📌  相关文章
📜  使N被25整除所需的最小给定移动次数

📅  最后修改于: 2021-06-26 22:50:11             🧑  作者: Mango

给定数字N (1≤N≤10 18 )而没有前导零。任务是找到使N被25整除所需的最小移动数。每次移动时，您都可以交换任意两个相邻的数字，并确保数字在任何时候都不得包含任何前导零。如果无法将N除以25，则打印-1 。

例子：

方法：遍历数字中的所有数字对。令该对中的第一个数字位于位置i，第二个数字位于位置j。让我们将这些数字放在数字的最后两个位置。但是，现在数字可以包含前导零。找到最左边的非零数字并将其移到第一个位置。然后，如果当前数字可被25整除，请尝试使用掉期数更新答案。所有这些操作之间的最小交换数是必需的答案。

下面是上述方法的实现：

C++
// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum number 
// of moves required to make n divisible 
// by 25
int minMoves(long long n)
{
  
    // Convert number into string
    string s = to_string(n);
  
    // To store required answer
    int ans = INT_MAX;
  
    // Length of the string
    int len = s.size();
  
    // To check all possible pairs
    for (int i = 0; i < len; ++i) {
        for (int j = 0; j < len; ++j) {
            if (i == j)
                continue;
  
            // Make a duplicate string
            string t = s;
            int cur = 0;
  
            // Number of swaps required to place
            // ith digit in last position
            for (int k = i; k < len - 1; ++k) {
                swap(t[k], t[k + 1]);
                ++cur;
            }
  
            // Number of swaps required to place
            // jth digit in 2nd last position
            for (int k = j - (j > i); k < len - 2; ++k) {
                swap(t[k], t[k + 1]);
                ++cur;
            }
  
            // Find first non zero digit
            int pos = -1;
            for (int k = 0; k < len; ++k) {
                if (t[k] != '0') {
                    pos = k;
                    break;
                }
            }
  
            // Place first non zero digit
            // in the first position
            for (int k = pos; k > 0; --k) {
                swap(t[k], t[k - 1]);
                ++cur;
            }
  
            // Convert string to number
            long long nn = atoll(t.c_str());
  
            // If this number is divisible by 25
            // then cur is one of the possible answer
            if (nn % 25 == 0)
                ans = min(ans, cur);
        }
    }
  
    // If not possible
    if (ans == INT_MAX)
        return -1;
  
    return ans;
}
  
// Driver code
int main()
{
    long long n = 509201;
    cout << minMoves(n);
    return 0;
}

Java
// Java implementation of the approach
class GFG
{
      
    // Function to return the minimum number 
    // of moves required to make n divisible 
    // by 25
    static int minMoves(int n)
    {
      
        // Convert number into string
        String s = Integer.toString(n);
          
        // To store required answer
        int ans = Integer.MAX_VALUE;
      
        // Length of the string
        int len = s.length();
      
        // To check all possible pairs
        for (int i = 0; i < len; ++i) 
        {
            for (int j = 0; j < len; ++j) 
            {
                if (i == j)
                    continue;
      
                // Make a duplicate string
                char t [] = s.toCharArray();
                int cur = 0;
      
                // Number of swaps required to place
                // ith digit in last position
                for (int k = i; k < len - 1; ++k) 
                {
                    swap(t, k, k + 1);
                    ++cur;
                }
      
                // Number of swaps required to place
                // jth digit in 2nd last position
                for (int k = j - ((j > i)? 1 : 0 ); k < len - 2; ++k) 
                {
                    swap(t, k, k + 1);
                    ++cur;
                }
      
                // Find first non zero digit
                int pos = -1;
                for (int k = 0; k < len; ++k)
                {
                    if (t[k] != '0') 
                    {
                        pos = k;
                        break;
                    }
                }
      
                // Place first non zero digit
                // in the first position
                for (int k = pos; k > 0; --k) 
                {
                    swap(t, k, k - 1);
                    ++cur;
                }
      
                // Convert string to number
                long nn = Integer.parseInt(String.valueOf(t));
      
                // If this number is divisible by 25
                // then cur is one of the possible answer
                if (nn % 25 == 0)
                    ans = Math.min(ans, cur);
            }
        }
      
        // If not possible
        if (ans == Integer.MAX_VALUE)
            return -1;
      
        return ans;
    }
      
    static void swap(char t [], int i, int j) 
    { 
        char temp = t[i]; 
        t[i] = t[j]; 
        t[j] = temp; 
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int n = 509201;
        System.out.println(minMoves(n));
    }
}
  
// This code is contributed by Archana_Kumari

Python3
# Python3 implementation of the approach
import sys
  
# Function to return the minimum number 
# of moves required to make n divisible 
# by 25
def minMoves(n):
  
    # Convert number into string
    s = str(n);
  
    # To store required answer
    ans = sys.maxsize;
  
    # Length of the string
    len1 = len(s);
  
    # To check all possible pairs
    for i in range(len1): 
        for j in range(len1): 
            if (i == j):
                continue;
  
