使用C ++将N表示为总和所需的最小回文数。
问题陈述
给定数字N，我们必须找到将N表示为总和所需的最小回文数
如果N=15，则需要2个回文，即8和7。
算法
1. Generate all the palindromes up to N in a sorted fashion 2. Find the size of the smallest subset such that its sum is N
示例
#include <iostream>
#include <vector>
#include <climits>
#include <algorithm>
using namespace std;
vector<vector<long long>> table;
int createPalindrome(int input, bool isOdd){
int n = input;
int palindrome = input;
if (isOdd)
n /= 10;
while (n > 0) {
palindrome = palindrome * 10 + (n % 10);
n /= 10;
}
return palindrome;
}
vector<int>generatePalindromes(int n){
vector<int> palindromes;
int number;
for (int j = 0; j < 2; j++) {
int i = 1;
while ((number = createPalindrome(i++, j)) <= n)
palindromes.push_back(number);
}
return palindromes;
}
long long minSubsetSize(vector<int>& vec, int i, int j, int n){
if (n == 0)
return 0;
if (i > j || vec[i] > n)
return INT_MAX;
if (table[i][n])
return table[i][n];
table[i][n] = min(1 + minSubsetSize(vec, i + 1, j, n - vec[i]), minSubsetSize(vec, i + 1, j, n));
return table[i][n];
}
int requiredPalindromes(int n){
vector<int> palindromes = generatePalindromes(n);
sort(palindromes.begin(), palindromes.end());
table = vector<vector<long long>>(palindromes.size(),
vector<long long>(n + 1, 0));
return minSubsetSize(palindromes, 0, palindromes.size() - 1, n);
}
int main(){
int n = 15;
cout << "Minimum required palindromes = " <<
requiredPalindromes(n) << endl;
return 0;
}输出结果
当您编译并执行上述程序时。它生成以下输出-
Minimum required palindromes = 2