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For each of the following code segments: -  identify the statements that are “mice”. -  identify the Big-0 run-time (show how you arrived at the answer ) a) sum=0; for( ctr = 0; ctr < 3; ++ctr ) for( ctc = 0; ctc < n; ++ctc ) sum += data[ctr] [ctc];   b) total =0; for( ctr = 0; ctr < n; ++ctr ) for( ctc = ctr + 1; ctc < n; ++ctc ) total += data[ctr] [ctc]; c) for ( ctr = 0; ctr < n; ++ ctr ) { sum[ctr] = 0; for ( ctc = 0; ctc < n; ++ctc ) sum[ctr] += data[ctr] [ctc]; } for ( ctr = 0; ctr < n; ++ctr ) cout << “Sum for row ”, ctr, “ is:   ”; ”, sum[ctr] << endl;

For each of the following code segments: -  identify the statements that are “mice”. -  identify the Big-0 run-time (show how you arrived at the answer ) a) sum=0; for( ctr = 0; ctr < 3; ++ctr ) for( ctc = 0; ctc < n; ++ctc ) sum += data[ctr] [ctc];   b) total =0; for( ctr = 0; ctr < n; ++ctr ) for( ctc = ctr + 1; ctc < n; ++ctc ) total += data[ctr] [ctc]; c) for ( ctr = 0; ctr < n; ++ ctr ) { sum[ctr] = 0; for ( ctc = 0; ctc < n; ++ctc ) sum[ctr] += data[ctr] [ctc]; } for ( ctr = 0; ctr < n; ++ctr ) cout << “Sum for row ”, ctr, “ is:   ”; ”, sum[ctr] << endl;

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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  1. For each of the following code segments:

    • -  identify the statements that are “mice”.

    • -  identify the Big-0 run-time (show how you arrived at the answer )

      a) sum=0;
      for( ctr = 0; ctr < 3; ++ctr )

      for( ctc = 0; ctc < n; ++ctc ) sum += data[ctr] [ctc];

       

      b) total =0;
      for( ctr = 0; ctr < n; ++ctr )

      for( ctc = ctr + 1; ctc < n; ++ctc ) total += data[ctr] [ctc];

      c) for ( ctr = 0; ctr < n; ++ ctr ) { sum[ctr] = 0;

      for ( ctc = 0; ctc < n; ++ctc ) sum[ctr] += data[ctr] [ctc];

      }
      for ( ctr = 0; ctr < n; ++ctr )

      cout << “Sum for row ”, ctr, “ is:

 

”;

”, sum[ctr] << endl;

 
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