查找升为y的x的单位数字

给定两个数字x和y，找到x ^y的单位数字。

例子 ：

Input  : x = 2, y = 1
Output : 2
Explanation
2^1 = 2 so units digit is 2.

Input : x = 4, y = 2
Output : 6
Explanation
4^2 = 16 so units digit is 6.

方法1(简单)计算x ^y的值并找到其最后一位。此方法导致x和y值稍大的溢出。
方法2(高效)
1)查找x的最后一位。
2)在模10下计算x ^ y并返回其值。

C++

// Efficient C++ program to
// find unit digit of x^y.
#include 
using namespace std;
 
// Returns unit digit of x
// raised to power y
int unitDigitXRaisedY(int x, int y)
{
    // Initialize result as 1 to
    // handle case when y is 0.
    int res = 1;
 
    // One by one multiply with x
    // mod 10 to avoid overflow.
    for (int i = 0; i < y; i++)
        res = (res * x) % 10;
 
    return res;
}
 
// Driver program
int main()
{ 
   cout << unitDigitXRaisedY(4, 2);
   return 0;
}

Java

// Efficient Java program to find
// unit digit of x^y.
import java.io.*;
 
class GFG {
    // Returns unit digit of x raised to power y
    static int unitDigitXRaisedY(int x, int y)
    {
        // Initialize result as 1 to
        // handle case when y is 0.
        int res = 1;
     
        // One by one multiply with x
        // mod 10 to avoid overflow.
        for (int i = 0; i < y; i++)
            res = (res * x) % 10;
     
        return res;
    }
     
    // Driver program
    public static void main(String args[])throws IOException
    {
    System.out.println(unitDigitXRaisedY(4, 2));
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3

# Python3 code to find
# unit digit of x^y.
 
# Returns unit digit of
# x raised to power y
def unitDigitXRaisedY( x , y ):
 
    # Initialize result as 1 to
    # handle case when y is 0.
    res = 1
     
    # One by one multiply with x
    # mod 10 to avoid overflow.
    for i in range(y):
        res = (res * x) % 10
     
    return res
     
# Driver program
print( unitDigitXRaisedY(4, 2))
 
 
# This code is contributed by Abhishek Sharma44.

C#

// Efficient Java program to find
// unit digit of x^y.
using System;
 
class GFG
{
    // Returns unit digit of x raised to power y
    static int unitDigitXRaisedY(int x, int y)
    {
        // Initialize result as 1 to
        // handle case when y is 0.
        int res = 1;
     
        // One by one multiply with x
        // mod 10 to avoid overflow.
        for (int i = 0; i < y; i++)
            res = (res * x) % 10;
     
        return res;
    }
     
    // Driver program
    public static void Main()
    {
    Console.WriteLine(unitDigitXRaisedY(4, 2));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// Efficient Javascript program to
// find unit digit of x^y.
 
// Returns unit digit of x
// raised to power y
function unitDigitXRaisedY(x, y)
{
     
    // Initialize result as 1 to
    // handle case when y is 0.
    let res = 1;
 
    // One by one multiply with x
    // mod 10 to avoid overflow.
    for(let i = 0; i < y; i++)
        res = (res * x) % 10;
 
    return res;
}
 
// Driver Code
document.write(unitDigitXRaisedY(4, 2));
 
// This code is contributed by _saurabh_jaiswal.
 

输出 ：

6

进一步优化：我们可以在Log y中计算模块化功率。

方法3(基于最后一位的循环性质直接进行)
此方法取决于x的最后一位的周期性

x   |  power 2  |  power 3  |   power 4  | Cyclicity  
0   | .................................. |  .... repeat with 0
1   | .................................. |  .... repeat with 1
2   |     4     |     8     |      6     | .... repeat with 2
3   |     9     |     7     |      1     | .... repeat with 3
4   |     6     |....................... |  .... repeat with 4
5   | .................................. |  .... repeat with 5
6   | .................................. |  .... repeat with 6
7   |     9     |     3     |      1     | .... repeat with 7
8   |     4     |     2     |      6     | .... repeat with 8
9   |     1     | ...................... |  .... repeat with 9 

