Print Given Money or Value in Words

PROGRAM : 

#include<stdio.h>

int main()
{
         int i,j,k=0,value,count,a[10],dummy;
         char ones[10][6]={"zero","one","two","three","four","five","six","seven","eight","nine"};
         char temp[10][10]={"ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};
         char tens[11][8]={"-","-","twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"};

 //hundreds are same as ones
         printf("Enter Value upto 6 digits to get in Words\n");
         dummy=value;
         scanf("%d",&value);
           while(value>0)
           {
               i=value%10;
               a[k]=i;
               k++;
               value/=10;
            }
 //Here if 12345 is given then a[0]=5,a[1]=4 and soon then last but one digit will be at a[1]
           int totalDigits=k-1;
           if(totalDigits<6)
           {
                    for(i=totalDigits;i>=0;i--)
                    {
                          if(i==5 && a[i]!=0)
                               printf("%s lakh ",ones[a[i]]);
                           if(i==4 || i==3)
                          {
       //Start
                               if(i==4 && a[i]==0)
                             {
                                 printf("%s thousand ",ones[a[i-1]]);
                                   i--;
                             }
                              else if(i==4 && a[i]==1)
                               {
                                 printf("%s thousand ",temp[a[i-1]]);
                                 i--;
                             }
                               else if(i==4)
                                {
                                 printf("%s ",tens[a[i]]);
                             }
                             else
                             {
                                 printf("%s thousand ",ones[a[i]]);
                             }
          //end
                          }
                        
                          if(i==2 && a[i]!=0)
                          {
                                  printf("%s hundred and ",ones[a[i]]);
                          }
                           if(i==2 && a[i]==0)
                           {
                                   printf("and ");
                          }
                            if(i==1 || i==0)
                          {
                               if(i==1 && a[i]==0)
                                 {
                                      printf("%s only\n",ones[a[i-1]]);
                                   i--;
                                  break;
                                  }
                               if(i==1 && a[i]==1)
                                 {
                                   printf("%s only\n",temp[a[i-1]]);
                                    i--;
                                 break;
                                }
                               if(i==1 && a[i]!=1 && a[i]!=0)
                                 {
                                    printf("%s ",tens[a[i]]);
                                 }
                                 if(i==0 && a[i]!=0)
                                 {
                                printf("%s only\n",ones[a[i]]);
                                  }
                             }
                    }
         }
        else
        {
             printf("Enter only upto 6 digits\n");
         }
return 0;
}

OUTPUT :

Number of Characters need to make a String Palindrome

#include<stdio.h>

#include<string.h>

int main(){

   

    int i=0,j,totChar=0,flag=0,c;

    char str[100];

    printf("Enter a String\n");

    scanf("%s",str);

    int len=strlen(str);

    j=len-1;

   

    while(i<j)

    {

           if(str[i]==str[j])

           {

               i++;

               j--;

              flag=1;

           }

        else

        {

            totChar++;

            i++;

            flag=0;

        }

    }

    if(flag!=1)

    totChar=len-1;

    c=totChar-1;

    printf("Total Characters required=%d\n",totChar);

   

    for(i=0;i<totChar;i++)

        str[len+i]=str[c--];

    str[len+i]='\0';

    printf("Final String of Palindrome=%s\n",str);

    return 0;

}

Input    : mada

Output : Total Characters required =1

               Final String of Palindrome = madam

Find Subsets of Given String

Method I:

       Can find substring for any length of string because we are taking binary number as string while Conversion itself

#include<stdio.h>

#include<string.h>

#include<math.h>

char *GetBinary(int dec);

int main()

{

 int i,j,len,count=0,no_of_zeros;

 char str[20],temp[20],binary[20];

 printf("Enter a string\n");

 gets(str);

 len=strlen(str);

 printf("Subsets of Given String are");

 for(i=0;i<(pow(2,len));i++)//2^len is the number of subsets for given length.

 {

  strcpy(binary,GetBinary(i));

  no_of_zeros=len-strlen(binary);//Since zeros will not be there before single digits and double etc depending on length of binary number.

 //To compensate them we used here

  for(j=0;j<len;j++)

  {

         if(j>=noofzeros)

        {

          if(binary[j-no_of_zeros]!='0')

           printf("%c",str[j]);

        }

  }

        printf("\n");

 }

 return(0);

}

char *GetBinary(int dec)

{

 int temp,i,j=1,k=0;

 char binary[20];

 temp=dec;

 while(temp!=0)

 {

   i=temp%2;

      binary[k]=i+'0';

   temp=temp/2;

      j=j*10;

      k++;

 }

 binary[k]='\0';

 return(strrev(binary));

}

Method II:

       Can find subsets for smaller strings as we are taking binary number as integer

 

#include<stdio.h>

#include<string.h>

#include<math.h>

int GetBinary(int dec);

int main()

{

 int i,j,binary,len,count=0,tempLen;

 char str[20],temp[20];

 printf("Enter a string\n");

 gets(str);

 len=strlen(str);

 for(i=1;i<(pow(2,len));i++)//2^len is the number of subsets for given length

 {

  binary=GetBinary(i);

  sprintf(temp, "%d", binary);//To convert integer to string

  tempLen=len-strlen(temp);

  for(j=0;j<len;j++)

  {

   if(j>=tempLen)

   {

    if(temp[j-tempLen]!='0')

       printf("%c",str[j]);

   }

  }

  printf("\n");

 }
return 0;

}

int GetBinary(int dec)

{

 int temp,i,j=1,binary=0;

 temp=dec;

 while(temp!=0)

 {

   i=temp%2;

      binary=binary+(i*j);

   temp=temp/2;

      j=j*10;

 }

 return(binary);

}

Input :

Tea

Output:

T
e 
a
Te
Ta
ea
Tea

Input :
Help
Output :
H
e
l
p
He
Hl
Hp
el
ep
lp
Hel
Hep
Hlp
elp
Help