给你两个整数，基数a（数字d的数量，1<=d<=1000）和索引b（0<=b<=922*10^15）。你必须找到a^b的最后一位。 例如：
null
Input : 3 10Output : 9Input : 6 2Output : 6Input : 150 53Output : 0
举几个例子之后，我们可以注意到下面的模式。
Number | Last digits that repeat in cycle 1 | 1 2 | 4, 8, 6, 2 3 | 9, 7, 1, 3 4 | 6, 4 5 | 5 6 | 6 7 | 9, 3, 1, 7 8 | 4, 2, 6, 8 9 | 1, 9
在给定的表格中，我们可以看到循环重复的最大长度为4。 例子： 2*2=4*2=8*2=16*2=32，32中的最后一个数字是2，这意味着在乘以4次数字后，重复它自己。所以算法非常简单。 资料来源： 非常棒。组织 算法：
- 由于数字非常大，我们将它们存储为字符串。
- 取a底的最后一位数字。
- 现在计算b%4。这里b很大。
- 如果b%4==0，这意味着b完全可以被4整除，所以我们的指数现在将是exp=4，因为通过将数字乘以4倍，我们可以根据上图中的循环表得到最后一位数字。
- 如果b%4=这意味着b不能完全被4整除，所以我们的指数现在将是exp=b%4，因为通过乘以数字指数，我们可以根据上图中的循环表得到最后一位数字。
- 现在计算ldigit=pow（基数中的最后一位，exp）。
- a^b的最后一位数字将是ldigit%10。
下面是上述算法的实现。
C++
// C++ code to find last digit of a^b #include <bits/stdc++.h> using namespace std; // Function to find b % a int Modulo( int a, char b[]) { // Initialize result int mod = 0; // calculating mod of b with a to make // b like 0 <= b < a for ( int i = 0; i < strlen (b); i++) mod = (mod * 10 + b[i] - '0' ) % a; return mod; // return modulo } // function to find last digit of a^b int LastDigit( char a[], char b[]) { int len_a = strlen (a), len_b = strlen (b); // if a and b both are 0 if (len_a == 1 && len_b == 1 && b[0] == '0' && a[0] == '0' ) return 1; // if exponent is 0 if (len_b == 1 && b[0] == '0' ) return 1; // if base is 0 if (len_a == 1 && a[0] == '0' ) return 0; // if exponent is divisible by 4 that means last // digit will be pow(a, 4) % 10. // if exponent is not divisible by 4 that means last // digit will be pow(a, b%4) % 10 int exp = (Modulo(4, b) == 0) ? 4 : Modulo(4, b); // Find last digit in 'a' and compute its exponent int res = pow (a[len_a - 1] - '0' , exp ); // Return last digit of result return res % 10; } // Driver program to run test case int main() { char a[] = "117" , b[] = "3" ; cout << LastDigit(a, b); return 0; } |
JAVA
// Java code to find last digit of a^b import java.io.*; import java.math.*; class GFG { // Function to find b % a static int Modulo( int a, char b[]) { // Initialize result int mod = 0 ; // calculating mod of b with a to make // b like 0 <= b < a for ( int i = 0 ; i < b.length; i++) mod = (mod * 10 + b[i] - '0' ) % a; return mod; // return modulo } // Function to find last digit of a^b static int LastDigit( char a[], char b[]) { int len_a = a.length, len_b = b.length; // if a and b both are 0 if (len_a == 1 && len_b == 1 && b[ 0 ] == '0' && a[ 0 ] == '0' ) return 1 ; // if exponent is 0 if (len_b == 1 && b[ 0 ] == '0' ) return 1 ; // if base is 0 if (len_a == 1 && a[ 0 ] == '0' ) return 0 ; // if exponent is divisible by 4 that means last // digit will be pow(a, 4) % 10. // if exponent is not divisible by 4 that means last // digit will be pow(a, b%4) % 10 int exp = (Modulo( 4 , b) == 0 ) ? 4 : Modulo( 4 , b); // Find last digit in 'a' and compute its exponent int res = ( int )(Math.pow(a[len_a - 1 ] - '0' , exp)); // Return last digit of result return res % 10 ; } // Driver program to run test case public static void main(String args[]) throws IOException { char a[] = "117" .toCharArray(), b[] = { '3' }; System.out.println(LastDigit(a, b)); } } // This code is contributed by Nikita Tiwari. |
# Python 3 code to find last digit of a ^ bimport math# Function to find b % adef Modulo(a, b) : # Initialize result mod = 0 # calculating mod of b with a to make # b like 0 <= b < a for i in range(0, len(b)) : mod = (mod * 10 + (int)(b[i])) % a return mod # return modulo# function to find last digit of a ^ bdef LastDigit(a, b) : len_a = len(a) len_b = len(b) # if a and b both are 0 if (len_a == 1 and len_b == 1 and b[0] == '0' and a[0] == '0') : return 1 # if exponent is 0 if (len_b == 1 and b[0]=='0') : return 1 # if base is 0 if (len_a == 1 and a[0] == '0') : return 0 # if exponent is divisible by 4 that means last # digit will be pow(a, 4) % 10. # if exponent is not divisible by 4 that means last # digit will be pow(a, b % 4) % 10 if((Modulo(4, b) == 0)) : exp = 4 else : exp = Modulo(4, b) # Find last digit in 'a' and compute its exponent res = math.pow((int)(a[len_a - 1]), exp) # Return last digit of result return res % 10 # Driver program to run test casea = ['1', '1', '7']b = ['3']print(LastDigit(a, b))# This code is contributed to Nikita Tiwari.
