dsal 除数与 X 的除数不同的超过 X 的最小值

除数与 X 的除数不同的超过 X 的最小值

原文:https://www . geesforgeks . org/最小值超出-x-谁的除数与-x 的除数具有不同的奇偶性/

给定一个整数 X ，任务是确定大于 X 的 Y 的最小值，使得 X 和 Y 的除数具有不同的重围。

示例:

输入: X = 5 输出: 9 说明:5 和 9 的除数分别为 2 和 3，两者的 parities 不同。

输入: X = 9 输出: 10 说明:9 和 10 的除数分别为 3 和 4，它们的重围不同。

天真法:解决问题最简单的方法是从 X + 1 开始迭代每个数，直到得到一个除数与奇偶性相反的元素 X 。

时间复杂度:O((1+√X)²) 辅助空间: O(1)

下面是上述方法的实现:

C++

// C++ program for the above approach

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

// Function to count divisors of n
int divisorCount(int n)
{
    int x = 0;
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            if (i == n / i)
                x++;
            else
                x += 2;
        }
    }
    return x;
}

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
int minvalue_y(int x)
{
    // Divisor count of x
    int a = divisorCount(x);
    int y = x + 1;

    // Iterate from x + 1 and
    // check for each element
    while ((a & 1)
           == (divisorCount(y) & 1))
        y++;
    return y;
}

// Driver Code
int main()
{
    // Given X
    int x = 5;

    // Function call
    cout << minvalue_y(x) << endl;
    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program for the above approach
import java.util.*;

class GFG{

// Function to count divisors of n
static int divisorCount(int n)
{
    int x = 0;
    for(int i = 1; i <= Math.sqrt(n); i++)
    {
        if (n % i == 0)
        {
            if (i == n / i)
                x++;
            else
                x += 2;
        }
    }
    return x;
}

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
static int minvalue_y(int x)
{

    // Divisor count of x
    int a = divisorCount(x);
    int y = x + 1;

    // Iterate from x + 1 and
    // check for each element
    while ((a & 1) == (divisorCount(y) & 1))
        y++;

    return y;
}

// Driver Code
public static void main(String[] args)
{

    // Given X
    int x = 5;

    // Function call
    System.out.println(minvalue_y(x));
}
}

// This code is contributed by chitranayal

C

// C# program for the above approach
using System;

class GFG{

// Function to count divisors of n
static int divisorCount(int n)
{
    int x = 0;
    for(int i = 1; i <= Math.Sqrt(n); i++)
    {
        if (n % i == 0)
        {
            if (i == n / i)
                x++;
            else
                x += 2;
        }
    }
    return x;
}

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
static int minvalue_y(int x)
{

    // Divisor count of x
    int a = divisorCount(x);
    int y = x + 1;

    // Iterate from x + 1 and
    // check for each element
    while ((a & 1) == (divisorCount(y) & 1))
        y++;

    return y;
}

// Driver Code
public static void Main()
{

    // Given X
    int x = 5;

    // Function call
    Console.WriteLine(minvalue_y(x));
}
}

// This code is contributed by susmitakundugoaldanga

Python 3

# Python program for the above approach

# Function to count divisors of n
def divisorCount(n):
    x = 0;
    for i in range(1, n):
        if (n % i == 0):
            if (i == n // i):
                x += 1;
            else:
                x += 2;
        if(i * i > n):
            break;

    return x;

# Function to find the minimum
# value exceeding x whose count
# of divisors has different parity
# with count of divisors of X
def minvalue_y(x):

    # Divisor count of x
    a = divisorCount(x);
    y = x + 1;

    # Iterate from x + 1 and
    # check for each element
    while ((a & 1) == (divisorCount(y) & 1)):
        y += 1;

    return y;

# Driver Code
if __name__ == '__main__':

    # Given X
    x = 5;

    # Function call
    print(minvalue_y(x));

# This code is contributed by 29AjayKumar

java 描述语言

<script>

// javascript program of the above approach

// Function to count divisors of n
function divisorCount(n)
{
    let x = 0;
    for(let i = 1; i <= Math.sqrt(n); i++)
    {
        if (n % i == 0)
        {
            if (i == n / i)
                x++;
            else
                x += 2;
        }
    }
    return x;
}

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
function minvalue_y(x)
{

    // Divisor count of x
    let a = divisorCount(x);
    let y = x + 1;

    // Iterate from x + 1 and
    // check for each element
    while ((a & 1) == (divisorCount(y) & 1))
        y++;

    return y;
}

    // Driver Code

     // Given X
    let x = 5;

    // Function call
    document.write(minvalue_y(x));

// This code is contributed by target_2.
</script>

Output:

有效方法:可以根据以下观察结果解决问题:

如果 X 是一个完美的正方形，那么 X + 1 就是答案。
否则(1+√X)²T3 将是答案。

按照以下步骤解决问题:

检查 X 是否是一个完美的正方形。如果发现是真的，打印 X + 1 。
否则，打印 (1 +楼层(√X)) ² ) 。

下面是上述方法的实现:

C++

// C++ program for the above approach

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

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
int minvalue_y(int x)
{
    // Check if x is
    // perfect square
    int n = sqrt(x);
    if (n * n == x)
        return x + 1;

    return pow(n + 1, 2);
}

// Driver Code
int main()
{
    int x = 5;
    cout << minvalue_y(x) << endl;
    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
static int minvalue_y(int x)
{

    // Check if x is
    // perfect square
    int n = (int)Math.sqrt(x);
    if (n * n == x)
        return x + 1;

    return (int)Math.pow(n + 1, 2);
}

// Driver Code
public static void main(String[] args)
{
    int x = 5;

    System.out.print(minvalue_y(x));
}
}

// This code is contributed by sanjoy_62

Python 3

# Python3 program for the above approach

# Function to find the minimum
# value exceeding x whose count
# of divisors has different parity
# with count of divisors of X
def minvalue_y(x):

    # Check if x is
    # perfect square
    n = int(pow(x, 1 / 2))

    if (n * n == x):
        return x + 1

    return(pow(n + 1, 2))

# Driver Code
if __name__ == '__main__':

    x = 5

    print(minvalue_y(x))

# This code is contributed by Rajput-Ji

C

// C# program for the above approach
using System;

class GFG{

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
static int minvalue_y(int x)
{

    // Check if x is
    // perfect square
    int n = (int)Math.Sqrt(x);
    if (n * n == x)
        return x + 1;

    return (int)Math.Pow(n + 1, 2);
}

// Driver Code
public static void Main()
{
    int x = 5;

    Console.WriteLine(minvalue_y(x));
}
}

// This code is contributed by code_hunt

java 描述语言

<script>

// JavaScript program to implement
// the above approach

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
function minvalue_y(x)
{

    // Check if x is
    // perfect square
    let n = Math.floor(Math.sqrt(x));
    if (n * n == x)
        return x + 1;

    return Math.floor(Math.pow(n + 1, 2));
}

// Driver code

    let x = 5;

    document.write(minvalue_y(x));

</script>

Output:

时间复杂度:O(1) T5辅助空间:** O(1)

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dsal 除数与 X 的除数不同的超过 X 的最小值

除数与 X 的除数不同的超过 X 的最小值

C++

Java 语言(一种计算机语言，尤用于创建网站)

C

Python 3

java 描述语言

C++

Java 语言(一种计算机语言，尤用于创建网站)

Python 3

C

java 描述语言

留言

联系客服

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