제출 #59149

#제출 시각아이디문제언어결과실행 시간메모리
59149SpeedOfMagicPalindrome-Free Numbers (BOI13_numbers)C++17
100 / 100
5 ms960 KiB
/** MIT License Copyright (c) 2018 Vasilyev Daniil **/ #include <bits/stdc++.h> using namespace std; #pragma GCC optimize("Ofast") template<typename T> using v = vector<T>; #define int long long typedef string str; typedef vector<int> vint; #define rep(a, l, r) for(int a = (l); a < (r); a++) #define pb push_back #define sz(a) ((int) a.size()) const long long inf = 4611686018427387903; //2^62 - 1 #if 0 //FileIO const string fileName = ""; ifstream fin ((fileName == "" ? "input.txt" : fileName + ".in" )); ofstream fout((fileName == "" ? "output.txt" : fileName + ".out")); #define get fin>> #define put fout<< #else #define get cin>> #define put cout<< #endif #define eol put endl void read() {} template<typename Arg,typename... Args> void read (Arg& arg,Args&... args){get (arg) ;read(args...) ;} void print(){} template<typename Arg,typename... Args> void print(Arg arg,Args... args){put (arg)<<" ";print(args...);} void debug(){eol;} template<typename Arg,typename... Args> void debug(Arg arg,Args... args){put (arg)<<" ";debug(args...);} int getInt(){int a; get a; return a;} //code goes here int solve(int r) { if (r <= 9) return r + 1; int ans = 0; int dd = 1; while (dd * 10 <= r) dd *= 10; ans += solve(dd - 1); vint digits; while (r) { digits.pb(r % 10); r /= 10; } reverse(digits.begin(), digits.end()); int n = sz(digits); int dp[n][10][11]; //dp[i][j][k] - i-th digit; j-th cur digit; k-th cur previous digit. rep(i, 0, n) { int prob1 = (i ? digits[i - 1] : 10), prob2 = (i > 1 ? digits[i - 2] : 10); bool cantGoFurther = 0; if (digits[i] == prob1 || digits[i] == prob2) cantGoFurther = 1; rep(j, 0, 10) { rep(k, 0, 11) dp[i][j][k] = 0; if (digits[i] < j || (i + 1 < n && digits[i] == j) || (i == 0 && j == 0)) continue; if (j != prob1 && j != prob2) dp[i][j][prob1] = 1; } rep(p, i + 1, n) rep(j, 0, 10) rep(k, 0, 11) { dp[p][j][k] = 0; if (k == 10) break; if (j != k) rep(l, 0, 11) if (l != j && l != k) dp[p][j][k] += dp[p - 1][k][l]; } rep(j, 0, 10) rep(k, 0, 10) ans += dp[n - 1][j][k]; if (cantGoFurther) break; } return ans; } void run() { int a, b; read(a, b); put solve(b) - solve(a - 1); } int32_t main() {srand(time(0)); ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); put fixed; put setprecision(15); run(); return 0;}
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