题目给出一个n ，求所有a/b=n 要求ab中数字不一样 枚举分母或者分分子就行了
注意输出格式
题目：
| Division |
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
代码：
1 #include <iostream> 2 #include <iomanip> 3 #include <memory.h> 4 #include <cstdio> 5 using namespace std; 6 7 int flag =0; 8 int n; 9 //int B[5]; 10 int judge(int i,long long k) 11 { 12 int vis[10]; 13 memset(vis,0,sizeof(vis)); 14 15 int ti=i,tk=k; 16 while(ti) 17 { 18 if(vis[ti%10]) 19 return 0; 20 else 21 vis[ti%10]=1; 22 23 ti/=10; 24 } 25 while(tk) 26 { 27 if(vis[tk%10])return 0; 28 else vis[tk%10]=1; 29 30 tk/=10; 31 32 } 33 for(int g=1;g<10;g++) 34 if(!vis[g]){return 0;} 35 36 cout<<setw(5)<<setfill('0')<<i<<" / "<<setw(5)<<setfill('0')<<k<<" = "<<n<<endl; 37 38 return 1; 39 } 40 41 int main() 42 { 43 freopen("out","w",stdout); 44 int tst=0; 45 46 while(cin>>n) 47 { 48 if(n==0)break; 49 if(tst>1)cout<<endl; 50 for(int i=10000;i<=99999;i++) 51 { 52 if(int(i/n)*n==i) 53 { 54 if(judge(i,i/n)) 55 { 56 flag = 1; 57 } 58 } 59 } 60 if(!flag) 61 { 62 cout<<"There are no solutions for "<<n<<"."<<endl; 63 } 64 flag =0; 65 if(!tst){cout<<endl;} 66 tst++; 67 } 68 69 return 0; 70 }