检查是否存在一个X因子除以Y的复合数

给定两个整数X和Y分别表示除数的总数和复合除数的数量，任务是检查是否存在整数N ，该整数正好具有X个除数， Y是复合数。

例子：

 
Input: X = 6, Y = 3 
Output: YES 
Explanation: 
N = 18 is such a number.
The divisors of 18 are 1, 2, 3, 6, 9 and 18. 
The composite divisors of 18 are 6, 9 and 18.
Input: X = 7, Y = 3 
Output: NO 
Explanation: 
We see that no such number exists that has 7 positive divisors out of which 3 are composite divisors. 
 

方法：

1. 首先计算一个数的质数除数，它等于：

主除数的数量=除数的总数–复合除数的数量– 1

1. 因此，素数除数C = X – Y – 1
2. 由于每个数字都有1作为因数，而1既不是质数也不是复合数，我们必须将其排除在素数的数量之外。
3. 如果复合除数的数量小于主除数的数量，则根本找不到这样的数量。
4. 因此，如果X的素数分解至少包含C个不同的整数，则可能有解决方案。否则，我们将找不到满足给定条件的数字N。
5. 找到最大数量的值X可以分解为每个值都大于1 。换句话说，我们可以找出X的素因式分解。
6. 如果该素数分解的项数大于或等于C ，那么这样的数是可能的。

下面是上述方法的实现：

C++

// C++ program to check if a number
// exists having exactly X positive
// divisors out of which Y are
// composite divisors
#include 
using namespace std;
 
int factorize(int N)
{
    int count = 0;
    int cnt = 0;
 
    // Count the number of
    // times 2 divides N
    while ((N % 2) == 0) {
        N = N / 2;
        count++;
    }
 
    cnt = cnt + count;
 
    // check for all possible
    // numbers that can divide it
    for (int i = 3; i <= sqrt(N); i += 2) {
        count = 0;
        while (N % i == 0) {
            count++;
            N = N / i;
        }
        cnt = cnt + count;
    }
 
    // if N at the end
    // is a prime number.
    if (N > 2)
        cnt = cnt + 1;
    return cnt;
}
 
// Function to check if any
// such number exists
void ifNumberExists(int X, int Y)
{
    int C, dsum;
 
    C = X - Y - 1;
    dsum = factorize(X);
    if (dsum >= C)
        cout << "YES \n";
    else
        cout << "NO \n";
}
 
// Driver Code
int main()
{
 
    int X, Y;
    X = 6;
    Y = 4;
 
    ifNumberExists(X, Y);
    return 0;
}

Java

// Java program to check if a number
// exists having exactly X positive
// divisors out of which Y are
// composite divisors
import java.lang.Math;
class GFG{
     
public static int factorize(int N)
{
    int count = 0;
    int cnt = 0;
     
    // Count the number of
    // times 2 divides N
    while ((N % 2) == 0)
    {
        N = N / 2;
        count++;
    }
     
    cnt = cnt + count;
     
    // Check for all possible
    // numbers that can divide it
    for(int i = 3; i <= Math.sqrt(N); i += 2)
    {
       count = 0;
       while (N % i == 0)
       {
           count++;
           N = N / i;
       }
       cnt = cnt + count;
    }
     
    // If N at the end
    // is a prime number.
    if (N > 2)
        cnt = cnt + 1;
    return cnt;
}
     
// Function to check if any
// such number exists
public static void ifNumberExists(int X, int Y)
{
    int C, dsum;
    C = X - Y - 1;
    dsum = factorize(X);
         
    if (dsum >= C)
        System.out.println("YES");
    else
        System.out.println("NO");
}
 
// Driver code   
public static void main(String[] args)
{
    int X, Y;
    X = 6;
    Y = 4;
     
    ifNumberExists(X, Y);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to check if a number exists
# having exactly X positive divisors out of
#  which Y are composite divisors
import math
 
def factorize(N):
 
    count = 0
    cnt = 0
 
    # Count the number of
    # times 2 divides N
    while ((N % 2) == 0):
        N = N // 2
        count+=1
 
    cnt = cnt + count
 
    # Check for all possible
    # numbers that can divide it
    sq = int(math.sqrt(N))
    for i in range(3, sq, 2):
        count = 0
         
        while (N % i == 0):
            count += 1
            N = N // i
         
        cnt = cnt + count
 
    # If N at the end
    # is a prime number.
    if (N > 2):
        cnt = cnt + 1
    return cnt
 
# Function to check if any
# such number exists
def ifNumberExists(X, Y):
 
    C = X - Y - 1
    dsum = factorize(X)
     
    if (dsum >= C):
        print ("YES")
    else:
        print("NO")
 
# Driver Code
if __name__ == "__main__":
     
    X = 6
    Y = 4
 
    ifNumberExists(X, Y)
 
# This code is contributed by chitranayal

C#

// C# program to check if a number
// exists having exactly X positive
// divisors out of which Y are
// composite divisors
using System;
class GFG{
 
public static int factorize(int N)
{
    int count = 0;
    int cnt = 0;
     
    // Count the number of
    // times 2 divides N
    while ((N % 2) == 0)
    {
        N = N / 2;
        count++;
    }
     
    cnt = cnt + count;
     
    // Check for all possible
    // numbers that can divide it
    for(int i = 3; i <= Math.Sqrt(N); i += 2)
    {
        count = 0;
        while (N % i == 0)
        {
            count++;
            N = N / i;
        }
        cnt = cnt + count;
    }
     
    // If N at the end
    // is a prime number.
    if (N > 2)
        cnt = cnt + 1;
    return cnt;
}
     
// Function to check if any
// such number exists
public static void ifNumberExists(int X, int Y)
{
    int C, dsum;
    C = X - Y - 1;
    dsum = factorize(X);
         
    if (dsum >= C)
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
 
// Driver code
public static void Main(string[] args)
{
    int X, Y;
    X = 6;
    Y = 4;
     
    ifNumberExists(X, Y);
}
}
 
// This code is contributed by AnkitRai01

Javascript

// javascript program to check if a number
// exists having exactly X positive
// divisors out of which Y are
// composite divisors
 
    function factorize(N) {
        var count = 0;
        var cnt = 0;
 
        // Count the number of
        // times 2 divides N
        while ((N % 2) == 0) {
            N = N / 2;
            count++;
        }
 
        cnt = cnt + count;
 
        // Check for all possible
        // numbers that can divide it
        for (i = 3; i <= Math.sqrt(N); i += 2) {
            count = 0;
            while (N % i == 0) {
                count++;
                N = N / i;
            }
            cnt = cnt + count;
        }
 
        // If N at the end
        // is a prime number.
        if (N > 2)
            cnt = cnt + 1;
        return cnt;
    }
 
    // Function to check if any
    // such number exists
    function ifNumberExists(X , Y) {
        var C, dsum;
        C = X - Y - 1;
        dsum = factorize(X);
 
        if (dsum >= C)
            document.write("YES");
        else
            document.write("NO");
    }
 
    // Driver code
     
        var X, Y;
        X = 6;
        Y = 4;
 
        ifNumberExists(X, Y);
 
// This code is contributed by todaysgaurav

输出：

YES

时间复杂度： O(N ^1/2 )
辅助空间： O(1)