Counting the number of valid bracket permutations for a given length

Example:
Input: 3
Output: 5
Explanation: The following permutations are valid:
((())),(())(),()(()),(()()),()()()
The following permutations are invalid:
)()()(,())(() etc.
Program:
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int main()
{
 int fun(char[],int,int,int);
 int n;
 cin>>n;
 char A[100]="";
 cout<<fun(A,0,0,n)<<endl;
}
int fun(char A[],int open,int close,int n)
{
 static int count=0;
 if(open-close<0 || open-close>n || open>n)
 {
  return count;
 }
 else if(strlen(A)==2*n)
 {
  //cout<<A<<endl;
  count++;
  //cout<<count<<" "<<endl;
  return count;
 }
 else
 {
  A[strlen(A)]='(';
  A[strlen(A)+1]=0;
  fun(A,open+1,close,n);
  A[strlen(A)-1]=0;
  A[strlen(A)]=')';
  A[strlen(A)+1]=0;
  fun(A,open,close+1,n);
  A[strlen(A)-1]=0;
  return count;
 }
}
Sample Input: 15
Sample Output: 9694845

Directory Path Parsing

Input: "C:\\a\\b\\c\\..\\.\\..\\g\\."
Output: "C:\\a\\g"

#include<iostream>
#include<string.h>
#include<signal.h>
using namespace std;
int main()
{
 void fun();
 char A[100]="C:\\a\\b\\c\\..\\.\\..\\g\\.\\..";
 //char A[100]="C:\\a\\b";
 char B[100]="";
 char temp[100]="";
 int top=-1;
 char C[20][100];
 int i=0;
 while(i<strlen(A) && A[i]!='\\')
 {
  B[i]=A[i];
  i++;
 }
 B[i]=0;
 while(i<strlen(A))
 {
  int j=0;
  while(i<strlen(A) && A[i]!='\\')
  {
   temp[j]=A[i];
   i++;
   j++;
  }
  temp[j]=0;
  if(top>=0 && strcmp(temp,"..")==0)
  {
   top--;
  }
  else if(strcmp(temp,".")==0)
  {
  }
  else
  {
   top++;
   strcpy(C[top],temp);
  }
  i++;
 }
 for(int j=0;j<=top;j++)
 {
  strcat(B,C[j]);
  strcat(B,"\\");
 }
 cout<<B<<endl;
}

Grid Traversal

There is a rectangular grid of size m * n . Bob is in location ( x, y ) inside grid. He can move in 4 directions up, down, right and left. He will die if he steps outside of rectangular grid. Find the probability that bob is alive given initial position of bob as ( x, y ) and number of steps he moves as N. (Given that Bob moves only one step at a time).

#include<iostream>
using namespace std;
int main()
{
    void fun(int,int,int,int,float*,float*,int,int);
    int m,n,x,y,N;
    float a=0.0,b=0.0,*p=&a,*q=&b;
    cin>>m>>n>>x>>y>>N;
    fun(m,n,x,y,p,q,0,N);
    cout<<(*q)/((*q)+(*p));
}
void fun(int m,int n,int x,int y,float *p,float *q,int S,int N)
{    
    if(S==N && x>=0 && y>=0 && x<m && y<n)
    {
        (*q)++;
        return;
    }
    else if((S<=N) && (x<0 || y<0 || x>=m || y>=n))
    {
        (*p)++;
        return;
    }
    else
    {
        fun(m,n,x+1,y,p,q,S+1,N);
        fun(m,n,x-1,y,p,q,S+1,N);
        fun(m,n,x,y+1,p,q,S+1,N);
        fun(m,n,x,y-1,p,q,S+1,N);
    }
}

Excel column header generator

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
        char out[27]={'Z','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y',0};
        char ans[100]={0};
        int t;
        cin>>t;
        while(t--)
        {
                int n;
                cin>>n;
                int j=0;
                while(n>0)
                {
                        int c=n%strlen(out);
                        //Using the remainder as the index of the character array.
                        ans[j]=out[c];
                        if(c==0)
                        {
                                n--;
                        }
                        n/=strlen(out);
                        j++;
                }
                //Reversing the answer string.
                for(int i=0;i<strlen(ans)/2;i++)
                {
                        char t=ans[i];
                        ans[i]=ans[strlen(ans)-i-1];
                        ans[strlen(ans)-i-1]=t;
                }
                cout<<ans<<endl;
        }
        return 0;
}

