# Find the first and last M digits from K-th power of N

Given a two integers N and K, the task is to find the first M and last M digits of the number NK .
Examples:

Input: N = 2345, K = 3, M = 3
Output: 128 625
Explanation:
2345 3 = 12895213625
Therefore, the first M(= 3) digits are 128 and the last three digits are 625.

Input: N = 12, K = 12, M = 4
Output: 8916 8256
Explanation:
12 12 = 8916100448256

Naive Approach:
The simplest approach to solve the problem is to calculate the value of NK and iterate to first M digits and find the last M digits by NK mod 10M
Time Complexity: O(K)
Auxiliary Space: O(1)

Efficient Approach:
The above approach can be optimized by the following observation:

Let us consider a number x which can be written as 10y Where y is a decimal number.
Let x = NK
NK = 10 y
Taking log10 on both sides of the above expression, we get:
K * log10(N) = log10(10 y
=> K * log10(N) = y * (log1010)
=> y = K * log10(N)
Now y will be a decimal number of form abc—.xyz—
Therefore,
NK = 10abc—.xyz—
=> NK = 10abc— + 0.xyz—
=> NK = 10abc— * 100.xyz—
In the above equation, 10abc— only moves the decimal point forward.
By calculating 100.xyz—, the first M digits can be figured out by moving the decimal point forward.

Follow the steps below to solve the problem:

• Find the first M digits of NK by calculating (NK)mod(10M).
• Calculate K * log10(N).
• Calculate 10K * log10(N).
• Find the first M digits by calculating 10K * log10(N) * 10M – 1 .
• Print the first and last M digits obtained.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement ` `// the above approach ` `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Function to find a^b modulo M ` `ll modPower(ll a, ll b, ll M) ` `{ ` `    ``ll res = 1; ` `    ``while` `(b) { ` `        ``if` `(b & 1) ` `            ``res = res * a % M; ` `        ``a = a * a % M; ` `        ``b >>= 1; ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Function to find the first and last ` `// M digits from N^K ` `void` `findFirstAndLastM(ll N, ll K, ll M) ` `{ ` `    ``// Calculate Last M digits ` `    ``ll lastM ` `        ``= modPower(N, K, (1LL) * ``pow``(10, M)); ` ` `  `    ``// Calculate First M digits ` `    ``ll firstM; ` ` `  `    ``double` `y = (``double``)K * ``log10``(N * 1.0); ` ` `  `    ``// Extract the number after decimal ` `    ``y = y - (ll)y; ` ` `  `    ``// Find 10 ^ y ` `    ``double` `temp = ``pow``(10.0, y); ` ` `  `    ``// Move the Decimal Point M - 1 digits forward ` `    ``firstM = temp * (1LL) * ``pow``(10, M - 1); ` ` `  `    ``// Print the result ` `    ``cout << firstM << ``" "` `<< lastM << endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``ll N = 12, K = 12, M = 4; ` ` `  `    ``findFirstAndLastM(N, K, M); ` `    ``return` `0; ` `} `

## Java

 `// Java program to implement ` `// the above approach ` `class` `GFG{ ` ` `  `// Function to find a^b modulo M ` `static` `long` `modPower(``long` `a, ``long` `b, ``long` `M) ` `{ ` `    ``long` `res = ``1``; ` `    ``while` `(b > ``0``)  ` `    ``{ ` `        ``if` `(b % ``2` `== ``1``) ` `            ``res = res * a % M; ` `             `  `        ``a = a * a % M; ` `        ``b >>= ``1``; ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Function to find the first and last ` `// M digits from N^K ` `static` `void` `findFirstAndLastM(``long` `N, ``long` `K, ` `                              ``long` `M) ` `{ ` `     `  `    ``// Calculate Last M digits ` `    ``long` `lastM = modPower(N, K, (1L) * ` `                 ``(``long``)Math.pow(``10``, M)); ` ` `  `    ``// Calculate First M digits ` `    ``long` `firstM; ` ` `  `    ``double` `y = (``double``)K * Math.log10(N * ``1.0``); ` ` `  `    ``// Extract the number after decimal ` `    ``y = y - (``long``)y; ` ` `  `    ``// Find 10 ^ y ` `    ``double` `temp = Math.pow(``10.0``, y); ` ` `  `    ``// Move the Decimal Point M - 1 digits forward ` `    ``firstM = (``long``)(temp * (1L) * ` `             ``Math.pow(``10``, (M - ``1``))); ` ` `  `    ``// Print the result ` `    ``System.out.print(firstM + ``" "` `+ ` `                      ``lastM + ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``long` `N = ``12``, K = ``12``, M = ``4``; ` ` `  `    ``findFirstAndLastM(N, K, M); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## C#

 `// C# program to implement ` `// the above approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG{ ` ` `  `// Function to find a^b modulo M ` `static` `long` `modPower(``long` `a, ``long` `b, ``long` `M) ` `{ ` `    ``long` `res = 1; ` `    ``while` `(b > 0)  ` `    ``{ ` `        ``if` `(b % 2 == 1) ` `            ``res = res * a % M; ` ` `  `        ``a = a * a % M; ` `        ``b >>= 1; ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Function to find the first and last ` `// M digits from N^K ` `static` `void` `findFirstAndLastM(``long` `N, ``long` `K, ` `                              ``long` `M) ` `{ ` ` `  `    ``// Calculate Last M digits ` `    ``long` `lastM = modPower(N, K, (1L) *  ` `                 ``(``long``)Math.Pow(10, M)); ` ` `  `    ``// Calculate First M digits ` `    ``long` `firstM; ` ` `  `    ``double` `y = (``double``)K * Math.Log10(N * 1.0); ` ` `  `    ``// Extract the number after decimal ` `    ``y = y - (``long``)y; ` ` `  `    ``// Find 10 ^ y ` `    ``double` `temp = Math.Pow(10.0, y); ` ` `  `    ``// Move the Decimal Point M - 1 digits forward ` `    ``firstM = (``long``)(temp * (1L) *  ` `             ``Math.Pow(10, (M - 1))); ` ` `  `    ``// Print the result ` `    ``Console.Write(firstM + ``" "` `+ ` `                   ``lastM + ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``long` `N = 12, K = 12, M = 4; ` ` `  `    ``findFirstAndLastM(N, K, M); ` `} ` `} ` ` `  `// This code is contributed by gauravrajput1 `

Output:

```8916 8256
```

Time Complexity: O(log K)
Auxiliary Space: O(1)

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