Bytelandian gold coins

COINS
#include<iostream>
#include<map>

typedef unsigned long long int int64;
using namespace std;
map<int64,int64> a;

int64 max(int64 x,int64 y)
{
      return x>y?x:y;
}


int64 f(int64 b)
{
 if(a[b]!=0)return a[b];
 else a[b] = max(b,(f(b/2)+f(b/3)+f(b/4)));     
 return a[b];     
}


int main()
{
a[0] = 0;
a[1] =1;
a[2] = 2;
a[3] = 3;
a[4] = 4;
a[5] = 5;
   
    int64 n;
    while(cin>>n)
    {
                 if(n==0||n==1||n==2||n==3)cout<<n<<endl;
                 else
       cout<<max(n,(f(n/2)+f(n/3)+f(n/4)))<<endl;        
    }           
return 0;   
}

Street parade

 STPAR
#include<iostream>
#include<stack>
#include<queue>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t)
    {
    queue<int> num;
    stack<int> rest;
    stack<int> pass;
    int temp;
        for(int i=0;i<t;i++)
            {
                cin>>temp;
                num.push(temp);
            }
        while(num.front()!=1)
            {
            //    cout<<num.front()<<endl;
            //    system("pause");
                rest.push(num.front());
                num.pop();
            }
    //    cout<<"out of first"<<endl;
        int start = 1;
        int flag;
    while(!num.empty()){
            while(!num.empty() && num.front()==start)
            {
            //    cout<<num.front();
            //    system("pause");               
                pass.push(num.front());
                num.pop();           
                start++;
            }
        //    cout<<"out of second"<<endl;
            while(!num.empty() && (num.front()!=start))
            {
                //cout<<num.front();
            //    /system("pause");
                //cout<<"out of third"<<endl;
                if(!(rest.empty()) && rest.top()==start)
                {
                //    cout<<rest.top();
                //    system("pause");                   
                    pass.push(rest.top());
                    rest.pop();
                    start++;
                }
                else if(!num.empty())
                {
                //    cout<<num.front()<<endl;
                //    system("pause");
                    rest.push(num.front());
                    num.pop();
                }   
            //    cout<<"out of fourth"<<endl;
            }
        }    while(!rest.empty())
            {
                //cout<<rest.top()<<" "<<pass.top()+1<<endl;
                if(rest.top()==pass.top()+1)
                    {pass.push(rest.top());rest.pop();}
                else {flag=1;goto ny;}
            }
           
        ny:
        if(flag==1){cout<<"no"<<endl;flag=0;}
        else cout<<"yes"<<endl;
        cin>>t;       
    }
return 0;
}

Algorithm

C program for prime number

Prime number program in c: c program for prime number, this code prints prime numbers using c programming language. To check whether a number is prime or not see another code below. Prime number logic: a number is prime if it is divisible only by one and itself. Remember two is the only even and also the smallest prime number. First few prime numbers are 2, 3, 5, 7, 11, 13, 17....etc. Prime numbers have many applications in computer science and mathematics. A number greater than one can be factorized into prime numbers, For example 540 = 2²*3³*5¹

Prime number program in c language

#include<stdio.h>
 
int main()
{
   int n, i = 3, count, c;
 
   printf("Enter the number of prime numbers required\n");
   scanf("%d",&n);
 
   if ( n >= 1 )
   {
      printf("First %d prime numbers are :\n",n);
      printf("2\n");
   }
 
   for ( count = 2 ; count <= n ;  )
   {
      for ( c = 2 ; c <= i - 1 ; c++ )
      {
         if ( i%c == 0 )
            break;
      }
      if ( c == i )
      {
         printf("%d\n",i);
         count++;
      }
      i++;
   }
 
   return 0;
}

Download Prime number program.
Output of program:

C program for prime number or not

#include<stdio.h>
 
main()
{
   int n, c = 2;
 
   printf("Enter a number to check if it is prime\n");
   scanf("%d",&n);
 
   for ( c = 2 ; c <= n - 1 ; c++ )
   {
      if ( n%c == 0 )
      {
         printf("%d is not prime.\n", n);
  break;
      }
   }
   if ( c == n )
      printf("%d is prime.\n", n);
 
   return 0;
}

C program for prime number using function

#include<stdio.h>
 
int check_prime(int);
 
main()
{
   int n, result;
 
   printf("Enter an integer to check whether it is prime or not.\n");
   scanf("%d",&n);
 
   result = check_prime(n);
 
   if ( result == 1 )
      printf("%d is prime.\n", n);
   else
      printf("%d is not prime.\n", n);
 
   return 0;
}
 
int check_prime(int a)
{
   int c;
 
   for ( c = 2 ; c <= a - 1 ; c++ )
   { 
      if ( a%c == 0 )
  return 0;
   }
   if ( c == a )
      return 1;
}

There are many logic to check prime numbers, one given below is more efficient then above method.
for ( c = 2 ; c <= (int)sqrt(n) ; c++ )
Only checking from 2 to square root of number is sufficient.
There are many more efficient logic available.

