找出任意K个不同的奇数整数，使它们的总和等于N

给定两个整数N和K ，任务是找到任意K个不同的奇数整数，以使它们的总和等于N。如果不存在这样的整数，则打印-1。
例子：

Input: N = 10, K = 2 
Output: 1, 9 
Explanation: 
There are two possible distinct odd integers, such that their sum is equal to N. 
Possible K integers can be – {(1, 9), (3, 7)}
Input: N = 5, K = 4 
Output: -1 
Explanation: 
There are no such 4 distinct odd integers such that their sum is 5. 
 

方法：

• 这个问题的主要观察结果是，如果N和K具有不同的奇偶校验，则不可能找到K这样的不同整数，以使它们的总和等于N，
• 否则，此类K – 1整数将由前K-1个奇数正整数组成
• 第K^个奇数等于(N –第一个(K-1)个奇整数的总和)

Kth Odd number  = N - sum of first K-1 integer

下面是上述方法的实现：

C++

// C++ implementation to find k
// odd integers such that their sum is N
 
#include 
using namespace std;
 
// Function to find K odd integers
// such that their sum is N
void oddIntegers(int n, int k)
{
    // Condition to check if there
    // exist such K integers
    if (n % 2 != k % 2) {
        cout << "-1"
             << "\n";
        return;
    }
 
    int sum = 0;
    int i = 1;
    int j = 1;
 
    // Loop to find first K-1
    // distinct odd integers
    while (j < k) {
        sum = sum + i;
        cout << i << " ";
        i = i + 2;
        j++;
    }
 
    // Final Kth odd number
    int finalOdd = n - sum;
 
    cout << finalOdd << "\n";
}
 
// Driver code
int main()
{
    int n = 10;
    int k = 2;
 
    oddIntegers(n, k);
 
    return 0;
}

Java

// Java implementation to find k
// odd integers such that their sum is N
class GFG
{
 
// Function to find K odd integers
// such that their sum is N
static void oddIntegers(int n, int k)
{
    // Condition to check if there
    // exist such K integers
    if (n % 2 != k % 2) {
        System.out.println("-1");
        return;
    }
 
    int sum = 0;
    int i = 1;
    int j = 1;
 
    // Loop to find first K-1
    // distinct odd integers
    while (j < k) {
        sum = sum + i;
        System.out.print(i+" ");
        i = i + 2;
        j++;
    }
 
    // Final Kth odd number
    int finalOdd = n - sum;
     
    System.out.println(finalOdd);
}
 
// Driver code
public static void main (String[] args)
{
    int n = 10;
    int k = 2;
 
    oddIntegers(n, k);
}
}
 
// This code is contributed by shubhamsingh

Python3

# Python3 implementation to find k
# odd integers such that their sum is N
 
# Function to find K odd integers
# such that their sum is N
def oddIntegers(n, k) :
 
    # Condition to check if there
    # exist such K integers
    if (n % 2 != k % 2) :
        print("-1");
         
        return;
 
    sum = 0;
    i = 1;
    j = 1;
 
    # Loop to find first K-1
    # distinct odd integers
    while (j < k) :
        sum += i;
        print(i,end= " ");
        i += 2;
        j += 1;
 
    # Final Kth odd number
    finalOdd = n - sum;
 
    print(finalOdd);
 
# Driver code
if __name__ == "__main__" :
 
    n = 10;
    k = 2;
 
    oddIntegers(n, k);
     
# This code is contributed by AnkitRai01

C#

// C# implementation to find k
// odd integers such that their sum is N
using System;
 
class GFG
{
  
// Function to find K odd integers
// such that their sum is N
static void oddints(int n, int k)
{
    // Condition to check if there
    // exist such K integers
    if (n % 2 != k % 2) {
        Console.WriteLine("-1");
        return;
    }
  
    int sum = 0;
    int i = 1;
    int j = 1;
  
    // Loop to find first K-1
    // distinct odd integers
    while (j < k) {
        sum = sum + i;
        Console.Write(i+" ");
        i = i + 2;
        j++;
    }
  
    // Final Kth odd number
    int finalOdd = n - sum;
      
    Console.WriteLine(finalOdd);
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 10;
    int k = 2;
  
    oddints(n, k);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// JavaScript implementation to find k
// odd integers such that their sum is N   
 
// Function to find K odd integers
// such that their sum is N
    function oddIntegers(n , k)
    {
        // Condition to check if there
        // exist such K integers
        if (n % 2 != k % 2) {
            document.write("-1");
            return;
        }
 
        var sum = 0;
        var i = 1;
        var j = 1;
 
        // Loop to find first K-1
        // distinct odd integers
        while (j < k) {
            sum = sum + i;
            document.write(i + " ");
            i = i + 2;
            j++;
        }
 
        // Final Kth odd number
        var finalOdd = n - sum;
 
        document.write(finalOdd);
    }
 
    // Driver code
     
        var n = 10;
        var k = 2;
 
        oddIntegers(n, k);
 
// This code contributed by Rajput-Ji
 

输出：

1 9

性能分析：

• 时间复杂度：与上述方法一样，在最坏的情况下，存在一个循环来查找这样的K个奇数整数，这需要O(K)时间。因此，时间复杂度将为O(K) 。
• 辅助空间复杂度：与上述方法一样，没有使用额外的空间。因此，辅助空间复杂度将为O(1) 。