所以在这里我们直接用4修改幂y，因为这是所有数字重复开始后的最后一个幂
在此之后，我们简单地用数字x最后一位加幂，然后得到产生的数字的单位位数。

C++

// C++ code to find the unit digit of x
// raised to power y.
#include
#include
using namespace std;
 
// find unit digit
int unitnumber(int x, int y)
{
    // Get last digit of x
    x = x % 10;
       
    // Last cyclic modular value
    if(y!=0)
        y = y % 4 + 4;
 
    // here we simply return the
    // unit digit or the power
    // of a number
    return (((int)(pow(x, y))) % 10);
}
 
int main()
{
    int x = 133, y = 5;
     
    // get unit digit number here we pass
    // the unit  digit of x and the last
    // cyclicity number that is y%4
    cout << unitnumber(x, y);
  
    return 0;
}

Java

// Java code to find the unit
// digit of x  raised to power y.
import java.io.*;
import java.util.*;
 
class GFG {
     
    // find unit digit
    static int unitnumber(int x, int y)
    {
        // Get last digit of x
        x = x % 10;
             
        // Last cyclic modular value
        if(y!=0)
             y = y % 4 + 4;
     
        // here we simply return the
        // unit digit or the power
        // of a number
        return (((int)(Math.pow(x, y))) % 10);
    }
     
     
    public static void main (String[] args)
    {
        int x = 133, y = 5;
     
        // get unit digit number here we pass
        // the unit digit of x and the last
        // cyclicity number that is y%4
        System.out.println(unitnumber(x, y));
     
    }
}
 
// This code is contributed by Gitanjali.

Python3

# Python3 code to find the unit  
# digit of x raised to power y.
import math
 
# Find unit digit
def unitnumber(x, y):
 
    # Get last digit of x
    x = x % 10
         
    # Last cyclic modular value
    if y!=0:
         y = y % 4 + 4
 
    # Here we simply return 
    # the unit digit or the 
    # power of a number
    return (((int)(math.pow(x, y))) % 10)
 
 
# Driver code
x = 133; y = 5
     
# Get unit digit number here we pass
# the unit digit of x and the last
# cyclicity number that is y%4
print(unitnumber(x, y))
 
 
# This code is contributed by Gitanjali.

C＃

// C# code to find the unit
// digit of x raised to power y.
using System;
 
class GFG {
     
    // find unit digit
    static int unitnumber(int x, int y)
    {
        // Get last digit of x
        x = x % 10;
             
        // Last cyclic modular value
        if(y!=0)
             y = y % 4 + 4;
     
        // here we simply return the
        // unit digit or the power
        // of a number
        return (((int)(Math.Pow(x, y))) % 10);
    }
     
    // Driver code
    public static void Main ()
    {
        int x = 133, y = 5;
     
        // get unit digit number here we pass
        // the unit digit of x and the last
        // cyclicity number that is y%4
        Console.WriteLine(unitnumber(x, y));
     
    }
}
 
// This code is contributed by vt_m.

Java脚本

 
// Javascript code to find the unit
// digit of x raised to power y.
 
// find unit digit
function unitnumber(x, y)
{
     
    // Get last digit of x
    x = x % 10;
     
    // Last cyclic modular value
    if (y != 0)
        y = y % 4 + 4;
 
    // here we simply return the
    // unit digit or the power
    // of a number
    return ((parseInt(Math.pow(x, y))) % 10);
}
 
// Driver code
let x = 133;
let y = 5;
     
// get unit digit number here we pass
// the unit digit of x and the last
// cyclicity number that is y%4
document.write(unitnumber(x, y));
 
// This code is contributed by _saurabh_jaiswal.
 

输出 ：

3

感谢DevanshuAgarwal建议上述解决方案。
如何处理大量数字?
大数a ^ b的最后一位数字的有效方法