元素的位置等于所有先前元素的总和

给定N个正整数的数组Arr [] 。任务是找到等于所有先前元素之和的所有元素的位置。如果不存在这样的元素，则打印-1。

例子：

Input : Arr[] = {1, 2, 3, 6, 3, 15, 5}
Output :3 4 6

Here, the element at index “3” i.e. 3 is equal to the sum of preceding elements (1 + 2).
Similarly, at index 4, 6 = 1+2+3 (sum of all preceding elements).
And element at index 6 i.e. 15 = 1 + 2 + 3 + 6 + 3.

Input: Arr[] = {7, 5, 17, 25}
Output: -1

方法：
在遍历数组Arr []时，请维护一个sum变量，该变量存储元素的总和，直到i – 1为止。将总和与当前元素Arr [i]进行比较。如果相等，则将该元素的索引推入答案向量。

下面是上述方法的实现：

C++

// C++ implementation
#include 
using namespace std;
  
// function to return valid indexes
vector find_idx(int ar[], int n)
{
  
    // Vector to store the answer
    vector answer;
  
    // Initial sum would always
    // be first element
    int sum = ar[0];
  
    for (int i = 1; i < n; i++) {
  
        // Check if sum till now
        // is equal to current element
        if (sum == ar[i]) {
            answer.push_back(i + 1);
        }
  
        // Updating the sum by
        // adding the current
        // element in each
        // iteration.
        sum += ar[i];
    }
  
    return answer;
}
  
// Driver code
int main()
{
    int ar[] = { 1, 2, 3, 6, 3, 15, 5 };
    int n = sizeof(ar) / sizeof(int);
  
    vector ans = find_idx(ar, n);
  
    if (ans.size() != 0) {
        for (int i : ans) {
            cout << i << " ";
        }
    }
    else {
        cout << "-1";
    }
  
    cout << endl;
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
      
// function to return valid indexes
static Vector find_idx(int ar[], int n)
{
  
    // Vector to store the answer
    Vector answer = new Vector();
  
    // Initial sum would always
    // be first element
    int sum = ar[0];
  
    for (int i = 1; i < n; i++) 
    {
  
        // Check if sum till now
        // is equal to current element
        if (sum == ar[i]) 
        {
            answer.add(i + 1);
        }
  
        // Updating the sum by adding the 
        // current element in each iteration.
        sum += ar[i];
    }
    return answer;
}
  
// Driver code
public static void main(String[] args) 
{
    int ar[] = { 1, 2, 3, 6, 3, 15, 5 };
    int n = ar.length;
  
    Vector ans = find_idx(ar, n);
  
    if (ans.size() != 0) 
    {
        for (int i : ans) 
        {
            System.out.print(i + " ");
        }
    }
    else 
    {
        System.out.println("-1");
    }
}
} 
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the above approach
  
# function to return valid indexes 
def find_idx(ar, n) : 
  
    # Vector to store the answer 
    answer = []; 
  
    # Initial sum would always 
    # be first element 
    sum = ar[0]; 
  
    for i in range(1, n) :
  
        # Check if sum till now 
        # is equal to current element 
        if (sum == ar[i]) :
            answer.append(i + 1); 
  
        # Updating the sum by 
        # adding the current 
        # element in each 
        # iteration. 
        sum += ar[i];
  
    return answer; 
  
# Driver code 
if __name__ == "__main__" : 
  
    ar = [ 1, 2, 3, 6, 3, 15, 5 ]; 
    n = len(ar); 
  
    ans = find_idx(ar, n); 
  
    if (len(ans) != 0) :
          
        for i in ans :
            print(i, end = " "); 
              
    else :
          
        print("-1"); 
  
    print(); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
      
class GFG 
{
      
// function to return valid indexes
static List find_idx(int []ar, int n)
{
  
    // Vector to store the answer
    List answer = new List();
  
    // Initial sum would always
    // be first element
    int sum = ar[0];
  
    for (int i = 1; i < n; i++) 
    {
  
        // Check if sum till now
        // is equal to current element
        if (sum == ar[i]) 
        {
            answer.Add(i + 1);
        }
  
        // Updating the sum by adding the 
        // current element in each iteration.
        sum += ar[i];
    }
    return answer;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []ar = { 1, 2, 3, 6, 3, 15, 5 };
    int n = ar.Length;
  
    List ans = find_idx(ar, n);
  
    if (ans.Count != 0) 
    {
        foreach (int i in ans) 
        {
            Console.Write(i + " ");
        }
    }
    else
    {
        Console.WriteLine("-1");
    }
}
}
  
// This code is contributed by Princi Singh

输出：

3 4 6