给定 sum 和 product 值可能的最小数组大小

给定两个正整数S和P ，任务是找到数组的最小可能大小，使得元素之和为S并且元素的乘积为P 。如果不存在任何这样的数组，则打印“-1” 。

例子：

Input: S = 5, P = 6
Output: 2
Explanation: The valid array can be {2, 3}, which is of minimum size.

Input: S = 5, P = 100
Output: -1

方法：可以根据以下观察解决给定的问题：

• 使用N个数字，可以组成一个数组 大小N总和S 。
• 当P的值介于[0, (S/N) ^N ]之间时，可以实现任何乘积值。

请按照以下步骤解决给定的问题：

• 首先检查S和P的值是否相同，然后返回 1，因为S值本身用于制作最小大小的数组。
• 使用变量i迭代范围[2, S] ，如果(S/i) >= pow(P, 1/i)的值，则打印i的值作为形成的数组的最小大小。
• 完成上述步骤后，如果没有任何可能的值i满足上述条件，则打印“-1” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum size
// of array  with sum S and product P
int minimumSizeArray(int S, int P)
{
    // Base Case
    if (S == P) {
 
        return 1;
    }
 
    // Iterate through all values of S
    // and check the mentioned condition
    for (int i = 2; i <= S; i++) {
 
        double d = i;
        if ((S / d) >= pow(P, 1.0 / d)) {
            return i;
        }
    }
 
    // Otherwise, print "-1"
    return -1;
}
 
// Driver Code
int main()
{
    int S = 5, P = 6;
    cout << minimumSizeArray(S, P);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to find the minimum size
// of array  with sum S and product P
static int minimumSizeArray(int S, int P)
{
    // Base Case
    if (S == P) {
 
        return 1;
    }
 
    // Iterate through all values of S
    // and check the mentioned condition
    for (int i = 2; i <= S; i++) {
 
        double d = i;
        if ((S / d) >= Math.pow(P, 1.0 / d)) {
            return i;
        }
    }
 
    // Otherwise, print "-1"
    return -1;
}
 
// Driver Code
public static void main(String args[])
{
    int S = 5, P = 6;
       System.out.println(minimumSizeArray(S, P));
}
}
 
// This code is contributed by AnkThon

Python3

# python program for the above approach
 
# Function to find the minimum size
# of array with sum S and product P
def minimumSizeArray(S, P):
 
    # Base Case
    if (S == P):
        return 1
 
    # Iterate through all values of S
    # and check the mentioned condition
    for i in range(2, S+1):
 
        d = i
        if ((S / d) >= pow(P, 1.0 / d)):
            return i
 
    # Otherwise, print "-1"
    return -1
 
# Driver Code
if __name__ == "__main__":
    S = 5
    P = 6
    print(minimumSizeArray(S, P))
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the minimum size
// of array  with sum S and product P
static int minimumSizeArray(int S, int P)
{
    // Base Case
    if (S == P) {
 
        return 1;
    }
 
    // Iterate through all values of S
    // and check the mentioned condition
    for (int i = 2; i <= S; i++) {
 
        double d = i;
        if ((S / d) >= Math.Pow(P, 1.0 / d)) {
            return i;
        }
    }
 
    // Otherwise, print "-1"
    return -1;
}
 
// Driver Code
public static void Main()
{
    int S = 5, P = 6;
    Console.Write(minimumSizeArray(S, P));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find the minimum size
        // of array  with sum S and product P
        function minimumSizeArray(S, P)
        {
         
            // Base Case
            if (S == P) {
 
                return 1;
            }
 
            // Iterate through all values of S
            // and check the mentioned condition
            for (let i = 2; i <= S; i++) {
 
                let d = i;
                if ((S / d) >= Math.pow(P, 1.0 / d)) {
                    return i;
                }
            }
 
            // Otherwise, print "-1"
            return -1;
        }
 
        // Driver Code
        let S = 5, P = 6;
        document.write(minimumSizeArray(S, P));
 
// This code is contributed by Potta Lokesh
    

输出：

2

时间复杂度： O(log P)
辅助空间： O(1)