检查是否可以从给定的 N 创建回文字符串

给定一个数字 N。任务是从该数字创建一个小写字母字符串，并判断该字符串是否为回文。 a = 0，b = 1……等等。

例如：如果数字是 61，则子字符串“gb”将打印到 7 (6+1) 个字符，即“gbgbgbg”，并检查是否为回文。

注意：没有数字会以零开头。仅考虑字母“a 到 j”，即从 0 到 9 的单个数字。

例子：

Input: N = 61
Output: YES
Numbers 6, 1 represent letters ‘g’, ‘b’ respectively. So the substring is ‘gb’ and the sum is 7(6+1). Thus the alphabetical string formed is ‘gbgbgbg’, and is a palindrome.

Input: N = 1998
Output: NO
Numbers 1, 9, 8 represent letters ‘b’, ‘j’ and ‘i’ respectively. So the substring is ‘bjji’ and sum is 27(1+9+9+8). Thus the alphabetical string formed is bjjibjjibjjibjjibjjibjjibjj’, and is not a palindrome.

编程需要懂一点英语

方法：

获取与给定数字 N 对应的子字符串并保持其数字的总和。
追加子串，直到其长度等于 N 的位数之和。
检查获得的字符串是否为回文。
如果是回文，打印 YES。
否则，打印 NO。

下面是上述方法的实现：

C++

// C++ implementation of the
// above approach
#include
using namespace std;
  
// Function to check if a string 
// is palindrome or not
bool isPalindrome(string s)
{
    // String that stores characters
    // of s in reverse order
    string s1 = "";
  
    // Length of the string s
    int N = s.length();
  
    for (int i = N - 1; i >= 0; i--)
        s1 += s[i];
  
    if (s == s1)
        return true;
    return false;
}
  
bool createString(int N)
{
    string str = "";
    string s = to_string(N);
  
    // String used to form substring 
    // using N
    string letters = "abcdefghij";
      
    // Variable to store sum 
    // of digits of N
    int sum = 0;
    string substr = "";
  
    // Forming the substring 
    // by traversing N
    for (int i = 0; i < s.length(); i++)
    {
        int digit = s[i] - '0';
        substr += letters[digit];
        sum += digit;
    }
  
    // Appending the substr to str till 
    // it's length becomes equal to sum
    while (str.length() <= sum)
    {
        str += substr;
    }
  
    // Trimming the string str so that 
    // it's length becomes equal to sum
    str = str.substr(0, sum);
  
    return isPalindrome(str);
}
  
// Driver code
int main()
{
    int N = 61;
  
    // Calling function isPalindrome to 
    // check if str is Palindrome or not
    bool flag = createString(N);
    if (flag)
        cout << "YES";
    else
        cout << "NO";
}
  
// This code is contributed by ihritik

Java

// Java implementation of the above approach
import java.io.*;
import java.util.*;
  
public class GFG {
  
    // Function to check if a string is palindrome or not
    static boolean isPalindrome(String s)
    {
        // String that stores characters
        // of s in reverse order
        String s1 = "";
  
        // Length of the string s
        int N = s.length();
  
        for (int i = N - 1; i >= 0; i--)
            s1 += s.charAt(i);
  
        if (s.equals(s1))
            return true;
        return false;
    }
  
    static boolean createString(int N)
    {
        String str = "";
        String s = "" + N;
  
        // String used to form substring using N
        String letters = "abcdefghij";
        // Variable to store sum of digits of N
        int sum = 0;
        String substr = "";
  
        // Forming the substring by traversing N
        for (int i = 0; i < s.length(); i++) {
            int digit = s.charAt(i) - '0';
            substr += letters.charAt(digit);
            sum += digit;
        }
  
        // Appending the substr to str 
        // till it's length becomes equal to sum
        while (str.length() <= sum) {
            str += substr;
        }
  
        // Trimming the string str so that 
        // it's length becomes equal to sum
        str = str.substring(0, sum);
  
        return isPalindrome(str);
    }
  
    // Driver code
    public static void main(String args[])
    {
        int N = 61;
  
        // Calling function isPalindrome to 
        // check if str is Palindrome or not
        boolean flag = createString(N);
        if (flag)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

Python3

# Python3 implementation of 
# the above approach
  
# Function to check if a string 
# is palindrome or not
def isPalindrome(s):
  
    # String that stores characters
    # of s in reverse order
    s1 = ""
  
    # Length of the string s
    N = len(s)
    i = (N - 1)
    while(i >= 0):
        s1 += s[i]
        i = i - 1
  
    if (s == s1):
        return True
    return False
  
def createString(N):
  
    s2 = ""
    s = str(N)
  
    # String used to form
    # substring using N
    letters = "abcdefghij"
      
    # Variable to store sum 
    # of digits of N
    sum = 0
    substr = ""
  
    # Forming the substring 
    # by traversing N
    for i in range(0, len(s)) :
        digit = int(s[i])
        substr += letters[digit]
        sum += digit
      
    # Appending the substr to str till
    # it's length becomes equal to sum
    while (len(s2) <= sum):
        s2 += substr
  
    # Trimming the string str so that 
    # it's length becomes equal to sum
    s2 = s2[:sum]
  
    return isPalindrome(s2)
  
# Driver code
N = 61;
  
# Calling function isPalindrome to 
# check if str is Palindrome or not
flag = createString(N)
if (flag):
    print("YES")
else:
    print("NO")
  
# This code is contributed by ihritik

C#

// C# implementation of the
// above approach
using System;
  
class GFG 
{
  
// Function to check if a string 
// is palindrome or not
static bool isPalindrome(String s)
{
    // String that stores characters
    // of s in reverse order
    String s1 = "";
  
    // Length of the string s
    int N = s.Length;
  
    for (int i = N - 1; i >= 0; i--)
        s1 += s[i];
  
    if (s.Equals(s1))
        return true;
    return false;
}
  
static bool createString(int N)
{
    String str = "";
    String s = "" + N;
  
    // String used to form substring 
    // using N
    String letters = "abcdefghij";
      
    // Variable to store sum 
    // of digits of N
    int sum = 0;
    String substr = "";
  
    // Forming the substring 
    // by traversing N
    for (int i = 0; i < s.Length; i++) 
    {
        int digit = s[i] - '0';
        substr += letters[digit];
        sum += digit;
    }
  
    // Appending the substr to str till 
    // it's length becomes equal to sum
    while (str.Length <= sum) 
    {
        str += substr;
    }
  
    // Trimming the string str so that 
    // it's length becomes equal to sum
    str = str.Substring(0, sum);
  
    return isPalindrome(str);
}
  
// Driver code
public static void Main()
{
    int N = 61;
  
    // Calling function isPalindrome to 
    // check if str is Palindrome or not
    bool flag = createString(N);
    if (flag)
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}
  
// This code is contributed 
// by ihritik

输出：

YES