求所有 I 和 j 对上 a[i] % a[j]的最大可能值
原文:https://www . geesforgeks . org/find-最大可能值-ai-aj-over-all-pair-I-and-j/
给定一个正整数数组arr[]N。任务是在所有对 i 和 j 上找到 a[i] % a[j] 的最大可能值。 举例:
输入: arr[] = {4，5，1，8} 输出: 5 如果我们选择对(5，8)，那么 5 % 8 给我们 5 这是最大可能。 输入: arr[] = {7，7，8，8，1} 输出: 7
方法:既然我们可以选择任何一对，因此 arr[i] 应该是数组的第二个最大值， arr[j] 应该是最大元素，以便最大化所需值。因此，数组的第二个最大值将是我们的答案。如果不存在任何第二大数字，那么 0 将是答案。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns the second largest
// element in the array if exists, else 0
int getMaxValue(int arr[], int arr_size)
{
int i, first, second;
// There must be at least two elements
if (arr_size < 2) {
return 0;
}
// To store the maximum and the second
// maximum element from the array
first = second = INT_MIN;
for (i = 0; i < arr_size; i++) {
// If current element is greater than first
// then update both first and second
if (arr[i] > first) {
second = first;
first = arr[i];
}
// If arr[i] is in between first and
// second then update second
else if (arr[i] > second && arr[i] != first)
second = arr[i];
}
// No second maximum found
if (second == INT_MIN)
return 0;
else
return second;
}
// Driver code
int main()
{
int arr[] = { 4, 5, 1, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << getMaxValue(arr, n);
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
{
// Function that returns the second largest
// element in the array if exists, else 0
static int getMaxValue(int arr[], int arr_size)
{
int i, first, second;
// There must be at least two elements
if (arr_size < 2)
{
return 0;
}
// To store the maximum and the second
// maximum element from the array
first = second = Integer.MIN_VALUE;
for (i = 0; i < arr_size; i++)
{
// If current element is greater than first
// then update both first and second
if (arr[i] > first)
{
second = first;
first = arr[i];
}
// If arr[i] is in between first and
// second then update second
else if (arr[i] > second && arr[i] != first)
{
second = arr[i];
}
}
// No second maximum found
if (second == Integer.MIN_VALUE)
{
return 0;
}
else
{
return second;
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = {4, 5, 1, 8};
int n = arr.length;
System.out.println(getMaxValue(arr, n));
}
}
// This code has been contributed by 29AjayKumar
Python 3
import sys
# Python 3 implementation of the approach
# Function that returns the second largest
# element in the array if exists, else 0
def getMaxValue(arr,arr_size):
# There must be at least two elements
if (arr_size < 2):
return 0
# To store the maximum and the second
# maximum element from the array
first = -sys.maxsize-1
second = -sys.maxsize-1
for i in range(arr_size):
# If current element is greater than first
# then update both first and second
if (arr[i] > first):
second = first
first = arr[i]
# If arr[i] is in between first and
# second then update second
elif (arr[i] > second and arr[i] != first):
second = arr[i]
# No second maximum found
if (second == -sys.maxsize-1):
return 0
else:
return second
# Driver code
if __name__ == '__main__':
arr = [4, 5, 1, 8]
n = len(arr)
print(getMaxValue(arr, n))
# This code is contributed by
# Surendra_Gangwar
C
// C# implementation of the approach
using System;
class GFG
{
// Function that returns the second largest
// element in the array if exists, else 0
static int getMaxValue(int []arr,
int arr_size)
{
int i, first, second;
// There must be at least two elements
if (arr_size < 2)
{
return 0;
}
// To store the maximum and the second
// maximum element from the array
first = second = int.MinValue;
for (i = 0; i < arr_size; i++)
{
// If current element is greater than first
// then update both first and second
if (arr[i] > first)
{
second = first;
first = arr[i];
}
// If arr[i] is in between first and
// second then update second
else if (arr[i] > second &&
arr[i] != first)
{
second = arr[i];
}
}
// No second maximum found
if (second == int.MinValue)
{
return 0;
}
else
{
return second;
}
}
// Driver code
public static void Main()
{
int []arr = {4, 5, 1, 8};
int n = arr.Length;
Console.Write(getMaxValue(arr, n));
}
}
// This code is contributed by Akanksha Rai
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function that returns the second largest
// element in the array if exists, else 0
function getMaxValue($arr, $arr_size)
{
// There must be at least two elements
if ($arr_size < 2)
{
return 0;
}
// To store the maximum and the second
// maximum element from the array
$first = $second = -(PHP_INT_MAX - 1);
for ($i = 0; $i < $arr_size; $i++)
{
// If current element is greater than first
// then update both first and second
if ($arr[$i] > $first)
{
$second = $first;
$first = $arr[$i];
}
// If arr[i] is in between first and
// second then update second
else if ($arr[$i] > $second &&
$arr[$i] != $first)
$second = $arr[$i];
}
// No second maximum found
if ($second == -(PHP_INT_MAX-1))
return 0;
else
return $second;
}
// Driver code
$arr = array(4, 5, 1, 8);
$n = count($arr);
echo getMaxValue($arr, $n);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// JavaScript implementation of the approach
// Function that returns the second largest
// element in the array if exists, else 0
function getMaxValue(arr, arr_size)
{
let i, first, second;
// There must be at least two elements
if (arr_size < 2)
{
return 0;
}
// To store the maximum and the second
// maximum element from the array
first = second = Number.MIN_VALUE;
for (i = 0; i < arr_size; i++)
{
// If current element is greater than first
// then update both first and second
if (arr[i] > first)
{
second = first;
first = arr[i];
}
// If arr[i] is in between first and
// second then update second
else if (arr[i] > second && arr[i] != first)
{
second = arr[i];
}
}
// No second maximum found
if (second == Number.MIN_VALUE)
{
return 0;
}
else
{
return second;
}
}
// Driver Code
let arr = [4, 5, 1, 8];
let n = arr.length;
document.write(getMaxValue(arr, n));
</script>
Output:
5
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