# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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
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.

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