给定大小为N的数组A [] ，任务是检查是否有可能使用数组中的K个不同元素来获得奇数和。
例子：
Input: N = 4, K = 2, A = {10, 19, 14, 14}
Output: YES
Explanation:
19 + 14 = 33, which is odd
Input: N = 3, K = 3, A = {101, 102, 103}
Output: NO
Explanation:
101 + 102 + 103 = 306, which is even
方法：
- 让我们看一些基本的数学属性：
EVEN + EVEN = EVEN
ODD + ODD = EVEN
EVEN + ODD = ODD
- 从以上属性，我们可以得出结论，当将奇数个奇数整数添加到任意数量的偶数整数时，结果将始终为奇数。
Examples:
- 3 + 2 + 6 = 11 (odd)
- 1 + 3 + 7 + 4 = 15 (odd)
因此，请按照以下步骤解决问题：
- 计算数组中存在的不同的奇数和偶数元素。
- 如果数组中存在K个或更多奇数元素，则打印“是”。
- 否则，对于奇数的每个奇数计数，检查数组中是否存在足够的偶数。
- 如果出现这种情况，请打印“是”。否则，打印“否”。
下面是上述方法的实现：
C++
// C++ program for above approach
#include
using namespace std;
// Function to return if
// odd sum is possible or not
bool oddSum(vector& A,
int N, int K)
{
// Stores distinct odd elements
set Odd;
// Stores distinct even elements
set Even;
// Iterating through given array
for (int i = 0; i < N; i++) {
// If element is even
if (A[i] % 2 == 0) {
Even.insert(A[i]);
}
// If element is odd
else {
Odd.insert(A[i]);
}
}
// If atleast K elements
// in the array are odd
if (Odd.size() >= K)
return true;
bool flag = false;
// Check for all odd frequencies
// of odd elements whether
// sufficient even numbers
// are present or not
for (int i = 1; i < K; i += 2) {
// Count of even numbers
// required
int needed = K - i;
// If required even numbers
// are present in the array
if (needed <= Even.size()) {
return true;
}
}
return flag;
}
// Driver Program
int main()
{
int K = 5;
vector A = { 12, 1, 7, 7, 26, 18 };
int N = 3;
if (oddSum(A, N, K))
cout << "YES";
else
cout << "NO";
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to return if
// odd sum is possible or not
static boolean oddSum(int []A,
int N, int K)
{
// Stores distinct odd elements
HashSet Odd = new HashSet();
// Stores distinct even elements
HashSet Even = new HashSet();
// Iterating through given array
for(int i = 0; i < N; i++)
{
// If element is even
if (A[i] % 2 == 0)
{
Even.add(A[i]);
}
// If element is odd
else
{
Odd.add(A[i]);
}
}
// If atleast K elements
// in the array are odd
if (Odd.size() >= K)
return true;
boolean flag = false;
// Check for all odd frequencies
// of odd elements whether
// sufficient even numbers
// are present or not
for(int i = 1; i < K; i += 2)
{
// Count of even numbers
// required
int needed = K - i;
// If required even numbers
// are present in the array
if (needed <= Even.size())
{
return true;
}
}
return flag;
}
// Driver code
public static void main(String[] args)
{
int K = 5;
int []A = { 12, 1, 7, 7, 26, 18 };
int N = 3;
if (oddSum(A, N, K))
System.out.print("YES");
else
System.out.print("NO");
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 program for
# the above approach
# Function to return if
# odd sum is possible or not
def oddSum(A, N, K):
# Stores distinct odd elements
Odd = set([])
# Stores distinct even elements
Even = set([])
# Iterating through given array
for i in range (N):
# If element is even
if (A[i] % 2 == 0):
Even.add(A[i])
# If element is odd
else:
Odd.add(A[i])
# If atleast K elements
# in the array are odd
if (len(Odd) >= K):
return True
flag = False
# Check for all odd frequencies
# of odd elements whether
# sufficient even numbers
# are present or not
for i in range (1, K, 2):
# Count of even numbers
# required
needed = K - i
# If required even numbers
# are present in the array
if (needed <= len(Even)):
return True
return flag
# Driver Program
if __name__ == "__main__":
K = 5
A = [12, 1, 7,
7, 26, 18]
N = 3
if (oddSum(A, N, K)):
print ("YES")
else:
print("NO")
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to return if
// odd sum is possible or not
static bool oddSum(int []A, int N, int K)
{
// Stores distinct odd elements
HashSet Odd = new HashSet();
// Stores distinct even elements
HashSet Even = new HashSet();
// Iterating through given array
for(int i = 0; i < N; i++)
{
// If element is even
if (A[i] % 2 == 0)
{
Even.Add(A[i]);
}
// If element is odd
else
{
Odd.Add(A[i]);
}
}
// If atleast K elements
// in the array are odd
if (Odd.Count >= K)
return true;
bool flag = false;
// Check for all odd frequencies
// of odd elements whether
// sufficient even numbers
// are present or not
for(int i = 1; i < K; i += 2)
{
// Count of even numbers
// required
int needed = K - i;
// If required even numbers
// are present in the array
if (needed <= Even.Count)
{
return true;
}
}
return flag;
}
// Driver code
public static void Main(String[] args)
{
int K = 5;
int []A = { 12, 1, 7, 7, 26, 18 };
int N = 3;
if (oddSum(A, N, K))
Console.Write("YES");
else
Console.Write("NO");
}
}
// This code is contributed by PrinciRaj1992 输出：
NO
时间复杂度： O(N)
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