从N的K次幂中查找前M个数字和后M个数字

给定两个整数N和K ，任务是找到数字N ^K的前M个和后M个数字。
例子：

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 
 

天真的方法：
解决该问题的最简单方法是计算N ^K的值，然后迭代N ^K mod 10 ^M到前M个数字，然后找到最后M个数字。
时间复杂度： O(K)
辅助空间： O(1)

高效方法：
可以通过以下观察来优化上述方法：

Let us consider a number x which can be written as 10^y Where y is a decimal number. 
Let x = N^K
N^K = 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, 
N^K = 10^{abc—.xyz—} 
=> N^K = 10^{abc— + 0.xyz—} 
=> N^K = 10^abc— * 10^0.xyz— 
In the above equation, 10^abc— only moves the decimal point forward. 
By calculating 10^0.xyz—, the first M digits can be figured out by moving the decimal point forward. 

请按照以下步骤解决问题：

• 通过计算(N ^K )mod(10 ^M )找出N ^K的前M个数字。
• 计算K * log ₁₀ (N) 。
• 计算10 ^{K * log ₁₀ (N)} 。
• 通过计算10 ^{K * log ₁₀ (N)} * 10 ^{M – 1来}找到前M个数字。
• 打印获得的前M位和后M位。

下面是上述方法的实现：

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

Python3

# Python3 program to implement
# the above approach
from math import *
  
# Function to find a^b modulo M
def modPower(a, b, M):
  
    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
def findFirstAndLastM(N, K, M):
  
    # Calculate Last M digits
    lastM = modPower(N, K, int(pow(10, M)))
  
    # Calculate First M digits
    firstM = 0
  
    y = K * log10(N * 1.0)
  
    # Extract the number after decimal 
    y = y - int(y)
  
    # Find 10 ^ y 
    temp = pow(10.0, y)
  
    # Move the Decimal Point M - 1
    # digits forward 
    firstM = int(temp * pow(10, M - 1))
  
    # Print the result
    print(firstM, lastM)
  
# Driver Code
N = 12
K = 12
M = 4
  
findFirstAndLastM(N, K, M) 
  
# This code is contributed by himanshu77

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

输出：

8916 8256

时间复杂度： O(log K)
辅助空间： O(1)