ABX01 - EDITORIAL

PROBLEM LINK:

Practice
Contest

Setter : Mohammad Salik

Tester : Hasan Jaddouh

Editorialist : Mohammad Salik

Difficulty

Simple/Easy

Prerequisites

Fast modular exponentiation, Observation skills

Problem

You are given two integers $A$ and $N$. The task is to compute $F(A^N)$ where $F(X)$ is a function which results in a non-negative single digit integer obtained by a process of summing digits,and on each iteration using the result from the previous iteration to compute a digit sum untill single digit number is reached.

Explanation

For Subtask 1

We can simply find A^N as it doesn’t exceed 10^{15} and then evaluate F(A^N) by the recursive process mentioned in the definition of this function.Note that we can evaluate F(A^N) in less than 5 iterations.

For Subtask 2 and 3

Let us first observe the function $F$ first.What will be $F(X)$ for a single digit integer $X$ $?$ It will simply be the number itself.Or we can say for a single digit integer $X$ ,$F(X)=X\%9$ for $X!=9$ and $F(X)=9$ for $X=9$.Combining these two we can write $F(X)=X\%9 + 9*(X\%9==0)$ for a single digit integer $X$ where $X\%9==0$ will return $1$ only when $X$ is $9$. Now what will be $F(X)$ for a two digit integer $X$. For a two digit integer $X$ which is a multiple of $9$ observe that $F(X)=9$ and for $X$ not a multiple of $9$ , $F(X)= X\%9$. Generalising this we can define $F(X)$ as follows : $F(X)= X\%9 + 9*(X\%9==0)$ for any non negative integer $X$ where $(X\%9==0)$ will return $1$ only when $X$ is a multiple of $9$.

Now lets try to find out F(X*X). Consider two Cases :

  • When X is a multiple of 9 i.e F(X)=9 then X*X is also a multiple of 9 and consequently F(X*X)=9. Also see here that F(F(X)*F(X))= F(9*9)=F(81)=9.Hence when X is a multiple of 9 then F(X*X)=F(F(X)*F(X)).

  • When X is not a multiple of 9 then F(X)=X\%9 . Hence, we have F(F(X)*F(X))=F((X\%9)*(X\%9))=((X\%9)*(X\%9))\%9 = (X*X)\%9 = F(X*X) .Hence in this case also F(X*X)=F(F(X)*F(X)).

So lets generalise this. F(X^2)=F(F(X)*F(X)). Similarly, F(X^4)=F( (F(F(X)*F(X)))^2 ).

Now for SUBTASK #2 where N<=100 we can evaluate F(A^N) by iterating from 1 to N and taking F at each step while multiplying.The following code evaluates this in N steps :

    ll Res=1;
    for(int i=1;i<=N;i++)
    {
            Res=Res*A;
            Res=F(Res);
    }

For Subtask #3 we have N<=10^{18} and hence we cannot take N steps as it will timeout. we can write a function similar to a fast modular exponentiation to evaluate F(A^N) which evaluates this in logN steps.The following code evaluates this :

int solve(long long A,long long N)
{
   long long res=1;
   while(N)
   {
        if(N%2==1)
        {
            res=res*F(A);
            res=F(res);
        }
        A=F(F(A)*F(A));
        N/=2;
   }
   return res;
}

The function F(A) for an integer A can be computed in O(1) easily as we have already defined this function.

Time Complexity

O(logN) per testcase

Space Complexity

O(1)

AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.
Tester’s solution can be found here.

4 Likes

i have done this question in O(1)
link:-https://www.codechef.com/viewsolution/16717619

1 Like

When you realize that F(A)\equiv A \mod 9 all you need to do is compute A^N \mod 9.

Solution.

And that’s also what the tester’s solution does.

4 Likes

Can anyone tell me the mistake in this solution? It passed for the last test case but failed in the first two.
I am not getting the mistake.

https://www.codechef.com/viewsolution/16722563

My this code is passing for 70 pts and getting WA for first two subtasks. I don’t know why it is happening atleast in first subtask because i have matched my output of all the possible test cases for first subtask with ouput of an AC code. If anyone could please help. @abx_2109

I have solved this in O(no_of_digits_in_N) but not sure whether my approach is fully correct or not. I have observed that after certain number of steps there will be a cycle and thus i just sum the digits of N untill it becomes single digit and then just print the answer for that single digit for that power.

my solution : https://www.codechef.com/viewsolution/16708827

I don’t know whether it fully correct or not and if it is and someone has some mathematical stuff regarding it’s proof then please share :slight_smile:

This is my solution.

