给定一个整数N ，任务是找到第N个偶数长度的偶数回文数，该数仅由数字X和Y组成，其中X，Y> 0 。
例子：
Input: N = 9, X = 4, Y = 5
Output: 454454
Explanation:
Even length palindromic numbers using 4 & 5 are
44, 55, 4444, 4554, 5445, 5555, 444444, 445544, 454454, …
9th term of the above series = 454454
Input: N = 6, X = 1, Y = 2
Output: 2222
Explanation:
Even length palindromic numbers using 1 & 2 are
11, 22, 1111, 1221, 2112, 2222, 111111, 112211, 121121, …
6th term of the above series = 2222
方法：
- 使用X＆Y的偶数回文数为
XX, YY, XXXX, XYYX, YXXY, YYYY, XXXXXX, XXYYXX, ... - 上面的序列可以观察为：
XX, -> Length (L) = 2 YY, -> Length (L) = 2 XXXX, -> Length (L) = 4 XYYX, -> Length (L) = 4 YXXY, -> Length (L) = 4 YYYY, -> Length (L) = 4 XXXXXX, -> Length (L) = 6 XXYYXX, -> Length (L) = 6 XYXXYX, -> Length (L) = 6 XYYYYX, -> Length (L) = 6 YXXXXY, -> Length (L) = 6 YXYYXY, -> Length (L) = 6 YYXXYY, -> Length (L) = 6 YYYYYY, -> Length (L) = 6 XXXXXXXX, -> Length (L) = 8 ... - 如果我们将任何一项分为两半，则下半部分与上半部分恰好相反
例子：Taking the term XXYYXX Dividing this into 2 halves XXYYXX = XXY | YXX So YXX is just the reverse of XXY - 仅取项的左半部分，将X = 0和Y = 1得到二进制字符串，可以看到长度为L的数字形成一个从0到(2 L / 2 – 1)的整数序列，作为等级(R) 。因此0＆leq; R＆leq; 2升/ 2 – 1
因此，可以观察到以下顺序：
L -> Left Half -> Binary String -> Rank (in Decimal) 2 -> X -> 0 -> 0 2 -> Y -> 1 -> 1 4 -> XX -> 00 -> 0 4 -> XY -> 01 -> 1 4 -> YX -> 10 -> 2 4 -> YY -> 11 -> 3 6 -> XXX -> 000 -> 0 6 -> XXY -> 001 -> 1 6 -> XYX -> 010 -> 2 6 -> XYY -> 011 -> 3 6 -> YXX -> 100 -> 4 6 -> YXY -> 101 -> 5 6 -> YYX -> 110 -> 6 6 -> YYY -> 111 -> 7 8 -> XXXX -> 0000 -> 0 ... - 因此，对于所需项N：
- 所需的第N个项的长度(L) ：
- 所需第N个词的等级(R) ：
- 所需的第N个项的前一半= R在L / 2位中的二进制表示，方法是将0替换为X，将1替换为Y
- 所需的第n项的下半年=前半叶的反
例子：
If N = 9, -> L = 2(ceil(log2(N + 2)) - 1) = 6 -> R = N - 2L/2 + 1 = 2 -> First Half = Binary (2) = 010 = XYX -> Second Half = Reverse (First Half) = XYX -> Complete Nth term = First Half + Second Half = XYXXYX which is the required Nth term. - 所需的第N个项的长度(L) ：
下面是上述方法的实现：
C++
// C++ program to find nth even
// palindromic number of only even
// length composing of 4's and 5's.
#include
using namespace std;
// Utility function to compute
// n'th palindrome number
string solve(int n, char x, char y)
{
// Calculate the length from above
// formula as discussed above
int length = ceil(log2(n + 2)) - 1;
// Calculate rank for length L
int rank = n - (1 << length) + 1;
string left = "", right = "";
for (int i = length - 1; i >= 0; i--) {
// Mask to check if i't bit
// is set or not
int mask = 1 << i;
// If bit is set append '5' else append '4'
bool bit = mask & rank;
if (bit) {
left += y;
right += y;
}
else {
left += x;
right += x;
}
}
reverse(right.begin(), right.end());
return left + right;
}
// Driver Code
int main()
{
int n = 23;
char x = '4', y = '5';
string ans = solve(n, x, y);
cout << ans << '\n';
return 0;
} Java
// Java program to find nth even
// palindromic number of only even
// length composing of 4's and 5's.
import java.util.*;
class GFG
{
// Utility function to compute
// n'th palindrome number
static String solve(int n, char x, char y)
{
// Calculate the length from above
// formula as discussed above
int length = (int)Math.ceil(Math.log(n + 2) /
Math.log(2)) - 1;
// Calculate rank for length L
int rank = n - (1 << length) + 1;
String left = "", right = "";
for (int i = length -1 ; i >= 0; i--)
{
// Mask to check if i't bit
// is set or not
int mask = (1 << i);
// If bit is set append '5' else append '4'
int bit = mask & rank;
if (bit > 0)
{
left += y;
right += y;
}
else
{
left += x;
right += x;
}
}
StringBuilder sb = new StringBuilder(right);
sb.reverse();
right = sb.toString();
String res = left + right;
return res;
}
// Driver Code
public static void main (String[] args)
{
int n = 23;
char x = '4', y = '5';
String ans = solve(n, x, y);
System.out.println(ans);
}
}
// This code is contributed by AnkitRai01Python3
# Python3 program to find nth even
# palindromic number of only even
# length composing of 4's and 5's.
from math import ceil, log2
# Utility function to compute
# n'th palindrome number
def solve(n, x, y) :
# Calculate the length from above
# formula as discussed above
length = ceil(log2(n + 2)) - 1;
# Calculate rank for length L
rank = n - (1 << length) + 1;
left = ""; right = "";
for i in range(length - 1 , -1, -1):
# Mask to check if i't bit
# is set or not
mask = (1 << i);
# If bit is set append '5'
# else append '4'
bit = (mask & rank);
if (bit) :
left += y;
right += y;
else :
left += x;
right += x;
right = right[::-1];
res = left + right;
return res;
# Driver Code
if __name__ == "__main__" :
n = 23;
x = '4';
y = '5';
ans = solve(n, x, y);
print(ans);
# This code is contributed by kanugargngC#
// C# program to find nth even
// palindromic number of only even
// length composing of 4's and 5's.
using System;
class GFG
{
// Utility function to compute
// n'th palindrome number
static String solve(int n, char x, char y)
{
// Calculate the length from above
// formula as discussed above
int length = (int)Math.Ceiling(Math.Log(n + 2) /
Math.Log(2)) - 1;
// Calculate rank for length L
int rank = n - (1 << length) + 1;
String left = "", right = "";
for (int i = length -1; i >= 0; i--)
{
// Mask to check if i't bit
// is set or not
int mask = (1 << i);
// If bit is set append '5'
// else append '4'
int bit = mask & rank;
if (bit > 0)
{
left += y;
right += y;
}
else
{
left += x;
right += x;
}
}
right = reverse(right);
String res = left + right;
return res;
}
static String reverse(String input)
{
char[] a = input.ToCharArray();
int l, r = 0;
r = a.Length - 1;
for (l = 0; l < r; l++, r--)
{
// Swap values of l and r
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("", a);
}
// Driver Code
public static void Main (String[] args)
{
int n = 23;
char x = '4', y = '5';
String ans = solve(n, x, y);
Console.WriteLine(ans);
}
}
// This code is contributed by Rajput-Ji时间复杂度： 
其中n是字符串的长度