最小偶数个数不少于N

给定一个数字N，我们需要编写一个程序以查找不小于N的最小数字，该数字具有全数字。
例子：

Input: N = 1345  
Output: 2000
Explanation: 2000 is the smallest number not 
less than N, whose all digits are even. 

Input : N = 2397 
Output : 2400 
Explanation: 2400 is the smallest number not 
less than N, whose all digits are even.

幼稚的方法：幼稚的方法是保持从N进行迭代，直到找到一个全数字甚至偶数为止。
下面是上述方法的实现：

C++

// CPP program to print the smallest
// integer not less than N with all even digits
#include 
using namespace std;
 
// function to check if all digits
// are even of a given number
int check_digits(int n)
{
    // iterate for all digits
    while (n) {
        if ((n % 10) % 2) // if digit is odd
            return 0;
 
        n /= 10;
    }
 
    // all digits are even
    return 1;
}
 
// function to return the smallest number
// with all digits even
int smallest_number(int n)
{
    // iterate till we find a
    // number with all digits even
    for (int i = n;; i++)
        if (check_digits(i))
            return i;
}
 
// Driver Code
int main()
{
    int N = 2397;
    cout << smallest_number(N);
    return 0;
}

Java

// Java program to print the smallest
// integer not less than N with all
// even digits
class GFG {
     
    // function to check if all digits
    // are even of a given number
    static int check_digits(int n)
    {
         
        // iterate for all digits
        while (n != 0) {
             
            // if digit is odd
            if ((n % 10) % 2 != 0)
                return 0;
 
            n /= 10;
        }
 
        // all digits are even
        return 1;
    }
 
    // function to return the smallest
    // number with all digits even
    static int smallest_number(int n)
    {
         
        // iterate till we find a
        // number with all digits even
        for (int i = n; ; i++)
            if (check_digits(i) != 0)
                return i;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 2397;
         
        System.out.println(smallest_number(N));
    }
}
 
// This code is contributed by
// Smitha Dinesh Semwal

Python3

# Python3 program to print the smallest
# integer not less than N with
# all even digits
 
# function to check if all digits
# are even of a given number
def check_digits(n) :
     
    # iterate for all digits
    while (n) :
         
        # if digit is odd
        if ((n % 10) % 2) :
            return 0
 
        n = int(n / 10)
         
    # all digits are even
    return 1
 
# function to return the
# smallest number with
# all digits even
def smallest_number(n) :
     
    # iterate till we find a
    # number with all digits even
    for i in range(n, 2401) :
        if (check_digits(i) == 1) :
            return (i)
 
# Driver Code
N = 2397
print (str(smallest_number(N)))
 
# This code is contributed by
# Manish Shaw (manishshaw1)

C#

// C# program to print the smallest
// integer not less than N with all
// even digits
using System;
class GFG {
     
    // function to check if all digits
    // are even of a given number
    static int check_digits(int n)
    {
         
        // iterate for all digits
        while (n != 0) {
             
            // if digit is odd
            if ((n % 10) % 2 != 0)
                return 0;
 
            n /= 10;
        }
 
        // all digits are even
        return 1;
    }
 
    // function to return the smallest
    // number with all digits even
    static int smallest_number(int n)
    {
         
        // iterate till we find a
        // number with all digits even
        for (int i = n; ; i++)
            if (check_digits(i) != 0)
                return i;
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 2397;
         
    Console.WriteLine(smallest_number(N));
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
// Javascript program to prvar the smallest
// integer not less than N with all
// even digits
 
     
// function to check if all digits
// are even of a given number
function check_digits(n)
{
     
    // iterate for all digits
    while (n != 0) {
         
        // if digit is odd
        if ((n % 10) % 2 != 0)
            return 0;
 
        n = parseInt(n/10);
    }
 
    // all digits are even
    return 1;
}
 
// function to return the smallest
// number with all digits even
function smallest_number(n)
{
     
    // iterate till we find a
    // number with all digits even
    for (i = n; ; i++)
        if (check_digits(i) != 0)
            return i;
}
 
// Driver Code
var N = 2397;
 
document.write(smallest_number(N));
 
// This code is contributed by 29AjayKumar
 

输出：

2400

时间复杂度： O(N)
高效的方法：我们可以通过将N中的第一个奇数位加1并用最小的偶数位(即0)替换该奇数位右边的所有数字来找到数字。如果N中没有奇数位，则N本身就是最小的数字。例如，考虑N =213。将N中的第一个奇数数字递增，即1到2，然后将其右边的所有数字都替换为0。因此，我们所需的数字将是220。
棘手的案例：

