从给定的数组中选择两个元素，使得它们的和不出现在任何数组中
原文:https://www . geeksforgeeks . org/从给定数组中选择两个元素，这样它们的总和就不会出现在任何数组中/
给定两个数组A【】和B【】，任务是选择两个元素 X 和 Y ，使得 X 属于A【】而 Y 属于B【】且 (X + Y) 不得出现在任何数组中。 示例:
输入: A[] = {3，2，2}，B[] = {1，5，7，7，9} 输出: 3 9 3 + 9 = 12 和 12 不存在于 任何给定的数组中。 输入: A[] = {1，3，5，7}，B[] = {7，5，3，1} 输出: 7 7
进场:从 A[] 中选择 X 作为最大元素，从 B[] 中选择 Y 作为最大元素。现在，很明显 (X + Y) 将大于两个阵列的最大值，即它不会出现在任何阵列中。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the numbers from
// the given arrays such that their
// sum is not present in any
// of the given array
void findNum(int a[], int n, int b[], int m)
{
// Find the maximum element
// from both the arrays
int x = *max_element(a, a + n);
int y = *max_element(b, b + m);
cout << x << " " << y;
}
// Driver code
int main()
{
int a[] = { 3, 2, 2 };
int n = sizeof(a) / sizeof(int);
int b[] = { 1, 5, 7, 7, 9 };
int m = sizeof(b) / sizeof(int);
findNum(a, n, b, m);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// find maximum element in an array
static int max_element(int a[], int n)
{
int m = Integer.MIN_VALUE;
for(int i = 0; i < n; i++)
m = Math.max(m, a[i]);
return m;
}
// Function to find the numbers from
// the given arrays such that their
// sum is not present in any
// of the given array
static void findNum(int a[], int n,
int b[], int m)
{
// Find the maximum element
// from both the arrays
int x = max_element(a, n);
int y = max_element(b, m);
System.out.print(x + " " + y);
}
// Driver code
public static void main(String args[])
{
int a[] = { 3, 2, 2 };
int n = a.length;
int b[] = { 1, 5, 7, 7, 9 };
int m = b.length;
findNum(a, n, b, m);
}
}
// This code is contributed by Arnub Kundu
Python 3
# Python3 implementation of the approach
# Function to find the numbers from
# the given arrays such that their
# sum is not present in any
# of the given array
def findNum(a, n, b, m) :
# Find the maximum element
# from both the arrays
x = max(a);
y = max(b);
print(x, y);
# Driver code
if __name__ == "__main__" :
a = [ 3, 2, 2 ];
n = len(a);
b = [ 1, 5, 7, 7, 9 ];
m = len(b);
findNum(a, n, b, m);
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// find maximum element in an array
static int max_element(int []a, int n)
{
int m = int.MinValue;
for(int i = 0; i < n; i++)
m = Math.Max(m, a[i]);
return m;
}
// Function to find the numbers from
// the given arrays such that their
// sum is not present in any
// of the given array
static void findNum(int []a, int n,
int []b, int m)
{
// Find the maximum element
// from both the arrays
int x = max_element(a, n);
int y = max_element(b, m);
Console.Write(x + " " + y);
}
// Driver code
public static void Main()
{
int []a = { 3, 2, 2 };
int n = a.Length;
int []b = { 1, 5, 7, 7, 9 };
int m = b.Length;
findNum(a, n, b, m);
}
}
// This code is contributed by kanugargng
java 描述语言
<script>
// Javascript implementation of the approach
// Function to find the numbers from
// the given arrays such that their
// sum is not present in any
// of the given array
function findNum(a, n, b, m)
{
// Find the maximum element
// from both the arrays
var x = a.reduce(function(a, b) { return Math.max(a, b); });
var y = b.reduce(function(a, b) { return Math.max(a, b); });
document.write(x + " " + y);
}
// Driver code
var a = [ 3, 2, 2 ];
var n = a.length;
var b = [ 1, 5, 7, 7, 9 ]
var m = b.length;
findNum(a, n, b, m);
// This code is contributed by rutvik_56.
</script>
Output:
3 9
时间复杂度: O(n)
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