非平凡波动数
原文:https://www.geeksforgeeks.org/nontrivial-undulant-numbers/
给定一个整数 N ，任务是检查 N 是否为非平凡波动数。
非平凡波动数是以 10 为基数大于 100 的数，其形式为 aba，abab，亚的斯亚贝巴，…，其中 a！= b。
示例:
输入: N = 121 输出:是 说明: 121 为 aba 形式
输入:N = 123 T3】输出:否
方法:思路是把数字转换成字符串。如果字符串的长度是偶数，那么我们将检查字符串的前半部分是否等于后半部分。如果长度是奇数，我们将在字符串的最后添加字符串的第二个字符，使其长度为偶数。最后，检查字符串的前半部分是否等于后半部分。
下面是上述方法的实现:
C++
// C++ implementation to check if N
// is a Nontrivial undulant number
#include<bits/stdc++.h>
using namespace std;
// Function to check if a string
// is double string or not
bool isDouble(int num)
{
string s = to_string(num);
int l = s.length();
// a and b should not be equal
if(s[0] == s[1])
return false;
// Condition to check
// if length is odd
// make length even
if(l % 2 == 1)
{
s = s + s[1];
l++;
}
// first half of s
string s1 = s.substr(0, l/2);
// second half of s
string s2 = s.substr(l/2);
// Double string if first
// and last half are equal
return s1 == s2;
}
// Function to check if N is an
// Nontrivial undulant number
bool isNontrivialUndulant(int N)
{
return N > 100 && isDouble(N);
}
// Driver Code
int main()
{
int n = 121;
if (isNontrivialUndulant(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to check if N
// is a Nontrivial undulant number
class GFG{
// Function to check if a string
// is double string or not
static boolean isDouble(int num)
{
String s = Integer.toString(num);
int l = s.length();
// a and b should not be equal
if(s.charAt(0) == s.charAt(1))
return false;
// Condition to check if length
// is odd make length even
if(l % 2 == 1)
{
s = s + s.charAt(1);
l++;
}
// First half of s
String s1 = s.substring(0, l / 2);
// Second half of s
String s2 = s.substring(l / 2);
// Double string if first
// and last half are equal
return s1.equals(s2);
}
// Function to check if N is an
// Nontrivial undulant number
static boolean isNontrivialUndulant(int N)
{
return N > 100 && isDouble(N);
}
// Driver code
public static void main(String[] args)
{
int n = 121;
if (isNontrivialUndulant(n))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by shubham
Python 3
# Python3 implementation to check if N
# is a Nontrivial undulant number
# Function to check if a string
# is double string or not
def isDouble(num):
s = str(num)
l = len(s)
# a and b should not be equal
if(s[0] == s[1]):
return False
# Condition to check
# if length is odd
# make length even
if(l % 2 == 1):
s = s + s[1]
l += 1
# First half of s
s1 = s[:l // 2]
# Second half of s
s2 = s[l // 2:]
# Double string if first
# and last half are equal
return s1 == s2
# Function to check if N is an
# Nontrivial undulant number
def isNontrivialUndulant(N):
return N > 100 and isDouble(N)
# Driver Code
n = 121
if (isNontrivialUndulant(n)):
print("Yes")
else:
print("No")
# This code is contributed by vishu2908
C
// C# implementation to check if N
// is a Nontrivial undulant number
using System;
class GFG{
// Function to check if a string
// is double string or not
static bool isDouble(int num)
{
String s = num.ToString();
int l = s.Length;
// a and b should not be equal
if(s[0] == s[1])
return false;
// Condition to check if length
// is odd make length even
if(l % 2 == 1)
{
s = s + s[1];
l++;
}
// First half of s
String s1 = s.Substring(0, l / 2);
// Second half of s
String s2 = s.Substring(l / 2);
// Double string if first
// and last half are equal
return s1.Equals(s2);
}
// Function to check if N is an
// Nontrivial undulant number
static bool isNontrivialUndulant(int N)
{
return N > 100 && isDouble(N);
}
// Driver code
public static void Main(String[] args)
{
int n = 121;
if (isNontrivialUndulant(n))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by gauravrajput1
java 描述语言
<script>
// JavaScript implementation to check if N
// is a Nontrivial undulant number
// Function to check if a string
// is double string or not
function isDouble(num)
{
let s = num.toString();
let l = s.length;
// a and b should not be equal
if (s[0] == s.charAt[1])
return false;
// Condition to check if length
// is odd make length even
if (l % 2 == 1)
{
s = s + s[1];
l++;
}
// First half of s
let s1 = s.substr(0, l / 2);
// Second half of s
let s2 = s.substr(l / 2);
// Double string if first
// and last half are equal
return (s1 == s2);
}
// Function to check if N is an
// Nontrivial undulant number
function isNontrivialUndulant(N)
{
return N > 100 && isDouble(N);
}
// Driver Code
let n = 121;
if (isNontrivialUndulant(n))
{
document.write("Yes");
}
else
{
document.write("No");
}
// This code is contributed by susmitakundugoaldanga
</script>
Output:
Yes
时间复杂度: O(|N|)
参考文献:OEIST4】
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