对可以表示为至少两个连续数组元素之和的数组元素进行计数

给定一个数组A []由[1，N]范围内的N个整数组成，任务是计算可以表示为两个或多个连续数组元素之和的数组元素的数量(不明显)。

例子：

Input: a[] = {3, 1, 4, 1, 5, 9, 2, 6, 5}
Output: 5
Explanation:
The array elements satisfying the condition are: 
4 = 3 + 1 
5 = 1 + 4 or 4 + 1 
9 = 3 + 1 + 4 + 1 
6 = 1 + 5 or 1 + 4 + 1 
5 = 1 + 4 or 4 + 1

Input: a[] = {1, 1, 1, 1, 1}
Output: 0
Explanation:
No such array element exists that can be represented as the sum of two or more consecutive elements.

天真的方法：遍历每个元素的给定数组，找到所有可能的子数组的总和，并检查任何子数组的总和是否等于当前元素的总和。如果发现是真的，则增加计数。最后，打印获得的计数。
时间复杂度： O(n ³ )
辅助空间： O(1)

高效方法：请按照以下步骤优化上述方法：

• 初始化数组cnt []来存储每个数组元素的出现次数。
• 迭代长度至少为2的所有子数组，并保持当前子数组总和。
• 如果当前总和不超过N ，则将cnt [sum]添加到答案并设置cnt [sum] = 0以防止多次计数相同的元素。
• 最后，打印获得的总和。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find the number of
// array elements that can be
// represented as the sum of two
// or more consecutive array elements
int countElements(int a[], int n)
{
     
    // Stores the frequencies
    // of array elements
    int cnt[n + 1] = {0};
    memset(cnt, 0, sizeof(cnt));
     
    // Stores required count
    int ans = 0;
 
    // Update frequency of
    // each array element
    for(int i = 0; i < n; i++)
    {
        ++cnt[a[i]];
    }
     
    // Find sum of all subarrays
    for(int l = 0; l < n; ++l)
    {
        int sum = 0;
 
        for(int r = l; r < n; ++r)
        {
            sum += a[r];
 
            if (l == r)
                continue;
 
            if (sum <= n)
            {
                 
                // Increment ans by cnt[sum]
                ans += cnt[sum];
 
                // Reset cnt[sum] by 0
                cnt[sum] = 0;
            }
        }
    }
 
    // Return ans
    return ans;
}
 
// Driver Code
int main()
{
     
    // Given array
    int a[] = { 1, 1, 1, 1, 1 };
    int N = sizeof(a) / sizeof(a[0]);
 
    // Function call
    cout << countElements(a, N);
}
 
// This code is contributed by Amit Katiyar

Java

// Java Program for above approach
 
import java.util.*;
class GFG {
 
    // Function to find the number of array
    // elements that can be represented as the sum
    // of two or more consecutive array elements
    static int countElements(int[] a, int n)
    {
        // Stores the frequencies
        // of array elements
        int[] cnt = new int[n + 1];
 
        // Stores required count
        int ans = 0;
 
        // Update frequency of
        // each array element
        for (int k : a) {
            ++cnt[k];
        }
 
        // Find sum of all subarrays
        for (int l = 0; l < n; ++l) {
 
            int sum = 0;
 
            for (int r = l; r < n; ++r) {
                sum += a[r];
 
                if (l == r)
                    continue;
 
                if (sum <= n) {
 
                    // Increment ans by cnt[sum]
                    ans += cnt[sum];
 
                    // Reset cnt[sum] by 0
                    cnt[sum] = 0;
                }
            }
        }
 
        // Return ans
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array
        int[] a = { 1, 1, 1, 1, 1 };
 
        // Function call
        System.out.println(
            countElements(a, a.length));
    }
}

Python3

# Python3 program for above approach
 
# Function to find the number of array
# elements that can be represented as the sum
# of two or more consecutive array elements
def countElements(a, n):
     
    # Stores the frequencies
    # of array elements
    cnt = [0] * (n + 1)
 
    # Stores required count
    ans = 0
 
    # Update frequency of
    # each array element
    for k in a:
        cnt[k] += 1
 
    # Find sum of all subarrays
    for l in range(n):
        sum = 0
 
        for r in range(l, n):
            sum += a[r]
 
            if (l == r):
                continue
            if (sum <= n):
 
                # Increment ans by cnt[sum]
                ans += cnt[sum]
 
                # Reset cnt[sum] by 0
                cnt[sum] = 0
 
    # Return ans
    return ans
 
# Driver Code
if __name__ == '__main__':
 
    # Given array
    a = [ 1, 1, 1, 1, 1 ]
 
    # Function call
    print(countElements(a, len(a)))
 
# This code is contributed by mohit kumar 29

C#

// C# program for above approach
using System;
 
class GFG{
 
// Function to find the number of array
// elements that can be represented as the sum
// of two or more consecutive array elements
static int countElements(int[] a, int n)
{
     
    // Stores the frequencies
    // of array elements
    int[] cnt = new int[n + 1];
 
    // Stores required count
    int ans = 0;
 
    // Update frequency of
    // each array element
    foreach(int k in a)
    {
        ++cnt[k];
    }
 
    // Find sum of all subarrays
    for(int l = 0; l < n; ++l)
    {
        int sum = 0;
 
        for(int r = l; r < n; ++r)
        {
            sum += a[r];
            if (l == r)
                continue;
 
            if (sum <= n)
            {
                 
                // Increment ans by cnt[sum]
                ans += cnt[sum];
 
                // Reset cnt[sum] by 0
                cnt[sum] = 0;
            }
        }
    }
 
    // Return ans
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array
    int[] a = { 1, 1, 1, 1, 1 };
 
    // Function call
    Console.WriteLine(countElements(a, a.Length));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// Javascript program for above approach
 
// Function to find the number of array
// elements that can be represented as the sum
// of two or more consecutive array elements
function countElements(a, n)
{
     
    // Stores the frequencies
    // of array elements
    var cnt = Array(n + 1).fill(0);
 
    // Stores required count
    var ans = 0;
 
    // Update frequency of
    // each array element
    for(k = 0; k < n; k++)
    {
        cnt[a[k]]++;
    }
 
    // Find sum of all subarrays
    for(l = 0; l < n; ++l)
    {
        var sum = 0;
 
        for(r = l; r < n; ++r)
        {
            sum += a[r];
 
            if (l == r)
                continue;
 
            if (sum <= n)
            {
                 
                // Increment ans by cnt[sum]
                ans += cnt[sum];
 
                // Reset cnt[sum] by 0
                cnt[sum] = 0;
            }
        }
    }
 
    // Return ans
    return ans;
}
 
// Driver Code
 
// Given array
var a = [ 1, 1, 1, 1, 1 ];
 
// Function call
document.write(countElements(a, a.length));
 
// This code is contributed by todaysgaurav
  

输出：

0

时间复杂度： O(N ² )
辅助空间： O(N)