给定的N位数字集可以被2、3和5整除的最大数字

给定一组“ N”个数字。任务是找到我们可以从这些数字中得出的最大整数。结果数必须可被2、3和5整除。
注意：没有必要使用集合中的所有数字。另外，不允许前导零。
例子：

Input: N = 11, setOfDigits = {3, 4, 5, 4, 5, 3, 5, 3, 4, 4, 0} 
Output: 5554443330 
After arranging all the elements in a non-increasing order as 5, 5, 5, 4, 4, 4, 4, 3, 3, 3, 0. The sum of all the digit is 40. Thus when we found out that the remainder of 40, when divided by 3, is 1. Then we’ll start traversing from the end to the start and if we encounter any digit with the same remainder, which we got 4 at the position 7 will be erased it. Now the sum is 36 which is divisible 3 and the new largest number will be 5554443330, which is divisible by 2, 3, and 5.
Input: N = 1, setOfDigits = {0} 
Output: 0

方法：下面是解决此问题的分步算法：

1. 初始化向量中的数字集。
2. 仅当数字总和可被3整除且最后一位为0时，任何数字才能被2、3和5整除。
3. 检查向量中是否不存在0，因此无法创建数字，因为它不会被5整除。
4. 如果第一个元素之后为0，则以不递增的方式对向量进行排序，然后打印0。
5. 求出所有数字的总和模数乘以3，如果为1，则从末尾遍历时删除具有相同余数的第一个元素。
6. 如果没有元素具有相同的余数，则删除两个元素的余数为3-y 。
7. 将向量的所有剩余数字打印为单个整数。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
#define ll long long
 
// Function to find the largest
// integer with the given set
int findLargest(int n, vector& v)
{
 
    int flag = 0;
    ll sum = 0;
 
    // find sum of all the digits
    // look if any 0 is present or not
    for (int i = 0; i < n; i++) {
        if (v[i] == 0)
            flag = 1;
        sum += v[i];
    }
 
    // if 0 is not present, the resultant number
    // won't be divisible by 5
    if (!flag)
        cout << "Not possible" << endl;
 
    else {
        // sort all the elements in a non-decreasing manner
        sort(v.begin(), v.end(), greater());
 
        // if there is just one element 0
        if (v[0] == 0) {
            cout << "0" << endl;
            return 0;
        }
        else {
            int flag = 0;
 
            // find the remainder of the sum
            // of digits when divided by 3
            int y = sum % 3;
 
            // there can a remainder as 1 or 2
            if (y != 0) {
 
                // traverse from the end of the digits
                for (int i = n - 1; i >= 0; i--) {
 
                    // first element which has the same remainder
                    // remove it
                    if (v[i] % 3 == y) {
                        v.erase(v.begin() + i);
                        flag = 1;
                        break;
                    }
                }
                // if there is no element which
                // has a same remainder as y
                if (flag == 0) {
 
                    // subtract it by 3 ( could be one or two)
                    y = 3 - y;
 
                    int cnt = 0;
                    for (int i = n - 1; i >= 0; i--) {
 
                        // delete two minimal digits
                        // which has a remainder as y
                        if (v[i] % 3 == y) {
                            v.erase(v.begin() + i);
                            cnt++;
 
                            if (cnt >= 2)
                                break;
                        }
                    }
                }
            }
            if (*v.begin() == 0)
                cout << "0" << endl;
 
            // print all the digits as a single integer
            else
                for (int i : v) {
                    cout << i;
                }
        }
    }
}
 
// Driver code
int main()
{
    // initialize the number of set of digits
    int n = 11;
 
    // initialize all the set of digits in a vector
    vector v{ 3, 9, 9, 6, 4, 3, 6, 4, 9, 6, 0 };
 
    findLargest(n, v);
 
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
 
class GFG {
 
    // Function to find the largest
    // integer with the given set
    static int findLargest(int n, Vector v)
    {
 
        int flag = 0;
        long sum = 0;
 
        // find sum of all the digits
        // look if any 0 is present or not
        for (int i = 0; i < n; i++) {
            if (v.get(i) == 0)
                flag = 1;
            sum += v.get(i);
        }
 
        // if 0 is not present, the resultant number
        // won't be divisible by 5
        if (flag != 1)
            System.out.println("Not possible");
 
        else {
            // sort all the elements in a non-decreasing manner
            Collections.sort(v, Collections.reverseOrder());
 
            // if there is just one element 0
            if (v.get(0) == 0) {
                System.out.println("0");
                return 0;
            }
            else {
                int flags = 0;
 