C#
// C# code to find last digit of a^b. using System; class GFG { // Function to find b % a static int Modulo( int a, char [] b) { // Initialize result int mod = 0; // calculating mod of b with a // to make b like 0 <= b < a for ( int i = 0; i < b.Length; i++) mod = (mod * 10 + b[i] - '0' ) % a; // return modulo return mod; } // Function to find last digit of a^b static int LastDigit( char [] a, char [] b) { int len_a = a.Length, len_b = b.Length; // if a and b both are 0 if (len_a == 1 && len_b == 1 && b[0] == '0' && a[0] == '0' ) return 1; // if exponent is 0 if (len_b == 1 && b[0] == '0' ) return 1; // if base is 0 if (len_a == 1 && a[0] == '0' ) return 0; // if exponent is divisible by 4 // that means last digit will be // pow(a, 4) % 10. if exponent is //not divisible by 4 that means last // digit will be pow(a, b%4) % 10 int exp = (Modulo(4, b) == 0) ? 4 : Modulo(4, b); // Find last digit in 'a' and // compute its exponent int res = ( int )(Math.Pow(a[len_a - 1] - '0' , exp)); // Return last digit of result return res % 10; } // Driver program to run test case public static void Main() { char [] a = "117" .ToCharArray(), b = { '3' }; Console.Write(LastDigit(a, b)); } } // This code is contributed by nitin mittal. |
PHP
<?php // php code to find last digit of a^b // Function to find b % a function Modulo( $a , $b ) { // Initialize result $mod = 0; // calculating mod of b with a to make // b like 0 <= b < a for ( $i = 0; $i < strlen ( $b ); $i ++) $mod = ( $mod * 10 + $b [ $i ] - '0' ) % $a ; return $mod ; // return modulo } // function to find last digit of a^b function LastDigit( $a , $b ) { $len_a = strlen ( $a ); $len_b = strlen ( $b ); // if a and b both are 0 if ( $len_a == 1 && $len_b == 1 && $b [0] == '0' && $a [0] == '0' ) return 1; // if exponent is 0 if ( $len_b == 1 && $b [0] == '0' ) return 1; // if base is 0 if ( $len_a == 1 && $a [0] == '0' ) return 0; // if exponent is divisible by 4 that // means last digit will be pow(a, 4) // % 10. if exponent is not divisible // by 4 that means last digit will be // pow(a, b%4) % 10 $exp = (Modulo(4, $b ) == 0) ? 4 : Modulo(4, $b ); // Find last digit in 'a' and compute // its exponent $res = pow( $a [ $len_a - 1] - '0' , $exp ); // Return last digit of result return $res % 10; } // Driver program to run test case $a = "117" ; $b = "3" ; echo LastDigit( $a , $b ); // This code is contributed by nitin mittal. ?> |
Javascript
<script> // Javascript code to find last digit of a^b // Function to find b % a function Modulo(a, b) { // Initialize result let mod = 0; // calculating mod of b with a to make // b like 0 <= b < a for (let i = 0; i < b.length; i++) mod = (mod * 10 + b[i] - '0' ) % a; return mod; // return modulo } // function to find last digit of a^b function LastDigit(a, b) { let len_a = a.length; let len_b = b.length; // if a and b both are 0 if (len_a == 1 && len_b == 1 && b[0] == '0' && a[0] == '0' ) return 1; // if exponent is 0 if (len_b == 1 && b[0] == '0' ) return 1; // if base is 0 if (len_a == 1 && a[0] == '0' ) return 0; // if exponent is divisible by 4 that // means last digit will be pow(a, 4) // % 10. if exponent is not divisible // by 4 that means last digit will be // pow(a, b%4) % 10 exp = (Modulo(4, b) == 0) ? 4 : Modulo(4, b); // Find last digit in 'a' and compute // its exponent res = Math.pow(a[len_a - 1] - '0' , exp); // Return last digit of result return res % 10; } // Driver program to run test case let a = "117" ; let b = "3" ; document.write(LastDigit(a, b)); // This code is contributed by _saurabh_jaiswal </script> |
输出：
3
本文由 沙申克·米什拉（古卢） .本文由Geeksforgeks团队审阅。 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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