Submission #57741

#	Time	Username	Problem	Language	Result	Execution time	Memory
57741		Benq	Deda (COCI17_deda)	C++14	140 / 140	132 ms	17444 KiB

deda

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;


template<class T, int SZ> struct Seg {
    T seg[2*SZ], MN = 0;
    
    Seg() {
        F0R(i,2*SZ) seg[i] = MOD;
    }
    
    T comb(T a, T b) { return min(a,b); } // easily change this to min or max
    
    void upd(int p, T value) {  // set value at position p
        for (seg[p += SZ] = value; p > 1; p >>= 1)
            seg[p>>1] = comb(seg[(p|1)^1],seg[p|1]); // non-commutative operations
    }
    
    void build() {
        F0Rd(i,SZ) seg[i] = comb(seg[2*i],seg[2*i+1]);
    }
    
    int solve(int ind, int L, int R, int B, int Y) { 
        if (seg[ind] > Y) return MOD;
        if (L == R) return L;
        
        int M = (L+R)/2;
        if (B <= M) {
            int x = solve(2*ind,L,M,B,Y);
            if (x != MOD) return x;
        }
        if (B <= R) {
            int x = solve(2*ind+1,M+1,R,B,Y);
            if (x != MOD) return x;
        }
        return MOD;
    }
};

Seg<int,1<<18> S;
int N,Q;

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> Q;
    F0R(i,Q) {
        char c; cin >> c;
        if (c == 'M') {
            int X,A; cin >> X >> A;
            S.upd(A,X);
        } else {
            int Y,B; cin >> Y >> B;
            int ans = S.solve(1,0,(1<<18)-1,B,Y);
            if (ans == MOD) cout << "-1\n";
            else cout << ans << "\n";
        }
    }
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 140 / 140