不大于 N 的最大数字，所有数字都是奇数
给定一个数字 N ，任务是找到不大于 N 的最大数字，它的所有数字都是奇数。 示例:
输入: N = 23 输出: 19 19 是小于 23 的最大数字，它有所有奇数位。 输入: N = 7236 输出: 7199
天真逼近:从 N 迭代到 0 ，找到第一个所有数字都是奇数的数字。如果跳过偶数，这种方法仍然可以优化，因为每个偶数的右边都有一个偶数。 以下是上述方法的实现:
C++
// CPP program to print the largest integer
// not greater than N with all odd digits
#include <bits/stdc++.h>
using namespace std;
// Function to check if all digits
// of a number are odd
bool allOddDigits(int n)
{
// iterate for all digits
while (n) {
// if digit is even
if ((n % 10) % 2 == 0)
return false;
n /= 10;
}
// all digits are odd
return true;
}
// function to return the largest number
// with all digits odd
int largestNumber(int n)
{
if (n % 2 == 0)
n--;
// iterate till we find a
// number with all digits odd
for (int i = n;; i -= 2)
if (allOddDigits(i))
return i;
}
// Driver Code
int main()
{
int N = 23;
cout << largestNumber(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print the largest integer
// not greater than N with all odd digits
public class GFG{
// Function to check if all digits
// of a number are odd
static boolean allOddDigits(int n)
{
// iterate for all digits
while (n != 0) {
// if digit is even
if ((n % 10) % 2 == 0)
return false;
n /= 10;
}
// all digits are odd
return true;
}
// function to return the largest number
// with all digits odd
static int largestNumber(int n)
{
if (n % 2 == 0)
n--;
// iterate till we find a
// number with all digits odd
for (int i = n;; i -= 2)
if (allOddDigits(i))
return i;
}
public static void main(String []args){
int N = 23;
System.out.println(largestNumber(N));
}
// This code is contributed by ANKITRAI1
}
Python 3
# Python 3 program to print the largest
# integer not greater than N with all
# odd digits
# Function to check if all digits
# of a number are odd
def allOddDigits(n):
# iterate for all digits
while (n):
# if digit is even
if ((n % 10) % 2 == 0):
return False
n = int(n / 10)
# all digits are odd
return True
# function to return the largest
# number with all digits odd
def largestNumber(n):
if (n % 2 == 0):
n -= 1
# iterate till we find a
# number with all digits odd
i = n
while(1):
if (allOddDigits(i)):
return i
i -= 2
# Driver Code
if __name__ =='__main__':
N = 23
print(largestNumber(N))
# This code is contributed by
# Shashank_Sharma
C
// C# program to print the largest
// integer not greater than N with
// all odd digits
using System;
class GFG
{
// Function to check if all
// digits of a number are odd
static bool allOddDigits(int n)
{
// iterate for all digits
while (n != 0)
{
// if digit is even
if ((n % 10) % 2 == 0)
return false;
n /= 10;
}
// all digits are odd
return true;
}
// function to return the largest
// number with all digits odd
static int largestNumber(int n)
{
if (n % 2 == 0)
n--;
// iterate till we find a
// number with all digits odd
for (int i = n;; i -= 2)
if (allOddDigits(i))
return i;
}
// Driver Code
public static void Main()
{
int N = 23;
Console.WriteLine(largestNumber(N));
}
}
// This code is contributed by anuj_67
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print the largest integer
// not greater than N with all odd digits
// Function to check if all digits
// of a number are odd
function allOddDigits($n)
{
// iterate for all digits
while ($n > 1)
{
// if digit is even
if (($n % 10) % 2 == 0)
return false;
$n = (int)$n / 10;
}
// all digits are odd
return true;
}
// function to return the largest
// number with all digits odd
function largestNumber($n)
{
if ($n % 2 == 0)
$n--;
// iterate till we find a
// number with all digits odd
for ($i = $n;; $i= ($i - 2))
if (allOddDigits($i))
return $i;
}
// Driver Code
$N = 23;
echo largestNumber($N);
// This code is contributed by ajit
?>
java 描述语言
<script>
// javascript program to print the largest integer
// not greater than N with all odd digits
// Function to check if all digits
// of a number are odd
function allOddDigits(n)
{
// iterate for all digits
while (n != 0)
{
// if digit is even
if ((n % 10) % 2 == 0)
return false;
n = parseInt(n / 10);
}
// all digits are odd
return true;
}
// function to return the largest number
// with all digits odd
function largestNumber(n)
{
if (parseInt(n % 2) == 0)
n--;
// iterate till we find a
// number with all digits odd
for (i = n; i > 0; i -= 2)
if (allOddDigits(i))
return i;
}
// Driver code
var N = 23;
document.write(largestNumber(N));
// This code is contributed by shikhasingrajput
</script>
Output:
19
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