给定两个数组，计算所有和为奇数的对
给定 N 和 M 个整数的两个数组。任务是从两个数组中找出由元素组成的无序对的数量，这样它们的和就是奇数。 注:一个元素只能是一对。 示例:
输入: a[] = {9，14，6，2，11}，b[] = {8，4，7，20} 输出: 3 {9，20}，{14，7}和{11，8} 输入: a[] = {2，4，6}，b[] = {8，10，12} 输出: 0
逼近:统计两个数组中奇数和偶数的个数，对个数的答案会是 min(odd1，even2) + min(odd2，even1)，因为奇数+偶数只是奇数。 以下是上述办法的实施情况:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns the number of pairs
int count_pairs(int a[], int b[], int n, int m)
{
// Count of odd and even numbers
int odd1 = 0, even1 = 0;
int odd2 = 0, even2 = 0;
// Traverse in the first array
// and count the number of odd
// and evene numbers in them
for (int i = 0; i < n; i++) {
if (a[i] % 2)
odd1++;
else
even1++;
}
// Traverse in the second array
// and count the number of odd
// and evene numbers in them
for (int i = 0; i < m; i++) {
if (b[i] % 2)
odd2++;
else
even2++;
}
// Count the number of pairs
int pairs = min(odd1, even2) + min(odd2, even1);
// Return the number of pairs
return pairs;
}
// Driver code
int main()
{
int a[] = { 9, 14, 6, 2, 11 };
int b[] = { 8, 4, 7, 20 };
int n = sizeof(a) / sizeof(a[0]);
int m = sizeof(b) / sizeof(b[0]);
cout << count_pairs(a, b, n, m);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
class GFG {
// Function that returns the number of pairs
static int count_pairs(int a[], int b[], int n, int m)
{
// Count of odd and even numbers
int odd1 = 0, even1 = 0;
int odd2 = 0, even2 = 0;
// Traverse in the first array
// and count the number of odd
// and evene numbers in them
for (int i = 0; i < n; i++) {
if (a[i] % 2 == 1) {
odd1++;
}
else {
even1++;
}
}
// Traverse in the second array
// and count the number of odd
// and evene numbers in them
for (int i = 0; i < m; i++) {
if (b[i] % 2 == 1) {
odd2++;
}
else {
even2++;
}
}
// Count the number of pairs
int pairs = Math.min(odd1, even2) + Math.min(odd2, even1);
// Return the number of pairs
return pairs;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 9, 14, 6, 2, 11 };
int b[] = { 8, 4, 7, 20 };
int n = a.length;
int m = b.length;
System.out.println(count_pairs(a, b, n, m));
}
}
// This code contributed by Rajput-Ji
Python 3
# Python 3 program to implement
# the above approach
# Function that returns
# the number of pairs
def count_pairs(a, b, n, m):
# Count of odd and even numbers
odd1 = 0
even1 = 0
odd2 = 0
even2 = 0
# Traverse in the first array
# and count the number of odd
# and evene numbers in them
for i in range(n):
if (a[i] % 2):
odd1 += 1
else:
even1 += 1
# Traverse in the second array
# and count the number of odd
# and evene numbers in them
for i in range(m):
if (b[i] % 2):
odd2 += 1
else:
even2 += 1
# Count the number of pairs
pairs = (min(odd1, even2) +
min(odd2, even1))
# Return the number of pairs
return pairs
# Driver code
if __name__ == '__main__':
a = [9, 14, 6, 2, 11]
b = [8, 4, 7, 20]
n = len(a)
m = len(b)
print(count_pairs(a, b, n, m))
# This code is contributed by
# Surendra_Gangwar
C
// C# program to implement
// the above approach
using System;
class GFG {
// Function that returns the number of pairs
static int count_pairs(int[] a, int[] b, int n, int m)
{
// Count of odd and even numbers
int odd1 = 0, even1 = 0;
int odd2 = 0, even2 = 0;
// Traverse in the first array
// and count the number of odd
// and evene numbers in them
for (int i = 0; i < n; i++) {
if (a[i] % 2 == 1) {
odd1++;
}
else {
even1++;
}
}
// Traverse in the second array
// and count the number of odd
// and evene numbers in them
for (int i = 0; i < m; i++) {
if (b[i] % 2 == 1) {
odd2++;
}
else {
even2++;
}
}
// Count the number of pairs
int pairs = Math.Min(odd1, even2) + Math.Min(odd2, even1);
// Return the number of pairs
return pairs;
}
// Driver code
static public void Main()
{
int[] a = { 9, 14, 6, 2, 11 };
int[] b = { 8, 4, 7, 20 };
int n = a.Length;
int m = b.Length;
Console.WriteLine(count_pairs(a, b, n, m));
}
}
// This code contributed by ajit.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to implement
// the above approach
// Function that returns the number of pairs
function count_pairs($a, $b, $n, $m)
{
// Count of odd and even numbers
$odd1 = 0; $even1 = 0;
$odd2 = 0; $even2 = 0;
// Traverse in the first array
// and count the number of odd
// and evene numbers in them
for ($i = 0; $i < $n; $i++)
{
if ($a[$i] % 2)
$odd1++;
else
$even1++;
}
// Traverse in the second array
// and count the number of odd
// and evene numbers in them
for ($i = 0; $i < $m; $i++)
{
if ($b[$i] % 2)
$odd2++;
else
$even2++;
}
// Count the number of pairs
$pairs = min($odd1, $even2) + min($odd2, $even1);
// Return the number of pairs
return $pairs;
}
// Driver code
$a = array( 9, 14, 6, 2, 11 );
$b = array( 8, 4, 7, 20 );
$n = count($a) ;
$m = count($b) ;
echo count_pairs($a, $b, $n, $m);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// JavaScript program to implement
// the above approach
// Function that returns the number of pairs
function count_pairs(a, b, n, m)
{
// Count of odd and even numbers
let odd1 = 0, even1 = 0;
let odd2 = 0, even2 = 0;
// Traverse in the first array
// and count the number of odd
// and even numbers in them
for (let i = 0; i < n; i++) {
if (a[i] % 2)
odd1++;
else
even1++;
}
// Traverse in the second array
// and count the number of odd
// and even numbers in them
for (let i = 0; i < m; i++) {
if (b[i] % 2)
odd2++;
else
even2++;
}
// Count the number of pairs
let pairs = Math.min(odd1, even2) + Math.min(odd2, even1);
// Return the number of pairs
return pairs;
}
// Driver code
let a = [ 9, 14, 6, 2, 11 ];
let b = [ 8, 4, 7, 20 ];
let n = a.length;
let m = b.length;
document.write(count_pairs(a, b, n, m));
// This code is contributed by Surbhi Tyagi.
</script>
Output:
3
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