Question

Given an integer N, subtract the first digit of the number N from it i. e if n is 245 then subtract 2 from it making it 245-2 = 243. Keep on doing this operation until the number becomes 0.
Your task is to calculate the number of operation required to make this number N zero

Answer 1

Answer:

In this case I am using C++ language here. Not sure if I am right or not... So here is my answer.

Explanation:

#include <iostream>

using namespace std;

int main() {

int n;

cout << "Enter number: ";

cin >> n;

cout << "The number of turns it took is " << n / 2 << endl;

return 0;

}

Answer 2

Concept:-

An expression set up of a subject, action word, item, and modifiers. A gathering of modifiers, a subject, an action word, and an item. A sentence that is developed with a subject, action word, item, and modifiers. a blend of subjects, action words, items, and modifiers.

Given:-

Given that an integer $N$, subtract the first digit of the number $N$ from it i. e if $n$ is $245$ then subtract $2$ from it making it $245-2 = 243$. Keep on doing this operation until the number becomes $0$.

Find:-

We need to find the number of operation required to make this number $N$ zero.

Solution:-

$// C++$ program to find the count of Steps to

$//$ reduce $N$ to zero by subtracting its most

$//$ significant digit at every step

#include $< bits/stdc++.h >$

using namespace std;

$//$ Function to count the number

$//$ of digits in a number m

int countdig(int m)

$\{$

if $(m == 0)$

return $0$;

else

return $1 + countdig(m / 10)$;

$\}$

$//$ Function to count the number of

$//$ steps to reach $0$

int countSteps(int x)

$\{$

$//$ count the total number of steps

int  $c= 0$;

int last $= x$;

$//$ iterate till we reach $0$

while (last) $\{$

$//$ count the digits in last

int digits $=$ countdig(last);

$//$ decrease it by $1$

digits $-= 1$;

$//$ find the number on whose division,

$//$ we get the first digit

int divisor $= pow(10$, digits$)$;

$//$ first digit in last

int first $=$ last$/$divisor;

$//$ find the first number less than

$//$ last where the first digit changes

int lastnumber $=$ first $\times$ divisor;

$//$ find the number of numbers

$//$ with same first digit that are jumped

int skipped $=$ (last $-$ lastnumber) $/$ first;

skipped$+= 1$;

$//$ count the steps

$c +=$ skipped;

$//$ the next number with a different

$//$ first digit

last $=$ last $-$ (first $\times$ skipped);

$\}$

return $c$;

$\}$

$//$ Driver code

int main$()$

{

int $n = 245$;

cout $< <$ countSteps$(n)$;

return $0$;

$\}$

Hence, the operation can be done with $N$ to be zero.

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