Question

Activity 16-02 With proper indentation and comments, With proper interface/menu on the monitor Write C program of the following problem. This question involve repetition: while, and for statements This question involve selection: Don't use switch, use IF Marks is based on comments, indentation and interface/menu. You must try Choose any one value of j to print screen the following i Three input: num less than j, 21 and any valid number three input: A, B and X. Question: 1.0 From number j=1,3..until 9, do the following step 2.0 until step 4.0. In j more than 9, go to step 6.0 2.0 Input an integer number, num (if num is less than j or greater than 20, display error "number must be greater or equal to j and less or equal than 20, enter input again) 3.0 Print the following menu: Choose one of the following menu: X: print '*' num times Y: print '#' num times Z: exit from program 4.0 Input a character X, Y or z. Otherwise, display message "Wrongly enter, input again" 5.0 Repeat 1.0 6.0 stop

Answer 1

CODE OF THE FOLLOWING PROGRAM:-

#include <stdio.h>
#include <stdlib.h>

int main()
{   
  
char choice;
int j=1,num;
//if J is less than equal to 9
while(j<=9){
printf("Input an integer number\n");
scanf("%d",&num);
//if number entered is less than j or greater than 20. User will enter again
while(num<j || num>20){
printf("Number must be greater or equal to j and less or equal than 20,enter input again\n");
scanf("%d",&num);
}
//Menu & asking user to enter a choice
printf("Choose one of the following menu\n");
printf("X: print '*' num times\n");
printf("Y: print '#' num times\n");
printf("Z: Exit from the program\n");
scanf(" %c",&choice);
  
//If the entered choice is not x,Y or Z enter again
while(choice!='X' && choice!='Y' && choice!='Z'){
printf("Wrongly enter,enter again\n");
scanf(" %c",&choice);
}
  
//If the entered choice is X,Y or Z do as per menu
if(choice=='X'){
//If user choice is X
for(int i=0;i<num;i++)
printf("*");
printf("\n");
}
else if(choice=='Y'){
//If user choice is Y
for(int i=0;i<num;i++)
printf("#");
printf("\n");
}
else if(choice=='Z'){
// Terminate if user choice is Z
exit(0);
}
}
return 0;
}

SCREENSHOT OF THE PROGRAM AND SAMPLE OUTPUT:-

9 #include <stdio.h> 10 #include <stdlib.h> 12 int main() 15 char choice; int j=1, num; //if j is less than equal to 9 while(j<=9){ printf("Input an integer number\n"); scanf("%d",&num); 1/if number entered is less than j or greater than 20. User will enter again while(num<j || num>20){ printf("Number must be greater or equal to j and less or equal than 20, enter input again\n"); scanf("%d",&num); //Menu & asking user to enter a choice printf("Choose one of the following menu\n"); printf("x: print '*' num times\n"); printf("Y: print '#' num times\n"); printf("Z: Exit from the program\n"); scanf("%c",&choice); //If the entered choice is not x, y or Z enter again while choice!='X' && choice!='Y' && choice!='Z' printf("Wrongly enter, enter again\n"); scanf(" %c",&choice);

//If the entered choice is x, y or Z do as per menu if(choice==''){ //If user choice is X for(int i=0;i<num;i++) printf("*"); printf("\n"); else if(choice=='Y'){ //If user choice is Y for(int i=0;i<num;i++)|| printf("#"); printf("\n"); else if(choice=='Z') { // Terminate if user choice is Z exit(); return ; 58 }
SAMPLE OUTPUT:

input Input an integer number Number must be greater or equal to j and less or equal than 20, enter input again Choose one of the following menu X: print "*' num times Y: print 'i' num times Z: Exit from the program Wrongly enter, enter again ##### Input an integer number Number must be greater or equal to j and less or equal than 20, enter input again Choose one of the following menu X: print '*' num times Y: print '' num times Z: Exit from the program ...Program finished with exit code 0 Press ENTER to exit console.
HAPPY LEARNING