不包含任何总和为 0 的子数组的子数组的计数

给定一个数组arr ，任务是找到给定数组的子数组的总数，这些子数组不包含任何元素总和为零的子数组。
例子：

Input: arr = {2, 4, -6} 
Output: 5 
Explanation: 
There are 5 subarrays which do not contain any subarray whose elements sum is equal to zero: [2], [4], [-6], [2, 4], [4, -6]
Input: arr = {10, -10, 10} 
Output: 3 
 

方法：

1. 首先将数组的所有元素存储为其前一个元素的总和。
2. 现在取两个指针，增加第二个指针并将值存储在映射中，而不会遇到相同的元素。
3. 如果遇到 map 中已经存在的元素，这意味着在两个元素总和等于 0 的指针之间存在一个子数组。
4. 现在增加第一个指针并从地图中删除元素，同时存在两个相同的元素。
5. 将答案存储在一个变量中并最终返回它。

下面是上述方法的实现：

C++

// C++ program to Count the no of subarray
// which do not contain any subarray
// whose sum of elements is equal to zero
 
#include 
using namespace std;
 
// Function that print the number of
// subarrays which do not contain any subarray
// whose elements sum is equal to 0
void numberOfSubarrays(int arr[], int n)
{
    vector v(n + 1);
    v[0] = 0;
 
    // Storing each element as sum
    // of its previous element
    for (int i = 0; i < n; i++) {
        v[i + 1] = v[i] + arr[i];
    }
 
    map mp;
 
    int begin = 0, end = 0, answer = 0;
 
    mp[0] = 1;
 
    while (begin < n) {
 
        while (end < n
               && mp.find(v[end + 1])
                      == mp.end()) {
            end++;
            mp[v[end]] = 1;
        }
 
        // Check if another same element found
        // this means a subarray exist between
        // end and begin whose sum
        // of elements is equal to 0
        answer = answer + end - begin;
 
        // Erase beginning element from map
        mp.erase(v[begin]);
 
        // Increase begin
        begin++;
    }
 
    // Print the result
    cout << answer << endl;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 2, 4, -6 };
    int size = sizeof(arr) / sizeof(arr[0]);
 
    numberOfSubarrays(arr, size);
 
    return 0;
}

Java

// Java program to Count the no of subarray
// which do not contain any subarray
// whose sum of elements is equal to zero
import java.util.*;
 
class GFG{
  
// Function that print the number of
// subarrays which do not contain any subarray
// whose elements sum is equal to 0
static void numberOfSubarrays(int arr[], int n)
{
    int []v = new int[n + 1];
    v[0] = 0;
  
    // Storing each element as sum
    // of its previous element
    for (int i = 0; i < n; i++) {
        v[i + 1] = v[i] + arr[i];
    }
  
    HashMap mp = new HashMap();
  
    int begin = 0, end = 0, answer = 0;
  
    mp.put(0, 1);
  
    while (begin < n) {
  
        while (end < n
               && !mp.containsKey(v[end + 1])) {
            end++;
            mp.put(v[end],  1);
        }
  
        // Check if another same element found
        // this means a subarray exist between
        // end and begin whose sum
        // of elements is equal to 0
        answer = answer + end - begin;
  
        // Erase beginning element from map
        mp.remove(v[begin]);
  
        // Increase begin
        begin++;
    }
  
    // Print the result
    System.out.print(answer +"\n");
}
  
// Driver Code
public static void main(String[] args)
{
  
    int arr[] = { 2, 4, -6 };
    int size = arr.length;
  
    numberOfSubarrays(arr, size);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python 3 program to Count the no of subarray
# which do not contain any subarray
# whose sum of elements is equal to zero
 
# Function that print the number of
# subarrays which do not contain any subarray
# whose elements sum is equal to 0
def numberOfSubarrays(arr, n):
 
    v = [0]*(n + 1)
 
    # Storing each element as sum
    # of its previous element
    for i in range( n):
        v[i + 1] = v[i] + arr[i]
 
    mp = {}
 
    begin, end, answer = 0 , 0 , 0
 
    mp[0] = 1
 
    while (begin < n):
 
        while (end < n
            and (v[end + 1]) not in mp):
            end += 1
            mp[v[end]] = 1
 
        # Check if another same element found
        # this means a subarray exist between
        # end and begin whose sum
        # of elements is equal to 0
        answer = answer + end - begin
 
        # Erase beginning element from map
        del mp[v[begin]]
 
        # Increase begin
        begin += 1
 
    # Print the result
    print(answer)
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 2, 4, -6 ]
    size = len(arr)
    numberOfSubarrays(arr, size)
 
# This code is contributed by chitranayal

C#

// C# program to Count the no of subarray
// which do not contain any subarray
// whose sum of elements is equal to zero
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function that print the number of
// subarrays which do not contain any subarray
// whose elements sum is equal to 0
static void numberOfSubarrays(int []arr, int n)
{
    int []v = new int[n + 1];
    v[0] = 0;
   
    // Storing each element as sum
    // of its previous element
    for (int i = 0; i < n; i++) {
        v[i + 1] = v[i] + arr[i];
    }
   
    Dictionary mp = new Dictionary();
   
    int begin = 0, end = 0, answer = 0;
   
    mp.Add(0, 1);
   
    while (begin < n) {
   
        while (end < n
               && !mp.ContainsKey(v[end + 1])) {
            end++;
            mp.Add(v[end],  1);
        }
   
        // Check if another same element found
        // this means a subarray exist between
        // end and begin whose sum
        // of elements is equal to 0
        answer = answer + end - begin;
   
        // Erase beginning element from map
        mp.Remove(v[begin]);
   
        // Increase begin
        begin++;
    }
   
    // Print the result
    Console.Write(answer +"\n");
}
   
// Driver Code
public static void Main(String[] args)
{
   
    int []arr = { 2, 4, -6 };
    int size = arr.Length;
   
    numberOfSubarrays(arr, size);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript program to Count the no of subarray
// which do not contain any subarray
// whose sum of elements is equal to zero
 
// Function that print the number of
// subarrays which do not contain any subarray
// whose elements sum is equal to 0
function numberOfSubarrays(arr, n)
{
    let v = new Array(n + 1);
    v[0] = 0;
 
    // Storing each element as sum
    // of its previous element
    for (let i = 0; i < n; i++) {
        v[i + 1] = v[i] + arr[i];
    }
 
    let mp = new Map();
 
    let begin = 0, end = 0, answer = 0;
 
    mp.set(0, 1);
 
    while (begin < n) {
 
        while (end < n && !mp.has(v[end + 1])) {
            end++;
            mp.set(v[end], 1);
        }
 
        // Check if another same element found
        // this means a subarray exist between
        // end and begin whose sum
        // of elements is equal to 0
        answer = answer + end - begin;
 
        // Erase beginning element from map
        mp.clear();
 
        // Increase begin
        begin++;
    }
 
    // Print the result
    document.write(answer + "<br>");
}
 
// Driver Code
 
let arr = [ 2, 4, -6 ];
let size = arr.length;
 
numberOfSubarrays(arr, size);
 
// This code is contributed by _saurabh_jaiswal
 

输出：

5

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