Programs in C & C++ with Example

c++ program to find prime numbers: The problem of finding prime number can be solved by checking all numbers, testing them for prime and then moving ahead. If you want to calculate nth prime. Then this can be done in a brutal way by checking the number one by one. This may sound odd, by there is no easy way then this  for prime numbers (Well Actually there are like Pollard's Rho Algorithm, Number Sieves or Shor's Quantum Algorithm, but we are talking about the one that most people may understand easily). There may be way to pre-calculate the prime numbers but that again is not sufficient. So how can we use c++ to create a program to find prime numbers.

Program for Matrix Multiplication in C++ using Operator Overloading : We will now try to multiply two matrix by using the concept of operator overloading. The described way is very very easy to understand. We will be overloading "*" operator for this purpose. Although this can be done by any binary operator.

Objective : To check if given number is palindrome or not in C. Programming logic : This is a very easy program if you know how to reverse a number. Once you have found the reverse of number just compare it with original, if both are same then the number is palindrome. Code: # include < stdio.h > int main ( ) { int k ; printf ( " \n enter number = " ) ; scanf ( " %d " , & k ) ; if ( palindrome ( k ) ) { printf ( " \n The number is palindrome " ) ; } else { printf ( " \n The number is not palindrome " ) ; } return ( 0 ) ; } int palindrome ( int x ) { int rev = 0 ; //to calculate reverse of number int k = x ; while ( k > 0 ) { rev = rev * 10 + k % 10 ; k = k / 10 ; } if ( x = = rev ) // check if reverse is equal to given number { return ( 1 ) ; //if yes then return 1 } else ...

Objective : To print given numbers of Fibonacci series. (1,1,2,3,5.....) non recursively. Programming Logic : Its easy to print the above Fibonacci series in C. the number is always sum of previous 2 terms. This series starts with numbers 1,1. so next term is 1+1=2. Then next term is 1+2=3 and so on. (It may be noted that series 1,3,4,7,11,18.... is also a Fibonacci series, but it starts with 1,3 ) Code : # include < stdio.h > int main ( ) { int k ; printf ( " \n enter number of terms to print= " ) ; scanf ( " %d " , & k ) ; fic ( k ) ; return ( 0 ) ; } void fic ( int x ) // x are number of terms to print { int a , b , sum , i ; a = b = 1 ; printf ( " 1,1, " , sum ) ; //first 2 terms for ( i = 0 ; i < x - 2 ; i + + ) { sum = a + b ; printf ( " %d , " , sum ) ; a = b ; b = sum ; } } It may be noted that using a non recu...

Objective : You are given a  number and you have to find the reverse of the given number in C. reverse of 1234 will be 4321. Programming logic : The logic lies in constructing the reverse numbers from the digits of the given number. So we extract each digit with % modulus operator and then build the number up.To build the number we simply multiply the reverse number by 10 and then add the next remainder to it. Like 1234: 1234%10=4 (remainder). rev=4. 123%10= 3. rev=4x10+3=43. 12%10=2 rev=43x10+2=432. 1%10=1 rev=432x10+1=4321. Code: # include < stdio.h > int main ( ) { int k ; printf ( " \n enter a number= " ) ; scanf ( " %d " , & k ) ; printf ( " \n Reverse of the number is= %d " , reverse ( k ) ) ; return 0 ; } int reverse ( int x ) { int ans = 0 , r ; while ( x > 0 ) { r = x % 10 ; //extract last digit ans = ans * 10 + r ; // multiply by 10 and then add the r...

Objective : To convert the given integer to binary in C language. Programming technique : The conversion of integer to binary can be done in many ways. I will demonstrate a very simple (possibly very unoptimised ) way which it is easy to understand. This way involves using an array to store the remainder of the number when we divide it by 2. The remainder will be saved in array. We will start filling the array from the last and at last we will print the array. This technique will convert integer number to binary using array, it may not be a true solution but you will get the picture. we will use % modules operator to extract remainder. Then will reduce the number till it reaches 0. The array will be initialized to 0. Code : # include < stdio.h > int main ( ) { int p ; printf ( " \n Enter a number= " ) ; scanf ( " %d " , & p ) ; bin ( p ) ; } void bin ( int x ) { long b = 0 ; int binary [ 20 ] , i ; ...