• 如果N中的第一个奇数位是9，那么我们必须立即用下一个偶数位替换该奇数位左边的数字。例如，如果N = 44934，则最小数= 46000。
• 另一个棘手的情况是，当第一个奇数位是9且直接在第一个奇数位左边的数字是8时。在这种情况下，我们必须将第一个奇数位左边的数字直接替换为0，并将左数位替换为该数字接下一个偶数数字，并继续执行此操作，直到找到8以外的其他数字。例如，如果N = 86891，则Y = 88000。最后，如果左边的所有数字继续为8，直到我们到达最左边的数字，或者N的第一个数字为9，则我们必须将最小的非零偶数数字(即2)添加为新的数字。左边。例如，如果N = 891或N = 910，则Y = 2000。

下面是有效方法的实现：

C++

// CPP program to print the smallest
// integer not less than N with all even digits
#include 
using namespace std;
 
// fucntion to return the answer when the
// first odd digit is 9
int trickyCase(string s, int index)
{
 
    int index1 = -1;
 
    // traverse towwars the left to find the non-8 digit
    for (int i = index - 1; i >= 0; i--) {
        // index digit
        int digit = s[i] - '0';
 
        // if digit is not 8, then break
        if (digit != 8) {
            index1 = i;
            break;
        }
    }
    // if on the left side of the '9', no 8
    // is found then we return by adding a 2 and 0's
    if (index1 == -1)
        return 2 * pow(10, s.length());
 
    int num = 0;
 
    // till non-8 digit add all numbers
    for (int i = 0; i < index1; i++)
        num = num * 10 + (s[i] - '0');
 
    // if non-8 is even or odd than add the next even.
    if (s[index1] % 2 == 0)
        num = num * 10 + (s[index1] - '0' + 2);
    else
        num = num * 10 + (s[index1] - '0' + 1);
 
    // add 0 to right of 9
    for (int i = index1 + 1; i < s.length(); i++)
        num = num * 10;
 
    return num;
}
 
// function to return the smallest number
// with all digits even
int smallestNumber(int n)
{
    int num = 0;
    string s = "";
 
    int duplicate = n;
    // convert the number to string to
    // perform operations
    while (n) {
        s = char(n % 10 + 48) + s;
        n /= 10;
    }
 
    int index = -1;
 
    // find out the first odd number
    for (int i = 0; i < s.length(); i++) {
        int digit = s[i] - '0';
        if (digit & 1) {
            index = i;
            break;
        }
    }
 
    // if no odd numbers are there, than n is the answer
    if (index == -1)
        return duplicate;
 
    // if the odd number is 9,
    // than tricky case handles it
    if (s[index] == '9') {
        num = trickyCase(s, index);
        return num;
    }
 
    // add all digits till first odd
    for (int i = 0; i < index; i++)
        num = num * 10 + (s[i] - '0');
 
    // increase the odd digit by 1
    num = num * 10 + (s[index] - '0' + 1);
 
    // add 0 to the right of the odd number
    for (int i = index + 1; i < s.length(); i++)
        num = num * 10;
 
    return num;
}
 
// Driver Code
int main()
{
    int N = 2397;
    cout << smallestNumber(N);
 
    return 0;
}

Java

// Java program to print the
// smallest integer not less
// than N with all even digits
import java.io.*;
import java.util.*;
import java.lang.*;
 
class GFG
{
// function to return
// the answer when the
// first odd digit is 9
static int trickyCase(String s,
                      int index)
{
 
    int index1 = -1;
 
    // traverse towwars the left
    // to find the non-8 digit
    for (int i = index - 1; i >= 0; i--)
    {
        // index digit
        int digit = s.charAt(i) - '0';
 
        // if digit is not 8,
        // then break
        if (digit != 8)
        {
            index1 = i;
            break;
        }
    }
     
    // if on the left side of the
    // '9', no 8 is found then we
    // return by adding a 2 and 0's
    if (index1 == -1)
        return 2 * (int)Math.pow(10, s.length());
 
    int num = 0;
 
    // till non-8 digit
    // add all numbers
    for (int i = 0; i < index1; i++)
        num = num * 10 + (s.charAt(i) - '0');
 
    // if non-8 is even or odd
    // than add the next even.
    if (s.charAt(index1) % 2 == 0)
        num = num * 10 +
            (s.charAt(index1) - '0' + 2);
    else
        num = num * 10 +
            (s.charAt(index1) - '0' + 1);
 
