仅由奇数组成的第 n 个数字
给定一个整数 N ，任务是找到仅由奇数位(1，3，5，7，9)组成的N号。 由奇数组成的前几个数字是 1，3，5，7，9，11，13，15，17，19，31，…
示例:
输入: N = 7 输出: 13 1、3、5、7、9、11、13 13 是系列中的第 7 个数字
输入:N = 10 T3】输出: 19
接近 1(简单):从 1 开始，继续检查号码是否仅由奇数(1、3、5、7、9)组成，当发现第 n 个这样的号码时停止。
下面是上述方法的实现:
C++
// C++ program to find nth number made up of odd digits only
#include <bits/stdc++.h>
using namespace std;
// Function to return nth number made up of odd digits only
int findNthOddDigitNumber(int n)
{
// Variable to keep track of how many
// such elements have been found
int count = 0;
for (int i = 1;; i++) {
int num = i;
bool isMadeOfOdd = true;
// Checking each digit of the number
while (num != 0) {
// If 0, 2, 4, 6 or 8 is found
// then the number is not made up of odd digits
if (num % 10 == 0
|| num % 10 == 2
|| num % 10 == 4
|| num % 10 == 6
|| num % 10 == 8) {
isMadeOfOdd = false;
break;
}
num = num / 10;
}
// If the number is made up of odd digits only
if (isMadeOfOdd == true)
count++;
// If it is the nth number
if (count == n)
return i;
}
}
// Driver Code
int main()
{
int n = 10;
cout << findNthOddDigitNumber(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find nth number
// made up of odd digits only
import java.io.*;
class GFG {
// Function to return nth number made up of odd digits only
static int findNthOddDigitNumber(int n)
{
// Variable to keep track of how many
// such elements have been found
int count = 0;
for (int i = 1;; i++) {
int num = i;
boolean isMadeOfOdd = true;
// Checking each digit of the number
while (num != 0) {
// If 0, 2, 4, 6 or 8 is found
// then the number is not made up of odd digits
if (num % 10 == 0
|| num % 10 == 2
|| num % 10 == 4
|| num % 10 == 6
|| num % 10 == 8) {
isMadeOfOdd = false;
break;
}
num = num / 10;
}
// If the number is made up of odd digits only
if (isMadeOfOdd == true)
count++;
// If it is the nth number
if (count == n)
return i;
}
}
// Driver Code
public static void main (String[] args) {
int n = 10;
System.out.println (findNthOddDigitNumber(n));
}
//This code is contributed by ajit
}
Python 3
# Python3 program to find nth number
# made up of odd digits only
# Function to return nth number made
# up of odd digits only
def findNthOddDigitNumber(n) :
# Variable to keep track of how many
# such elements have been found
count = 0
i = 1
while True :
num = i
isMadeOfOdd = True
# Checking each digit of the number
while num != 0 :
# If 0, 2, 4, 6 or 8 is found
# then the number is not made
# up of odd digits
if (num % 10 == 0 or num % 10 == 2 or
num % 10 == 4 or num % 10 == 6 or
num % 10 == 8) :
isMadeOfOdd = False
break
num /= 10
# If the number is made up of
# odd digits only
if isMadeOfOdd == True :
count += 1
# If it is the nth number
if count == n :
return i
i += 1
# Driver code
if __name__ == "__main__" :
n = 10
# Function call
print(findNthOddDigitNumber(n))
# This code is contributed by Ryuga
C
// C# program to find nth number
// made up of odd digits only
using System;
class GFG
{
// Function to return nth number
// made up of odd digits only
static int findNthOddDigitNumber(int n)
{
// Variable to keep track of
// how many such elements have
// been found
int count = 0;
for (int i = 1;; i++)
{
int num = i;
bool isMadeOfOdd = true;
// Checking each digit
// of the number
while (num != 0)
{
// If 0, 2, 4, 6 or 8 is found
// then the number is not made
// up of odd digits
if (num % 10 == 0 || num % 10 == 2 ||
num % 10 == 4 || num % 10 == 6 ||
num % 10 == 8)
{
isMadeOfOdd = false;
break;
}
num = num / 10;
}
// If the number is made up of
// odd digits only
if (isMadeOfOdd == true)
count++;
// If it is the nth number
if (count == n)
return i;
}
}
// Driver Code
static public void Main ()
{
int n = 10;
Console.WriteLine(findNthOddDigitNumber(n));
}
}
// This code is contributed
// by Ajit Deshpal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find nth number
// made up of odd digits only
// Function to return nth number
// made up of odd digits only
function findNthOddDigitNumber($n)
{
// Variable to keep track of how many
// such elements have been found
$count = 0;
$i = 1;
while (true)
{
$num = $i;
$isMadeOfOdd = true;
// Checking each digit of
// the number
while ($num != 0)
{
// If 0, 2, 4, 6 or 8 is found
// then the number is not made
// up of odd digits
if ($num % 10 == 0 or $num % 10 == 2 or
$num % 10 == 4 or $num % 10 == 6 or
$num % 10 == 8)
{
$isMadeOfOdd = false;
break;
}
$num = (int)($num / 10);
}
// If the number is made up of
// odd digits only
if ($isMadeOfOdd == true)
$count += 1;
// If it is the nth number
if ($count == $n)
return $i;
$i += 1;
}
}
// Driver code
$n = 10;
// Function call
print(findNthOddDigitNumber($n));
// This code is contributed by mits
?>
java 描述语言
<script>
// JavaScript program to find nth
// number made up of odd digits only
// Function to return nth number
// made up of odd digits only
function findNthOddDigitNumber(n)
{
// Variable to keep track of how many
// such elements have been found
let count = 0;
for(let i = 1;; i++)
{
let num = i;
let isMadeOfOdd = true;
// Checking each digit of the number
while (num != 0)
{
// If 0, 2, 4, 6 or 8 is found
// then the number is not made
// up of odd digits
if (num % 10 == 0 ||
num % 10 == 2 ||
num % 10 == 4 ||
num % 10 == 6 ||
num % 10 == 8)
{
isMadeOfOdd = false;
break;
}
num = Math.floor(num / 10);
}
// If the number is made up of
// odd digits only
if (isMadeOfOdd == true)
count++;
// If it is the nth number
if (count === n)
return i;
}
}
// Driver Code
let n = 10;
document.write(findNthOddDigitNumber(n));
// This code is contributed by Manoj.
</script>
Output:
19
方法 2(基于队列):想法是生成所有只包含奇数的数字(小于 n)。如何用奇数生成所有小于 n 的数字？我们用队列来做这个。最初，我们将“1”、“3”、“5”、“7”和“9”推送到队列中。然后我们运行一个循环，同时处理的元素数小于 n。我们逐个弹出一个项目，对于每个弹出的项目 x，我们生成下一个数字 x10 + 1、x10 + 3、x10 + 5、x10 + 7 和 x*10 + 9。我们把这些新数字排成队列。该方法的时间复杂度为 O(n) 请参考以下帖子了解该方法的实施情况。 小于 N 的二进制数字计数
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