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Flauer [41]
3 years ago
9

Add Try/Catch error checking To make sure that the user enters valid values in your program. To catch negative values passed to

constructor. To catch if not enough memory is available for list. (Hint: bad_alloc)
Computers and Technology
1 answer:
ryzh [129]3 years ago
4 0

Answer:

int main()

{

 cout<<"Enter the size of array\n";

 int n;

 try{

 cin>>n;//get size of array

 if (n > 0){

 List list;//make object of list class

 try{

     list.init(n);//initialize the list

     list.fillData();//fill the data

     list.print();//print the data

     list.sortData();//sort the  data

     cout<<"\n--- After sorting the data---\n";

     list.print();//print the data

     cout<<"Enter the element to search\n";

     int x;

     cin>>x;//get element to search

     list.searchElement(x);//call the search function

 }

 catch (std::bad_alloc){

   cout<<"Sorry, could not allocate memory for list object.";

 }

 }else{

   throw('Negative Number detected');

 }

 }

 catch (int n){

   cout<<"The number should be a positive integer.";

 }

   return 0;

}

Explanation:

The try and catch keywords come in pairs, as they are used to control exceptions in the C++ source code. The try keyword checks for error in the source code given the condition (if the n integer variable is greater than 0). If the condition is met, the code in the try code block runs otherwise the catch keyword catches the error of a negative number (if the n variable in less than 0).

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Answer:

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Explanation:

Solution

(1) By making use of the  dynamic programming, we can solve the problem in linear time.

Now,

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Thus, given that:

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Then,

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Using the above recurrence relation, we can compute the sum of the optimal sub sequence for array A, which would just be the maximum over G[i] for 0 <= i<= n-1.

However, we are required to output the starting and ending indices of an optimal sub-sequence, we would use another array V where V[i] would store the starting index for a maximum sum contiguous sub sequence ending at index i.

Now the algorithm would be:

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V[0] = 0;

max = G[0];

max_start = 0, max_end = 0;

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If ( G[i-1] > 0)

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V[i] = i;

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max = G[i];

EndFor.

Output max_start and max_end.

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