COP3502-Sum11-Exam1-solexp

COP3502-Sum11-Exam1-solexp - COP 3502 Summer 2011 Exam 1...

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COP 3502 – Summer 2011 – Exam 1 Solutions (with Instructor comments) Exam 1 Comments: As mentioned in class, my intention with this first exam is to have, at most, one hard question, which is almost always going to be the recursive coding question or the linked-list coding question, simply due to the fact that you have to actually think instead of regurgitate. Other than that, I wanted all other questions to be “workable”; meaning, every student who did an average amount of studying could successfully solve the problem. All test questions were straightforward, and anyone who truly studied and understood the PPT slides, lab questions, and programs would have done very well. Statistics: # of Exams Taken: 135 Time Stats: st student finished at the 25 minute mark 27 students finished with 15 minutes remaining (about 1/5 of class) 44 students finished with 10 minutes remaining (about 1/3 of class) Average Grade: 82.36 (which is a HIGH average) # of Grades >= 100: 16 # of Grades >= 90: 47 # of Grades >= 80: 87 So out of 135 students, 87 scored higher than 80. Yeah, that is HUGE. I’d love to say this is completely due to the awesomeness of (a) the students and (b) the instructor; that would be great. Most likely, there are other factors at play. Regarding the length, I think it was fair, and the stats “Time Stats” agree. Ideally, the first student to leave (assuming he/she did well) finishes around the midway point. A few more trickle in. And with ten minutes left, it is a good sign when one-third of the class has finished. Finally, when the time was up, only around 15-20 students (less than 10% of the class) were still working (or reading over their answers). These are good indicators as to the length and difficulty of an exam. So although some may feel the exam was long, I do feel the length was appropriate for this class and certainly workable by those students that invested enough time in studying. Now to the solutions… 1) (10 pts) Dynamic Memory Allocation. The following struct can be used to store information about a Super Market: struct SuperMarket { char SuperMarketName[30]; int *itemNumbers; double *itemPrices; }; The last two members of this struct are meant to be pointers to dynamically allocated arrays of ints and doubles, respectively. The first will be an array of item numbers, each stored as an int ,
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representing the item numbers for all items in the store. The second will be an array of prices, each stored as a double , representing the prices of those respective item numbers.
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This note was uploaded on 09/21/2011 for the course COP 3330 taught by Professor Staff during the Spring '08 term at University of Central Florida.

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COP3502-Sum11-Exam1-solexp - COP 3502 Summer 2011 Exam 1...

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