Objective : Your are given a number and you have to find the sum of its digits. Like 1234 so its digits sum 1+2+3+4=10. Programming technique : This problem  can be solved using the fact that modulus operator (%) gives the last digit when divided by 10. So we successively Divide it , till all numbers are exhausted. so The program runs like this >> 1234%10 gives last digit. that is 4. (we save it ) then divide the number by 10 to reduce it. 1234/10=123. and repeat the process till the number becomes 0. First run : 1234%10 gives 4, 1234/10=123. Second run: 123%10 gives 3 , 123/10=12. Third Run :12%10 gives 2 , 12/10=1 Fourth run: 1%10 gives 1, 1/10=0 (we stop here). pretty easy stuff. Code :- #include<stdio.h> int main() {     int k;     printf("\n enter any number=");     scanf("%d",&k);     printf("\n sum of digit=%d",sumdigit(k));         ...

Objective: To print any Prime number in C.  This task may sound easy but is a very good exercise to help optimize CPU intensive programs. Finding Prime number is a very CPU hogging task and it becomes even more dreaded as the number starts to grow. Lets say we want to find 30001th prime. how will we find it?? Here is the program. Programming logic : The easiest method is to check if the given number N is divisible by any number from 2 to N-1. But it may be noted that we don't need to go beyond the number that is square root of N.  why?? lets have a look. Say we are testing number 49 for prime number. Our objective is to find any number that may divide it. so we start from 2,3,4 etc. As soon as we cross 7 there will be no number that will divide 49, if there was we have got it earlier since it will be smaller then 7.  Get it? another example 64 : 2x32 4x16 8x8 16x4 32x2 as you can see when we pass number 8 we don't need to test as the number that will come ,...

Factorial: its a mathematical function defined as 4!=4x3x2x1=120. Yes just like that. It will multiply all numbers before the given numbers till 1(Worst definition ever).

LCM : least common multiplier, is a number to which given two numbers can divide exactly (and of course there are a lot of numbers like that , but LCM is the least one of them). E.g. given two numbers 12,15 these numbers can divide 60,120,180 but LCM is 60 as its the least.

Palindrome String : are those string which is same if you read it backwards."MADAM" or "AABBCCCCBBAA" are palindrome but "TATA" is not. Its a good practice for starters to check their recursion knowledge with this kind of problem. As we know this can be implemented in recursive way too. but a non recursive way will always compare the first letter of the string to the last letter. Method : checking if string is palindrome in C can be done by creating a function. The Function checkPalindromeStr uses 2 char pointers start and end . Assign first pointer to start char and other to last char. Then we start matching start pointer to end pointer, when they don't match the given string cant be palindrome so we return 0. else we continue by increasing start and decreasing end   till start crosses end pointer. At that point the string will be palindrome and we return 1. Code: # include < stdio.h > # include < string.h > //for strlen ...

HCF : Highest common factor can be calculated using long division method. HCF is also called as GCD (greatest commomn divisor). Method : Two numbers x and y and passed to the function HCF which calculates HCF and returns it. we keep on diviing x with y untill it completely divides x. If it does y becomes HCF, if not then we replace x with the remainder and y with x . Code : # include < stdio.h > int HCF ( int x , int y ) { int tmp ; if ( x < y ) { tmp = x ; x = y ; y = tmp ; //using tmp to swap x and y } while ( x % y ! = 0 ) { tmp = x % y ; //using tmp to store remainder y = x ; x = tmp ; } return ( y ) ; } int main ( ) { int k ; k = HCF ( 18 , 24 ) ; printf ( " \n HCF= %d " , k ) ; }