Starting C programming || Print text in c language

#include<stdio.h>
int main()
{
    printf("Hello, what is your name: \n");
    /* This will print the text in the double quote and by using \n the cursor move to
    the next line.*/
    return 0;
}

Comments

Post a Comment

Popular posts from this blog

Generating a Sequence: Exploring Loops in C Programming

#include <stdio.h> int main () {     int num ;     int a ;     // Prompt user to enter the number of digits to print     printf ( "How many digits you want to print: \n " );     scanf ( " %d " , & num );     // Loop to print the specified number of digits     for ( a = 1 ; num >= a ; a ++ )     {         printf ( " %d \n " , a );     }     return 0 ; } // Blog Title: "Generating a Sequence: Exploring Loops in C Programming" // Keywords: C programming, loops, sequence generation, number printing
Read more

C Programming: Finding Factors Using 'do-while' Loop and User Interaction

  #include <stdio.h> int main () {     int a , b ;     do     {         // Initialize 'b' to 2 for each iteration of the loop         b = 2 ;         // Prompt the user to enter a digit         printf ( "Enter a digit: \n " );         // Read the input from the user         scanf ( " %d " , & a );         // Display message for searching factors         printf ( "Searching: \n " );         // Loop to find factors of the entered number         while ( a > b )         {             if ( a % b == 0 )             {                 // If 'b' is a factor of 'a', print 'b' and increment 'b' to search for the next factor...
Read more

Weird Algorithm || Introductory Problem || CSES

#include <bits/stdc++.h> using namespace std ; #define MOD 1000000007 ; #define int long long int int32_t main (){     ios_base :: sync_with_stdio ( false );     cin . tie ( NULL );     #ifndef ONLINE_JUDGE     freopen ( "input.txt" , "r" , stdin );     #endif     //----Solution-----//     int n ;     cin >> n ;     while ( n != 1 ){         cout << n << " " ;         if ( n & 1 ){ // if n is not divisible by 3 then             n = n * 3 + 1 ; // multiply it by 3 and add 1         } else {             n /= 2 ; // else divide it by 2         }     }     cout << n ;     return 0 ; }
Read more