                // find the remainder of the sum
                // of digits when divided by 3
                int y = (int)(sum % 3);
 
                // there can a remainder as 1 or 2
                if (y != 0) {
 
                    // traverse from the end of the digits
                    for (int i = n - 1; i >= 0; i--) {
 
                        // first element which has the same remainder
                        // remove it
                        if (v.get(i) % 3 == y) {
                            v.remove(i);
                            flags = 1;
                            break;
                        }
                    }
 
                    // if there is no element which
                    // has a same remainder as y
                    if (flags == 0) {
 
                        // subtract it by 3 ( could be one or two)
                        y = 3 - y;
 
                        int cnt = 0;
                        for (int i = n - 1; i >= 0; i--) {
 
                            // delete two minimal digits
                            // which has a remainder as y
                            if (v.get(i) % 3 == y) {
                                v.remove(i);
                                cnt++;
 
                                if (cnt >= 2)
                                    break;
                            }
                        }
                    }
                }
                if (v.get(0) == 0)
                    System.out.println("0");
 
                // print all the digits as a single integer
                else
                    for (Integer i : v) {
                        System.out.print(i);
                    }
            }
        }
        return Integer.MIN_VALUE;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        // initialize the number of set of digits
        int arr[] = { 3, 9, 9, 6, 4, 3, 6, 4, 9, 6, 0 };
        int n = 11;
 
        Vector v = new Vector();
 
        // initialize all the set of digits in a vector
        for (int i = 0; i < n; i++)
            v.add(i, arr[i]);
 
        findLargest(n, v);
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python 3 implementation of above approach
 
# Function to find the largest
# integer with the given set
def findLargest(n, v):
    flag = 0
    sum = 0
     
    # find sum of all the digits
    # look if any 0 is present or not
    for i in range(n):
        if (v[i] == 0):
            flag = 1
        sum += v[i]
 
    # if 0 is not present, the resultant number
    # won't be divisible by 5
    if (flag == 0):
        print("Not possible")
 
    else:
         
        # sort all the elements in a
        # non-decreasing manner
        v.sort(reverse = True)
 
        # if there is just one element 0
        if (v[0] == 0):
            print("0")
            return 0
         
        else:
            flag = 0
 
            # find the remainder of the sum
            # of digits when divided by 3
            y = sum % 3
 
            # there can a remainder as 1 or 2
            if (y != 0):
                 
                # traverse from the end of the digits
                i = n - 1
                while(i >= 0):
                     
                    # first element which has the same
                    # remainder, remove it
                    if (v[i] % 3 == y):
                        v.remove(v[i])
                        flag = 1
                        break
                    i -= 1
                 
                # if there is no element which
                # has a same remainder as y
                if (flag == 0):
                     
                    # subtract it by 3 ( could be one or two)
                    y = 3 - y
 
                    cnt = 0
                    i = n - 1
                    while(i >= 0):
                         
                        # delete two minimal digits
                        # which has a remainder as y
                        if (v[i] % 3 == y):
                            v.remove(v[i])
                            cnt += 1
 
                            if (cnt >= 2):
                                break
                         
                        i -= 1
                 
    
 
            # print all the digits as a single integer
            for i in (v):
               print(i, end = "")
         
# Driver code
if __name__ == '__main__':
     
    # initialize the number of set of digits
    n = 11
 
    # initialize all the set of
    # digits in a vector
    v = [3, 9, 9, 6, 4, 3,
            6, 4, 9, 6, 0]
 
    findLargest(n, v)
     
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach
using System;
using System.Collections;
 
class GFG {
 
    // Function to find the largest
    // integer with the given set
    static int findLargest(int n, ArrayList v)
    {
 
        int flag = 0;
        long sum = 0;
 
        // find sum of all the digits
        // look if any 0 is present or not
        for (int i = 0; i < n; i++) {
            if ((int)v[i] == 0)
                flag = 1;
            sum += (int)v[i];
        }
 
        // if 0 is not present, the resultant number
        // won't be divisible by 5
        if (flag != 1)
            Console.WriteLine("Not possible");
 
        else {
            // sort all the elements in a non-decreasing manner
            v.Sort();
            v.Reverse();
 
            // if there is just one element 0
            if ((int)v[0] == 0) {
                Console.WriteLine("0");
                return 0;
            }
            else {
                int flags = 0;
 
                // find the remainder of the sum
                // of digits when divided by 3
                int y = (int)(sum % 3);
 
