检查左半部分的数字总和是否可以在N的最大排列中被右半部分的数字总和整除

给定一个正整数N ，任务是通过重新排列数字来最大化整数N并检查左半数的总和是否可以被右半数的总和整除。如果发现是真的，则打印“是” 。否则，打印“否” 。

If the number of digits(say D) in the given number N is odd, then consider any of the two possibilities:

Consider the first (D/2) digit in the left half.
Consider the first (D/2 + 1) digit in the left half.

例子：

Input: N = 43729
Output: Yes
Explanation:
Maximum value of N possible by rearranging the digits is 97432.
Case 1:
Sum of digits in the left half = 9 + 7 = 16.
Sum of digits in the right half = 4 + 3 + 2 = 9.
Since, 16 is not divisible by 9, the condition is not satisfied.
Case 2:
Sum of digits in the left half = 9 + 7 + 4 = 20.
Sum of digits in the right half = 3 + 2 = 5.
Since, 20 is divisible by 5, the condition is satisfied.

Input: N = 3746
Output: No

编程需要懂一点英语

方法：给定的问题可以通过将数字N的给定数字按降序排序，最大化N的值，然后检查左右半数之和的可分性来解决。请按照以下步骤解决问题：

初始化一个变量，比如D ，它存储给定数字N的位数。
将给定的数字作为字符串存储在变量中，比如temp 。
按降序对字符串temp进行排序。
现在检查前(D/2)位数字的总和是否可以被最后(D/2)位数字的总和整除，然后打印“Yes” 。否则，打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// Function to check if sum of digits
// in left half is divisible by the sum
// of digits in the right half or not
void checkDivisible(int N)
{
 
    // Convert N to its equivalent string
    string temp = to_string(N);
 
    // Count of digits
    int D = temp.length();
 
    // Sort the digits in decreasing order
    sort(temp.begin(), temp.end(),
         greater());
 
    int leftSum = 0, rightSum = 0;
 
    // Find the sum of digits
    for (int i = 0; temp[i]; i++) {
        rightSum += (temp[i] - '0');
    }
 
    for (int i = 0; i < D / 2; i++) {
 
        // Update the leftSum and the
        // rightSum
        leftSum += (temp[i] - '0');
        rightSum -= (temp[i] - '0');
    }
 
    if (D & 1) {
 
        // Consider Case 1: Check divisibility
        if (leftSum % rightSum == 0) {
            cout << "Yes";
        }
        else {
 
            leftSum += (temp[D / 2] - '0');
            rightSum -= (temp[D / 2] - '0');
 
            // Consider Case 2: Check divisibility
            if (leftSum % rightSum == 0) {
                cout << "Yes";
            }
            else {
                cout << "No";
            }
        }
    }
 
    else {
 
        // Check divisibility
        if (leftSum % rightSum == 0) {
            cout << "Yes";
        }
        else {
            cout << "No";
        }
    }
}
 
// Driver Code
int main()
{
    int N = 43729;
    checkDivisible(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
   
// Utility function to sort array in
// descending order
static void descOrder(char[] s)
{
    Arrays.sort(s);
    reverse(s);
}
 
// Utility function to reverse an array
static void reverse(char[] a)
{
    int i, n = a.length;
    char t;
    for (i = 0; i < n / 2; i++)
    {
    t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
}
     
// Function to check if sum of digits
// in left half is divisible by the sum
// of digits in the right half or not
static void checkDivisible(int N)
{
 
    // Convert N to its equivalent string
    String temp = Integer.toString(N);
    char []arr = temp.toCharArray();
 
    // Count of digits
    int D = arr.length;
 
    // Sort the digits in decreasing order
    descOrder(arr);
 
    int leftSum = 0, rightSum = 0;
 
    // Find the sum of digits
    for (int i = 0; i < D; i++) {
        rightSum += (arr[i] - '0');
    }
 
    for (int i = 0; i < D / 2; i++) {
 
        // Update the leftSum and the
        // rightSum
        leftSum += (arr[i] - '0');
        rightSum -= (arr[i] - '0');
    }
 
    if (D%2 == 1) {
 
        // Consider Case 1: Check divisibility
        if (leftSum % rightSum == 0) {
            System.out.print("Yes");
        }
        else {
 
            leftSum += (arr[D / 2] - '0');
            rightSum -= (arr[D / 2] - '0');
 
