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  1.  
  2. #include <iostream>
  3.  
  4. using namespace std;
  5.  
  6. int n, m, a[10][10], b[100], k;
  7.  
  8. void nhapMang(int a[10][10], int n, int m)
  9. {
  10.     for (int i = 0 ; i < n; i++)
  11.         for (int j = 0; j < m; j++) cin >> a[i][j];
  12. }
  13.  
  14. void mangB (int a[10][10], int n, int m, int b[100])
  15. {
  16.     int k = 0;
  17.     for (int i = 0 ; i < n ; i++)
  18.     {
  19.         for (int j = 0 ; j < m ; j++)
  20.         {
  21.             b[k] = a[i][j];
  22. cout << b[k] << "\t";
  23.             k++;
  24.         }
  25.  
  26.     }
  27. }
  28.  
  29. void sapxep (int b[100], int n, int m)
  30. {
  31.     for (int i = m*n ; i > 0 ; i--)
  32.     {
  33.         for (int j = 0 ; j < i ; j++)
  34.             if (b[j] < b[j + 1])
  35.             {
  36.                 int tmp;
  37.                 tmp = b[j];
  38.                 b[j] = b[j + 1];
  39.                 b[j + 1] = tmp;
  40.             }
  41.     }
  42. }
  43.  
  44. void xuat(int b[100], int n, int m){
  45. for(int i = 0; i < n*m; i++){
  46. cout << b[i] << "\t";
  47. if(i%m==m-1) cout << endl;
  48. }
  49. }
  50.  
  51. int main()
  52. {
  53.     cin >> n;
  54.     cin >> m;
  55.  
  56.     nhapMang(a, n, m);
  57.  
  58.     cout << "chua sx: " << endl;
  59.     mangB(a, n, m, b);
  60.  
  61.     cout << "\nda sx: " << endl;
  62.     sapxep(b, n, m);
  63. xuat(b,n,m);
  64.     return 0;
  65. }
Success #stdin #stdout 0s 4344KB
comments (?)
stdin
copy 
2 3
3 5 9
2 6 8
stdout
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chua sx: 
3	5	9	2	6	8	
da sx: 
9	8	6	
5	3	2
https://ideone.com/5uWI8m
language:
C++ (gcc 8.3)
created:
4 years ago
visibility:
public

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