检查前半部分的数组总和是否可以被另一半的总和整除，反之亦然

给定一个大小为N的数组arr[] ，任务是检查左子数组的总和是否可以被右子数组的总和整除，反之亦然。如果是则打印Yes ，否则打印No 。这里，左子数组将包含从0到mid=(N-1)/2的字符串，右子数组将包含从mid+1到N-1的字符串。

例子：

Input: arr[] = [1, 2, 3, 4, 5]
Output: No
Explanation:
Sum of left subarray: 1+2+3=6
Sum of right subarray: 4+5=9
So, the sum of neither of them is divisible by the other one.

Input: arr[] = [4, 5, 6, 1, 2, 2]
Output: Yes
Explanation:
Sum of left subarray: 4+5+6=15
Sum of right subarray: 1+2+2=5
So, the sum of left subarray is divisible by the sum of right subarray

方法：按照以下步骤，解决这个问题：

1. 创建两个变量sumL和sumR分别存储左子数组和右子数组的和。用0初始化它们。
2. 现在，从0到N-1运行一个循环，对于0到(N-1)/2范围内的迭代，将元素添加到sumL中。对于(N-1)/2之后的迭代，添加sumR中的元素。
3. 循环结束后，检查 sumL 是否可以被sumR整除，或者sumR是否可以被sumL整除。如果满足这些条件中的任何一个，则打印Yes否则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 the sum of left subarray
// is divisible by the sum of right or vice - versa
bool isDivisible(int* arr, int N)
{
 
    // Variables to store the sum of
    // left and right subarrays
    int sumL = 0, sumR = 0;
 
    for (int i = 0; i < N; ++i) {
        if (i <= (N - 1) / 2) {
            sumL += arr[i];
        }
        else {
            sumR += arr[i];
        }
    }
 
    // If Divisible
    if (sumL % sumR == 0
        or sumR % sumL == 0) {
        return 1;
    }
 
    return 0;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 5, 6, 1, 2, 2 };
    int N = sizeof(arr) / sizeof(int);
 
    if (isDivisible(arr, N))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program for the above approach
class GFG {
 
    // Function to check if the sum of left subarray
    // is divisible by the sum of right or vice - versa
    public static boolean isDivisible(int[] arr, int N)
    {
 
        // Variables to store the sum of
        // left and right subarrays
        int sumL = 0, sumR = 0;
 
        for (int i = 0; i < N; ++i) {
            if (i <= (N - 1) / 2) {
                sumL += arr[i];
            } else {
                sumR += arr[i];
            }
        }
 
        // If Divisible
        if (sumL % sumR == 0 || sumR % sumL == 0) {
            return true;
        }
 
        return false;
    }
 
    // Driver Code
    public static void main(String args[]) {
        int[] arr = { 4, 5, 6, 1, 2, 2 };
        int N = arr.length;
 
        if (isDivisible(arr, N))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# Python3 program for the above approach
 
# Function to check if the sum of left subarray
# is divisible by the sum of right or vice - versa
def isDivisible(arr, N) :
 
    # Variables to store the sum of
    # left and right subarrays
    sumL = 0; sumR = 0;
 
    for i in range(N) :
        if (i <= (N - 1) // 2) :
            sumL += arr[i];
         
        else :
            sumR += arr[i];
 
    # If Divisible
    if (sumL % sumR == 0 or sumR % sumL == 0) :
        return 1;
 
    return 0;
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 4, 5, 6, 1, 2, 2 ];
    N = len(arr);
 
    if (isDivisible(arr, N)) :
        print("Yes");
    else :
        print("No");
 
    # This code is contributed by AnkThon

C#

// C# program for the above approach
using System;
class GFG {
 
    // Function to check if the sum of left subarray
    // is divisible by the sum of right or vice - versa
    public static bool isDivisible(int[] arr, int N)
    {
 
        // Variables to store the sum of
        // left and right subarrays
        int sumL = 0, sumR = 0;
 
        for (int i = 0; i < N; ++i) {
            if (i <= (N - 1) / 2) {
                sumL += arr[i];
            } else {
                sumR += arr[i];
            }
        }
 
        // If Divisible
        if (sumL % sumR == 0 || sumR % sumL == 0) {
            return true;
        }
 
        return false;
    }
 
    // Driver Code
    public static void Main() {
        int[] arr = { 4, 5, 6, 1, 2, 2 };
        int N = arr.Length;
 
        if (isDivisible(arr, N))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
 
// This code is contributed by gfgking.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to check if the sum of left subarray
        // is divisible by the sum of right or vice - versa
        function isDivisible(arr, N) {
 
            // Variables to store the sum of
            // left and right subarrays
            let sumL = 0, sumR = 0;
 
            for (let i = 0; i < N; ++i) {
                if (i <= (N - 1) / 2) {
                    sumL += arr[i];
                }
                else {
                    sumR += arr[i];
                }
            }
 
            // If Divisible
            if (sumL % sumR == 0
                || sumR % sumL == 0) {
                return 1;
            }
 
            return 0;
        }
 
        // Driver Code
 
        let arr = [4, 5, 6, 1, 2, 2];
        let N = arr.length;
 
        if (isDivisible(arr, N))
            document.write("Yes");
        else
            document.write("No");
 
    // This code is contributed by Potta Lokesh
    

输出

Yes

时间复杂度： O(N)
辅助空间： O(1)