Given a number n, find the first k digits of n^n

Given a number n, find the first k digits of nn, where k is a value less than the number of digits in nn

 Input :  n = 10
          k = 2
 Output : 10
The first 2 digits in 1010 are 10.

 Input :  n = 144
          k = 6
 Output : 637087 

 Input:  n = 1250
         k = 5
 Output:  13725 

The problem can be solved in multiple ways, of which two are:

Method 1 (Simple): A naive method which involves calculating the actual value and then dividing by 10 until we get the required answer. However this method cannot take input greater than n = 15 as it would cause overflow.

C/C++

// C++ program to find the first k digits of n^n
#include <bits/stdc++.h>
using namespace std;

// function that manually calculates n^n and then
// removes digits until k digits remain
unsigned long long firstkdigits(int n, int k)
{
   unsigned long long product = 1;

   for (int i = 0 ; i < n ; i++)
      product *= n;

   // loop will terminate when there are only
   // k digits left
   while ((int)(product / pow(10, k)) != 0)
      product = product / 10;

   return product;
}

//driver function
int main()
{
   int n = 15;
   int k = 4;
   cout << firstkdigits(n, k);
   return 0;
}

Java

// Java program to find the first k digits of n^n
public class Digits
{
    // function that manually calculates n^n and then
    // removes digits until k digits remain
    static long firstkdigits(int n, int k)
    {
        long product = 1;
        for (int i = 0 ; i < n ; i++)
           product *= n;
    
       // loop will terminate when there are only
        // k digits left
       while ((int)(product / Math.pow(10, k)) != 0)
            product = product / 10;
        return product;
    }
    
    public static void main(String[] args)
    {
      int n = 15;
      int k = 4;
      System.out.println(firstkdigits(n, k));
    }
}

//This code is contributed by Saket Kumar


Output:

4378

 

Method 2: The next method involves using logarithms to calculate the first k digits. The method and steps are explained below:

  1. Let product = nn. Take logarithm base 10 on both sides of the equation. We get log10(product) = log10(nn), which we can also write as n*log10(n)
  2. In this example, we get log10(product) = 3871.137516. We can split the RHS as 3871 + 0.137516, so our equation can now be written as log10(product) = 3871 + 0.137516
  3. Raise both sides with base 10, and using the above example, we get product = 103871 x 100.137516. 103871 will not make a difference to our first k digits as it only shifts decimal points. We are interested in the next part, 100.137516, as this will determine the first few digits.
    In this case, the value of 100.137516 is 1.37251.
  4. Hence our required first 5 digits would be 13725.

C

//C++ program to generate first k digits of
// n ^ n
#include <bits/stdc++.h>
using namespace std;

// function to calculate first k digits
// of n^n
long long firstkdigits(int n,int k)
{

   //take log10 of n^n. log10(n^n) = n*log10(n)
   long double product = n * log10(n);

   // We now try to separate the decimal and
   // integral part of the /product. The floor
   // function returns the smallest integer
   // less than or equal to the argument. So in
   // this case, product - floor(product) will
   // give us the decimal part of product
   long double decimal_part = product - floor(product);

   // we now exponentiate this back by raising 10
   // to the power of decimal part
   decimal_part = pow(10, decimal_part);

   // We now try to find the power of 10 by which
   // we will have to multiply the decimal part to
   // obtain our final answer
   long long digits = pow(10, k - 1), i = 0;

   return decimal_part * digits;
}

// driver function
int main()
{
   int n = 1450;
   int k = 6;
   cout << firstkdigits(n, k);
   return 0;
}

Java

// Java  program to find the first k digits of n^n

import java.util.*;
import java.lang.*;
import java.io.*;

class KDigitSquare
{
      /* function that manually calculates 
         n^n and then removes digits until 
         k digits remain */
    public static long  firstkdigits(int n, int k)
    {
        //take log10 of n^n. 
        // log10(n^n) = n*log10(n)
        double product = n * Math.log10(n);
     
       /* We will now try to separate the decimal 
          and integral part of the /product. The 
          floor function returns the smallest integer
          less than or equal to the argument. So in
          this case, product - floor(product) will
          give us the decimal part of product */
        double decimal_part = product - Math.floor(product);
     
        // we will now exponentiate this back by 
        // raising 10 to the power of decimal part
        decimal_part = Math.pow(10, decimal_part);
     
        /* We now try to find the power of 10 by 
           which we will have to multiply the decimal 
           part to obtain our final answer*/
        double digits = Math.pow(10, k - 1), i = 0;
        
        return ((long)(decimal_part * digits));
    }

    // driver function
    public static void main (String[] args)
    {
        int n = 1450;
        int k = 6;
        System.out.println(firstkdigits(n,k));
    }
}

/* This code is contributed by Mr. Somesh Awasthi */


Output:

962948

This code runs in constant time and can handle large input values of n

This article is contributed by Deepak Srivatsav. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice





Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.