求 N 个元素数组中所有可能对的和
给定一个由 N 个整数组成的数组 arr[] ，任务是从给定的数组中找出所有可能的对的和。请注意，
- (arr[i]，arr[i]) 也被认为是有效对。
- (arr[i]、arr[j]) 和 (arr[j]、arr[i]) 被认为是两个不同的对。
示例:
输入: arr[] = {1，2} 输出: 12 所有有效对为(1，1)、(1，2)、(2，1)和(2，2)。 1 + 1 + 1 + 2 + 2 + 1 + 2 + 2 = 12
输入: arr[] = {1，2，3，1，4} 输出: 110
天真法:找出所有可能的对，计算每对元素的和。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the sum of the elements
// of all possible pairs from the array
int sumPairs(int arr[], int n)
{
// To store the required sum
int sum = 0;
// Nested loop for all possible pairs
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Add the sum of the elements
// of the current pair
sum += (arr[i] + arr[j]);
}
}
return sum;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << sumPairs(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
import java.util.*;
class GFG
{
// Function to return the sum of the elements
// of all possible pairs from the array
static int sumPairs(int arr[], int n)
{
// To store the required sum
int sum = 0;
// Nested loop for all possible pairs
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// Add the sum of the elements
// of the current pair
sum += (arr[i] + arr[j]);
}
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3};
int n = arr.length;
System.out.println(sumPairs(arr, n));
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 implementation of the approach
# Function to return the summ of the elements
# of all possible pairs from the array
def summPairs(arr, n):
# To store the required summ
summ = 0
# Nested loop for all possible pairs
for i in range(n):
for j in range(n):
# Add the summ of the elements
# of the current pair
summ += (arr[i] + arr[j])
return summ
# Driver code
arr = [1, 2, 3]
n = len(arr)
print(summPairs(arr, n))
# This code is contributed by Mohit Kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the sum of the elements
// of all possible pairs from the array
static int sumPairs(int []arr, int n)
{
// To store the required sum
int sum = 0;
// Nested loop for all possible pairs
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// Add the sum of the elements
// of the current pair
sum += (arr[i] + arr[j]);
}
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {1, 2, 3};
int n = arr.Length;
Console.WriteLine(sumPairs(arr, n));
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the sum of the elements
// of all possible pairs from the array
function sumPairs(arr, n)
{
// To store the required sum
var sum = 0;
var i, j;
// Nested loop for all possible pairs
for(i = 0; i < n; i++)
{
for(j = 0; j < n; j++)
{
// Add the sum of the elements
// of the current pair
sum += (arr[i] + arr[j]);
}
}
return sum;
}
// Driver code
var arr = [ 1, 2, 3 ];
var n = arr.length;
document.write(sumPairs(arr, n));
// This code is contributed by ipg2016107
</script>
Output:
36
时间复杂度: O(N 2 )
高效方法:可以观察到，每个元素作为一对 (x，y) 的元素之一，恰好出现 (2 * N) 次。精确地 N 倍于 x ，精确地 N 倍于 y 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the sum of the elements
// of all possible pairs from the array
int sumPairs(int arr[], int n)
{
// To store the required sum
int sum = 0;
// For every element of the array
for (int i = 0; i < n; i++) {
// It appears (2 * n) times
sum = sum + (arr[i] * (2 * n));
}
return sum;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << sumPairs(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
import java.util.*;
class GFG
{
// Function to return the sum of the elements
// of all possible pairs from the array
static int sumPairs(int arr[], int n)
{
// To store the required sum
int sum = 0;
// For every element of the array
for (int i = 0; i < n; i++)
{
// It appears (2 * n) times
sum = sum + (arr[i] * (2 * n));
}
return sum;
}
// Driver code
static public void main(String []arg)
{
int arr[] = { 1, 2, 3 };
int n = arr.length;
System.out.println(sumPairs(arr, n));
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
# Function to return the sum of the elements
# of all possible pairs from the array
def sumPairs(arr, n) :
# To store the required sum
sum = 0;
# For every element of the array
for i in range(n) :
# It appears (2 * n) times
sum = sum + (arr[i] * (2 * n));
return sum;
# Driver code
if __name__ == "__main__" :
arr = [ 1, 2, 3 ];
n = len(arr);
print(sumPairs(arr, n));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the sum of the elements
// of all possible pairs from the array
static int sumPairs(int []arr, int n)
{
// To store the required sum
int sum = 0;
// For every element of the array
for (int i = 0; i < n; i++)
{
// It appears (2 * n) times
sum = sum + (arr[i] * (2 * n));
}
return sum;
}
// Driver code
static public void Main(String []arg)
{
int []arr = { 1, 2, 3 };
int n = arr.Length;
Console.WriteLine(sumPairs(arr, n));
}
}
// This code contributed by Rajput-Ji
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the sum of the elements
// of all possible pairs from the array
function sumPairs(arr, n)
{
// To store the required sum
let sum = 0;
// For every element of the array
for (let i = 0; i < n; i++) {
// It appears (2 * n) times
sum = sum + (arr[i] * (2 * n));
}
return sum;
}
// Driver code
let arr = [ 1, 2, 3 ];
let n = arr.length;
document.write(sumPairs(arr, n));
</script>
Output:
36
时间复杂度: O(N)
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