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Oct 19th, 2018
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  1. #include <stdio.h>
  2. #include <conio.h>
  3.  
  4.  
  5. void main ()
  6. {
  7. int N, M, i, j, A[15][15], S=0;
  8. printf ("vvedite N i M\n");
  9. scanf ("%d % d", &N, &M);
  10. printf ("vvedite matrizu\n");
  11. for (i = 1; i <= N; i++)
  12. for (j = 1; j<=M; j++)
  13. scanf ("%d", &A[i][j]);
  14.  
  15. //*printf ("vvedite i i j\n");
  16. //*scanf ("%d %d", &i, &j);
  17.  
  18. printf("\n i\t j\t summa\n");
  19. for (i = 1; i <= N; i++)
  20. for (j = 1; j <=M; j++)
  21. {
  22. S=obrabotka(i,j,N,M,A, S);
  23.  
  24. printf ("%d\t%d\t%d\n",i, j, S); }
  25. getch();
  26.  
  27. }
  28.  
  29.  
  30.  
  31.  
  32. int obrabotka (int i, int j, int N, int M, int A[15][15], int S)
  33. {
  34. int ii, jj, j1, B[10];
  35.  
  36. int n=1;
  37.  
  38. if ((i+1) <= N)
  39. { for (j1=1; j1 <= M; j1++)
  40. {
  41. B[n]=A[i+1][j1];
  42. n++;
  43. }
  44. for (jj=j, ii=i; jj<=M, ii>=1; jj++,ii--)
  45. {
  46. B[n]=A[ii][jj];
  47. n++;
  48. }
  49. for (jj=j, ii=i; jj>=1, ii>=1; jj--, ii--)
  50. {
  51. B[n]=A[ii][jj];
  52. n++;
  53. }
  54. }
  55.  
  56. for (n=1;n<=(n-1);n++)
  57. if (B[n]%B[n]/2==0)
  58. {
  59. S=S+B[n];
  60.  
  61. }
  62. return S;
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