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Nov 17th, 2019
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  1. #include <stdio.h>
  2. #include <stdlib.h>
  3.  
  4. void wypisz (int n, int tab [n] [n])
  5. {
  6. printf (" ");
  7. for(int i=0; i<10; i++)
  8. {
  9. printf("%d",i);
  10. }
  11. printf("\n\n");
  12.  
  13. for (int x=0; x<10; x++)
  14. {
  15. printf("%d ",x);
  16. for (int y=0; y<10; y++)
  17. {
  18.  
  19. if (tab [x] [y] == 1)
  20. {
  21. printf("X");
  22. }
  23. else
  24. {
  25. printf ( "%d",tab[x] [y]);
  26. }
  27. }
  28. printf("\n");
  29. }
  30. }
  31. int zamien (int kolumna, int wiersz, int m, int tab [m][m])
  32. {
  33. if (kolumna+1 < m)
  34. {
  35. tab[kolumna+1][wiersz] = !tab[kolumna+1][wiersz];
  36. }
  37. if (wiersz+1 < m)
  38. {
  39. tab[kolumna][wiersz+1] = !tab[kolumna][wiersz+1];
  40. }
  41. if (kolumna-1 >= 0)
  42. {
  43. tab[kolumna-1][wiersz] = !tab[kolumna-1][wiersz];
  44. }
  45. if (wiersz-1 >= 0)
  46. {
  47. tab[kolumna][wiersz-1] = !tab[kolumna][wiersz-1];
  48. }
  49.  
  50. int suma = 0;
  51. for (int r=0; r<10; r++)
  52. {
  53. for (int t=0; t<10; t++)
  54. {
  55. if (tab [r][t]==1)
  56. {
  57. suma ++;
  58. }
  59. }
  60. }
  61. return suma;
  62. }
  63. int main (int argc, char ** argv)
  64. {
  65. int tab [10][10] = {0};
  66.  
  67. int n = 10;
  68. int level = 1;
  69. printf ("[*] Wybierz poziom trudnosci: 1:easy, 2:medium, 3:hard : ");
  70. int steps_remaining;
  71. scanf ("%d",&level);
  72. if (level == 1)
  73. {
  74. steps_remaining = 150;
  75. }
  76. else if (level == 2)
  77. {
  78. steps_remaining = 75;
  79. }
  80. else if (level == 1)
  81. {
  82. steps_remaining = 30;
  83. }
  84. int actual_steps = 0;
  85. int xksy = 0;
  86. do
  87. {
  88.  
  89. int a,b;
  90. wypisz(n, tab);
  91. printf ("[*] Pozostalo Ci %d krokow\n",steps_remaining-actual_steps);
  92. printf("[*] Zostalo %d pol do zamienienia\n", n*n-xksy);
  93. printf("[*] Podaj wiersz i kolumne ");
  94. scanf("%d %d", &a, &b);
  95. xksy = zamien(a,b,n,tab);
  96.  
  97. actual_steps++;
  98.  
  99. }
  100. while (xksy < n*n || actual_steps < steps_remaining);
  101.  
  102. return 0;
  103. }
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