找到最小的X，使X！至少包含Y个尾随零。

给定一个整数Y，找到最小的X使得X！至少包含Y个尾随零。
先决条件–计算数字阶乘中的尾随零

例子：

Input : Y = 2 
Output : 10 
10! = 3628800, which has 2 trailing zeros. 9! = 362880, which has 1 trailing zero. Hence, 10 is the correct answer.

Input : Y = 6 
Output : 25 
25! = 15511210043330985984000000, which has 6 trailing zeros. 24! = 620448401733239439360000, which has 4 trailing zeros. Hence, 25 is the correct answer. 
 

方法：使用Binary Search可以轻松解决问题。 N中的尾随零数！由N！中的因子5的计数给出。请阅读本文以了解先决条件。 countFactor(5，N)函数以N为单位返回因子5的计数！等于N！中尾随零的计数。 X的最小数就是X！通过使用此函数在范围[0，5 * Y]上进行二进制搜索，可以快速计算出至少包含Y个结尾的零。

下面是上述方法的实现。

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to count the number
// of factors P in X!
int countFactor(int P, int X)
{
    if (X < P)
        return 0;
 
    return (X / P + countFactor(P, X / P));
}
 
// Function to find the smallest X such
// that X! contains Y trailing zeros
int findSmallestX(int Y)
{
    int low = 0, high = 5 * Y;
    int N = 0;
    while (low <= high) {
        int mid = (high + low) / 2;
 
        if (countFactor(5, mid) < Y) {
            low = mid + 1;
        }
 
        else {
            N = mid;
            high = mid - 1;
        }
    }
 
    return N;
}
 
// Driver code
int main()
{
    int Y = 10;
 
    cout << findSmallestX(Y);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
     
    // Function to count the number
    // of factors P in X!
    static int countFactor(int P, int X)
    {
        if (X < P)
            return 0;
     
        return (X / P + countFactor(P, X / P));
    }
     
    // Function to find the smallest X such
    // that X! contains Y trailing zeros
    static int findSmallestX(int Y)
    {
        int low = 0, high = 5 * Y;
        int N = 0;
        while (low <= high)
        {
            int mid = (high + low) / 2;
     
            if (countFactor(5, mid) < Y)
            {
                low = mid + 1;
            }
     
            else
            {
                N = mid;
                high = mid - 1;
            }
        }
     
        return N;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int Y = 10;
     
        System.out.println(findSmallestX(Y));
    }
}
 
// This code is contributed by Ryuga

Python3

# Python3 implementation of the approach
 
# Function to count the number
# of factors P in X!
def countFactor(P, X):
    if (X < P):
        return 0;
 
    return (X // P + countFactor(P, X // P));
 
# Function to find the smallest X such
# that X! contains Y trailing zeros
def findSmallestX(Y):
 
    low = 0;
    high = 5 * Y;
    N = 0;
    while (low <= high):
        mid = (high + low) // 2;
 
        if (countFactor(5, mid) < Y):
            low = mid + 1;
 
        else:
            N = mid;
            high = mid - 1;
 
    return N;
 
# Driver code
Y = 10;
 
print(findSmallestX(Y));
 
# This code is contributed by mits

C#

// C# implementation of the approach
class GFG
{
     
// Function to count the number
// of factors P in X!
static int countFactor(int P, int X)
{
    if (X < P)
        return 0;
 
    return (X / P + countFactor(P, X / P));
}
 
// Function to find the smallest X such
// that X! contains Y trailing zeros
static int findSmallestX(int Y)
{
    int low = 0, high = 5 * Y;
    int N = 0;
    while (low <= high)
    {
        int mid = (high + low) / 2;
 
        if (countFactor(5, mid) < Y)
        {
            low = mid + 1;
        }
 
        else
        {
            N = mid;
            high = mid - 1;
        }
    }
 
    return N;
}
 
// Driver code
static void Main()
{
    int Y = 10;
 
    System.Console.WriteLine(findSmallestX(Y));
}
}
 
// This code is contributed by mits

Javascript

 
// Javascript implementation of the approach
 
// Function to count the number
// of factors P in X!
function countFactor(P, X)
{
    if (X < P)
        return 0;
 
    return (parseInt(X / P) +
    countFactor(P, parseInt(X / P)));
}
 
// Function to find the smallest X such
// that X! contains Y trailing zeros
function findSmallestX(Y)
{
    let low = 0, high = 5 * Y;
    let N = 0;
     
    while (low <= high)
    {
        let mid = parseInt((high + low) / 2);
 
        if (countFactor(5, mid) < Y)
        {
            low = mid + 1;
        }
        else
        {
            N = mid;
            high = mid - 1;
        }
    }
    return N;
}
 
// Driver code
let Y = 10;
 
document.write(findSmallestX(Y));
 
// This code is contributed by subhammahato348
 

输出：

45

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