📜  C / C++程序,用于计算数字阶乘中的尾随零

📅  最后修改于: 2021-05-28 05:25:48             🧑  作者: Mango

给定整数n,编写一个函数,返回n!中尾随零的计数。

例子 :

Input: n = 5
Output: 1 
Factorial of 5 is 120 which has one trailing 0.

Input: n = 20
Output: 4
Factorial of 20 is 2432902008176640000 which has
4 trailing zeroes.

Input: n = 100
Output: 24
Trailing 0s in n! = Count of 5s in prime factors of n!
                  = floor(n/5) + floor(n/25) + floor(n/125) + ....
C++
// C++ program to count
// trailing 0s in n!
#include 
using namespace std;
  
// Function to return trailing
// 0s in factorial of n
int findTrailingZeros(int n)
{
    // Initialize result
    int count = 0;
  
    // Keep dividing n by powers of
    // 5 and update count
    for (int i = 5; n / i >= 1; i *= 5)
        count += n / i;
  
    return count;
}
  
// Driver Code
int main()
{
    int n = 100;
    cout << "Count of trailing 0s in " << 100
         << "! is " << findTrailingZeros(n);
    return 0;
}


输出:
Count of trailing 0s in 100! is 24

有关更多详细信息,请参阅完整的关于计数因数的尾随零的文章!

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