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vlada-n [284]
3 years ago
10

Consider the series of alternating 1s and 0s 1 0 1 0 1 0 1 0 1 0 ...... The user will enter a number n . And you have to output

the first n numbers of this sequence (separated by spaces) You may assume the user will enter n > 0 and an integer using two if statements

Computers and Technology
1 answer:
Vika [28.1K]3 years ago
8 0

Answer:

# user is prompted to enter the value of n

n = int(input("Enter your number: "))

# if statement to check if n > 0

if (n > 0):

   # for-loop to loop through the value of n

   for digit in range(1, (n+1)):

       # if digit is odd, print 1

       if((digit % 2) == 1):

           print("1", end=" ")

       # else if digit is even print 0

       else:

           print("0", end=" ")

Explanation:

The code is written in Python and is well commented.

A sample output of the code execution is attached.

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How to write a function that counts the letters in a string in C?
stiv31 [10]

Answer:

Here is the C function that counts the letters in a string:

int LetterCount(char string[]){  //function to count letters in a string passed as parameter

 string[100];  // char type string with size 100

  int i, letters;  // letter variable stores the count for no. of letters in string

   i = letters = 0;  //initialize variables i and letters to 0

  while (string[i] != '\0')  { // loop iterates through the entire string until end of string is reached

    if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  { // if condition checks the letters in the string

     letters++;    }  // increments 1 to the count of letters variable each time a letter is found in the string

    i++;  }  //increments value of i to move one character forward in string

   printf("Number of Letters in this String = %d", letters);   // displays the number of letters in the string

   return 0; }                              

Explanation:

Here the question means that the function should count the letters in a string. So the letters are the alphabets from a to z and A to Z.

Here is the complete program:

#include <stdio.h>   // to use input output functions

int LetterCount(char string[]){  // method that takes a character string as parameter and counts the letters in the string

string[100];  

int i, letters;

i = letters = 0;

while (string[i] != '\0')   {

 if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z'))

   {letters++;  }

   i++; }

   printf("Number of Alphabets in this String = %d", letters);  

   return 0;}  

int main(){  // start of main function

  char string[100];  //declares a char array of string

   printf("Please Enter a String : ");   //prompts user to enter a string

  fgets(string,100,stdin);  //get the input string from user

   LetterCount(string); } // calls method to count letters in input string

I will explain this function with an example

Lets suppose the user input "abc3" string

string[100] = "abc3"

Now the function has a while loop that while (string[i] != '\0')  that checks if string character at i-th position is not '\0' which represents the end of the character string. As value of i = 0 so this means i is positioned at the first character of array i.e. 'a'

At first iteration:

i = 0

letter = 0

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )   if condition checks if the character at i-th position of string is a letter. As the character at 0-th position of string is 'a' which is a letter so this condition evaluates to true. So the statement letter++ inside if condition executes which increments the letter variable to 1. So the value of letter becomes 1. Next statement i++ increments the value of i to 1. So i becomes 1. Hence:

i = 1

letter = 1

At second iteration:

i = 1

letter = 1

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 1st position of string is 'b' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 2

letter = 2

At third iteration:

i = 2

letter = 2

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 2nd position of string is 'c' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 3

letter = 3

At fourth iteration:

i = 3

letter = 3

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 3rd position of string is '3' which is not a letter but a digit so this condition evaluates to false. So the statement letter++ inside if condition does not execute. Now i++ executes which increments this variable to 1. Hence:

i = 4

letter = 3

Now the loop breaks because while (string[i] != '\0') condition valuates to false as it reaches the end of the string.

So the statement: printf("\n Number of Letters in this String = %d", letters); executes which prints the value of letters on the output screen. Hence the output is:

Number of Letters in this String = 3

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