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_sqr = 0; for( ct = 0; ct < size; ++ ct ) sum_sqr += data[ct] * data[ct]; mean_sqr = sum_sqr / size; b) sum = 0 for( ct = 1; ct <= size; ++ct ) sum += ct; product = 1; for( ct = 1; ct <= size; +
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For each of the following code segments:
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- identify the statements that are “mice”.
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- identify the Big-0 run-time (show how you arrived at the answer )
a) sum_sqr = 0;
for( ct = 0; ct < size; ++ ct )sum_sqr += data[ct] * data[ct]; mean_sqr = sum_sqr / size;
b) sum = 0
for( ct = 1; ct <= size; ++ct )sum += ct; product = 1;
for( ct = 1; ct <= size; ++ct ) product *= ct;
difference = product – sum;
c) max = data[0, 0];
for( ctr = 0; ctr < n; ++ctr )for( ctc = 0; ctc < n; ++ctc
if ( data[ctr] [ctc] > max )max = data[ctr] [ctc];
d) max = data[0, 0];
for( ctr =0; ctr < 5; ++ctr )for( ctc = 0; ctc < 5; ++ctc )
if ( data[ctr] [ctc] > max )max = data[ctr] [ctc];
e)sum=0;
for( ctr = 0; ctr < 3; ++ctr )for( ctc = 0; ctc < n; ++ctc ) sum += data[ctr] [ctc];
f) total =0;
for( ctr = 0; ctr < n; ++ctr )for( ctc = ctr + 1; ctc < n; ++ctc ) total += data[ctr] [ctc];
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