結果
問題 | No.911 ラッキーソート |
ユーザー |
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提出日時 | 2019-10-28 21:15:20 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 74 ms / 2,000 ms |
コード長 | 3,519 bytes |
コンパイル時間 | 1,669 ms |
コンパイル使用メモリ | 174,448 KB |
実行使用メモリ | 6,528 KB |
最終ジャッジ日時 | 2024-09-14 21:19:00 |
合計ジャッジ時間 | 5,071 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 46 |
ソースコード
#include <bits/stdc++.h>using namespace std;/*#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>using namespace __gnu_pbds;template<typename T> using gpp_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;template<typename T, typename L> using gpp_map = tree<T, L, less<T>, rb_tree_tag, tree_order_statistics_node_update>;template<typename T> using gpp_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;*/struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)#define REP(i, n) FOR(i,0,n)#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)#define IREP(i, n) IFOR(i,0,n)#define Sort(v) sort(v.begin(), v.end())#define Reverse(v) reverse(v.begin(), v.end())#define all(v) v.begin(),v.end()#define SZ(v) ((int)v.size())#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))#define Max(a, b) a = max(a, b)#define Min(a, b) a = min(a, b)#define bit(n) (1LL<<(n))#define bit_exist(x, n) ((x >> n) & 1)#define debug(x) cout << #x << "=" << x << endl;#define vdebug(v) cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl;#define mdebug(m) cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout <<endl;}#define pb push_back#define f first#define s second#define int long long#define INF 1000000000000000000template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout <<endl; }; return os; }template<typename T> void Out(T x) { cout << x << endl; }template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }using vec = vector<int>;using mat = vector<vec>;using Pii = pair<int, int>;using PiP = pair<int, Pii>;using PPi = pair<Pii, int>;using bools = vector<bool>;using pairs = vector<Pii>;//int dx[4] = {1,0,-1,0};//int dy[4] = {0,1,0,-1};//char d[4] = {'D','R','U','L'};const int mod = 1000000007;//const int mod = 998244353;//#define Add(x, y) x = (x + (y)) % mod//#define Mult(x, y) x = (x * (y)) % modint calcans(mat c, int R){int ans = 0, tmp = 0;mat dp(62, vec(2, 0));dp[61][0] = 1;IREP(k, 61){int b0 = (R >> k) & 1;for(int b: c[k]){dp[k][1] += dp[k + 1][1];if(b == b0) dp[k][0] += dp[k + 1][0];else if(b < b0) dp[k][1] += dp[k + 1][0];}}return dp[0][1] + dp[0][0];}signed main(){int N, L, R; cin >> N >> L >> R;vec A(N); cin >> A;mat c(61);vec a(N);IREP(k, 61){REP(b, 2){bool f = true;REP(i, N) a[i] = (A[i] >> k) ^ b;REP(i, N - 1) if(a[i] > a[i + 1]){f = false;break;}if(f) c[k].pb(b);}if(SZ(c[k]) == 0){Out(0);return 0;}else if(c[k][0] == 1){REP(i, N) A[i] ^= bit(k);}}int ans = calcans(c, R);if(L > 0) ans -= calcans(c, L - 1);Out(ans);return 0;}