10^x 的任意正除数是 10^Y 的整数倍的概率

原文:https://www . geesforgeks . org/10x 的任意正除数是 10y 的整数倍的概率/

给定两个数字 X 和 Y ，任务是求10XT7】的任意正除数是10YT11】的整数倍的概率。 注: Y 应为< = X. 例:****

输入: X = 2，Y = 1 输出: 4/9 说明: 102的正因子为 1，2，4，5，10，20，25，50，100。一共 9 个。 101到 10 2 的倍数为 10，20，50，100。一共 4 个。 P(102的除数是 10 1 的倍数)= 4/9。 输入: X = 99，Y = 88 输出: 9/625 解释:T30】A = 1099，B = 1088T35】P(10 的除数 99 是 10 的倍数 88 ) = 9/625

先决条件: 一个数的除数总数 方法: 为了解决问题，我们需要观察以下步骤:

  • 10 X 的所有除数将为:

    (2P 5Q*)，其中 0 < = P，Q<= X

    = T14】

  • 求**10XT3 T5

    10X= 2X* 5X

    T14】的质因数数**

  • 因此，10 X 的除数总数将为: ( X + 1 ) * ( X + 1 ) 。

  • 现在，考虑 10 Y 的所有倍数，它们可以是 10 X 的潜在约数。它们的形式还有:

    (2A* 5B，其中 Y < = A，B<= X

    = T12】

  • 所以10XT3【也是10YT7】的倍数的潜在约数为(X–Y+1)(X–Y+1)**。*

  • 因此，所需概率为((X–Y+1)(X–Y+1))/((X+1)(X+1))。计算给定值 X 和 Y 的表达式的值，给出了所需的概率。

以下是上述方法的实现:

C++

// C++ program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10

#include <bits/stdc++.h>
using namespace std;
#define int long long int

// Function to calculate and print
// the required probability
void prob(int x, int y)
{
    // Count of potential divisors
    // of X-th power of 10 which are
    // also multiples of Y-th power
    // of 10
    int num = abs(x - y + 1)
              * abs(x - y + 1);

    // Count of divisors of X-th
    // power of 10
    int den = (x + 1) * (x + 1);

    // Calculate GCD
    int gcd = __gcd(num, den);

    // Print the reduced
    // fraction probability
    cout << num / gcd << "/"
         << den / gcd << endl;
}

// Driver Code
signed main()
{
    int X = 2, Y = 1;
    prob(X, Y);
    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10
import java.util.*;

class GFG{

// Function to calculate and print
// the required probability
static void prob(int x, int y)
{

    // Count of potential divisors
    // of X-th power of 10 which are
    // also multiples of Y-th power
    // of 10
    int num = Math.abs(x - y + 1) * 
              Math.abs(x - y + 1);

    // Count of divisors of X-th
    // power of 10
    int den = (x + 1) * (x + 1);

    // Calculate GCD
    int gcd = __gcd(num, den);

    // Print the reduced
    // fraction probability
    System.out.print(num / gcd + "/" + 
                     den / gcd + "\n");
}

static int __gcd(int a, int b) 
{ 
    return b == 0 ? a : __gcd(b, a % b);     
} 

// Driver code
public static void main(String[] args)
{
    int X = 2, Y = 1;

    prob(X, Y);
}
}

// This code is contributed by amal kumar choubey

Python 3

# Python3 program to find the probability
# of an arbitrary positive divisor of
# Xth power of 10 to be a multiple of
# Yth power of 10
from math import *

# Function to calculate and print
# the required probability
def prob(x, y):

    # Count of potential divisors
    # of X-th power of 10 which are
    # also multiples of Y-th power
    # of 10
    num = abs(x - y + 1) * abs(x - y + 1)

    # Count of divisors of X-th
    # power of 10
    den = (x + 1) * (x + 1)

    # Calculate GCD
    gcd1 = gcd(num, den)

    # Print the reduced
    # fraction probability
    print(num // gcd1, end = "")
    print("/", end = "")
    print(den // gcd1)

# Driver Code
if __name__ == '__main__':

    X = 2
    Y = 1

    prob(X, Y)

# This code is contributed by Surendra_Gangwar

C

// C# program to find the probability 
// of an arbitrary positive divisor of 
// Xth power of 10 to be a multiple of 
// Yth power of 10 
using System;
class GFG{ 

// Function to calculate and print 
// the required probability 
static void prob(int x, int y) 
{ 

    // Count of potential divisors 
    // of X-th power of 10 which are 
    // also multiples of Y-th power 
    // of 10 
    int num = Math.Abs(x - y + 1) * 
              Math.Abs(x - y + 1); 

    // Count of divisors of X-th 
    // power of 10 
    int den = (x + 1) * (x + 1); 

    // Calculate GCD 
    int gcd = __gcd(num, den); 

    // Print the reduced 
    // fraction probability 
    Console.Write(num / gcd + "/" + 
                  den / gcd + "\n"); 
} 

static int __gcd(int a, int b) 
{ 
    return b == 0 ? a : __gcd(b, a % b);     
} 

// Driver code 
public static void Main(string[] args) 
{ 
    int X = 2, Y = 1; 

    prob(X, Y); 
} 
} 

// This code is contributed by AnkitRai01 

java 描述语言

<script>

// JavaScript program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10

// Function to calculate and print
// the required probability
function prob(x, y)
{

    // Count of potential divisors
    // of X-th power of 10 which are
    // also multiples of Y-th power
    // of 10
    var num = Math.abs(x - y + 1) * 
              Math.abs(x - y + 1);

    // Count of divisors of X-th
    // power of 10
    var den = (x + 1) * (x + 1);

    // Calculate GCD
    var gcd = __gcd(num, den);

    // Print the reduced
    // fraction probability
    document.write(num / gcd + "/" + 
                   den / gcd + "\n");
}

function __gcd(a, b) 
{ 
    return b == 0 ? a : __gcd(b, a % b);     
} 

// Driver Code
var X = 2, Y = 1;

prob(X, Y);

// This code is contributed by Khushboogoyal499

</script>

Output: 

4/9

时间复杂度: O(log(N))

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