    // add 0 to right of 9
    for (int i = index1 + 1;
            i < s.length(); i++)
        num = num * 10;
 
    return num;
}
 
// function to return
// the smallest number
// with all digits even
static int smallestNumber(int n)
{
    int num = 0;
    String s = "";
 
    int duplicate = n;
     
    // convert the number to
    // string to perform operations
    while (n > 0)
    {
        s = (char)(n % 10 + 48) + s;
        n /= 10;
    }
 
    int index = -1;
 
    // find out the
    // first odd number
    for (int i = 0; i < s.length(); i++)
    {
        int digit = s.charAt(i) - '0';
        int val = digit & 1;
        if (val == 1)
        {
            index = i;
            break;
        }
    }
 
    // if no odd numbers are there,
    // than n is the answer
    if (index == -1)
        return duplicate;
 
    // if the odd number is 9,
    // than tricky case handles it
    if (s.charAt(index) == '9')
    {
        num = trickyCase(s, index);
        return num;
    }
 
    // add all digits till first odd
    for (int i = 0; i < index; i++)
        num = num * 10 +
             (s.charAt(i) - '0');
 
    // increase the
    // odd digit by 1
    num = num * 10 +
        (s.charAt(index) - '0' + 1);
 
    // add 0 to the right
    // of the odd number
    for (int i = index + 1;
             i < s.length(); i++)
        num = num * 10;
 
    return num;
}
 
// Driver Code
public static void main(String args[])
{
    int N = 2397;
    System.out.print(smallestNumber(N));
}
}
 
// This code is contributed
// by Akanksha rai(Abby_akku)

Python3

# Python3 program to print the smallest
# integer not less than N with all even digits
 
# Function to return the answer when the
# first odd digit is 9
def trickyCase(s, index):
    index1 = -1;
 
    # traverse towwars the left to find
    # the non-8 digit
    for i in range(index - 1, -1, -1):
 
        # index digit
        digit = s[i] - '0';
 
        # if digit is not 8, then break
        if (digit != 8):
            index1 = i;
            break;
     
    # if on the left side of the '9',
    # no 8 is found then we return by
    # adding a 2 and 0's
    if (index1 == -1):
        return 2 * pow(10, len(s));
 
    num = 0;
 
    # till non-8 digit add all numbers
    for i in range(index1):
        num = num * 10 + (s[i] - '0');
 
    # if non-8 is even or odd
    # than add the next even.
    if (s[index1] % 2 == 0):
        num = num * 10 + (s[index1] - '0' + 2);
    else:
        num = num * 10 + (s[index1] - '0' + 1);
 
    # add 0 to right of 9
    for i in range(index1 + 1, len(s)):
        num = num * 10;
 
    return num;
 
# function to return the smallest
# number with all digits even
def smallestNumber(n):
 
    num = 0;
    s = "";
 
    duplicate = n;
     
    # convert the number to string to
    # perform operations
    while (n):
        s = chr(n % 10 + 48) + s;
        n = int(n / 10);
 
    index = -1;
 
    # find out the first odd number
    for i in range(len(s)):
        digit = ord(s[i]) - ord('0');
        if (digit & 1):
            index = i;
            break;
 
    # if no odd numbers are
    # there, than n is the answer
    if (index == -1):
        return duplicate;
 
    # if the odd number is 9, than
    # tricky case handles it
    if (s[index] == '9'):
        num = trickyCase(s, index);
        return num;
 
    # add all digits till first odd
    for i in range(index):
        num = num * 10 + ord(s[i]) - ord('0');
 
    # increase the odd digit by 1
    num = num * 10 + (ord(s[index]) -
                      ord('0') + 1);
 
    # add 0 to the right of the odd number
    for i in range(index + 1, len(s)):
        num = num * 10;
 
    return num;
 
# Driver Code
N = 2397;
print(smallestNumber(N));
 
# This code is contributed
# by mits

C＃

// C# program to print the smallest integer
// not less than N with all even digits
using System;
 
class GFG
{
// function to return the answer when
// the first odd digit is 9
static int trickyCase(string s,
                      int index)
{
    int index1 = -1;
 
    // traverse towwars the left
    // to find the non-8 digit
    for (int i = index - 1; i >= 0; i--)
    {
        // index digit
        int digit = s[i] - '0';
 
        // if digit is not 8, then break
        if (digit != 8)
        {
            index1 = i;
            break;
        }
    }
     
    // if on the left side of the
    // '9', no 8 is found then we
    // return by adding a 2 and 0's
    if (index1 == -1)
        return 2 * (int)Math.Pow(10, s.Length);
 
    int num = 0;
 
