给定一个由n个正数组成的数组，任务是计算带有偶数和奇数和的有序对的数量。
例子：
Input: arr[] = {1, 2, 4}
Output: Even sum Pairs = 2, Odd sum Pairs = 4
The ordered pairs are (1, 2), (1, 4), (2, 1), (4, 1), (2, 4), (4, 2)
Pairs with Even sum: (2, 4), (4, 2)
Pairs with Odd sum: (1, 2), (1, 4), (2, 1), (4, 1)
Input: arr[] = {2, 4, 5, 9, 1, 8}
Output: Even sum Pairs = 12, Odd sum Pairs = 18
方法：
如果一个数字为奇数而另一个数字为偶数，则两个数字的总和为奇数。因此，现在我们必须计算偶数和奇数。正如(a，b)和(b，a)对一样，它们都被视为不同的对，因此
Number of odd sum pairs = (count of even numbers) * (count of odd numbers) * 2
这是因为每个偶数都可以与每个奇数配对，并且每个奇数都可以与每个偶数配对。因此，答案是乘以2。
并且偶数和对的数量将是奇数和对的数量的倒数。所以：
Number of even sum pairs = Total Number of pairs – Number of odd sum pairs
下面是上述方法的实现：
CPP
// C++ implementation of the above approach
#include
using namespace std;
// function to count odd sum pair
int count_odd_pair(int n, int a[])
{
int odd = 0, even = 0;
for (int i = 0; i < n; i++) {
// if number is even
if (a[i] % 2 == 0)
even++;
// if number is odd
else
odd++;
}
// count of ordered pairs
int ans = odd * even * 2;
return ans;
}
// function to count even sum pair
int count_even_pair(int odd_sum_pairs, int n)
{
int total_pairs = (n * (n - 1));
int ans = total_pairs - odd_sum_pairs;
return ans;
}
// Driver code
int main()
{
int n = 6;
int a[] = { 2, 4, 5, 9, 1, 8 };
int odd_sum_pairs = count_odd_pair(n, a);
int even_sum_pairs = count_even_pair(
odd_sum_pairs, n);
cout << "Even Sum Pairs = "
<< even_sum_pairs
<< endl;
cout << "Odd Sum Pairs= "
<< odd_sum_pairs
<< endl;
return 0;
} Java
// Java implementation of the above approach
class GFG
{
// function to count odd sum pair
static int count_odd_pair(int n, int a[])
{
int odd = 0, even = 0;
for (int i = 0; i < n; i++) {
// if number is even
if (a[i] % 2 == 0)
even++;
// if number is odd
else
odd++;
}
// count of ordered pairs
int ans = odd * even * 2;
return ans;
}
// function to count even sum pair
static int count_even_pair(int odd_sum_pairs, int n)
{
int total_pairs = (n * (n - 1));
int ans = total_pairs - odd_sum_pairs;
return ans;
}
// Driver code
public static void main(String []args)
{
int n = 6;
int []a = { 2, 4, 5, 9, 1, 8 };
int odd_sum_pairs = count_odd_pair(n, a);
int even_sum_pairs = count_even_pair( odd_sum_pairs, n);
System.out.println("Even Sum Pairs = " + even_sum_pairs);
System.out.println("Odd Sum Pairs= " + odd_sum_pairs);
}
}
// This code is contributed by ihritikPython3
# Pytho3 implementation of the above approach
# function to count odd sum pair
def count_odd_pair( n, a):
odd = 0
even = 0
for i in range(0,n):
# if number is even
if ( a[ i] % 2 == 0):
even=even+1
# if number is odd
else:
odd=odd+1
# count of ordered pairs
ans = odd * even * 2
return ans
# function to count even sum pair
def count_even_pair( odd_sum_pairs, n):
total_pairs = ( n * ( n - 1))
ans = total_pairs - odd_sum_pairs
return ans
# Driver code
n = 6
a = [2, 4, 5, 9, 1, 8]
odd_sum_pairs = count_odd_pair( n, a)
even_sum_pairs = count_even_pair( odd_sum_pairs, n)
print("Even Sum Pairs =", even_sum_pairs)
print("Odd Sum Pairs=", odd_sum_pairs)
# This code is contributed by ihritikC#
// C# implementation of the above approach
using System;
class GFG
{
// function to count odd sum pair
static int count_odd_pair(int n, int []a)
{
int odd = 0, even = 0;
for (int i = 0; i < n; i++) {
// if number is even
if (a[i] % 2 == 0)
even++;
// if number is odd
else
odd++;
}
// count of ordered pairs
int ans = odd * even * 2;
return ans;
}
// function to count even sum pair
static int count_even_pair(int odd_sum_pairs, int n)
{
int total_pairs = (n * (n - 1));
int ans = total_pairs - odd_sum_pairs;
return ans;
}
// Driver code
public static void Main()
{
int n = 6;
int []a = { 2, 4, 5, 9, 1, 8 };
int odd_sum_pairs = count_odd_pair(n, a);
int even_sum_pairs = count_even_pair( odd_sum_pairs, n);
Console.WriteLine("Even Sum Pairs = " + even_sum_pairs);
Console.WriteLine("Odd Sum Pairs= " + odd_sum_pairs);
}
}
// This code is contributed by ihritikPHP
输出：
Even Sum Pairs = 12
Odd Sum Pairs= 18