                // there can a remainder as 1 or 2
                if (y != 0) {
 
                    // traverse from the end of the digits
                    for (int i = n - 1; i >= 0; i--) {
 
                        // first element which has the same remainder
                        // remove it
                        if ((int)v[i] % 3 == y) {
                            v.RemoveAt(i);
                            flags = 1;
                            break;
                        }
                    }
 
                    // if there is no element which
                    // has a same remainder as y
                    if (flags == 0) {
 
                        // subtract it by 3 ( could be one or two)
                        y = 3 - y;
 
                        int cnt = 0;
                        for (int i = n - 1; i >= 0; i--) {
 
                            // delete two minimal digits
                            // which has a remainder as y
                            if ((int)v[i] % 3 == y) {
                                v.RemoveAt(i);
                                cnt++;
 
                                if (cnt >= 2)
                                    break;
                            }
                        }
                    }
                }
                if ((int)v[0] == 0)
                    Console.WriteLine("0");
 
                // print all the digits as a single integer
                else
                    for (int i = 0; i < v.Count; i++) {
                        Console.Write(v[i]);
                    }
            }
        }
        return int.MinValue;
    }
 
    // Driver code
    static void Main()
    {
        // initialize the number of set of digits
        int[] arr = { 3, 9, 9, 6, 4, 3, 6, 4, 9, 6, 0 };
        int n = 11;
 
        ArrayList v = new ArrayList();
 
        // initialize all the set of digits in a vector
        for (int i = 0; i < n; i++)
            v.Add(arr[i]);
 
        findLargest(n, v);
    }
}
 
// This code contributed by mits

输出：

999666330

替代解决方案：
以下是由华盛顿州西雅图市的Keegan Fisher实施的

C++

#include 
using namespace std;
 
// return a String representing the largest value that is
// both a combination of the values from the parameter array
// "vals" and divisible by 2, 3, and 5.
string findLargest(int vals[], int N)
{
   
    // sort the array in ascending order
    sort(vals, vals + N);
    string sb;
 
    // index of the lowest value divisible by 3 in "vals"
    // if not present, no possible value
    int index_div_3 = -1;
 
    // if a zero is not found, no possible value
    bool zero = false;
 
    // find minimum multiple of 3 and check for 0
    for (int i = 0; i < N; i++)
    {
 
        // break when finding the first multiple of 3
        // which is minimal due to sort in asc order
        if (vals[i] % 3 == 0 && vals[i] != 0)
        {
            index_div_3 = i;
            break;
        }
        if (vals[i] == 0)
        {
            zero = true;
        }
    }
 
    // if no multiple of 3 or no zero, then no value
    if (index_div_3 == -1 || !zero)
    {
        return sb;
    }
 
    // construct the output StringBuilder
    // adding the values to the string from highest
    // to lowest, not adding the val[index_div_3]
    // until reaching the 0s in the array, at which
    // point add the multiple of 3 followed by
    // any 0s in the array
    for (int i = N - 1; i >= 0; i--)
    {
        if (i != index_div_3)
        {
            if (vals[i] == 0 && index_div_3 != -1)
            {
                sb = sb + to_string(vals[index_div_3]);
                sb = sb + to_string(vals[i]);
                index_div_3 = -1;
            }
            else
            {
                sb = sb + to_string(vals[i]);
            }
        }
    }
    return sb;
}
     
int main()
{
    int vals[] = { 0, 0, 0, 2, 3, 4, 2, 7, 6, 9 };
    int N = sizeof(vals) / sizeof(vals[0]);
    cout << "Output = " << findLargest(vals, N) << endl;
 
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

import java.util.Arrays;
 
class GFG {
    // Driver
    public static void main()
    {
        int[] vals = { 0, 0, 0, 2, 3, 4, 2, 7, 6, 9 };
        System.out.println("Output = " + findLargest(vals));
    }
 
    // return a String representing the largest value that is
    // both a combination of the values from the parameter array
    // "vals" and divisible by 2, 3, and 5.
    public static String findLargest(int[] vals)
    {
        // sort the array in ascending order
        Arrays.sort(vals);
        StringBuilder sb = new StringBuilder();
 
        // index of the lowest value divisible by 3 in "vals"
        // if not present, no possible value
        int index_div_3 = -1;
 
        // if a zero is not found, no possible value
        boolean zero = false;
 
        // find minimum multiple of 3 and check for 0
        for (int i = 0; i < vals.length; i++) {
 
