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  1. // Latinski kvadrat
  2. #include <iostream>
  3.  
  4. using namespace std;
  5.  
  6. int latinski_kvadrat[6][6];
  7. int n;
  8.  
  9. bool moze_broj(int red, int kolona, int broj)
  10. {
  11.     for (int x = 0; x < n; x++)
  12.         if (latinski_kvadrat[red][x] == broj or latinski_kvadrat[x][kolona] == broj)
  13.             return false;
  14.     return true;
  15. }
  16.  
  17. void resi(int red, int kolona)
  18. {
  19.     if (red == n)
  20.     {
  21.         for (int i = 0; i < n; i++)
  22.         {
  23.             for (int j = 0; j < n; j++)
  24.                 cout << latinski_kvadrat[i][j];
  25.             cout << endl;
  26.         }
  27.         cout << endl;
  28.         return;
  29.     }
  30.     int sledeci_red = red;
  31.     if (kolona == n - 1)
  32.         sledeci_red++;
  33.     int sledeca_kolona;
  34.     if (kolona == n - 1)
  35.         sledeca_kolona = 0;
  36.     else
  37.         sledeca_kolona = kolona + 1;
  38.     if (latinski_kvadrat[red][kolona] != 0)
  39.         resi(sledeci_red, sledeca_kolona);
  40.     else
  41.         for (int broj = 1; broj <= n; broj++)
  42.             if (moze_broj(red, kolona, broj))
  43.             {
  44.                 latinski_kvadrat[red][kolona] = broj;
  45.                 resi(sledeci_red, sledeca_kolona);
  46.                 latinski_kvadrat[red][kolona] = 0;
  47.             }
  48. }
  49.  
  50. int main()
  51. {
  52.     cin >> n;
  53.     for (int i = 0; i < n; i++)
  54.     {
  55.     	string linija;
  56.     	cin >> linija;
  57.     	for (int j = 0; j < n; j++)
  58.             latinski_kvadrat[i][j] = linija[j] - '0';
  59.     }
  60.     resi(0, 0);
  61.     return 0;
  62. }
Success #stdin #stdout 0s 5308KB
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C++14 (gcc 8.3)

					

					

						
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