📌  相关文章
📜  将N分成K个唯一的部分,以使这些部分的gcd最大

📅  最后修改于: 2021-06-26 13:38:09             🧑  作者: Mango

给定正整数N ,任务是将其划分为K个唯一的部分,以使这些部分的总和等于原始数,并且所有部分的gcd最大。如果存在这样的除法,则打印最大gcd ,否则打印-1

例子:

天真的方法:找到N的所有可能除法,并计算它们的最大gcd。但是这种方法将花费指数时间和空间。
高效的方法:N的除法数为A 1 ,A 2 ……..A K
现在,已知gcd(a,b)= gcd(a,b,a + b) ,因此gcd(A 1 ,A 2 ….A K )= gcd(A 1 ,A 2 ….A K ,A 1 + A 2 …。+ A K )
但是A 1 + A 2 …。 + A K = N ,因此,除法的gcd将成为N的因素之一。
aN的因子,这可能是答案:
由于是GCD,所有的分裂将是倍数,因此,对于K独特B I,
a * B 1 + a * B 2 ……。 + a * B K = N
a *(B 1 + B 2 ……. + B K )= N
由于所有的B i都是唯一的,
B 1 + B 2 ……。+ B K≥1 + 2 + 3….. + K
1 + B 2 ……。+ b k的序列≥K *(K + 1)/ 2
因此,互补因子大于或等于K *(K +1)/ 2的所有N个因子都可以作为可能的答案之一,而我们已将所有可能的答案最大化。

下面是上述方法的实现:

CPP
// C++ implementation of the approach
#include 
using namespace std;
 
// Function to calculate maximum GCD
int maxGCD(int N, int K)
{
 
    // Minimum possible sum for
    // K unique positive integers
    int minSum = (K * (K + 1)) / 2;
 
    // It is not possible to divide
    // N into K unique parts
    if (N < minSum)
        return -1;
 
    // All the factors greater than sqrt(N)
    // are complementary of the factors less
    // than sqrt(N)
    int i = sqrt(N);
    int res = 1;
    while (i >= 1) {
 
        // If i is a factor of N
        if (N % i == 0) {
            if (i >= minSum)
                res = max(res, N / i);
 
            if (N / i >= minSum)
                res = max(res, i);
        }
        i--;
    }
 
    return res;
}
 
// Driver code
int main()
{
    int N = 18, K = 3;
 
    cout << maxGCD(N, K);
 
    return 0;
}


Java
// Java implementation of the approach
import java.io.*;
import java.lang.Math;
 
class GFG
{
    // Function to calculate maximum GCD
    static int maxGCD(int N, int K)
    {
 
        // Minimum possible sum for
        // K unique positive integers
        int minSum = (K * (K + 1)) / 2;
 
        // It is not possible to divide
        // N into K unique parts
        if (N < minSum)
            return -1;
 
        // All the factors greater than sqrt(N)
        // are complementary of the factors less
        // than sqrt(N)
        int i = (int) Math.sqrt(N);
        int res = 1;
        while (i >= 1)
        {
 
            // If i is a factor of N
            if (N % i == 0)
            {
                if (i >= minSum)
                    res = Math.max(res, N / i);
 
                if (N / i >= minSum)
                    res = Math.max(res, i);
            }
            i--;
        }
 
        return res;
    }
 
    // Driver code
    public static void main (String[] args)
    {
        int N = 18, K = 3;
 
        System.out.println(maxGCD(N, K));
    }
}
 
// This code is contributed by ApurvaRaj


Python
# Python3 implementation of the approach
from math import sqrt,ceil,floor
 
# Function to calculate maximum GCD
def maxGCD(N, K):
 
    # Minimum possible sum for
    # K unique positive integers
    minSum = (K * (K + 1)) / 2
 
    # It is not possible to divide
    # N into K unique parts
    if (N < minSum):
        return -1
 
    # All the factors greater than sqrt(N)
    # are complementary of the factors less
    # than sqrt(N)
    i = ceil(sqrt(N))
    res = 1
    while (i >= 1):
 
        # If i is a factor of N
        if (N % i == 0):
            if (i >= minSum):
                res = max(res, N / i)
 
            if (N / i >= minSum):
                res = max(res, i)
 
        i-=1
 
    return res
 
# Driver code
 
N = 18
K = 3
 
print(maxGCD(N, K))
 
# This code is contributed by mohit kumar 29


C#
// C# implementation of the approach
using System;
 
class GFG
{
    // Function to calculate maximum GCD
    static int maxGCD(int N, int K)
    {
 
        // Minimum possible sum for
        // K unique positive integers
        int minSum = (K * (K + 1)) / 2;
 
        // It is not possible to divide
        // N into K unique parts
        if (N < minSum)
            return -1;
 
        // All the factors greater than sqrt(N)
        // are complementary of the factors less
        // than sqrt(N)
        int i = (int) Math.Sqrt(N);
        int res = 1;
        while (i >= 1)
        {
 
            // If i is a factor of N
            if (N % i == 0)
            {
                if (i >= minSum)
                    res = Math.Max(res, N / i);
 
                if (N / i >= minSum)
                    res = Math.Max(res, i);
            }
            i--;
        }
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 18, K = 3;
 
        Console.WriteLine(maxGCD(N, K));
    }
}
 
// This code is contributed by AnkitRai01


Javascript


输出:
3

如果您希望与行业专家一起参加现场课程,请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。