UVA

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 integerN , where 2 N 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 integerN . 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 forN .’. Separate the output for two different values of N by a blank line.

Sample Input

61620

Sample Output

There are no solutions for 61.

79546 / 01283 = 62

94736 / 01528 = 62

Sample Code

//版本1 #include<cstdio>#include<cstring>int arr[10];//0~9int n,div,dived;bool solution;bool judge(int num){if(num>98765)return false;int temp[10];memcpy(temp,arr,sizeof(arr));while(num>0){temp[num%10]++;num/=10;}for(int i=0;i<10;i++)if(temp[i]!=1)return false;return true;}void get_digit(int cnt){if(cnt==0){dived=n*div;if(judge(dived)==true){solution=true;printf("%5d / %05d = %d\n",dived,div,n);}return;}for(int i=0;i<=9;i++){if(arr[i]==0){arr[i]=1;cnt--;div=div*10+i;get_digit(cnt);arr[i]=0;cnt++;div=(div-i)/10;}}}int main(){int kase=0;while(scanf("%d",&n)==1 && n){if(kase++)printf("\n");//要求两个n间有一个空行。 div=dived=0;solution=false;memset(arr,0,sizeof(arr));get_digit(5);if(solution==false)printf("There are no solutions for %d.\n",n);}return 0;}

//版本2 #include<cstdio>#include<cstring>int arr[10];//0~9void part_digit(int num){while(num>0){arr[num%10]++;num/=10;}}bool judge(){for(int i=0;i<10;i++)if(arr[i]!=1)return false;return true;}int main(){int n;int kase=0;while(scanf("%d",&n)==1 && n){if(kase++)printf("\n");//要求两个n间有一个空行。 int div,dived;bool solution=false;for(div=1234;div<98765;div++){dived=div*n;if(dived>98765)break;memset(arr,0,sizeof(arr));if(div<10000)arr[0]++;part_digit(div);part_digit(dived);if(judge()){solution=true;printf("%05d / %05d = %d\n",dived,div,n);}} if(solution==false)printf("There are no solutions for %d.\n",n);}return 0;}