Print Diamond

C program to print diamond pattern

Diamond pattern in c: This code print diamond pattern of stars. Diamond shape is as follows:

  *
 ***
*****
 ***
  *

C programming code

#include <stdio.h>
 
int main()
{
  int n, c, k, space = 1;
 
  printf("Enter number of rows\n");
  scanf("%d", &n);
 
  space = n - 1;
 
  for (k = 1; k <= n; k++)
  {
    for (c = 1; c <= space; c++)
      printf(" ");
 
    space--;
 
    for (c = 1; c <= 2*k-1; c++)
      printf("*");
 
    printf("\n");
  }
 
  space = 1;
 
  for (k = 1; k <= n - 1; k++)
  {
    for (c = 1; c <= space; c++)
      printf(" ");
 
    space++;
 
    for (c = 1 ; c <= 2*(n-k)-1; c++)
      printf("*");
 
    printf("\n");
  }
 
  return 0;
}

Download Diamond program.
Output of program:

Print Pattern in C

C program to print patterns of numbers and stars

These program prints various different patterns of numbers and stars. These codes illustrate how to create various patterns using c programming. Most of these c programs involve usage of nested loops and space. A pattern of numbers, star or characters is a way of arranging these in some logical manner or they may form a sequence. Some of these patterns are triangles which have special importance in mathematics. Some patterns are symmetrical while other are not. Please see the complete page and look at comments for many different patterns.

    *
   ***
  *****
 *******
*********

We have shown five rows above, in the program you will be asked to enter the numbers of rows you want to print in the pyramid of stars.

C programming code

#include <stdio.h>
 
int main()
{
   int row, c, n, temp;
 
   printf("Enter the number of rows in pyramid of stars you wish to see ");
   scanf("%d",&n);
 
   temp = n;
 
   for ( row = 1 ; row <= n ; row++ )
   {
      for ( c = 1 ; c < temp ; c++ )
         printf(" ");
 
      temp--;
 
      for ( c = 1 ; c <= 2*row - 1 ; c++ )
         printf("*");
 
      printf("\n");
   }
 
   return 0;
}

Download Stars pyramid program.
Output of program:

For more patterns or shapes on numbers and characters see comments below and also see codes on following pages:
Floyd triangle
Pascal triangle
Consider the pattern
*
**
***
****
*****
to print above pattern see the code below:

#include <stdio.h>
 
int main()
{
    int n, c, k;
 
    printf("Enter number of rows\n");
    scanf("%d",&n);
 
    for ( c = 1 ; c <= n ; c++ )
    {
        for( k = 1 ; k <= c ; k++ )
            printf("*");
 
        printf("\n");
    }
 
    return 0;
}

Using these examples you are in a better position to create your desired pattern for yourself. Creating a pattern involves how to use nested loops properly, some pattern may involve alphabets or other special characters. Key aspect is knowing how the characters in pattern changes.

C pattern programs

Pattern:

   
   *
  *A*
 *A*A*
*A*A*A*

C pattern program of stars and alphabets:

#include<stdio.h>
 
main()
{
    int n, c, k, space, count = 1;
 
    printf("Enter number of rows\n");
    scanf("%d",&n);
 
    space = n;
 
    for ( c = 1 ; c <= n ; c++)
    {
        for( k = 1 ; k < space ; k++)
           printf(" ");
 
        for ( k = 1 ; k <= c ; k++)
        {
            printf("*");
 
            if ( c > 1 && count < c)
            {
                 printf("A");    
                 count++; 
            }      
        }    
 
        printf("\n");
        space--;
        count = 1;
    }
    return 0;
}

Pattern:

        1
       232
      34543
     4567654
    567898765

C program:

#include<stdio.h>
 
main()
{
      int n, c, d, num = 1, space;
 
      scanf("%d",&n);
 
      space = n - 1;
 
      for ( d = 1 ; d <= n ; d++ )
      {
          num = d;
 
          for ( c = 1 ; c <= space ; c++ )
              printf(" ");
 
          space--;
 
          for ( c = 1 ; c <= d ; c++ )
          {
              printf("%d", num);
              num++;
          }
          num--;
          num--;
          for ( c = 1 ; c < d ; c++)
          {
              printf("%d", num);
              num--;
          }
          printf("\n");
 
      }
 
      return 0;
}

Palindrome Numbers

Palindrome number in c: A palindrome number is a number such that if we reverse it, it will not change. For example some palindrome numbers examples are 121, 212, 12321, -454. To check whether a number is palindrome or not first we reverse it and then compare the number obtained with the original, if both are same then number is palindrome otherwise not. C program for palindrome number is given below.

