给定数字N，我们必须找到将N表示为总和所需的回文数最少。
例子：
Input : N = 11
Output : 1
Explanation: 11 is itself a palindrome.
Input : N = 65
Output : 3
Explanation: 65 can be expressed as a sum of three palindromes (55, 9, 1).
在上一篇文章中，我们讨论了针对该问题的动态编程方法，该方法的时空复杂度为O(N 3/2 )。
Cilleruelo，Luca和Baxter在2016年的一篇研究论文中证明，每个数字都可以表示为在任何b> = 5的基数中最多三个回文的总和(此下限后来提高到3)。有关该定理的证明，请参阅原始论文。如果数字N本身不是回文并且不能表示为两个回文的总和，我们可以通过安全地假定答案为3来利用该定理。
下面是上述方法的实现：
C++
// C++ program to find the minimum number of
// palindromes required to express N as a sum
#include
using namespace std;
// A utility for creating palindrome
int createPalindrome(int input, bool isOdd)
{
int n = input;
int palin = input;
// checks if number of digits is odd or even
// if odd then neglect the last digit of input in
// finding reverse as in case of odd number of
// digits middle element occur once
if (isOdd)
n /= 10;
// Creates palindrome by just appending revers
// of number to itself
while (n > 0) {
palin = palin * 10 + (n % 10);
n /= 10;
}
return palin;
}
// Function to generate palindromes
vector generatePalindromes(int N)
{
vector palindromes;
int number;
// Run two times for odd and even
// length palindromes
for (int j = 0; j < 2; j++) {
// Creates palindrome numbers with first half as i.
// Value of j decides whether we need an odd length
// or even length palindrome.
int i = 1;
while ((number = createPalindrome(i++, j)) <= N)
palindromes.push_back(number);
}
return palindromes;
}
// Function to find the minimum
// number of palindromes required
// to express N as a sum
int minimumNoOfPalindromes(int N)
{
// Checking if the number is a palindrome
string a, b = a = to_string(N);
reverse(b.begin(), b.end());
if (a == b)
return 1;
// Checking if the number is a
// sum of two palindromes
// Getting the list of all palindromes upto N
vector palindromes = generatePalindromes(N);
// Sorting the list of palindromes
sort(palindromes.begin(), palindromes.end());
int l = 0, r = palindromes.size() - 1;
while (l < r) {
if (palindromes[l] + palindromes[r] == N)
return 2;
else if (palindromes[l] + palindromes[r] < N)
++l;
else
--r;
}
// The answer is three if the
// control reaches till this point
return 3;
}
// Driver code
int main()
{
int N = 65;
cout << minimumNoOfPalindromes(N);
return 0;
} Java
// Java program to find the minimum number of
// palindromes required to express N as a sum
import java.util.*;
class GFG
{
// A utility for creating palindrome
static int createPalindrome(int input, int isOdd)
{
int n = input;
int palin = input;
// checks if number of digits is odd or even
// if odd then neglect the last digit of input in
// finding reverse as in case of odd number of
// digits middle element occur once
if (isOdd % 2 == 1)
{
n /= 10;
}
// Creates palindrome by just appending revers
// of number to itself
while (n > 0)
{
palin = palin * 10 + (n % 10);
n /= 10;
}
return palin;
}
// Function to generate palindromes
static Vector generatePalindromes(int N)
{
Vector palindromes = new Vector<>();
int number;
// Run two times for odd and even
// length palindromes
for (int j = 0; j < 2; j++)
{
// Creates palindrome numbers with first half as i.
// Value of j decides whether we need an odd length
// or even length palindrome.
