Programming and Data Structures

Transcription

1 Programming and Data Structures Tutorial sheet: 2 Topics: Decision Making, Looping and Branching Q 1. (a) What is wrong with the following loop while ( n <= 100) sum += n*n; Ans : It is an infinite loop. The control variable n doesn t change (b) If s is a compound statement, and e1, e2 and e2 are expressions, then what is the difference between the program fragment: for ( e1; e2; e3 ) s; and the fragment: e1; while ( e2) s; e3; Ans : There is no difference between these two fragments, unless s contains a continue statement (tricky). For example the following for statement will iterate 4 times and then terminate normally, but the while statement will be an infinite loop for (i=0; i < 4; i++) if(i == 2) continue; i=0; while (i < 4) if(i == 2) continue; i++; Q 2. Convert the following for loop into a while loop: for ( int i = 1; i <= n; i++) printf ( %d,i*i); Ans : int i= 1; while ( i <= n) printf ( %d,i*i); i++;

2 Q 3. Write the output of the following code a) main( ) Ans : for( int i = 0; i < 8; i++) b) main() int n=100, s; Ans : 68 if ( i%2 == 0) printf( %d\n,i+1); else if ( i%3 == 0) printf( %d\n,i*1); else if ( i%5 == 0) printf( %d\n,.2*i); else printf( %d,i); for(s=2; n>2; --n) s +=2; n-=2; printf( %d,s); Q 4. (a) What is wrong with the following code? printf( Enter n: ; scanf( %d,n); if(n<0) printf( That is negative. Try again.\n ); scanf( %d,n); else printf( o.k. n = %d \n,n); Ans: There is more than one statement between the if clause and the else clause. They need to be made into a compound statement by enclosing them in braces. (b) What does the variable sum compute for any value of N in the following code? int i, sum = 0; for ( i = 1 ; i < N ; i *=2) sum += i ; Ans: It computes the sum of a G.P. starting with 1 and common ratio 2 upto N terms.

3 Q 5. WAP that reads a positive integer n and then prints the following pattern ********* _******** ******* ****** ***** **** *** ** * where n is the number of lines void main() int i=0,j=0,n; printf( Enter no. of lines - ); for(i = 0; i < n ;i++) for(j = 0; j < n; j++) if( i > j) printf( _ ); else printf( * ); printf( \n ); Q 6. Find the output of the following C program int i = 20, k = 0; for (int j = 1; j < i; j = i + 4*( i / j) ) printf( %d, k ); Ans: 4 k += j < 10? 4 : 3;

4 Q 7 Write C code which reads a floating point number x and an integer n from keyboard and computes the following. x + (x+1) + (x+2) + (x+3) (x+n) Ans : #include<stdio.h> void main() float x, sum; int n,i; printf("enter value of n : "); scanf("%d",&n); printf("enter value of x : "); scanf("%f",&x); for( i = 0; i <= n; i++) sum += x+i; printf( %f, sum); Q 8. WAP to compute the sine of an angle using the following sine infinite series upto 5 decimal places sin(x) = x x 3 /3! + x 5 /5! x 7 /7! + where x is in radian. void main ( ) float x, sum, t1, t2; int i printf( Enter angle in radians : ); scanf( %d,&x); t1 = x; t2 = -1*x*x*x/3; for (sum = t1, i = 5;(t2-t1)<.00001; i += 2) sum += t1; t1=t2; t2 = -1*t2*x*x/(i*(i-1)); Q 9. Consider the evaluation of the following function: 1.5x 3, x 2 y 2x, x>2 and z 4x 100, x<40 300, x x 150, x>40 How can y and z be evaluated using conditional operator (?:). Ans: y = ( x <= 2 )? (1.5*x + 3) : 2*x ; z = ( x <= 40 )? ( ( x < 40 )? ( 4*x ) : 300 ) : ( 4.5*x );

5 Q 10. WAP to take a number from the user and check whether it is a palindrome or not. A number is a palindrome if it is the same as its reverse. Ex is a palindrome. void main( ) int n,i,sum; printf( Enter a number : ); sum=0; i = n; while(i>0) sum = ( sum*10 ) + i%10; i = i/10; if(n == sum) printf( \n%d is a palindrome.,n); else printf( \n%d is not a palindrome.,n); Q 11.WAP that plays the game Rock, Paper, Scissors. In this game, two players simultaneously say (or display a hand symbol representing) either rock, paper or scissors. The winner is the one whose choice dominates the other. The rules are: paper dominates (wraps) rock, rock dominates (breaks) scissors, and scissors dominates (cut) paper. enum Choice rock, paper, scissors ; enum Result player1, player2, tie ; void main( ) int n; Choice choice1, choice2; Result result; printf( Choose rock(0), paper(1), or scissors(2) :\n ); printf( Player #1 : ); choice1 = n; printf( Player #2 : ); choice2 = n; if( choice1 == choice2) result = tie; else if( choice1 == rock ) if( choice2 == paper) result = player2; else result = player1; else if( choice1 == paper ) if( choice2 == rock) result = player1; else result = player2; else // ( choice1 == scissors ) if( choice2 == rock ) result = player2; else result = player1; if( result == tie ) printf( \nyou tied. ); else if( result == player1 ) printf( \nplayer #1 wins. ); else if( result == player2 ) printf( \nplayer #2 wins. ); ** Alternatively one can use normal integer data type for choice and result instead of enum.

6 Q 12. Amicable numbers are found in pairs. A given pair of numbers is Amicable if the sum of the proper divisors (not including itself) of one number is equal to the other number and vice versa. For example 220 & 284 are amicable numbers First we find the proper divisors of 220: 220:1, 2, 4, 5, 10, 11, 20, 22, 44, 55, = 284 Now, 284: 1, 2, 4, 71, = 220 WAP to check that the input pair of numbers is amicable. void main( ) int n1, n2, sum1, sum2, i; printf( Enter 2 numbers : ); scanf( %d%d,&n1,&n2); sum1 = sum2 = 0; for(i = 1; i <= n1; i++) if( (n1 % i) == 0 ) sum1 += i; for(i = 1; i <= n2; i++) if( (n2 % i) == 0 ) sum2 += i; if( ( sum2 == n1 ) && ( sum1 == n2) ) printf( \n%d and %d is an amicable pair.,n1,n2); else printf( \n%d and %d is not an amicable pair.,n1,n2);