[1, N] 范围内不具有任何子集总和为 K 的不同整数的计数

给定两个正整数N和K ，使得K≤N，任务是找到范围[1, N]中不存在总和等于K的子集的不同整数的最大数量。如果有多个解决方案，请打印任何一个。

例子：

Input: N = 5, K = 3
Output: 1 4 5
Explanation: There are two sets of distinct numbers of size 3 which don’t have any subset sum of 3.
These are {1, 4, 5} and {2, 4, 5}. So, print any of them in any order.

Input: N = 1, K =1
Output: 0

方法：这个想法基于以下观察：

• 可以选择任何大于K的数字，因为它们永远不会对总和为K的子集做出贡献。
• 无法选择K。
• 对于小于K的数字，最多可以选择K/2 个数字。例如：
  • 如果K =5、1+4=5、2+3=5，那么可以选择1或者4，同样可以选择2或者3。因此，最多可以选择 (5/2=2) 个数字。
  • 如果K =6, 1+5=6, 2+4=6 和 3+3=6。同样，可以选择 3 个数字，使得没有子集和等于 6。始终可以选择 3，因为只选择不同的数字，并且可以选择 1 或 5，类似地可以选择 2 或 4。因此，最多可以选择（6/3=3）个数字。
• 因此，可以选择的不同数字的最大数量是(NK)+(K/2) 。

请按照以下步骤解决问题：

• 可以选择的不同数字的最大数量是(NK)+(K/2) 。
• 需要选择所有大于K的数字，即从末尾开始的NK 个数字。还需要选择小于K的K/2 个元素。
• 因此，一种可能的解决方案是从N开始选择(NK)+(K/2)个连续数字，不包括K 。
• 最简单的方法是创建一个数组，存储从1到N的所有值，除了K ，反转数组，并从头开始打印(NK)+(K/2) 个元素。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find maximum number of distinct integers
// in [1, N] having no subset with sum equal to K
void findSet(int N, int K)
{
    // Declare a vector to store
    // the required numbers
    vector a;
 
    // Store all the numbers in [1, N] except K
    for (int i = 1; i <= N; i++) {
        if (i != K)
            a.push_back(i);
    }
 
    // Store the maximum number
    // of distinct numbers
    int MaxDistinct = (N - K) + (K / 2);
 
    // Reverse the array
    reverse(a.begin(), a.end());
 
    // Print the required numbers
    for (int i = 0; i < MaxDistinct; i++)
        cout << a[i] << " ";
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 5, K = 3;
 
    // Function Call
    findSet(N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find maximum number of distinct
// integers in [1, N] having no subset with
// sum equal to K
static void findSet(int N, int K)
{
     
    // Declare a vector to store
    // the required numbers
    ArrayList a = new ArrayList();
 
    // Store all the numbers in [1, N] except K
    for(int i = 1; i <= N; i++)
    {
        if (i != K)
            a.add(i);
    }
 
    // Store the maximum number
    // of distinct numbers
    int MaxDistinct = (N - K) + (K / 2);
 
    // Reverse the array
    Collections.reverse(a);
 
    // Print the required numbers
    for(int i = 0; i < MaxDistinct; i++)
        System.out.print(a.get(i) + " ");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int N = 5, K = 3;
     
    // Function Call
    findSet(N, K);
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to find maximum number of distinct
# integers in [1, N] having no subset with
# sum equal to K
def findSet(N, K):
     
    # Declare a vector to store
    # the required numbers
    a = []
 
    # Store all the numbers in [1, N] except K
    for i in range(1, N + 1):
        if (i != K):
            a.append(i)
 
    # Store the maximum number
    # of distinct numbers
    MaxDistinct = (N - K) + (K // 2)
 
    # Reverse the array
    a = a[::-1]
     
    # Print the required numbers
    for i in range(MaxDistinct):
        print(a[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    N = 5
    K = 3
 
    # Function Call
    findSet(N, K)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find maximum number of distinct
// integers in [1, N] having no subset with
// sum equal to K
static void findSet(int N, int K)
{
     
    // Declare a vector to store
    // the required numbers
    List a = new List();
 
    // Store all the numbers in [1, N] except K
    for(int i = 1; i <= N; i++)
    {
        if (i != K)
            a.Add(i);
    }
 
    // Store the maximum number
    // of distinct numbers
    int MaxDistinct = (N - K) + (K / 2);
 
    // Reverse the array
    a.Reverse();
 
    // Print the required numbers
    for(int i = 0; i < MaxDistinct; i++)
        Console.Write(a[i] + " ");
}
 
// Driver Code
public static void Main()
{
    // Given Input
    int N = 5, K = 3;
     
    // Function Call
    findSet(N, K);
}
}
 
// This code is contributed by avijitmondal1998.

Javascript

   
// JavaScript program for the above approach
 
// Function to find maximum number of distinct integers
// in [1, N] having no subset with sum equal to K
function findSet( N,  K)
{
    // Declare a vector to store
    // the required numbers
    let a = [];
 
    // Store all the numbers in [1, N] except K
    for (let i = 1; i <= N; i++) {
        if (i != K)
            a.push(i);
    }
 
    // Store the maximum number
    // of distinct numbers
    let MaxDistinct = (N - K) + parseInt(K / 2);
 
    // Reverse the array
    a.reverse();
    // Print the required numbers
    for (let i = 0; i < MaxDistinct; i++)
        document.write(a[i]+" ");
}
 
// Driver Code
// Given Input
    let N = 5, K = 3;
 
    // Function Call
    findSet(N, K);
 
  // This code is contributed by Potta Lokesh
   

输出

5 4 2 

时间复杂度： O(N)
辅助空间： O(N)