
        else visited[temp]=true;
    }
    return false;
}

int main(){
    int casen=0,flag;
    while(scanf("%d",&n)!=EOF&&n){
        if(casen>0) printf("\n");
        casen++;flag=0;
        for(int den=1234;den<100000;den++){//分子控制在5个宽度内，所以要小于100000.
            if(den%n==0){//之前用已知分子求分母，耗时太多，改用已知分母求分子后，因为只有整除时才进入子函数，所以效率提高了许多.
                num=den/n;
                if(!haveCommonElement(num,den)){
                    printf("%05d / %05d = %d\n",den,num,n);
                    flag=1;
                }
            }
        }
        if(flag==0)
            printf("There are no solutions for %d.\n",n);
    }
    return 0;
}