int i = 1;
while ((number = createPalindrome(i++, j)) <= N)
{
palindromes.add(number);
}
}
return palindromes;
}
static String reverse(String input)
{
char[] temparray = input.toCharArray();
int left, right = 0;
right = temparray.length - 1;
for (left = 0; left < right; left++, right--)
{
// Swap values of left and right
char temp = temparray[left];
temparray[left] = temparray[right];
temparray[right] = temp;
}
return String.valueOf(temparray);
}
// Function to find the minimum
// number of palindromes required
// to express N as a sum
static int minimumNoOfPalindromes(int N)
{
// Checking if the number is a palindrome
String a = String.valueOf(N);
String b = String.valueOf(N);
b = reverse(b);
if (a.equals(b))
{
return 1;
}
// Checking if the number is a
// sum of two palindromes
// Getting the list of all palindromes upto N
Vector palindromes = generatePalindromes(N);
// Sorting the list of palindromes
Collections.sort(palindromes);
int l = 0, r = palindromes.size() - 1;
while (l < r)
{
if (palindromes.get(l) + palindromes.get(r) == N)
{
return 2;
}
else if (palindromes.get(l) + palindromes.get(r) < N)
{
++l;
}
else
{
--r;
}
}
// The answer is three if the
// control reaches till this point
return 3;
}
// Driver code
public static void main(String[] args)
{
int N = 65;
System.out.println(minimumNoOfPalindromes(N));
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 program to find the minimum number of
# palindromes required to express N as a sum
# A utility for creating palindrome
def createPalindrome(_input, isOdd):
n = palin = _input
# checks if number of digits is odd or even
# if odd then neglect the last digit of _input in
# finding reverse as in case of odd number of
# digits middle element occur once
if isOdd:
n //= 10
# Creates palindrome by just appending revers
# of number to itself
while n > 0:
palin = palin * 10 + (n % 10)
n //= 10
return palin
# Function to generate palindromes
def generatePalindromes(N):
palindromes = []
# Run two times for odd and even
# length palindromes
for j in range(0, 2):
# Creates palindrome numbers with first half as i.
# Value of j decides whether we need an odd length
# or even length palindrome.
i = 1
number = createPalindrome(i, j)
while number <= N:
palindromes.append(number)
i += 1
number = createPalindrome(i, j)
return palindromes
# Function to find the minimum
# number of palindromes required
# to express N as a sum
def minimumNoOfPalindromes(N):
# Checking if the number is a palindrome
b = a = str(N)
b = b[::-1]
if a == b:
return 1
# Checking if the number is a
# sum of two palindromes
# Getting the list of all palindromes upto N
palindromes = generatePalindromes(N)
# Sorting the list of palindromes
palindromes.sort()
l, r = 0, len(palindromes) - 1
while l < r:
if palindromes[l] + palindromes[r] == N:
return 2
elif palindromes[l] + palindromes[r] < N:
l += 1
else:
r -= 1
# The answer is three if the
# control reaches till this point
return 3
# Driver code
if __name__ == "__main__":
N = 65
print(minimumNoOfPalindromes(N))
# This code is contributed by Rituraj JainC#
// C# program to find the
// minimum number of palindromes
// required to express N as a sum
using System;
using System.Collections.Generic;
class GFG{
// A utility for creating palindrome
static int createPalindrome(int input,
int isOdd)
{
int n = input;
int palin = input;
// checks if number of digits
// is odd or even if odd then
// neglect the last digit of
// input in finding reverse
// as in case of odd number of
// digits middle element occur once
if (isOdd % 2 == 1)
{
n /= 10;
}
// Creates palindrome by
// just appending revers
// of number to itself
while (n > 0)
{
palin = palin * 10 + (n % 10);
n /= 10;
}
return palin;
}
// Function to generate palindromes
static List generatePalindromes(int N)
{
List palindromes = new List();
int number;
// Run two times for
// odd and even length
// palindromes
for (int j = 0; j < 2; j++)
{
// Creates palindrome numbers
// with first half as i. Value
// of j decides whether we need
// an odd length or even length
// palindrome.
int i = 1;
while ((number = createPalindrome(i++,
j)) <= N)
{
palindromes.Add(number);
}
}
return palindromes;
}
static String reverse(String input)
{
char[] temparray = input.ToCharArray();
int left, right = 0;
right = temparray.Length - 1;
for (left = 0;
left < right; left++, right--)
{
// Swap values of left and right
char temp = temparray[left];
temparray[left] = temparray[right];
temparray[right] = temp;
}
return String.Join("", temparray);
}
// Function to find the minimum
// number of palindromes required
// to express N as a sum
static int minimumNoOfPalindromes(int N)
{
// Checking if the number
// is a palindrome
String a = String.Join("", N);
String b = String.Join("", N);
b = reverse(b);
if (a.Equals(b))
{
return 1;
}
// Checking if the number is
// a sum of two palindromes
// Getting the list of all
// palindromes upto N
List palindromes =
generatePalindromes(N);
// Sorting the list
// of palindromes
palindromes.Sort();
int l = 0,
r = palindromes.Count - 1;
while (l < r)
{
if (palindromes[l] +
palindromes[r] == N)
{
return 2;
}
else if (palindromes[l] +
palindromes[r] < N)
{
++l;
}
else
{
--r;
}
}
// The answer is three if the
// control reaches till this point
return 3;
}
// Driver code
public static void Main(String[] args)
{
int N = 65;
Console.WriteLine(minimumNoOfPalindromes(N));
}
}
// This code is contributed by Rajput-Ji 输出：
3
时间复杂度：O(√(N)log N)。