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CS 2073 Computer Programming with Engineering Applications
Fall 2008 – Final — May 9, 2008 1:30pm - 4:00 pm
Name:……………………… Score: ……./ Sequence number:…………. This exam has six questions. You have 120 minutes. Good luck…
- (5 points) Complete the following C program that first reads a, b, c, where a and b are two degrees and c is increment. The program then generates a table of conversions from degrees to radians. Start degrees at a and increment by c until the degree is greater than b. Recall that we say 180 o^ is equal to π radian. You can assume that π is 3.14.
#include <stdio.h> #define PI 3. int main(void) { /* Declare variables. If needed, you can declare more*/ int a,b,c, degree;
/* Enter the start degree, end degree, and increment */ printf("Enter a b c : "); scanf("%d %d %d", &a, &b, &c);
system("pause");
/* Exit program. */ return 0; }
Name:…………..
- (5 points) Suppose a data file (say student.txt) contains student ID followed by nine numbers indicating the grades she/he got for each hw. As we will do in this class, we want to drop the lowest hw and compute the average for each student based on the best eight hw. For example, here is a sample file containing three students:
1 30 30 10 30 30 30 30 30 30 2 40 40 30 40 40 40 40 40 40 3 50 50 40 50 50 50 50 50 50
Write a program that can read the above file and output the student ID, and the average grade based on the best eight hw after dropping the lowest one.
(Hint: in a loop find the total of nine while finding the minimum of nine for each student, then you can take (total-min)/8 as the average).
For the above file, your program should generate the following output on the screen:
Student Id Average
1 30 2 40 3 50
Complete the code in the next page.
- (5 points) In this question you will trace the following program (i.e., run it by hand) and show how values in variables and arrays change in memory.
x[0] 5
x[1]
x[2] 3
y[0] 0
y[1] 0
y[2] 0
j a b n i j a i j
sum
Name:……………………
main()
int x[3]={5, 2, 3};
int y[3] = {0};
int j;
F1(x, y, 3);
for(j = 0; j < 3; j++){
printf(“%d ”, x[j]);
return 0;
void F1 (int a[], int b[], int n)
int i, j;
for(i=0; i < n; i++){
b[i] = F2(a, i);
return;
int F2 (int a[], int i)
int j, sum=0;
for(j = 0; j < i+1; j++){
sum = sum + a[j];
return sum
OUTPUT:
- (5 points) Suppose we declare int a[N], b[M], c[N] , where N and M are predefined global constants, somewhat we initialize a and b such that they contain two sets and assume that the values in a and b are already sorted in ascending order (so you don’t need to anything else, but use them as is). NOW, We are interested in finding: a. the difference between a and b , and put the elements of a-b into c and b. the number of elements in c. Write a function that will take a, b, c as parameters and determine the difference set c and the number of elements in c. The function will return the number of elements in c while storing the difference between a and b into c.
For example, if a = {2, 5, 7, 8, 13} and b = {1, 5, 7, 9, 12} , then your function will return 3 as the number of elements in c , and we will have c = {2, 8, 13} as a-b.
int set_difference(int a[N], int b[M], int c[N]) {
return …………….
}
Name:……………………
- (5points) Trace the following program. Using the below table show how variables change in the and give the output.
name Address Content/Value
x 11
y 12
z 13
p1 14
p2 15
16
a 59
b 60
c 61
d 61
x 63
y 64
65
Name:……………………
main() {
**int x, y, z, p1, p2;
p1 = &x; p2 = p1; x=4; y=5;
*z = f1(x, p1, p2, &y);
**printf(“x=%d y=%d z=%d “ “ and *p1=%d *p2=%d “, x, y, z, p1, p2);
}
**int f1(int a, int b, int c, int d)
{
int x, y;
**x = *c + a; y = *d – b; c = x; d = y; return x-y;
}
MEMORY
OUTPUT:
