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deff fn [24]
3 years ago
9

A medium-sized highway construction company CST has chosen to constrain its investment in computing resources to laptop and desk

top computers and the necessary networking hardware and software to access the Internet efficiently. All of CST's data is to be stored off-site through an Internet-enabled storage provider that is used by many different companies and organizations. The service model employed for the data storage capability of only CST is:
Computers and Technology
1 answer:
Valentin [98]3 years ago
5 0

Answer:

The service model employed for the data storage capability of only CST is:

SaaS (Software as a Service).

Explanation:

The three cloud computing categories are Infrastructure as a Service (IaaS), Software as a Service (SaaS), and Platform as a Service (PaaS).  The three cloud computing services provide different virtual computing resources.  In the case of CST, it utilizes only the data storage software of a provider, who provides the data storage infrastructure, including cloud-based servers, while CST uses its own "laptop and desktop computers and other networking hardware and software to access the Internet efficiently."

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Answer:

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#include <iostream>

using namespace std;  

int main()  

{

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   int input;

   int found = 0;

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   for( int k = 0; k < 10; k++ )

   {

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   }

   if( found == 0 )

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   return 0;

}

OUTPUT

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Enter any number  

7

Found

Explanation:

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User is prompted to input any number. No validation is implemented for user input since it is not mentioned. The user input is taken in an integer variable, input.

int input;

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Another integer variable found is declared and initialized to 0.

int found = 0;

Inside for loop, each element is compared with the user input.

If the user input is present in the given array, variable found is incremented by 1.

for( int k = 0; k < 10; k++ )

   {

       if( input == sequence[k] )

           found = found + 1;

   }

If value of variable found is 0, program displays “ Not Found ”.

If the value of variable found is greater than 0, program displays “ Found ”.

if( found == 0 )

       cout << " Not Found " << endl;

   else

       cout << " Found " << endl;

This program can be tested for both positive and negative numbers.

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