Question

write C++ program using functions

(separate function for each bottom)

Write a program to find if a number is large word for two given bottom base - bottom1 and bottom2. You can predict that a number, when converted to any given base shall not exceed 10 digits. .

the program should ask from user to enter a number that it should ask to enter the base ranging from 2 to 16 after that it should check if the number is palindrom or not
Sample Input:

A number is called a word if it's represented in its bottom. e.g. Let  bottom1 = 6 and bottom2 = 2 as it is not a word in base 5(1010)(as the reciprocal of the number isnt a palindrome).
bottom is base of a number
take bases as input from user
bases can go from decimal-hexadecimal.

for bottom1 = 3 & bottom2 = 4, then the number 130 (in base 10) will be called a large_word, as it is word in bottom 3 (11211) as well as in bottom 4 (2002). However, it is not a large_word for bottom1 = 3 and bottom2 = 5 as it is not a word in bottom 5(1010).

Number: 51

bottom(base) 1: 6

bottom(base) 2: 2

Sample Output:
51 is not large word

hint:
bottom is basically base of a number

A large _word is a word, phrase, number, or other sequence of characters which reads the same reverse or straightward.

word==palindrome

use function in the program at all time

Answer 1

Code-

#include <stdio.h>

#include <iostream>

#include <string.h>

using namespace std;

char val(int n)// to return the charcter of the number accordingly

{

if(n>=0 && n<=9)

return (char)(n+'0'); // returns the normal charcter from 0-9

else

return (char)(n-10+'A');// returns char from A-F accordingly

}

string reverse(string s)// to reverse the string

{

string rev="";

for(int i=s.length()-1; i>=0; i--)

rev+=s[i];

return rev;

}

int base1(int b1,int num)// to convert to base 1

{

string s1;

while(num>0)

{

s1+=val(num%b1);

num=num/b1;

}

if(s1==reverse(s1))// compare reverse with original

return 1;

else

return 0;

}

int base2(int b2,int num)// to convert to base 2

{

string s2;

while(num>0)

{

s2+=val(num%b2);

num=num/b2;

}

if(s2==reverse(s2))//compare reverse with original

return 1;

else

return 0;

}

int main()

{

cout<<"Number: ";

int num,b1,b2;

cin>>num;

cout<<"bottom base 1: ";

cin>>b1;

cout<<"bottom base 2: ";

cin>>b2;

if(base1(b1,num)==1 && base2(b2,num)==1)// checking for large word

cout<<num<<" is a large word";

else

cout<<num<<" is not a large word";

return 0;

}

Indentation-

Output-