- Translate each rule into one explicit state update.
- Maintain the invariant after every processed item.
- Return the accumulated state once all relevant input has been handled.
Code notes
- 63 lines of C++ from the credited upstream file abc414_c.cpp.
- The implementation visibly relies on sequence storage.
- 8 loop blocks detected.
Complexity
Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.
Check the problem constraints before deciding whether this complexity will pass.
Use this to learn the idea, then write your own version.
1#include <iostream>2#include <set>3#include <vector>4 5using namespace std;6using ull = unsigned long long;7 8bool isPalindrome(ull num, int base) {9 vector<int> digits;10 ull temp = num;11 while (temp > 0) {12 digits.push_back(temp % base);13 temp /= base;14 }15 16 int len = digits.size();17 for (int i = 0; i < len / 2; i++)18 if (digits[i] != digits[len - 1 - i]) return false;19 20 return true;21}22 23void generatePalindromes(vector<ull>& palindromes, ull maxN) {24 for (int len = 1; len <= 12; len++) {25 if (len == 1) {26 for (ull i = 1; i <= 9; i++)27 if (i <= maxN) palindromes.push_back(i);28 } else {29 int halfLen = (len + 1) / 2;30 ull start = 1;31 for (int i = 1; i < halfLen; i++) start *= 10;32 33 ull end = start * 10;34 35 for (ull half = start; half < end; half++) {36 string halfStr = to_string(half);37 string palindrome = halfStr;38 39 int startIdx = (len % 2 == 0) ? halfLen - 1 : halfLen - 2;40 for (int i = startIdx; i >= 0; i--) palindrome += halfStr[i];41 42 ull num = stoull(palindrome);43 if (num <= maxN) palindromes.push_back(num);44 }45 }46 }47}48 49int main() {50 int a;51 ull n;52 cin >> a >> n;53 54 vector<ull> palindromes;55 generatePalindromes(palindromes, n);56 57 ull sum = 0;58 for (ull num : palindromes)59 if (isPalindrome(num, a)) sum += num;60 61 cout << sum << endl;62 return 0;63}