Palindrome number algorithm

1. Get the number from user.
2. Reverse it.
3. Compare it with the number entered by the user.
4. If both are same then print palindrome number
5. Else print not a palindrome number.

Palindrome number program c

#include <stdio.h>
 
int main()
{
   int n, reverse = 0, temp;
 
   printf("Enter a number to check if it is a palindrome or not\n");
   scanf("%d",&n);
 
   temp = n;
 
   while( temp != 0 )
   {
      reverse = reverse * 10;
      reverse = reverse + temp%10;
      temp = temp/10;
   }
 
   if ( n == reverse )
      printf("%d is a palindrome number.\n", n);
   else
      printf("%d is not a palindrome number.\n", n);
 
   return 0;
}

Download Palindrome number program.
Output of program:

Reversing a Number.

C program to reverse a number

C Program to reverse a number :- This program reverse the number entered by the user and then prints the reversed number on the screen. For example if user enter 123 as input then 321 is printed as output. In our program we use modulus(%) operator to obtain the digits of a number. To invert number look at it and write it from opposite direction or the output of code is a number obtained by writing original number from right to left. To reverse or invert large numbers use long data type or long long data type if your compiler supports it, if you still have large numbers then use strings or other data structure.

C programming code

 
#include <stdio.h>
 
int main()
{
   int n, reverse = 0;
 
   printf("Enter a number to reverse\n");
   scanf("%d",&n);
 
   while (n != 0)
   {
      reverse = reverse * 10;
      reverse = reverse + n%10;
      n = n/10;
   }
 
   printf("Reverse of entered number is = %d\n", reverse);
 
   return 0;
}

Download Reverse number program.
Output of program:

Swapping two numbers

C program to swap two numbers

C program to swap two numbers with and without using third variable, swapping in c using pointers, functions (Call by reference) and using bitwise XOR operator, swapping means interchanging. For example if in your c program you have taken two variable a and b where a = 4 and b = 5, then before swapping a = 4, b = 5 after swapping a = 5, b = 4
In our c program to swap numbers we will use a temp variable to swap two numbers.

Swapping of two numbers in c

#include <stdio.h>
 
int main()
{
   int x, y, temp;
 
   printf("Enter the value of x and y\n");
   scanf("%d%d", &x, &y);
 
   printf("Before Swapping\nx = %d\ny = %d\n",x,y);
 
   temp = x;
   x    = y;
   y    = temp;
 
   printf("After Swapping\nx = %d\ny = %d\n",x,y);
 
   return 0;
}

Download Swap numbers program.
Output of program:

Swapping of two numbers without third variable

You can also swap two numbers without using temp or temporary or third variable. In that case c program will be as shown :-

#include <stdio.h>
 
int main()
{
   int a, b;
 
   printf("Enter two integers to swap\n");
   scanf("%d%d", &a, &b);
 
   a = a + b;
   b = a - b;
   a = a - b;
 
   printf("a = %d\nb = %d\n",a,b);
   return 0;
}

To understand above logic simply choose a as 7 and b as 9 and then do what is written in program. You can choose any other combination of numbers as well. Sometimes it's a good way to understand a program.

Swap two numbers using pointers

#include <stdio.h>
 
int main()
{
   int x, y, *a, *b, temp;
 
   printf("Enter the value of x and y\n");
   scanf("%d%d", &x, &y);
 
   printf("Before Swapping\nx = %d\ny = %d\n", x, y);
 
   a = &x;
   b = &y;
 
   temp = *b;
   *b   = *a;
   *a   = temp;
 
   printf("After Swapping\nx = %d\ny = %d\n", x, y);
 
   return 0;
}

Swapping numbers using call by reference

In this method we will make a function to swap numbers.