Sample Input: The first line is the number of test cases.
7
1
25
26
27
51
52
676

Sample Output:
A
Y
Z
AA
AY
AZ
YZ

C++ Program for Quicksort

#include<iostream>
using namespace std;
int main()
{
        int A[7]={3,6,1,2,7,5,4};
        void quicksort(int[],int,int);
        quicksort(A,0,6);
        for(int i=0;i<7;i++)
        {
                cout<<A[i]<<" ";
        }
        cout<<endl;         
        return 0; 
} 
void quicksort(int A[7],int low,int high) 
{         
        if(low>=high)
        {
                return;
        }
        int p=A[low];
        int i=low+1;
        int j=high;
        while(j>=i)
        {
                while(A[i]<=p)
                {
                         i++;                 
                }
                while(A[j]>p)
                {
                        j--;
                }
                if(j>i)
                {
                        int t=A[i];
                        A[i]=A[j];
                        A[j]=t;
                }
        }
        int t=A[j];
        A[j]=A[low];
        A[low]=t;
        quicksort(A,low,j-1);
        quicksort(A,j+1,high);
}

C++ Program for Mergesort

#include<iostream>
using namespace std;
int main()
{
        void mergesort(int[],int,int);
        int A[7]={1,2,5,3,7,6,4};
        int low=0,high=6;
        mergesort(A,0,6);
        for(int i=low;i<=high;i++)
        {
                cout<<A[i]<<" ";
        }
        cout<<endl;
        return 0;
}
void mergesort(int A[7],int low,int high)
{
        if(low==high)
        {
                return;
        }
        int mid=(low+high)/2;
        mergesort(A,low,mid);
        mergesort(A,mid+1,high);
        int a=low;
        int b=mid+1;
        int B[100],j=0;
        while(a<=mid && b<=high)
        {
                if(A[a]<A[b])
                {
                        B[j]=A[a];
                        j++;
                        a++;
                }
                else if(A[b]<A[a])
                {
                        B[j]=A[b];
                        j++;
                        b++;
                }
        }
        while(a<=mid)
        {
                B[j]=A[a];
                j++;
                a++;
        }
        while(b<=high)
        {
                B[j]=A[b];
                j++;
                b++;
        }
        int k=0;
        for(int i=low;i<=high;i++)
        {
                A[i]=B[k];
                k++;
        }
}

Given an array of positive integers, print out all the numbers which are repeated an even number of times without using additional storage

using System;
using System.Collections.Generic;
using System.Collections;
using System.Linq;
using System.Text.RegularExpressions;

namespace ConsoleApplication3
{
    class Program
    {
        static void Main(string[] args)
        {
            //This part assumes that all the elements in the array lie in the range (min,min + arraylength -1).
            int[] arr1 = { 1, 2, 3, 4, 4, 5, 2, 1, 1, 1 };
            int count = 1;
            int min = arr1.Min();
            int max = arr1.Max();
            for (int i = 0; i < arr1.GetLength(0); i++)
            {
                arr1[Math.Abs(arr1[i])] *= -1;
            }
            for (int i = 0; i < arr1.GetLength(0); i++)
            {
                Console.Write(arr1[i] + " ");
            }
            Console.WriteLine();
            for (int i = min; i <= max; i++)             {                 if (arr1[i] > 0)
                {
                    Console.Write(i + " ");
                }
            }
            Console.WriteLine();
            //This part sorts the array and finds the frequency map and counts the frequency of each element.
            int[] arr2 = { 1, 2, 3, 4, 4, 5, 2, 1, 1, 1 };
            Array.Sort(arr2);
            for (int i = 0; i < arr2.GetLength(0) - 1; i++)
            {
                count = 1;
                while (arr2[i + 1] == arr2[i])
                {
                    i++;
                    count++;
                }
                if (count % 2 == 0)
                {
                    Console.WriteLine(arr2[i] + " " + count);
                }
            }
        }
    }
}