            // break when finding the first multiple of 3
            // which is minimal due to sort in asc order
            if (vals[i] % 3 == 0 && vals[i] != 0) {
                index_div_3 = i;
                break;
            }
            if (vals[i] == 0) {
                zero = true;
            }
        }
 
        // if no multiple of 3 or no zero, then no value
        if (index_div_3 == -1 || !zero) {
            return sb.toString();
        }
 
        // construct the output StringBuilder
        // adding the values to the string from highest
        // to lowest, not adding the val[index_div_3]
        // until reaching the 0s in the array, at which
        // point add the multiple of 3 followed by
        // any 0s in the array
        for (int i = vals.length - 1; i >= 0; i--) {
            if (i != index_div_3) {
                if (vals[i] == 0 && index_div_3 != -1) {
                    sb.append(vals[index_div_3]);
                    sb.append(vals[i]);
                    index_div_3 = -1;
                }
                else {
                    sb.append(vals[i]);
                }
            }
        }
        return sb.toString();
    }
}

Python3

# Return a String representing the largest
# value that is both a combination of the
# values from the parameter array
# "vals" and divisible by 2, 3, and 5.
def findLargest (vals):
 
    # Sort the array in ascending order
    vals.sort()
    sb = ""
 
    # Index of the lowest value divisible
    # by 3 in "vals" if not present, no
    # possible value
    index_div_3 = -1
 
    # If a zero is not found, no possible value
    zero = False
 
    # Find minimum multiple of 3 and check for 0
    for i in range(len(vals)):
 
        # Break when finding the first
        # multiple of 3 which is minimal
        # due to sort in asc order
        if (vals[i] % 3 == 0 and vals[i] != 0):
            index_div_3 = i
            break
 
        if (vals[i] == 0):
            zero = True
 
    # If no multiple of 3 or no zero, then no value
    if (index_div_3 == -1 or zero == False):
        return str(sb)
 
    # Construct the output String by
    # adding the values to the string from highest
    # to lowest, not adding the val[index_div_3]
    # until reaching the 0s in the array, at which
    # point add the multiple of 3 followed by
    # any 0s in the array
    for i in range(len(vals) - 1, -1, -1):
        if (i != index_div_3):
 
            if (vals[i] == 0 and index_div_3 != -1):
                sb += str(vals[index_div_3])
                sb += str(vals[i])
                index_div_3 = -1
            else:
                sb += str(vals[i])
 
    return str(sb)
 
# Driver code
if __name__ == '__main__':
 
    vals = [ 0, 0, 0, 2, 3, 4, 2, 7, 6, 9 ]
     
    print("Output =", findLargest(vals))
 
# This code is contributed by himanshu77

C＃

using System;
using System.Text;
class GFG
{
    // Driver
    public static void Main()
    {
        int[] vals = { 0, 0, 0, 2, 3, 4, 2, 7, 6, 9 };
        Console.WriteLine("Output = " + findLargest(vals));
    }
 
    // return a String representing the largest value that is
    // both a combination of the values from the parameter array
    // "vals" and divisible by 2, 3, and 5.
    public static string findLargest(int[] vals)
    {
        // sort the array in ascending order
        Array.Sort(vals);
        StringBuilder sb = new StringBuilder();
 
        // index of the lowest value divisible by 3 in "vals"
        // if not present, no possible value
        int index_div_3 = -1;
 
        // if a zero is not found, no possible value
        bool zero = false;
 
        // find minimum multiple of 3 and check for 0
        for (int i = 0; i < vals.Length; i++) {
 
            // break when finding the first multiple of 3
            // which is minimal due to sort in asc order
            if (vals[i] % 3 == 0 && vals[i] != 0) {
                index_div_3 = i;
                break;
            }
            if (vals[i] == 0) {
                zero = true;
            }
        }
 
        // if no multiple of 3 or no zero, then no value
        if (index_div_3 == -1 || !zero) {
            return sb.ToString();
        }
 
        // construct the output StringBuilder
        // adding the values to the string from highest
        // to lowest, not adding the val[index_div_3]
        // until reaching the 0s in the array, at which
        // point add the multiple of 3 followed by
        // any 0s in the array
        for (int i = vals.Length - 1; i >= 0; i--) {
            if (i != index_div_3) {
                if (vals[i] == 0 && index_div_3 != -1) {
                    sb.Append(vals[index_div_3]);
                    sb.Append(vals[i]);
                    index_div_3 = -1;
                }
                else {
                    sb.Append(vals[i]);
                }
            }
        }
        return sb.ToString();
    }
}
 
// This code is contributed by SoumikMondal

输出：

Output = 9764223000