結果

問題 No.3075 Mex Recurrence Formula
ユーザー MMRZ
提出日時 2025-03-28 22:03:54
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 143 ms / 2,000 ms
コード長 4,840 bytes
コンパイル時間 3,531 ms
コンパイル使用メモリ 284,928 KB
実行使用メモリ 9,740 KB
最終ジャッジ日時 2025-03-28 22:04:09
合計ジャッジ時間 8,453 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 46
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

# include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const double pi = acos(-1);
template<class T>constexpr T inf() { return ::std::numeric_limits<T>::max(); }
template<class T>constexpr T hinf() { return inf<T>() / 2; }
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); }
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); }
template<class T> bool chmin(T& a,T b) { if(a > b){a = b; return true;} return false; }
template<class T> bool chmax(T& a,T b) { if(a < b){a = b; return true;} return false; }
int popcnt(unsigned long long n) { int cnt = 0; for (int i = 0; i < 64; i++)if ((n >> i) & 1)cnt++; return cnt; }
int d_sum(ll n) { int ret = 0; while (n > 0) { ret += n % 10; n /= 10; }return ret; }
int d_cnt(ll n) { int ret = 0; while (n > 0) { ret++; n /= 10; }return ret; }
ll gcd(ll a, ll b) { if (b == 0)return a; return gcd(b, a%b); };
ll lcm(ll a, ll b) { ll g = gcd(a, b); return a / g*b; };
ll MOD(ll x, ll m){return (x%m+m)%m; }
ll FLOOR(ll x, ll m) {ll r = (x%m+m)%m; return (x-r)/m; }
template<class T> using dijk = priority_queue<T, vector<T>, greater<T>>;
# define all(qpqpq)           (qpqpq).begin(),(qpqpq).end()
# define UNIQUE(wpwpw)        (wpwpw).erase(unique(all((wpwpw))),(wpwpw).end())
# define LOWER(epepe)         transform(all((epepe)),(epepe).begin(),TL<char>)
# define UPPER(rprpr)         transform(all((rprpr)),(rprpr).begin(),TU<char>)
# define rep(i,upupu)         for(ll i = 0, i##_len = (upupu);(i) < (i##_len);(i)++)
# define reps(i,opopo)        for(ll i = 1, i##_len = (opopo);(i) <= (i##_len);(i)++)
# define len(x)                ((ll)(x).size())
# define bit(n)               (1LL << (n))
# define pb push_back
# define eb emplace_back
# define exists(c, e)         ((c).find(e) != (c).end())
struct INIT{
    INIT(){
        std::ios::sync_with_stdio(false);
        std::cin.tie(0);
        cout << fixed << setprecision(20);
    }
}INIT;
namespace mmrz {
    void solve();
}
int main(){
    mmrz::solve();
}
#define debug(...) (static_cast<void>(0))
using namespace mmrz;
template<typename T>struct segment_tree {
    using F = function<T(T, T)>;
    int offset;
    int n;
    vector<T> node;
    F combine;
    T identify;
    segment_tree(int _n, F _combine, T _identify) : segment_tree(vector<T>(_n, _identify), _combine, _identify) {}
    segment_tree(const vector<T> &v, F _combine, T _identify) : n((int)v.size()), combine(_combine), identify(_identify) {
        offset = 1;
        while(offset < n)offset <<= 1;
        node.resize(2*offset, identify);
        for(int i = 0;i < n;i++)node[i + offset] = v[i];
        for(int i = offset - 1;i >= 1;i--)node[i] = combine(node[2 * i + 0], node[2 * i + 1]);
    }
    T operator[](int x) {return node[x + offset]; }
    void set(int x, T val){
        assert(0 <= x && x < n);
        x += offset;
        node[x] = val;
        while(x >>= 1){
            node[x] = combine(node[2 * x + 0], node[2 * x + 1]);
        }
    }
    T fold(int l, int r){
        assert(0 <= l && l <= r && r <= n);
        if(l == r)return identify;
        T L = identify, R = identify;
        for(l += offset, r += offset; l < r;l >>= 1, r >>= 1){
            if(l&1)L = combine(L, node[l++]);
            if(r&1)R = combine(node[--r], R);
        }
        return combine(L, R);
    }
    T all_fold() { return node[1]; };
    int max_right(const function<bool(T)> f, int l = 0){
        assert(0 <= l && l <= n);
        assert(f(identify));
        if(l == n)return n;
        
        l += offset;
        T sum = identify;
        do{
            while(l%2 == 0)l >>= 1;
            if(not f(combine(sum, node[l]))){
                while(l < offset){
                    l <<= 1;
                    if(f(combine(sum, node[l]))){
                        sum = combine(sum, node[l]);
                        l++;
                    }
                }
                return l - offset;
            }
            sum = combine(sum, node[l]);
            l++;
        }while((l&-l) != l);
        return n;
    }
    int min_left(const function<bool(T)> f, int r = -1){
        if(r == 0)return 0;
        if(r == -1)r = n;
        r += offset;
        T sum = identify;
        do{
            --r;
            while(r > 1 && (r % 2))r >>= 1;
            if(not f(combine(node[r], sum))){
                while(r < offset){
                    r = r*2 + 1;
                    if(f(combine(node[r], sum))){
                        sum = combine(node[r], sum);
                        --r;
                    }
                }
                return r+1 - offset;
            }
            sum = combine(node[r], sum);
        }while((r&-r) != r);
        return 0;
    }
};
void SOLVE(){
    ll n, m;
    cin >> n >> m;
    vector<int> a(n);
    for(auto &x : a){
        cin >> x;
    }
    m--;
    if(m < n){
        cout << a[m] << '\n';
        return;
    }
    for(auto &x : a){
        if(x >= n+2){
            x = n+2;
        }
    }
    ll md = (m-n)%(n+1);
    if(md == 0)md = n+1;
    segment_tree<int> seg(n+10, [](int l, int r){return min(l, r); }, hinf<int>());
    rep(i, n+1)seg.set(i, 0);
    rep(i, n)seg.set(a[i], seg[a[i]]+1);
    for(int i = n;i <= n+n+1;i++){
        int g = seg.max_right([&](int val){return val >= 1;}, 0);
        debug(g);
        a.pb(g);
        debug(a);
        seg.set(a[i-n], seg[a[i-n]]-1);
        seg.set(a[i], seg[a[i]]+1);
    }
    cout << a[n+md] << '\n';
    debug(a);
}
void mmrz::solve(){
    int t = 1;
    //cin >> t;
    while(t--)SOLVE();
}
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