两个 1 到 N 值数组之间不同对和的计数

给定一个正整数N ，使得存在两个数组a[]和b[]，每个数组都包含值 {1, 2, 3, .., N}，任务是找到所有对 (a[i], b[j]) 使得 a[i] + b[j] 在所有对中是唯一的，即如果两对的总和相等，那么结果中只会计算一个。
例子：

Input: N = 2 
Output: 3 
Explanation: 
a[] = {1, 2}, b[] = {1, 2} 
The three possible pairs are (a[0], b[0]), (a[1], b[0]) and (a[1], b[1]). 
Pair 1: 1 + 1 = 2 
Pair 2: 2 + 1 = 3 
Pair 3: 2 + 2 = 4
Input: N = 3 
Output: 5 
a[] = {1, 2, 3}, b[] = {1, 2, 3} 
The possible pairs with distinct sum are: 
Pair 1: 1 + 1 = 2 
Pair 2: 2 + 1 = 3 
Pair 3: 2 + 2 = 4 
Pair 4: 3 + 2 = 5 
Pair 5: 3 + 3 = 6 
 

幼稚的方法：
为了解决上面提到的问题，最简单的方法是使用一个 Set 来存储 {1, 2, 3, .. N} 和 {1, 2, 3, .. N} 的不同和，使用两个循环。
下面是上述方法的实现：

C++

// C++ implementation to count
// of distinct pair sum between
// two Array with values 1 to N
 
#include 
using namespace std;
 
// Function to find the distinct sums
int findDistinctSums(int n)
{
    // Set to store distinct sums
    set s;
 
    for (int i = 1; i <= n; i++) {
        for (int j = i; j <= n; j++) {
 
            // Inserting every sum
            s.insert(i + j);
        }
    }
 
    // returning distinct sums
    return s.size();
}
 
// Driver code
int main()
{
 
    int N = 3;
 
    cout << findDistinctSums(N);
 
    return 0;
}

Java

// Java implementation to count
// of distinct pair sum between
// two Array with values 1 to N
import java.util.*;
 
class GFG{
 
// Function to find the distinct sums
static int findDistinctSums(int n)
{
     
    // Set to store distinct sums
    HashSet s = new HashSet<>();
 
    for(int i = 1; i <= n; i++)
    {
        for(int j = i; j <= n; j++)
        {
             
            // Inserting every sum
            s.add(i + j);
        }
    }
     
    // Returning distinct sums
    return s.size();
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
 
    System.out.print(findDistinctSums(N));
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 implementation to count
# of distinct pair sum between
# two Array with values 1 to N
 
# Function to find the distinct sums
def findDistinctSums(n):
 
    # Set to store distinct sums
    s = set()
 
    for i in range(1, n + 1):
        for j in range(i, n + 1):
 
            # Inserting every sum
            s.add(i + j)
         
    # Returning distinct sums
    return len(s)
 
# Driver code
N = 3
print(findDistinctSums(N))
     
# This code is contributed by divyamohan123

C#

// C# implementation to count
// of distinct pair sum between
// two Array with values 1 to N
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the distinct sums
static int findDistinctSums(int n)
{
     
    // Set to store distinct sums
    HashSet s = new HashSet();
 
    for(int i = 1; i <= n; i++)
    {
        for(int j = i; j <= n; j++)
        {
             
            // Inserting every sum
            s.Add(i + j);
        }
    }
     
    // Returning distinct sums
    return s.Count;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 3;
 
    Console.Write(findDistinctSums(N));
}
}
 
// This code is contributed by gauravrajput1

Javascript

 
// Javascript implementation to count
// of distinct pair sum between
// two Array with values 1 to N
 
// Function to find the distinct sums
function findDistinctSums(n)
{
    // Set to store distinct sums
    s = new Set();
 
    for (var i = 1; i <= n; i++) {
        for (var j = i; j <= n; j++) {
 
            // Inserting every sum
            s.add(i + j);
        }
    }
 
    // returning distinct sums
    return s.size;
}
 
// Driver code
var N = 3;
document.write( findDistinctSums(N));
 

输出：

5

高效方法：优化上述方法：

1. 观察到为 {1, 2, 3, .., N} 和 {1, 2, 3, .., N} 中的不同和的计数而形成的系列给出为1, 3, 5, 7, …
2. 因此上述系列的第 N 项 = 2 * N – 1
3. 因此不同对和的计数可以计算为2 * N – 1

下面是上述方法的实现：

C++

// C++ implementation to find count
// of distinct pair sum between
// two 1 to N value Arrays
 
#include 
using namespace std;
 
// Function to find the distinct sums
int findDistinctSums(int N)
{
    return (2 * N - 1);
}
 
// Driver code
int main()
{
 
    int N = 3;
 
    cout << findDistinctSums(N);
 
    return 0;
}

Java

// Java implementation to find count
// of distinct pair sum between
// two 1 to N value Arrays
import java.util.*;
class GFG{
 
// Function to find the distinct sums
static int findDistinctSums(int N)
{
    return (2 * N - 1);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
 
    System.out.print(findDistinctSums(N));
}
}
 
// This code is contributed by shivanisinghss2110

蟒蛇3

# Python3 implementation to find count
# of distinct pair sum between
# two 1 to N value Arrays
 
# Function to find the distinct sums
def findDistinctSums(N):
 
    return (2 * N - 1)
 
# Driver code
N = 3
print(findDistinctSums(N))
 
# This code is contributed by divyamohan123

C＃

// C# implementation to find count
// of distinct pair sum between
// two 1 to N value Arrays
using System;
class GFG{
 
// Function to find the distinct sums
static int findDistinctSums(int N)
{
    return (2 * N - 1);
}
 
// Driver code
public static void Main()
{
    int N = 3;
 
    Console.Write(findDistinctSums(N));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
    // Javascript implementation to find count
    // of distinct pair sum between
    // two 1 to N value Arrays
     
    // Function to find the distinct sums
    function findDistinctSums(N)
    {
        return (2 * N - 1);
    }
     
    let N = 3;
  
    document.write(findDistinctSums(N));
 

输出：

5

时间复杂度： O(1)
辅助空间复杂度： O(1)

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