检查数组中所有可能对 (i, j) 的 arr[i] / j 和是否为 0

给定一个由N 个整数组成的数组arr[] ，任务是检查所有对(i, j)的(arr[i] / j)的所有可能值的总和是否满足0 < i ≤ j < (N – 1)是否为0 。如果发现是真的，则打印“是” 。否则，打印“否” 。

例子：

Input: arr[] = {1, -1, 3, -2, -1}
Output: Yes
Explanation:
For all possible pairs (i, j), such that 0 < i <= j < (N – 1), required sum = 1/1 + -1/2 + 3/3 + -2/4 + -1/5 + -1/2 + 3/3 + -2/4 + -1/5 + 3/3 + -2/4 + -1/5 + -2/ 4 + -1/5 + -1/5 = 0.

Input: arr[] = {1, 2, 3, 4, 5}
Output: No

方法：根据以下观察可以解决给定的问题：

• 对于[0, N – 1]范围内i的每个可能值以及j 的每个可能值，以下是表达式：
  • j = 1： $\frac{a_1}{1} + \frac{a_2}{2} + \frac{a_3}{3} .... +\frac{a_n}{n}$
  • j = 2： $\frac{a_2}{2} + \frac{a_3}{3} .... +\frac{a_n}{n}$
  • j = 3： $\frac{a_3}{3} .... +\frac{a_n}{n}$第三行等等……
• 因此，上述所有表达式的总和由下式给出：

=> $\frac{a_1}{1} + \frac{2*a_2}{2} + \frac{3*a_3}{3} ...+ \frac{n*a_n}{n}$

=> $a_1 + a_2 + a_3 + ... + a_n$

根据上述观察，如果数组的总和为0 ，则打印Yes 。否则，打印No 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if sum of all
// values of (arr[i]/j) for all
// 0 < i <= j < (N - 1) is 0 or not
void check(int arr[], int N)
{
    // Stores the required sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++)
        sum += arr[i];
 
    // If the sum is equal to 0
    if (sum == 0)
        cout << "Yes";
 
    // Otherwise
    else
        cout << "No";
}
 
// Driver Code
int main()
{
    int arr[] = { 1, -1, 3, -2, -1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    check(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to check if sum of all
// values of (arr[i]/j) for all
// 0 < i <= j < (N - 1) is 0 or not
static void check(int arr[], int N)
{
     
    // Stores the required sum
    int sum = 0;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
        sum += arr[i];
 
    // If the sum is equal to 0
    if (sum == 0)
        System.out.println("Yes");
 
    // Otherwise
    else
        System.out.println("No");
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, -1, 3, -2, -1 };
    int N = arr.length;
     
    check(arr, N);
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
 
# Function to check if sum of all
# values of (arr[i]/j) for all
# 0 < i <= j < (N - 1) is 0 or not
def check(arr, N):
     
    # Stores the required sum
    sum = 0
 
    # Traverse the array
    for i in range(N):
        sum += arr[i]
 
    # If the sum is equal to 0
    if (sum == 0):
        print("Yes")
 
    # Otherwise
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, -1, 3, -2, -1 ]
    N = len(arr)
     
    check(arr, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG {
 
    // Function to check if sum of all
    // values of (arr[i]/j) for all
    // 0 < i <= j < (N - 1) is 0 or not
    static void check(int[] arr, int N)
    {
 
        // Stores the required sum
        int sum = 0;
 
        // Traverse the array
        for (int i = 0; i < N; i++)
            sum += arr[i];
 
        // If the sum is equal to 0
        if (sum == 0)
            Console.WriteLine("Yes");
 
        // Otherwise
        else
            Console.WriteLine("No");
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 1, -1, 3, -2, -1 };
        int N = arr.Length;
 
        check(arr, N);
    }
}
 
// This code is contributed by ukasp.

Javascript

// javascript program for the above approach
    // Function to check if sum of all
    // values of (arr[i]/j) for all
    // 0 < i <= j < (N - 1) is 0 or not
    function check(arr , N) {
 
        // Stores the required sum
        var sum = 0;
 
        // Traverse the array
        for (i = 0; i < N; i++)
            sum += arr[i];
 
        // If the sum is equal to 0
        if (sum == 0)
            document.write("Yes");
 
        // Otherwise
        else
            document.write("No");
    }
 
    // Driver Code
     
        var arr = [ 1, -1, 3, -2, -1 ];
        var N = arr.length;
 
        check(arr, N);
 
// This code contributed by umadevi9616

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live