I did it in a different way.
I took the sum of digits according to the question given in a query. And then found out the pattern in which result is repeating.
Finally, printed the result according to the condition satisfied by the exponential term given in the query.

Here is my solution.

2 Likes

DONE THIS USING OBSERVATION https://www.codechef.com/viewsolution/16726993

And here is O(1) solution.

1 Like

@abx_2109 u can do this question more efficiently when each time for updating res u should go for checking

if(res%9==0) res=9;else res%=9;

rather than again making an recursive function as given in definition

And similarly updating A with

     A = ((A%M)*(A%M))%M;
      
    if(A%9==0) A=9;else A%=9;

Problem Link
Contest
Practise

Setter : Mohammad Salik

Tester : Hasan Jaddouh

Editorialist : Ashwany Aggarwal

Difficulty:
Easy

Prerequisites:
Mathematics,Observation skills

Problem
You are given two integers A and N. The task is to compute F(A^N) where F(X) is a function which results in a non-negative single digit integer obtained by a process of summing digitsuntil single digit number is reached.

Explanation:

STEP 1: First of all we find the sum of integers of the given integer A until it reaches to a single integer (say y), using function :

int digsum(unsigned long long p)
{ unsigned long long x,sum=0;
     while(p>0)
               {
                 while(p != 0)
                 {
                      x = p%10;
                      sum = sum+x;
                      p=p/10;
                 }
                 if(sum > 9)
                 {
                         p = sum;
                        sum = 0;
                 }
               }
               return sum;
}

Now, by the property of exponents that sum of digits repeat after every 6th exponent.i.e, digsum(A^N)= digsum(A^(N%6)), (except for A=3 or A=6 , where A^N= A for N=1 and A^N=9 for N>1).

For eg. 2^1=2, 2^7=128=1+2+8=11=1+1=2

For eg: 3^1=3; 3^4=81=8+1=9, 6^3=216=2+1+6=9 (you can check for any N).

STEP 2:

Now we have y (the single digit sum for A). Now We have 2 cases either y=(3 or 6) or y= else than 3 or 6.

1st case: If y= 3 or 6 or 9

1: If N>1 output 9

2: If N=1 output Y

2nd case: Y is not 3 or 6 or 9.(I have taken 9 also because for 9 answer is always 9)

Then take remainder of the given large N when divided by 6 (say z).(The main LOGIC)
Now single digit raised to power maximum 5 will be 3 digit in worst case this reduces the complexity of problem.

Step 3:
Find y^z and apply the same function. You will get the required answer.

Here is my solution: https://www.codechef.com/viewsolution/16728889

As some wonder why there is a magic modulo 9, here is a short explanation:

Any number can be written as a sum of power of 10 and you can write it like this:

1543 = 1 * (999+1) + 5 * (99+1) + 4 * (9+1) + 3

If you rearrange:

1543 = (1 * 999 + 5 * 99 + 4 * 9) + (1 + 5 + 4 + 3)

And then you understand why mod 9 is the magic trick.

More details here: http://www.flyingcoloursmaths.co.uk/a-neat-number-trick-digital-roots-and-modulo-9-arithmetic/

7 Likes

We can use "set stl " also

I was trying to use recursive version of fast exponentiation algorithm but getting TLE on 2nd and 3rd subtask. Can anyone help me regarding same? Please upvote this answer so that I can ask queries :slight_smile:

https://www.codechef.com/viewsolution/16740009

1 Like

Superfast - 0s

And it works with arbitrarily large integers!

@eugalt u really squeezed up !!!

1 Like

Are you sure it’s O(1)? :wink:

3 Likes

Consider this TC:

2

18 0

18 1

Expected o/p:

1

9

Your code gives:

18

18

Hope this helps!

For A=18, N=0 and A=18, N=1 your code gives 9 and 0 respectively which should be 1 and 9.

1 Like