code for James found a love letter his friend Harry has written for his girlfriend. James is a prankster, so he decides to meddle with the letter. He changes all the words in the letter into palindromes. To do this, he follows 2 rules: (a) He can reduce the value of a letter, e.g. he can change 'd' to 'c', but he cannot change 'c' to 'd'. (b) In order to form a palindrome, if he has to repeatedly reduce the value of a letter, he can do it until the letter becomes 'a'. Once a letter has been changed to 'a', it can no longer be changed. Each reduction in the value of any letter is counted as a single operation. Find the minimum number of operations required to convert a given string into a palindrome.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include<string.h>
using namespace std;


int main() {
  char c[100000];
unsigned long long int a,b,i,j,k,l,temp,count=0;
cin>>a;
for(i=0;i<a;i++)
{
cin>>c;
k=strlen(c);
if(k%2==0)
{
j=k/2;
temp=j;
l=j-1;
for(;temp!=0;)
{
if(c[j]>=c[l])
{
if(c[l]==c[j])
{
l--;
j++;
temp--;
}
else
{
c[j]--;
count++;
}
}
else
{
if(c[l]==c[j])
{
l--;
j++;
temp--;
}
else
{
c[l]--;
count++;
}
}
}
}
else
{
j=k/2;
temp=j;
l=j-1;
j++;
for(;temp!=0;)
{
if(c[j]>=c[l])
{
if(c[l]==c[j])
{
l--;
j++;
temp--;
}
else
{
c[j]--;
count++;
}
}
else
{
if(c[l]==c[j])
{
l--;
j++;
temp--;
}
else
{
c[l]--;
count++;
}
}
}
}
cout<<count<<endl;
count=0;
}
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */  
    return 0;
}

program to print the pattern

#include<stdio.h>
#include<conio.h>
void main()
{
int a,b,c,i,j;
clrscr();
printf("enter the number of column");
scanf("%d",&a);
printf("enter the number of row");
scanf("%d",&b);
for(i=0;i<b;i++)
{
for(j=0;j<a;j++)
{
if(i==0||i==b-1)
{
printf("X");
continue;
}
if(j==0||j==a-1)
{
printf("X");
continue;
}
if(i!=0&&j!=0&&i!=b-1&&j!=a-1)
{
if(i==1||i==b-2||j==1||j==a-2)
{
printf("0");
continue;
}
}
printf("X");
}
printf("\n");
}
getch();
}