uva 725 Division 暴力求解入门

本是古典何须时尚 2022-07-10 09:58 230阅读 0赞

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 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 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

#include<iostream>
    #include<cstring>
    
    using namespace std;
    
    int main()
    {
        int sum,num,N,j;
        int a,b,c,d,e;
        int f,g,h,l,m;
        int x[10],ok = 0;
        while(cin>>N&& N)
        {
            if(ok == 0)
                ok = 1;
            else
                cout<<"\n";
            sum = 0;
            for(int i = 1234; i <= 98765; i++)
            {
                memset(x,0,sizeof(x));
                a = i%10,b = i/10%10,c = i/100%10,d = i/1000%10,e = i/10000%10;
                num = i*N;
                if(num>98765)
                    break;
                f = num%10,g = num/10%10,h = num/100%10,l = num/1000%10,m = num/10000%10;
                x[a]++,x[b]++,x[c]++,x[d]++,x[e]++;
                x[f]++,x[g]++,x[h]++,x[l]++,x[m]++;
                for(j = 0; j <= 9; j++)
                    if(x[j]>1)
                        break;
                if(j==10)
                {
                   cout<<m<<l<<h<<g<<f<<" / "<<e<<d<<c<<b<<a<<" = "<<N<<"\n";
                   sum++;
                }
            }
            if(sum==0)
               cout<<"There are no solutions for "<<N<<".\n";
        }
        return 0;
    }