给定数字n，请找到n！中的最后一个非零数字。
例子：
Input : n = 5
Output : 2
5! = 5 * 4 * 3 * 2 * 1 = 120
Last non-zero digit in 120 is 2.
Input : n = 33
Output : 4
一个简单的解决方案是首先找到n !，然后找到n的最后一个非零数字。由于算术溢出，该解决方案即使对于较大的数字也不起作用。
更好的解决方案基于以下递归公式
Let D(n) be the last non-zero digit in n!
If tens digit (or second last digit) of n is odd
D(n) = 4 * D(floor(n/5)) * D(Unit digit of n)
If tens digit (or second last digit) of n is even
D(n) = 6 * D(floor(n/5)) * D(Unit digit of n)
公式说明：
对于小于10的数字，我们可以通过上述简单的解决方案轻松地找到最后一个非零数字，即先计算n !，然后找到最后一个数字。
D(1)= 1，D(2)= 2，D(3)= 6，D(4)= 4，D(5)= 2
D(6)= 2，D(7)= 4，D(8)= 2，D(9)= 8。
D(1) to D(9) are assumed to be precomputed.
Example 1: n = 27 [Second last digit is even]:
D(27) = 6 * D(floor(27/5)) * D(7)
= 6 * D(5) * D(7)
= 6 * 2 * 4
= 48
Last non-zero digit is 8
Example 2: n = 33 [Second last digit is odd]:
D(33) = 4 * D(floor(33/5)) * D(3)
= 4 * D(6) * 6
= 4 * 2 * 6
= 48
Last non-zero digit is 8
以上公式如何运作?
下面的解释提供了公式背后的直觉。读者可以参考http://math.stackexchange.com/questions/130352/last-non-zero-digit-of-a-factorial以获得完整的证明。
14! = 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 *
6 * 5 * 4 * 3 * 2 * 1
Since we are asked about last non-zero digit,
we remove all 5's and equal number of 2's from
factors of 14!. We get following:
14! = 14 * 13 * 12 * 11 * 2 * 9 * 8 * 7 *
6 * 3 * 2 * 1
Now we can get last non-zero digit by multiplying
last digits of above factors!
在n！中，2的数量总是大于5的数量。要删除尾随的0，我们将删除5和相等的2。
令a = floor(n / 5)， b = n％5。在除去相等数量的5和2之后，我们可以从n减少问题！到2个*一个！ * b！
D(n)= 2 a * D(a)* D(b)
执行：
C++
// C++ program to find last non-zero digit in n!
#include
using namespace std;
// Initialize values of last non-zero digit of
// numbers from 0 to 9
int dig[] = {1, 1, 2, 6, 4, 2, 2, 4, 2, 8};
int lastNon0Digit(int n)
{
if (n < 10)
return dig[n];
// Check whether tens (or second last) digit
// is odd or even
// If n = 375, So n/10 = 37 and (n/10)%10 = 7
// Applying formula for even and odd cases.
if (((n/10)%10)%2 == 0)
return (6*lastNon0Digit(n/5)*dig[n%10]) % 10;
else
return (4*lastNon0Digit(n/5)*dig[n%10]) % 10;
}
// Driver code
int main()
{
int n = 14;
cout << lastNon0Digit(n);
return 0;
} Java
// Java program to find last
// non-zero digit in n!
class GFG
{
// Initialize values of last non-zero digit of
// numbers from 0 to 9
static int dig[] = {1, 1, 2, 6, 4, 2, 2, 4, 2, 8};
static int lastNon0Digit(int n)
{
if (n < 10)
return dig[n];
// Check whether tens (or second last)
// digit is odd or even
// If n = 375, So n/10 = 37 and
// (n/10)%10 = 7 Applying formula for
// even and odd cases.
if (((n / 10) % 10) % 2 == 0)
return (6 * lastNon0Digit(n / 5)
* dig[n % 10]) % 10;
else
return (4 * lastNon0Digit(n / 5)
* dig[n % 10]) % 10;
}
// Driver code
public static void main (String[] args)
{
int n = 14;
System.out.print(lastNon0Digit(n));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python program to find
# last non-zero digit in n!
# Initialize values of
# last non-zero digit of
# numbers from 0 to 9
dig= [1, 1, 2, 6, 4, 2, 2, 4, 2, 8]
def lastNon0Digit(n):
if (n < 10):
return dig[n]
# Check whether tens (or second last) digit
# is odd or even
# If n = 375, So n/10 = 37 and (n/10)%10 = 7
# Applying formula for even and odd cases.
if (((n//10)%10)%2 == 0):
return (6*lastNon0Digit(n//5)*dig[n%10]) % 10
else:
return (4*lastNon0Digit(n//5)*dig[n%10]) % 10
return 0
# driver code
n = 14
print(lastNon0Digit(n))
# This code is contributed
# by Anant Agarwal.C#
// C# program to find last
// non-zero digit in n!
using System;
class GFG {
// Initialize values of last non-zero
// digit of numbers from 0 to 9
static int []dig = {1, 1, 2, 6, 4, 2, 2, 4, 2, 8};
static int lastNon0Digit(int n)
{
if (n < 10)
return dig[n];
// Check whether tens (or second
// last) digit is odd or even
// If n = 375, So n/10 = 37 and
// (n/10)%10 = 7 Applying formula
// for even and odd cases.
if (((n / 10) % 10) % 2 == 0)
return (6 * lastNon0Digit(n / 5) *
dig[n % 10]) % 10;
else
return (4 * lastNon0Digit(n / 5) *
dig[n % 10]) % 10;
}
// Driver code
public static void Main ()
{
int n = 14;
Console.Write(lastNon0Digit(n));
}
}
// This code is contributed by Nitin Mittal.PHP
输出：
2