找出N个阶乘和的单位位数

给定数字N，任务是查找前N个自然数阶乘的单位位数，即1！+2！+3！+….N!。其中N <= 10e18。
例子：

Input: n = 2 
Output: 3
1! + 2! = 3
Last digit is 3

Input: n = 3
Output: 9
1! + 2! + 3! = 9
Last digit is 9

天真的方法：在这种方法中，只需计算每个数字的阶乘并求和。最后得到总和的单位位数。这将花费大量时间和不必要的计算。
高效的方法：在这种方法中，仅在[1，5]范围内计算N的单位数字，因为：
1！ = 1
2！ = 2
3！ = 6
4！ = 24
5！ = 120
6！ = 720
7！ = 5040
很快。
由于5！= 120，且大于5的阶乘具有尾随零。因此，在进行求和时，N> = 5不会在单位位置发生变化。
所以：

if (n < 5)
    ans = (1 ! + 2 ! +..+ n !) % 10;
else
    ans = (1 ! + 2 ! + 3 ! + 4 !) % 10;

Note : We know (1! + 2! + 3! + 4!) % 10 = 3
So we always return 3 when n is greater 
than 4.

下面是有效方法的实现：

C++

// C++ program to find the unit place digit
// of the first N natural numbers factorials
#include 
using namespace std;
 
// Function to find the unit's place digit
int get_unit_digit(long long int N)
{
 
    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if (N == 0 || N == 1)
       return 1;
    else if (N == 2)
       return 3;
    else  if (N == 3)
       return 9;
 
    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
       return 3;
}
 
// Driver code
int main()
{
    long long int N = 1;
 
    for (N = 0; N <= 10; N++)
        cout << "For N = " << N
             << " : " << get_unit_digit(N)
             << endl;
 
    return 0;
}

Java

// Java  program to find the unit place digit
// of the first N natural numbers factorials
 
import java.io.*;
 
class GFG {
     
     
// Function to find the unit's place digit
static int get_unit_digit(  int N)
{
 
    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if (N == 0 || N == 1)
    return 1;
    else if (N == 2)
    return 3;
    else if (N == 3)
    return 9;
 
    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
    return 3;
}
 
// Driver code
     
    public static void main (String[] args) {
         
      int N = 1;
 
    for (N = 0; N <= 10; N++)
            System.out.println ("For N = " + N
            + " : " + get_unit_digit(N));
    }
}
//This Code is Contributed by ajit

Python3

# Python3 program to find the unit
# place digit of the first N natural
# numbers factorials
 
# Function to find the unit's place digit
def get_unit_digit(N):
     
    # Let us write for cases when
    # N is smaller than or equal
    # to 4.
    if (N == 0 or N == 1):
        return 1
    elif (N == 2):
        return 3
    elif(N == 3):
        return 9
         
    # We know following
    # (1! + 2! + 3! + 4!) % 10 = 3
    else:
        return 3
 
# Driver code
N = 1
for N in range(11):
    print("For N = ", N, ":",
        get_unit_digit(N), sep = ' ')
 
# This code is contributed
# by sahilshelangia

C#

// C# program to find the unit
// place digit of the first N
// natural numbers factorials
using System;
 
class GFG
{
     
// Function to find the unit's
// place digit
static int get_unit_digit( int N)
{
 
    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if (N == 0 || N == 1)
    return 1;
    else if (N == 2)
    return 3;
    else if (N == 3)
    return 9;
 
    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
    return 3;
}
 
// Driver code
static public void Main ()
{
    int N = 1;
 
    for (N = 0; N <= 10; N++)
        Console.WriteLine ("For N = " + N +
                " : " + get_unit_digit(N));
}
}
 
// This Code is Contributed by akt_mit

PHP

= 4)
        return 3;
}
 
// Driver code
$N = 1;
 
for ($N = 0; $N <= 10; $N++)
    echo "For N = " . $N.
         " : " . get_unit_digit($N) . "\n";
 
// This code is contributed
// by ChitraNayal
?>

Javascript

 
// Javascript program to find the unit place digit
// of the first N natural numbers factorials
 
// Function to find the unit's place digit
function get_unit_digit(N)
{
 
    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if (N == 0 || N == 1)
      return 1;
    else if (N == 2)
      return 3;
    else if (N == 3)
      return 9;
    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
      return 3;
}
 
// Driver code
var N = 1;
for (N = 0; N <= 10; N++)
    document.write( "For N = " + N
        + " : " + get_unit_digit(N)+"<br>")
     
 

输出：

For N = 0 : 1
For N = 1 : 1
For N = 2 : 3
For N = 3 : 9
For N = 4 : 3
For N = 5 : 3
For N = 6 : 3
For N = 7 : 3
For N = 8 : 3
For N = 9 : 3
For N = 10 : 3