제출 #380159

#제출 시각아이디문제언어결과실행 시간메모리
380159nguyen31hoang08minh2003Palindrome-Free Numbers (BOI13_numbers)C++14
100 / 100
2 ms512 KiB
#include <iostream> #include <fstream> #include <sstream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cctype> #include <cmath> #include <math.h> #include <string.h> #include <string> #include <queue> #include <stack> #include <vector> #include <deque> #include <set> #include <bitset> #include <iterator> #include <functional> #include <algorithm> #include <iomanip> #include <map> #include <ctime> #define fore(i, a, b) for(ll i = (a), _b = (b); i < (_b); i++) #define fort(i, a, b) for(ll i = (a), _b = (b); i <= (_b); i++) #define ford(i, a, b) for(ll i = (a), _b = (b); i >= (_b); i--) #define all(x) (x).begin(),(x).end() #define rall(x) (x).rbegin(),(x).rend() #define sz(x) ((int)(x).size()) #define pb push_back #define pf push_front #define fi first #define se second using namespace std; template<class A,class B> bool mini(A &a,const B b){return a > b ? (a = b, 1) : 0; } template<class A,class B> bool maxi(A &a,const B b){return a < b ? (a = b, 1) : 0; } typedef unsigned long long ull; typedef long long ll; typedef long double ld; typedef vector<ll> vi; typedef pair<ll,ll> ii; typedef vector<ii> vii; typedef vector<vi> vvi; typedef vector<vii> vvii; ll n, a, b, T, dp[25][2][12][12][2], vis[25][2][12][12][2]; string t; ll DP(ll i, bool f, ll x, ll y, bool st) { if (i >= n) return 1; #define v i][f][x][y][st if (vis[v] != T) { vis[v] = T; dp[v] = 0; const int d = f ? 9 : (t[i] - '0'); fort(j, 0, d) if (!j) { if (st) dp[v] += DP(i + 1, f || (j < d), 10, 10, true); else if (x != j && y != j) dp[v] += DP(i + 1, f || (j < d), y, j, false); } else { if (x != j && y != j) dp[v] += DP(i + 1, f || (j < d), y, j, false); } } return dp[v]; #undef v } ll calc(ll x) { stringstream ss; ss << x; ss >> t; ++T; n = sz(t); return DP(0, false, 10, 10, true); } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); cin >> a >> b; if (!a) cout << calc(b) << '\n'; else cout << calc(b) - calc(a - 1) << '\n'; return 0; }
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