计算系列K，K + 1，K + 2，K + 3，K + 4，…，K + N的所有可能的唯一和

给定一个数N以及对于任何数K，对于形成为K，K + 1，K + 2，K + 3，K + 4，……，K + N的序列。任务是找到可以通过将N +1个整数序列中的两个或多个整数相加而获得的唯一和的数量。

例子：

Input: N = 3 
Output: 10 
Explanation: 
The possible unique combinations are: 
(K) + (K + 1) = 2*K + 1 
(K) + (K + 2) = 2*K + 2 
(K) + (K + 3) = 2*K + 3 
(K + 1) + (K + 3) = 2*K + 4 
(K + 2) + (K + 3) = 2*K + 5 
(K) + (K + 1) + (K + 2) = 3*K + 3 
(K) + (K + 1) + (K + 3) = 3*K + 4 
(K) + (K + 2) + (K + 3) = 3*K + 5 
(K + 1) + (K + 2) + (K + 3) = 3*K + 6 
(K) + (K + 1) + (K + 2) + (K + 3) = 4*K + 6 
So in total, there are 10 unique ways. 
Input: N = 4 
Output: 20 
Explanation: 
The possible unique combinations are 20 in number. 
 

处理方法：进行以下观察：

• 由于K是数字，所以它唯一具有的意义是大小不同的两个子集不能具有相同的总和。
• 可以从子集的最小可能总和到子集的最大可能总和获得唯一的总和，其中子集的大小为X ，其中(2≤X≤(N +1))。

For Example:N = 4
The Series is = {K, K + 1, K + 2, K + 3, K + 4}
For K = 2, minimum_sum = (K, K + 1) = 2*K + 1
and maximum_sum = (K + 3, K + 4) = 2*K + 7
The maximum distinct sums possible with K = 2 are (maximum_sum – minimum_sum + 1) = (7 – 1 + 1) = 7.

请按照以下步骤解决问题：

1. 初始化两个数组，数组fsum []和rsum []的大小均为N +1至0 。
2. 对于这两种阵列FSUM []和率R sum []，更新FSUM的每个元素[I]与AR [I] + FSUM [I – 1]用Ar和率R sum [I] [I] + FSUM第[i + 1]。
3. 将变量ans初始化为1，该变量存储给定序列的不同可能和的计数。
4. 对于每个可能的子集大小X ，其中( 2≤X≤ (N + 1))，将值1 + rsum [n +1-k] + fsum [k]加到ans 。
5. ans的值是必需的答案，因此将其返回。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the unique sum
int count_unique_sum(int n)
{
 
    int i, ar[n + 1], fsum[n + 1];
    int rsum[n + 1], ans = 1;
 
    // Initialize array fsum[] with 0
    memset(fsum, 0, sizeof fsum);
 
    // Initialize array rsum[] with 0
    memset(rsum, 0, sizeof rsum);
 
    for (i = 0; i <= n; i++) {
        ar[i] = i;
    }
 
    // Set fsum[0] as ar[0]
    fsum[0] = ar[0];
 
    // Set rsum[0] as ar[n]
    rsum[n] = ar[n];
 
    // For each i update fsum[i] with
    // ar[i] + fsum[i - 1]
    for (i = 1; i <= n; i++) {
        fsum[i] = ar[i] + fsum[i - 1];
    }
 
    // For each i from n-1, update
    // rsum[i] with ar[i] + fsum[i + 1]
    for (i = n - 1; i >= 0; i--) {
        rsum[i] = ar[i] + rsum[i + 1];
    }
 
    // K represent size of subset as
    // explained above
    for (int k = 2; k <= n; k++) {
 
        // Using above relation
        ans += 1 + rsum[n + 1 - k]
               - fsum[k - 1];
    }
 
    // Return the result
    return ans;
}
 
// Driver Code
int main()
{
    // Given a number N
    int N = 4;
 
    // Function Call
    cout << count_unique_sum(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
   
// Function to count the unique sum
static int count_unique_sum(int n)
{
    int i;
    int ar[] = new int[n + 1];
    int fsum[] = new int[n + 1];
    int rsum[] = new int[n + 1];
    int ans = 1;
   
    // Initialize array fsum[] with 0
    Arrays.fill(fsum, 0);
  
    // Initialize array rsum[] with 0
    Arrays.fill(rsum, 0);
  
    for (i = 0; i <= n; i++)
    {
        ar[i] = i;
    }
  
    // Set fsum[0] as ar[0]
    fsum[0] = ar[0];
  
    // Set rsum[0] as ar[n]
    rsum[n] = ar[n];
  
    // For each i update fsum[i] with
    // ar[i] + fsum[i - 1]
    for (i = 1; i <= n; i++)
    {
        fsum[i] = ar[i] + fsum[i - 1];
    }
  
