UVa 725 DIVISION 除法

题目描述：

   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, where2 ≤ 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 bezero.

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 nosolutions 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

一道枚举题，又由于每个数字的不能重复，启发我用回溯写了一个全排列。然后先求fghij,靠fghij和n的乘积来求abcde以此来减小枚举量。

不过因为没有看清楚输出的格式让我错了好几次，怪难受的。

#include <cstdio>#include <iostream>#include <cstring>using namespace std;int n;void print(long int y,long int x);bool pd(long int x);bool fd[11]={0};bool fdd[11]={0};long int sum=0;int ff=0;void hui(int x)     //回溯，来求每一种可能。{    if(x<=5)    {        for(int i=0;i<=9;i++)            if(fd[i]==0)            {                sum=sum*10+i;                fd[i]=1;                     hui(x+1);                     fd[i]=0;                sum=(sum-i)/10;              }    }    else    {        long int y=sum*n;               if(pd(y))        {            print(y,sum);        }        else return;    }}bool pd(long int y){    for(int i=0;i<=10;i++)fdd[i]=fd[i];    if(y<1234||y>98765)return 0;    else    {        int t=5,f=1;        while(t--)        {            int z=y%10;            y=y/10;            if(fdd[z]==1){f=0;break;}            fdd[z]=1;        }        if(f==1)return 1;        else return 0;    }}void print(long int y,long int x){        if(y<10000)printf("0%ld /",y);        else printf("%ld /",y);        if(x<10000)printf(" 0%ld = %d\n",x,n);        else printf(" %ld = %d\n",x,n);        ff=1;}int main(){    int fz=0;    while(cin>>n&&n)    {        ff=0;        if(fz==1)cout<<endl;        int y;        memset(fd,0,sizeof(fd));        hui(1);        if(ff==0)printf("There are no solutions for %d.\n",n);        fz=1;    }    return 0;        //虽然比较长，可是看起来还是挺可爱的}