Advertisement
Guest User

Untitled

a guest
Mar 21st, 2018
58
0
Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
text 1.69 KB | None | 0 0
  1. #include<stdio.h>
  2. #include<math.h>
  3. #include<color.h>
  4. #include <conio.h>
  5.  
  6. int i,p,u,rez,nr,l,alo,n,o,k[2][2],e,a1,m,j;
  7. char den;
  8.  
  9. int lol (int main1);
  10. int kek (int main1);
  11. int aga (int main1);
  12.  
  13. main()
  14.  
  15. {
  16. main1:{
  17. printf("dati litera noua pentru tabel\n");
  18. for (l = 0; l < 2; l++) {
  19. scanf("%c",&den); }
  20. printf("Dati nr de linii ale tabloului \n");
  21. scanf("%i",&m);
  22. printf("Dati nr de coloane ale tabloului \n");
  23. scanf("%i",&n);
  24.  
  25. for(i=0;i<m;i++)
  26. for(j=0;j<n;j++)
  27. { printf("\n k[%i,%i]= ",i,j);
  28. scanf("%i",&k[i][j]); }
  29.  
  30. for(i=0;i<m;i++)
  31. { printf("\n");
  32. for(j=0;j<n;j++)
  33. printf(" %i ",k[i][j]);
  34. }
  35. }
  36. printf("\n Alege: \n");
  37. printf("1.nr impare negative\n");
  38. printf("2.nr divizibile cu 3 si nu cu 5\n");
  39. printf("3.satisfac conditia j+i<k[i][j]<k[i][j]^2\n");
  40. scanf("%d",&e);
  41. switch(e)
  42. { case 1:
  43.  
  44. printf("%d",lol(k[i][j]));
  45.  
  46. break;
  47.  
  48.  
  49. case 2:
  50.  
  51. printf("%d",kek(k[i][j]));
  52. break;
  53.  
  54. break;
  55.  
  56. case 3:
  57.  
  58. printf("%d",aga(k[i][j]));
  59. break;
  60. default:printf("punctul %d nu exista",e);
  61. }
  62. printf("\nDoriti sa repetati programul? 0 - nu \n");
  63. scanf("%d",&p);
  64. if (p > 0) goto main1;
  65. if (p == 0) printf("Progr. sa finisat");
  66. }
  67. int lol (main1)
  68. { for(i=0;i<m;i++)
  69. for(j=0;j<n;j++){
  70. if((k[i][j] < 0)&&(k[i][j] % 2 == 1)) ; printf(" nr impare negative sunt=%d ",k[i][j]);
  71.  
  72. } }
  73.  
  74. int kek (main1)
  75. {
  76. for(i=0;i<m;i++)
  77. for(j=0;j<n;j++) {
  78. if((k[i][j]%3==0)&&(k[i][j]%5==1)); }
  79.  
  80. }
  81. int aga (main1)
  82. {
  83. for(i=0;i<m;i++)
  84. for(j=0;j<n;j++) {
  85. if( (j + i) < k[i][j] < (k[i][j] * k[i][j])) ; printf(" \n satisfac conditia j+i<k[i][j]<k[i][j]^2= %d",k[i]) ;
  86. }
  87.  
  88. }
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement