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  1. #include <iostream>
  2. #include <cstring>
  3. #include <string>
  4.  
  5. using namespace std;
  6.  
  7. typedef long long ll;
  8.  
  9. string N;
  10. ll memo[20][11][11][2][2];
  11.  
  12. ll dp(int pos, int prev1, int prev2, bool tight, bool leading_zero) {
  13.     if (pos == N.size()) {
  14.         return 1; // We have formed a valid number
  15.     }
  16.     if (memo[pos][prev1 + 1][prev2 + 1][tight][leading_zero] != -1) {
  17.         return memo[pos][prev1 + 1][prev2 + 1][tight][leading_zero];
  18.     }
  19.     ll res = 0;
  20.     int limit = tight ? N[pos] - '0' : 9;
  21.     for (int d = 0; d <= limit; ++d) {
  22.         bool new_tight = tight && (d == limit);
  23.         bool new_leading_zero = leading_zero && (d == 0);
  24.         int new_prev1 = prev1, new_prev2 = prev2;
  25.         if (new_leading_zero) {
  26.             // Leading zeros, prev1 and prev2 remain -1
  27.             res += dp(pos + 1, -1, -1, new_tight, true);
  28.         } else {
  29.             // Check for palindromic substrings
  30.             if (prev1 == d) continue; // Palindrome of length 2
  31.             if (prev2 == d) continue; // Palindrome of length 3
  32.             new_prev2 = prev1;
  33.             new_prev1 = d;
  34.             res += dp(pos + 1, new_prev1, new_prev2, new_tight, false);
  35.         }
  36.     }
  37.     return memo[pos][prev1 + 1][prev2 + 1][tight][leading_zero] = res;
  38. }
  39.  
  40. ll count(string num) {
  41.     N = num;
  42.     memset(memo, -1, sizeof(memo));
  43.     return dp(0, -1, -1, true, true);
  44. }
  45.  
  46. int main() {
  47.     ll a, b;
  48.     cin >> a >> b;
  49.     if (a > b) swap(a, b); // Ensure a <= b
  50.     ll ans = count(to_string(b)) - (a > 0 ? count(to_string(a - 1)) : 0);
  51.     cout << ans << endl;
  52.     return 0;
  53. }
  54.  
Success #stdin #stdout 0.01s 5276KB
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https://ideone.com/7O9qKi
language:
C++14 (gcc 8.3)
created:
7 days ago
visibility:
public

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