求 N 阶乘之和的单位位数
原文:https://www . geeksforgeeks . org/find-the-unit-place-digit-of-n-factories/
给定一个数字 N，任务是找到单位第一个 N 个自然数阶乘的位置数字，即 1！+2!+3!+….n！其中 N <=10e18. 例:
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
天真法:在这种方法中，只需计算每个数的阶乘，并求出这些数的和。最后得到和的单位位数。这将花费大量时间和不必要的计算。 有效方法:在这种方法中，只有单位的数字 N 要在范围[1，5]内计算，因为: 1！= 1 2！= 2 3！= 6 4！= 24 5！= 120 6！= 720 7！= 5040 以此类推。 As 5！=120，大于 5 的阶乘有尾随零。所以，N > =5 在做求和的时候不在单位位置做贡献。 因此:
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.
下面是高效方法的实现:
C++
// C++ program to find the unit place digit
// of the first N natural numbers factorials
#include <iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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
$N = 1;
for ($N = 0; $N <= 10; $N++)
echo "For N = " . $N.
" : " . get_unit_digit($N) . "\n";
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// 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>")
</script>
Output:
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
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