結果

問題 No.2901 Logical Sum of Substring
ユーザー 👑 binap
提出日時 2024-09-21 05:23:28
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,725 bytes
コンパイル時間 4,910 ms
コンパイル使用メモリ 266,776 KB
最終ジャッジ日時 2025-02-24 11:01:50
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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 * 2
other WA * 6 TLE * 1 -- * 23
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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>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;
ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << '('<< p.first << ',' << p.second << ')'; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return 
    os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); 
    return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return 
    true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder
    ::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n 
    - 1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
int MAX;
const int INF = 1001001001;
struct S{
    vector<pair<int, int>> left, right;
    int ans = INF;
    int len;
    S(int x = -1){
        len = 0;
        if(x == -1){
            
        }else{
            left = {{x, 1}};
            right = {{x, 1}};
            len++;
        }
        if(x == MAX) ans = 1;
    }
    int all(){
        if(len == 0) return 0;
        return (left.back()).first;
    }
};
S op(S a, S b){
    S res;
    res.left = a.left;
    res.right = b.right;
    res.ans = min(a.ans, b.ans);
    {
        int m = b.left.size();
        int len = a.len;
        rep(i, m){
            auto [val, time] = b.left[i];
            int prod = a.all() | val;
            if(prod == MAX) chmin(res.ans, len + 1);
            if(a.all() == prod){
                res.left.back().second += time;
            }else{
                res.left.emplace_back(prod, time);
            }
            len += time;
        }
    }
    
    {
        int m = a.right.size();
        int len = b.len;
        rep(i, m){
            auto [val, time] = a.right[i];
            int prod = val | b.all();
            if(prod == MAX) chmin(res.ans, len + 1);
            if(b.all() == prod){
                res.right.back().second += time;
            }else{
                res.right.emplace_back(prod, time);
            }
            len += time;
        }
    }
    return res;
};
S e(){
    S res;
    return res;
};
int main(){
    int n, k;
    cin >> n >> k;
    MAX = (1 << k) - 1;
    vector<int> a(n);
    cin >> a;
    int q;
    cin >> q;
    vector<S> init;
    
    rep(i, n) init.emplace_back(a[i]);
    
    segtree<S, op, e> seg(init);
    rep(_, q){
        int t;
        cin >> t;
        if(t == 1){
            int i, val;
            cin >> i >> val;
            i--;
            seg.set(i, S(val));
        }
        if(t == 2){
            int l, r;
            cin >> l >> r;
            l--; r--;
            auto res = seg.prod(l, r + 1);
            int ans = res.ans;
            if(ans == INF) cout << "-1\n";
            else cout << ans << "\n";
        }
    }
    return 0;
}
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