给定一个数字 N，我们必须找到将 N 表示为它们的总和所需的最小回文数。
例子：
Input: N = 11
Output: 1
11 is itself a palindrome.
Input: N = 65
Output: 3
65 can be expressed as a sum of three palindromes (55, 9, 1).
方法：
我们可以使用动态规划来解决这个问题。这个想法是首先以排序的方式生成最多 N 的所有回文，然后我们可以将此问题视为子集和问题的变体，我们必须找到最小子集的大小，使其总和为 N .
下面是上述方法的实现：
CPP
// C++ implementation of above approach
#include
using namespace std;
// Declaring the DP table as global variable
vector > dp;
// 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 elements in a sorted
// array A[i..j] such that their sum is N
long long minimumSubsetSize(vector& A, int i, int j, int N)
{
if (!N)
return 0;
if (i > j || A[i] > N)
return INT_MAX;
if (dp[i][N])
return dp[i][N];
dp[i][N] = min(1 + minimumSubsetSize(A, i + 1, j,
N - A[i]),
minimumSubsetSize(A, i + 1, j, N));
return dp[i][N];
}
// Function to find the minimum
// number of palindromes that N
// can be expressed as a sum of
int minimumNoOfPalindromes(int N)
{
// Getting the list of all palindromes upto N
vector palindromes = generatePalindromes(N);
// Sorting the list of palindromes
sort(palindromes.begin(), palindromes.end());
// Initializing the DP table
dp = vector >(palindromes.size(),
vector(N + 1, 0));
// Returning the required value
return minimumSubsetSize(palindromes, 0,
palindromes.size() - 1, N);
}
// Driver code
int main()
{
int N = 65;
cout << minimumNoOfPalindromes(N);
return 0;
} Python3
# Python3 implementation of above approach
# Declaring the DP table as global variable
dp = [[0 for i in range(1000)] for i in range(1000)]
# A utility for creating palindrome
def createPalindrome(input, isOdd):
n = input
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 = []
number = 0
# Run two times for odd and even length palindromes
for j in range(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:
number = createPalindrome(i, j)
palindromes.append(number)
i += 1
return palindromes
# Function to find the minimum
# number of elements in a sorted
# array A[i..j] such that their sum is N
def minimumSubsetSize(A, i, j, N):
if (not N):
return 0
if (i > j or A[i] > N):
return 10**9
if (dp[i][N]):
return dp[i][N]
dp[i][N] = min(1 + minimumSubsetSize(A, i + 1, j, N - A[i]),
minimumSubsetSize(A, i + 1, j, N))
return dp[i][N]
# Function to find the minimum
# number of palindromes that N
# can be expressed as a sum of
def minimumNoOfPalindromes(N):
# Getting the list of all palindromes upto N
palindromes = generatePalindromes(N)
# Sorting the list of palindromes
palindromes = sorted(palindromes)
# Returning the required value
return minimumSubsetSize(palindromes, 0, len(palindromes) - 1, N)
# Driver code
N = 65
print(minimumNoOfPalindromes(N))
# This code is contributed by mohit kumar 29输出：
3
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