#include <stdio.h>
 
void swap(int*, int*);
 
int main()
{
   int x, y;
 
   printf("Enter the value of x and y\n");
   scanf("%d%d",&x,&y);
 
   printf("Before Swapping\nx = %d\ny = %d\n", x, y);
 
   swap(&x, &y); 
 
   printf("After Swapping\nx = %d\ny = %d\n", x, y);
 
   return 0;
}
 
void swap(int *a, int *b)
{
   int temp;
 
   temp = *b;
   *b   = *a;
   *a   = temp;   
}

C programming code to swap using bitwise XOR

#include <stdio.h>
 
int main()
{
  int x, y;
 
  scanf("%d%d", &x, &y);
 
  printf("x = %d\ny = %d\n", x, y);
 
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
 
  printf("x = %d\ny = %d\n", x, y);
 
  return 0;
}

Swapping is used in sorting algorithms that is when we wish to arrange numbers in a particular order either in ascending order or in descending.

Add n Numbers

C program to add n numbers

This c program add n numbers which will be entered by the user. Firstly user will enter a number indicating how many numbers user wishes to add and then user will enter n numbers. In the first c program to add numbers we are not using an array, and using array in the second code.

C programming code

#include <stdio.h>
 
int main()
{
   int n, sum = 0, c, value;
 
   printf("Enter the number of integers you want to add\n");
   scanf("%d", &n);
 
   printf("Enter %d integers\n",n);
 
   for (c = 1; c <= n; c++)
   {
      scanf("%d",&value);
      sum = sum + value;
   }
 
   printf("Sum of entered integers = %d\n",sum);
 
   return 0;
}

Download Add n numbers program.
Output of program:

C programming code using array

#include <stdio.h>
 
int main()
{
   int n, sum = 0, c, array[100];
 
   scanf("%d", &n);
 
   for (c = 0; c < n; c++)
   {
      scanf("%d", &array[c]);
      sum = sum + array[c];
   }
 
   printf("Sum = %d\n",sum);
 
   return 0;
}

nPr and nCr

C program to find nCr and nPr: This code calculate nCr which is n!/(r!*(n-r)!) and nPr = n!/(n-r)!

C program to find nCr using function

#include<stdio.h>
 
long factorial(int);
long find_ncr(int, int);
long find_npr(int, int);
 
main()
{
   int n, r;
   long ncr, npr;
 
   printf("Enter the value of n and r\n");
   scanf("%d%d",&n,&r);
 
   ncr = find_ncr(n, r);
   npr = find_npr(n, r);
 
   printf("%dC%d = %ld\n", n, r, ncr);
   printf("%dP%d = %ld\n", n, r, npr);
 
   return 0;
}
 
long find_ncr(int n, int r)
{
   long result;
 
   result = factorial(n)/(factorial(r)*factorial(n-r));
 
   return result;
}
 
long find_npr(int n, int r)
{
   long result;
 
   result = factorial(n)/factorial(n-r);
 
   return result;
} 
 
long factorial(int n)
{
   int c;
   long result = 1;
 
   for( c = 1 ; c <= n ; c++ )
      result = result*c;
 
   return ( result );
}

Download NCR and NPR program.
Output of program:

Decimal to binary conversion

  C program to convert decimal to binary: c language code to convert an integer from decimal number system(base-10) to binary number system(base-2). Size of integer is assumed to be 32 bits. We use bitwise operators to perform the desired task. We right shift the original number by 31, 30, 29, ..., 1, 0 bits using a loop and bitwise AND the number obtained with 1(one), if the result is 1 then that bit is 1 otherwise it is 0(zero).

#include<stdio.h>

int main(){

    long int decimalNumber,remainder,quotient;

    int binaryNumber[100],i=1,j;

    printf("Enter any decimal number: ");

    scanf("%ld",&decimalNumber);

    quotient = decimalNumber;

    while(quotient!=0){

         binaryNumber[i++]= quotient % 2;

         quotient = quotient / 2;

    }

    printf("Equivalent binary value of decimal number %d: ",decimalNumber);

    for(j = i -1 ;j> 0;j--)

         printf("%d",binaryNumber[j]);

    return 0;

}

Sample output:

Enter any decimal number: 50

Equivalent binary value of decimal number 50: 110010

HCF and LCM of two Number

There are many ways to find the H.C.F and LCM  of two numbers is C programming. The codes below will take two integers from user and displays the H.C.F  and LCM of those two integers.

#include <stdio.h>
 
int main() {
  int a, b, x, y, t, gcd, lcm;
 
  printf("Enter two integers\n");
  scanf("%d%d", &x, &y);
 
  a = x;
  b = y;
 
  while (b != 0) {
    t = b;
    b = a % b;
    a = t;
  }
 
  gcd = a;
  lcm = (x*y)/gcd;
 
  printf("Greatest common divisor of %d and %d = %d\n", x, y, gcd);
  printf("Least common multiple of %d and %d = %d\n", x, y, lcm);
 
  return 0;
}