Name:


CS 30 Winter 2007 Midterm 1 Solutions to problems 2 through 4.


2. (20 pts) What does the following program print?

#include<stdio.h>

main()
{
int i;

i = 17;
while (i <= 28) {
  printf("%d ", i++);
  i = i + 3;
}
}

Answer:
17 21 25

3. (20 pts) Rewrite the program in problem 2 using a for loop instead of a while statement.


#include<stdio.h>

main()
{
int i;

i = 17;

for (i = 17; i <= 28; i = i+4) {
	printf("%d ", i);
}
}








4. (20 pts)  What does the following program print out?

#include<stdio.h>

main()
{
int i,j,k;


for (i = 1;   i <= 2 ;  i++) {
        for (j = i+1; j <= 4; j++) {
                for (k = j+2; k <= 7; k++) {
                printf("%d %d %d\n", i, j, k);
                }
        }
}
}

Answer:
1 2 4
1 2 5
1 2 6
1 2 7
1 3 5
1 3 6
1 3 7
1 4 6
1 4 7
2 3 5
2 3 6
2 3 7
2 4 6
2 4 7


5. (40 pts) Write a program that asks the user to input a list of
real numbers which can either be positive or negative or zero.
Inputing ends when the user enters the EOF symbol.  The program
counts the number of input numbers that are strictly positive, the
number that are zero and the number that are strictly negative, and
prints out those three counts, along with appropriate explanation of 
the output.  The program also computes the product of all
the strictly positive numbers and prints that number.

Answer:


/* CS30 midterm 1 problem 5 solution */
#include <stdio.h>

int main()
{
	float input, product;       /* "real" numbers are floats, not ints */
	int pos, neg, zero;         /* our counters */
	pos = 0;
	neg = 0;
	zero = 0;
	product = 1.0;

	printf("Please enter a number: ");
	while(scanf("%f",&input) != EOF)
	{
		if(input > 0)       /* positive input */
		{
			pos++;
			product *= input;
		}
		else if(input < 0)  /* negative input */
			neg++;
		else if(input == 0) /* zero input */
			zero++;
		else                /* we should never get here */
			printf("error: this number is not positive, negative, or zero!\n");

		printf("Please enter a number: ");
	}

	printf("\nPositive count: %d, Negative count: %d, Zero count: %d\n", pos, neg, zero);
	printf("Product of positive numbers: %f\n", product);

	return 0;

}