Solutions to Assessment: Basic Programming Constructs

Transcription

1 Solutions to Assessment: Basic Programming Constructs Question 1: Specify the minimum number of comparisons required to find the largest number among a set of 3 integers X, Y and Z. We already know how to find the largest number between X, Y and Z using 2 comparisons. Using 1 comparison, we can compare only two numbers; so there is no way of telling the largest among X, Y, Z using just 1 comparison. Therefore, the answer is 2 Ans: 2 Question 2: Specify the minimum number of comparisons required to find the largest number among a set of 4 integers W, X, Y and Z. We need N-1 comparisons in general to find the minimum of N numbers. Ans: 3 Question 3: What is the value of variable a at the end of the following program segment? 1. int a,b,c,d,e; 2. a = 6; 3. b = 3; 4. c = -2; 5. d = 2;e = 4; 6. a = a * b; 7. d = d * 2; 8. a = a + c * d / e; 9. a = a - 5; Value of variable a after execution of line 6 is 18 Value of variable d after execution of line 7 is 4 Line 8 can be re-written as a = (a + ((c*d)/e)), because operators * and / have an equal precedence, which is higher than + Value of variable a after execution of line 8 is 16 Value of variable a after execution of line 9 is 11 Ans: 11 Question 4: What is the value of variable a at the end of the following program segment?

2 b = 10; a = b++; The expression b++ increments the value of the variable b to 11, but the expression b++ evaluates to the value 10. This value is assigned to the variable a. Ans: 10 Note: ++ in the expression b++ is called post-increment operator. Operator ++ can also be written before a variable s name (Eg: ++b). In that case, it s called pre-increment operator and the expression ++b evaluates to 11 Question 5: Which among the following operators has the highest precedence? a.) * b) == c) & d) + The order of precedence for the operators in descending order is as follows: *, +, ==,& Ans: a Question 6: What is the output of the following code segment? int x=10; int y=5; printf("%d",(y,x)); In the expression (y, x) comma acts as an operator. The expression (y, x) evaluates to the second value, i.e. x. Thus the value of variable x is printed. Please see for more examples Ans: 10 Question 7: What is the output of the following code segment? int x=1000; int y=5000; printf("%d",y,x); In this question, comma acts only as separator. printf() statement prints the value of the variable y and ignores the variable x

3 Ans: 5000 Question 8: What is the output of the following code segment? int a=10,b=200; int c; c=a,b; printf("%d",c); c = a, b; is equivalent to (c = a), b; so the variable c is assigned the value of variable a. Therefore the answer is 10. Please see for more examples Ans: 10 Question 9: Consider the following code snippet. What is the output if x is 10? switch(x) default: printf( x ); case 10: printf( y ); case 20: printf( z ); In switch statement, code under a matching label is first executed, and the execution continues till the end of switch. When x = 10; case 10: block is executed and then case 20: Ans: yz Question 10: What is the output of the following code snippet? 1. int x=10; 2. if (x==5) 3. printf("5"); 4. if (x%5==0) 5. printf("10"); 6. else 7. printf("bye");

4 In line 2, x == 5 evaluates to 0 (or False), line 3 doesn t get executed. In line 4, x % 5 == 0 evaluates to 1(or True), statement in line 5 is executed. else in line 6 is not executed obviously because if statement was already true. Ans: 10 Question 11: What is the output of the following code snippet? 1. int x=5; 2. if ( x < 10 ) 3. printf( "<10" ); 4. else if ( x % 2 == 0 ) 5. printf( "even" ); 6. else if ( x >= 5) 7. printf(">=5" ); 8. else 9. printf( "odd and <=5"); In if else if else construct, only the first true statement will be executed. x < 10 in Line 2 evaluates to True, Line 3 will be executed. Ans: <10 Question 12: Is the program segment below valid? for(;;); The statement above is a valid C statement Ans: Yes Question 13: What is of the following is the output of the following code segment? int i; for(i=1;i++<=5;printf( %d,i)); for(variable initialization; condition; variable update) Code to execute while the condition is true For the first iteration, condition is executed and if it evaluates to True, the body of for loop is executed. For the subsequent iterations, statement(s) in variable update is/are executed, followed by statement(s)