    // till non-8 digit add all numbers
    for (int i = 0; i < index1; i++)
        num = num * 10 + (s[i] - '0');
 
    // if non-8 is even or odd
    // than add the next even.
    if (s[index1] % 2 == 0)
        num = num * 10 +
            (s[index1] - '0' + 2);
    else
        num = num * 10 +
             (s[index1] - '0' + 1);
 
    // add 0 to right of 9
    for (int i = index1 + 1;
            i < s.Length; i++)
        num = num * 10;
 
    return num;
}
 
// function to return the smallest number
// with all digits even
static int smallestNumber(int n)
{
    int num = 0;
    string s = "";
 
    int duplicate = n;
     
    // convert the number to
    // string to perform operations
    while (n > 0)
    {
        s = (char)(n % 10 + 48) + s;
        n /= 10;
    }
 
    int index = -1;
 
    // find out the first odd number
    for (int i = 0; i < s.Length; i++)
    {
        int digit = s[i] - '0';
        int val = digit & 1;
        if (val == 1)
        {
            index = i;
            break;
        }
    }
 
    // if no odd numbers are there,
    // than n is the answer
    if (index == -1)
        return duplicate;
 
    // if the odd number is 9,
    // than tricky case handles it
    if (s[index] == '9')
    {
        num = trickyCase(s, index);
        return num;
    }
 
    // add all digits till first odd
    for (int i = 0; i < index; i++)
        num = num * 10 +
            (s[i] - '0');
 
    // increase the odd digit by 1
    num = num * 10 +
        (s[index] - '0' + 1);
 
    // add 0 to the right of the odd number
    for (int i = index + 1;
            i < s.Length; i++)
        num = num * 10;
 
    return num;
}
 
// Driver Code
public static void Main()
{
    int N = 2397;
    Console.Write(smallestNumber(N));
}
}
 
// This code is contributed
// by Akanksha Rai
?>

的PHP

= 0; $i--)
    {
        // index digit
        $digit = $s[$i] - '0';
 
        // if digit is not
        // 8, then break
        if ($digit != 8)
        {
            $index1 = $i;
            break;
        }
    }
     
    // if on the left side
    // of the '9', no 8
    // is found then we
    // return by adding a 2
    // and 0's
    if ($index1 == -1)
        return 2 * pow(10,
              strlen($s));
 
    $num = 0;
 
    // till non-8 digit
    // add all numbers
    for ($i = 0; $i < $index1; $i++)
        $num = $num * 10 +
              ($s[$i] - '0');
 
    // if non-8 is even or
    // odd than add the next even.
    if ($s[$index1] % 2 == 0)
        $num = $num * 10 +
              ($s[$index1] - '0' + 2);
    else
        $num = $num * 10 +
              ($s[$index1] - '0' + 1);
 
    // add 0 to right of 9
    for ($i = $index1 + 1;
         $i < strlen($s); $i++)
        $num = $num * 10;
 
    return $num;
}
 
// function to return
// the smallest number
// with all digits even
function smallestNumber($n)
{
    $num = 0;
    $s = "";
 
    $duplicate = $n;
     
    // convert the number
    // to string to perform
    // operations
    while ($n)
    {
        $s = chr($n % 10 + 48) . $s;
        $n = (int)($n / 10);
    }
 
    $index = -1;
 
    // find out the
    // first odd number
    for ($i = 0;
         $i < strlen($s); $i++)
    {
        $digit = $s[$i] - '0';
        if ($digit & 1)
        {
            $index = $i;
            break;
        }
    }
 
    // if no odd numbers are
    // there, than n is the answer
    if ($index == -1)
        return $duplicate;
 
    // if the odd number
    // is 9, than tricky
    // case handles it
    if ($s[$index] == '9')
    {
        $num = trickyCase($s, $index);
        return $num;
    }
 
    // add all digits
    // till first odd
    for ($i = 0; $i < $index; $i++)
        $num = $num * 10 +
              ($s[$i] - '0');
 
    // increase the
    // odd digit by 1
    $num = $num * 10 +
          ($s[$index] - '0' + 1);
 
    // add 0 to the right
    // of the odd number
    for ($i = $index + 1;
         $i < strlen($s); $i++)
        $num = $num * 10;
 
    return $num;
}
 
// Driver Code
$N = 2397;
echo smallestNumber($N);
 
// This code is contributed
// by mits
?>

输出：

2400

时间复杂度： O(M)，其中M是N中的位数。