            # Make a duplicate string
            t = s;
            cur = 0;
  
            # Number of swaps required to place
            # ith digit in last position
            list1 = list(t);
            for k in range(i,len1 - 1):
                e = list1[k];
                list1[k] = list1[k + 1];
                list1[k + 1] = e;
                cur += 1;
            t = ''.join(list1);
  
            # Number of swaps required to place
            # jth digit in 2nd last position
            list1 = list(t);
            for k in range(j - (j > i),len1 - 2): 
                e = list1[k];
                list1[k] = list1[k + 1];
                list1[k + 1] = e;
                cur += 1;
            t = ''.join(list1);
  
            # Find first non zero digit
            pos = -1;
            for k in range(len1): 
                if (t[k] != '0'): 
                    pos = k;
                    break;
  
            # Place first non zero digit
            # in the first position
            for k in range(pos,0,-1):
                e = list1[k];
                list1[k] = list1[k + 1];
                list1[k + 1] = e;
                cur += 1;
            t = ''.join(list1);
  
  
            # Convert string to number
            nn = int(t);
  
            # If this number is divisible by 25
            # then cur is one of the possible answer
            if (nn % 25 == 0):
                ans = min(ans, cur);
  
    # If not possible
    if (ans == sys.maxsize):
        return -1;
  
    return ans;
  
# Driver code
n = 509201;
print(minMoves(n));
  
# This code is contributed
# by chandan_jnu

C#
// C# implementation of the approach
using System;
  
class GFG
{
      
    // Function to return the minimum number 
    // of moves required to make n divisible 
    // by 25
    static int minMoves(int n)
    {
      
        // Convert number into string
        string s = n.ToString();
          
        // To store required answer
        int ans = Int32.MaxValue;
      
        // Length of the string
        int len = s.Length;
      
        // To check all possible pairs
        for (int i = 0; i < len; ++i) 
        {
            for (int j = 0; j < len; ++j) 
            {
                if (i == j)
                    continue;
      
                // Make a duplicate string
                char[] t = s.ToCharArray();
                int cur = 0;
      
                // Number of swaps required to place
                // ith digit in last position
                for (int k = i; k < len - 1; ++k) 
                {
                    swap(t, k, k + 1);
                    ++cur;
                }
      
                // Number of swaps required to place
                // jth digit in 2nd last position
                for (int k = j - ((j > i)? 1 : 0 ); k < len - 2; ++k) 
                {
                    swap(t, k, k + 1);
                    ++cur;
                }
      
                // Find first non zero digit
                int pos = -1;
                for (int k = 0; k < len; ++k)
                {
                    if (t[k] != '0') 
                    {
                        pos = k;
                        break;
                    }
                }
      
                // Place first non zero digit
                // in the first position
                for (int k = pos; k > 0; --k) 
                {
                    swap(t, k, k - 1);
                    ++cur;
                }
      
                // Convert string to number
                int nn = Convert.ToInt32(new String(t));
      
                // If this number is divisible by 25
                // then cur is one of the possible answer
                if (nn % 25 == 0)
                    ans = Math.Min(ans, cur);
            }
        }
      
        // If not possible
        if (ans == Int32.MaxValue)
            return -1;
      
        return ans;
    }
      
    static void swap(char []t, int i, int j) 
    { 
        char temp = t[i]; 
        t[i] = t[j]; 
        t[j] = temp; 
    }
      
    // Driver code
    static void Main()
    {
        int n = 509201;
        Console.WriteLine(minMoves(n));
    }
}
  
// This code is contributed by mits

PHP
 $i); 
                 $k < $len - 2; ++$k) 
            {
                $e = $t[$k];
                $t[$k] = $t[$k + 1];
                $t[$k + 1] = $e;
                ++$cur;
            }
  
            // Find first non zero digit
            $pos = -1;
            for ($k = 0; $k < $len; ++$k) 
            {
                if ($t[$k] != '0') 
                {
                    $pos = $k;
                    break;
                }
            }
  
            // Place first non zero digit
            // in the first position
            for ($k = $pos; $k > 0; --$k) 
            {
                $e = $t[$k];
                $t[$k] = $t[$k + 1];
                $t[$k + 1] = $e;
                ++$cur;
            }
  
            // Convert string to number
            $nn = intval($t);
  
            // If this number is divisible by 25
            // then cur is one of the possible answer
            if ($nn % 25 == 0)
                $ans = min($ans, $cur);
        }
    }
  
    // If not possible
    if ($ans == PHP_INT_MAX)
        return -1;
  
    return $ans;
}
  
// Driver code
$n = 509201;
echo minMoves($n);
  
// This code is contributed
// by chandan_jnu
?>

输出： 
4

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。