    // For each i from n-1, update
    // rsum[i] with ar[i] + fsum[i + 1]
    for (i = n - 1; i >= 0; i--)
    {
        rsum[i] = ar[i] + rsum[i + 1];
    }
  
    // K represent size of subset as
    // explained above
    for (int k = 2; k <= n; k++)
    {
  
        // Using above relation
        ans += 1 + rsum[n + 1 - k] -
                     fsum[k - 1];
    }
  
    // Return the result
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    // Given a number N
    int N = 4;
  
    // Function Call
    System.out.print(count_unique_sum(N));
}
}
 
// This code is contributed by rock__cool

Python3

# Python3 program for the above approach
 
# Function to count the unique sum
def count_unique_sum(n):
 
    ar = [0] * (n + 1)
    fsum = [0] * (n + 1)
    rsum = [0] * (n + 1)
    ans = 1
 
    for i in range(0, n + 1):
        ar[i] = i
     
    # Set fsum[0] as ar[0]
    fsum[0] = ar[0]
 
    # Set rsum[0] as ar[n]
    rsum[n] = ar[n]
 
    # For each i update fsum[i] with
    # ar[i] + fsum[i - 1]
    for i in range(1, n + 1):
        fsum[i] = ar[i] + fsum[i - 1]
 
    # For each i from n-1, update
    # rsum[i] with ar[i] + fsum[i + 1]
    for i in range(n - 1, -1, -1):
        rsum[i] = ar[i] + rsum[i + 1]
     
    # K represent size of subset as
    # explained above
    for k in range(2, n + 1):
 
        # Using above relation
        ans += (1 + rsum[n + 1 - k] -
                    fsum[k - 1])
     
    # Return the result
    return ans
 
# Driver Code
 
# Given a number N
N = 4
 
# Function call
print(count_unique_sum(N))
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
class GFG{
   
// Function to count the unique sum
static int count_unique_sum(int n)
{
    int i;
    int []ar = new int[n + 1];
    int []fsum = new int[n + 1];
    int []rsum = new int[n + 1];
    int ans = 1;
  
    for (i = 0; i <= n; i++)
    {
        ar[i] = i;
    }
  
    // Set fsum[0] as ar[0]
    fsum[0] = ar[0];
  
    // Set rsum[0] as ar[n]
    rsum[n] = ar[n];
  
    // For each i update fsum[i] with
    // ar[i] + fsum[i - 1]
    for (i = 1; i <= n; i++)
    {
        fsum[i] = ar[i] + fsum[i - 1];
    }
  
    // For each i from n-1, update
    // rsum[i] with ar[i] + fsum[i + 1]
    for (i = n - 1; i >= 0; i--)
    {
        rsum[i] = ar[i] + rsum[i + 1];
    }
  
    // K represent size of subset as
    // explained above
    for (int k = 2; k <= n; k++)
    {
  
        // Using above relation
        ans += 1 + rsum[n + 1 - k] -
                   fsum[k - 1];
    }
  
    // Return the result
    return ans;
}
  
// Driver Code
public static void Main(String[] args)
{
    // Given a number N
    int N = 4;
  
    // Function Call
    Console.Write(count_unique_sum(N));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

    // Javascript program for the above approach
     
    // Function to count the unique sum
    function count_unique_sum(n)
    {
 
        let i;
        let ans = 1;
        let ar = new Array(n + 1);
        let fsum = new Array(n + 1);
        let rsum = new Array(n + 1);
 
        // Initialize array fsum[] with 0
        fsum.fill(0);
 
        // Initialize array rsum[] with 0
        rsum.fill(0);
 
        for (i = 0; i <= n; i++) {
            ar[i] = i;
        }
 
        // Set fsum[0] as ar[0]
        fsum[0] = ar[0];
 
        // Set rsum[0] as ar[n]
        rsum[n] = ar[n];
 
        // For each i update fsum[i] with
        // ar[i] + fsum[i - 1]
        for (i = 1; i <= n; i++) {
            fsum[i] = ar[i] + fsum[i - 1];
        }
 
        // For each i from n-1, update
        // rsum[i] with ar[i] + fsum[i + 1]
        for (i = n - 1; i >= 0; i--) {
            rsum[i] = ar[i] + rsum[i + 1];
        }
 
        // K represent size of subset as
        // explained above
        for (let k = 2; k <= n; k++) {
 
            // Using above relation
            ans += 1 + rsum[n + 1 - k]
                   - fsum[k - 1];
        }
 
        // Return the result
        return ans;
    }
     
    // Given a number N
    let N = 4;
   
    // Function Call
    document.write(count_unique_sum(N));
 
//This code is contribute by suresh07.

输出

20

时间复杂度： O(N)
辅助空间： O(N)