5 in condition. If condition evaluates to True, body of the for loop is executed. Also note that, if i = 5, the statement i++ evaluates to 5; and hence i++ <= 5 evaluates to True Ans: Question 14: What is the output of the following code segment? int i=1; while(i++<5); printf("%d",i); The while loop starts with the check (1 < 5) but it gets post-incremented to 2. This loop continues till i = 5. When you check (i++ < 5), it evaluates (5 <5) which is FALSE and the loop exits. However, because of i++, i gets incremented to 6. The result printed would be 6. Ans: 6 Question 15: What is the output of the following code segment? 1. int i=1; 2. do 3. while(i++<=5); 4. while(i++<=4); 5. while(i++<=3); 6. printf( %d,i); Just like in Q15, the while loop in Line 3, would leave i with the value 7. The check in line 4, checks (7<=4) but also increments it to 8. Finally, the check in line 9 fails but the value of i gets incremented to 9. Thus 9 is the final answer. Ans: 9 Solutions to Programming Assignments Question 1: Write a program that reads numbers which are in the range 0 to 100, till it encounters -1. Print the sum of all the integers that you have read before you encountered -1

6 INPUT: A sequence of integers separated by whitespace. There may be other integers following -1. OUTPUT: Sum of all integers in the sequence before you encounter -1. Any integer that is input after -1 in the sequence should be ignored. CONSTRAINTS: Atmost 10 integers will be given in the input. One of them is guaranteed to be a -1. s will be in the range 0 to 100 (both included). #include<stdio.h> int main() int sum=0; int number=0; do sum=sum+number; //update partial sum scanf("%d",&number); //read a number from the input while(number!=-1); //check whether recently added number is -1 or not printf("%d",sum); //print the final sum return 0; Public test cases: Private test cases:

7 Question 2: Solution described in the video. Given three points (x1, y1), (x2, y2) and (x3, y3), write a program to check if all the three points fall on one straight line. INPUT: Six integers x1, y1, x2, y2, x3, y3 separated by whitespace. OUTPUT: Print Yes if all the points fall on straight line, No otherwise. CONSTRAINTS: <= x1, y1, x2, y2, x3, y3 <= 1000 #include<stdio.h> int main() int x1,y1,x2,y2,x3,y3; //Reading all 6 integers from the input using scanf() scanf("%d %d %d %d %d %d", &x1,&y1,&x2,&y2,&x3,&y3); //Checking if the slopes (y2-y1)/(x2-x1) == (y3-y2)/(x3-x2). If they are equal, then all the points lie on the same line. //Instead of performing division in LHS and RHS, we cross multiply (x2-x1) and (x3-x2) to handle the case when either (x2-x1) = 0 or (x3-x2) = 0. if ((y2-y1)*(x3-x2) == (y3-y2)*(x2-x1)) printf("yes"); else printf("no"); return 0; PUBLIC TEST CASES:

8 Yes Yes No Yes PRIVATE TEST CASES: No Yes No Yes No Yes Yes Yes Question 3: The digital root (also called repeated digital sum) of a number is a single digit value obtained by an iterative process of summing digits. Digital sum of is 7, because =25 and 2+5 = 7. Write a program that takes an integer as input and prints its digital root. INPUT: A single integer N OUTPUT: Digital root of the number N. CONSTRAINTS: 1 <= N <= 10^7 PUBLIC TEST CASES:

9 PRIVATE TEST CASES: Check the video posted in the course webpage for a detailed explanation of the following code: #include<stdio.h> int main() int N, sum=0; scanf("%d",&n); while(n>9) sum=0; while(n>0) sum += N%10; N = N/10; N = sum;

10 printf("%d",n); return 0; Question 4: Write a program to print all the factors of a positive integer A. INPUT: A single integer A OUTPUT: Factors of the number A, in ascending order, separated by whitespace. 1 and A are also factors of A. CONSTRAINTS: 2 <= A <= #include<stdio.h> int main() int number; int i=1; scanf("%d",&number); printf("%d",i); for(i=2;i<=number;i++) //iterating over all numbers from 2 to number if(number%i==0) //checking whether 'i' is a factor of number or not printf(" %d",i); //printing a factor with a preceding whitespace return 0; PUBLIC TEST CASES:

11 PRIVATE TEST CASES: