给定两个整数N和K ，任务是找到一个大小为K的仅包含偶数或奇数元素的数组，其中该数组的所有元素之和为N。如果没有这样的数组，则打印“ No”。
例子：
Input: N = 18, K = 3
Output: 6 6 6
Input: N = 19, K = 5
Output: 3 3 3 3 7
方法：想法是选择K-1次最小的偶数或奇数，最后，在总和的帮助下计算最后一个数。如果最后一个数字对于偶数也是偶数，对于最小的奇数也是奇数。然后可以选择这样的阵列。否则，就不可能有这样的数组。
下面是上述方法的实现：
C++
// C++ implementation to find an
// array of size K with all the
// even or odd elements in the array
#include
using namespace std;
// Function to find the array with
// all the even / odd elements
void getArrayOfSizeK(int n, int k)
{
vector ans;
// If array could be constructed
// by adding odd elements
// we need to find the kth
// element is odd or even
// if array of odd elements
// would be the answer then
// k-1 elements would be 1
// First let's check
// kth is odd or even
int odd = n - ((k - 1) * 1);
// if last element is also
// an odd number then
// we can choose odd
// elements for our answer
if (odd > 0
&& odd % 2 != 0) {
// Add 1 in the array (k-1) times
for (int i = 0; i < k - 1; i++) {
ans.push_back(1);
}
// Add last odd element
ans.push_back(odd);
}
// If array of even elements
// would be the answer then
// k-1 elements would be 2
int even = n - ((k - 1) * 2);
// if last element is also
// an even number then
// we can choose even
// elements for our answer
if (even > 0
&& even % 2 == 0
&& ans.size() == 0) {
// Add 2 in the array (k-1) times
for (int i = 0; i < k - 1; i++) {
ans.push_back(2);
}
// Add last even element
ans.push_back(even);
}
// Printing the array
if (ans.size() > 0) {
for (int i = 0; i < k; i++) {
cout << ans[i] << " ";
}
}
else {
cout << "NO" << endl;
}
}
// Driver Code
int main()
{
int n = 10, k = 3;
getArrayOfSizeK(n, k);
return 0;
} Java
// Java implementation to find an
// array of size K with all the
// even or odd elements in the array
import java.util.*;
class GFG{
// Function to find the array with
// all the even / odd elements
static void getArrayOfSizeK(int n, int k)
{
Vector ans = new Vector();
// If array could be constructed
// by adding odd elements
// we need to find the kth
// element is odd or even
// If array of odd elements
// would be the answer then
// k-1 elements would be 1
// First let's check
// kth is odd or even
int odd = n - ((k - 1) * 1);
// If last element is also
// an odd number then
// we can choose odd
// elements for our answer
if (odd > 0 && odd % 2 != 0)
{
// Add 1 in the array (k-1) times
for(int i = 0; i < k - 1; i++)
{
ans.add(1);
}
// Add last odd element
ans.add(odd);
}
// If array of even elements
// would be the answer then
// k-1 elements would be 2
int even = n - ((k - 1) * 2);
// If last element is also
// an even number then
// we can choose even
// elements for our answer
if (even > 0 && even % 2 == 0 &&
ans.size() == 0)
{
// Add 2 in the array (k-1) times
for(int i = 0; i < k - 1; i++)
{
ans.add(2);
}
// Add last even element
ans.add(even);
}
// Printing the array
if (ans.size() > 0)
{
for(int i = 0; i < k; i++)
{
System.out.print(ans.get(i) + " ");
}
}
else
{
System.out.println("NO");
}
}
// Driver code
public static void main(String args[])
{
int n = 10, k = 3;
getArrayOfSizeK(n, k);
}
}
// This code is contributed by Surendra_Gangwar Python3
# Python 3 implementation to find an
# array of size K with all the
# even or odd elements in the array
# Function to find the array with
# all the even / odd elements
def getArrayOfSizeK(n, k):
ans = []
# If array could be constructed
# by adding odd elements
# we need to find the kth
# element is odd or even
# if array of odd elements
# would be the answer then
# k-1 elements would be 1
# First let's check
# kth is odd or even
odd = n - ((k - 1) * 1)
# if last element is also
# an odd number then
# we can choose odd
# elements for our answer
if (odd > 0
and odd % 2 != 0):
# Add 1 in the array (k-1) times
for i in range(k - 1):
ans.append(1)
# Add last odd element
ans.append(odd)
# If array of even elements
# would be the answer then
# k-1 elements would be 2
even = n - ((k - 1) * 2)
# if last element is also
# an even number then
# we can choose even
# elements for our answer
if (even > 0
and even % 2 == 0
and len(ans) == 0):
# Add 2 in the array (k-1) times
for i in range(k - 1):
ans.append(2)
# Add last even element
ans.append(even)
# Printing the array
if (len(ans) > 0):
for i in range( k):
print (ans[i], end = " ")
else :
print ("NO")
# Driver Code
if __name__=="__main__":
n, k = 10, 3
getArrayOfSizeK(n, k)
# This code is contributed by chitranayalC#
// C# implementation to find an
// array of size K with all the
// even or odd elements in the array
using System;
using System.Collections.Generic;
class GFG{
// Function to find the array with
// all the even / odd elements
static void getArrayOfSizeK(int n, int k)
{
List ans = new List();
// If array could be constructed
// by adding odd elements
// we need to find the kth
// element is odd or even
// If array of odd elements
// would be the answer then
// k-1 elements would be 1
// First let's check
// kth is odd or even
int odd = n - ((k - 1) * 1);
// If last element is also
// an odd number then
// we can choose odd
// elements for our answer
if (odd > 0 && odd % 2 != 0)
{
// Add 1 in the array (k-1) times
for(int i = 0; i < k - 1; i++)
{
ans.Add(1);
}
// Add last odd element
ans.Add(odd);
}
// If array of even elements
// would be the answer then
// k-1 elements would be 2
int even = n - ((k - 1) * 2);
// If last element is also
// an even number then
// we can choose even
// elements for our answer
if (even > 0 && even % 2 == 0 &&
ans.Count == 0)
{
// Add 2 in the array (k-1) times
for(int i = 0; i < k - 1; i++)
{
ans.Add(2);
}
// Add last even element
ans.Add(even);
}
// Printing the array
if (ans.Count > 0)
{
for(int i = 0; i < k; i++)
{
Console.Write(ans[i] + " ");
}
}
else
{
Console.WriteLine("NO");
}
}
// Driver code
public static void Main(String []args)
{
int n = 10, k = 3;
getArrayOfSizeK(n, k);
}
}
// This code is contributed by amal kumar choubey 输出：
2 2 6