Hex Factorial 是如何一个实现的

Question

Problem Description
The expression N!, reads as the factorial of N, denoting the product of the first N positive integers. If the factorial of N is written in hexadecimal without leading zeros, can you tell us how many zeros are there in it? Take 15! as an example, you should answer "3" because (15)10! = (13077775800)16, and there are 3 zeros in it.

Input
The input contains several cases. Each case has one line containing a non-negative decimal integer N (N ≤ 100). You need to count the zeros in N! in hexadecimal. A negative number terminates the input.

Output
For each non-negative integer N, output one line containing exactly one integer, indicating the number of
zeros in N!.

Sample Input
1
15
-1

Sample Output
0
3

Answer 1

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# 读取输入数据
input <- readLines()

# 定义变量
zero_count <- 0

for (line in input) {
  # 将输入字符串转换为整数
  num <- as.integer(strsplit(line, "")[[1]])
  
  # 计算n!
  fact <- factorial(num)
  
  # 计算fact在十六进制中的长度，并添加到零计数中
  zero_count <- zero_count + length(as.character(fact)) %% 16
  
  # 输出结果
  print(zero_count)
}

请注意，此代码可能不会正确地处理所有边缘情况。例如，如果输入包含负数或非数字字符，它将抛出错误。此外，由于Python和R之间的差异，某些函数（如factorial()）可能不适用。在这种情况下，您可能需要使用特定于R的语言库来解决这个问题。