            // Consider Case 2: Check divisibility
            if (leftSum % rightSum == 0) {
                System.out.print("Yes");
            }
            else {
                System.out.print("No");
            }
        }
    }
 
    else {
 
        // Check divisibility
        if (leftSum % rightSum == 0) {
            System.out.print("Yes");
        }
        else {
            System.out.print("No");
        }
    }
}
 
// Driver Code
public static void main(String args[])
{
    int N = 43729;
    checkDivisible(N);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
 
# Function to check if sum of digits
# in left half is divisible by the sum
# of digits in the right half or not
def checkDivisible(N):
 
    # Convert N to its equivalent string
    temp = str(N)
 
    # Count of digits
    D = len(temp)
 
    # Sort the digits in decreasing order
    temp = ''.join(sorted(temp, reverse = True))
 
    leftSum,rightSum = 0,0
 
    # Find the sum of digits
    for i in range(D):
        rightSum += ord(temp[i]) - ord('0')
 
    for i in range(D // 2):
 
        # Update the leftSum and the
        # rightSum
        leftSum += ord(temp[i]) - ord('0')
        rightSum -= ord(temp[i]) - ord('0')
 
    if (D % 2 == 1):
 
        # Consider Case 1: Check divisibility
        if (leftSum % rightSum == 0):
            print("Yes")
        else:
 
            leftSum += ord(temp[D // 2]) - ord('0')
            rightSum -= ord(temp[D // 2]) - ord('0')
 
            # Consider Case 2: Check divisibility
            if (leftSum % rightSum == 0):
                print("Yes")
            else:
                print("No")
 
    else:
 
        # Check divisibility
        if (leftSum % rightSum == 0):
            print("Yes")
        else:
            print("No")
 
# Driver Code
N = 43729
checkDivisible(N)
 
# This code is contributed by shinjanpatra

C#

// C# program for the above approach
using System;
using System.Collections;
 
public class GFG
{
// Utility function to sort array in
// descending order
static void descOrder(char []s)
{
    Array.Sort(s);
    reverse(s);
}
 
// Utility function to reverse an array
static void reverse(char []a)
{
    int i, n = a.Length;
    char t;
    for (i = 0; i < n / 2; i++)
    {
    t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
}
     
// Function to check if sum of digits
// in left half is divisible by the sum
// of digits in the right half or not
static void checkDivisible(int N)
{
 
    // Convert N to its equivalent string
    String temp = N.ToString();
    char []arr = temp.ToCharArray();
 
    // Count of digits
    int D = arr.Length;
 
    // Sort the digits in decreasing order
    descOrder(arr);
 
    int leftSum = 0, rightSum = 0;
 
    // Find the sum of digits
    for (int i = 0; i < D; i++) {
        rightSum += (arr[i] - '0');
    }
 
    for (int i = 0; i < D / 2; i++) {
 
        // Update the leftSum and the
        // rightSum
        leftSum += (arr[i] - '0');
        rightSum -= (arr[i] - '0');
    }
 
    if (D%2 == 1) {
 
        // Consider Case 1: Check divisibility
        if (leftSum % rightSum == 0) {
            Console.Write("Yes");
        }
        else {
 
            leftSum += (arr[D / 2] - '0');
            rightSum -= (arr[D / 2] - '0');
 
            // Consider Case 2: Check divisibility
            if (leftSum % rightSum == 0) {
                Console.Write("Yes");
            }
            else {
                Console.Write("No");
            }
        }
    }
 
    else {
 
        // Check divisibility
        if (leftSum % rightSum == 0) {
            Console.Write("Yes");
        }
        else {
            Console.Write("No");
        }
    }
}
 
// Driver Code
public static void Main()
{
    int N = 43729;
    checkDivisible(N);
}
}
// This code is contributed by Samim Hossain Mondal.

Javascript

输出：

Yes

时间复杂度： O(log ₁₀ N)
辅助空间： O(log ₁₀ N)