Submission #253967

#	Time	Username	Problem	Language	Result	Execution time	Memory
253967		ne4eHbKa	Nekameleoni (COCI15_nekameleoni)	C++17	42 / 140	3096 ms	102008 KiB

nekameleoni

//{ <defines>

#include <bits/stdc++.h>
using namespace std;

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,-O3")

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int((x).size())
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset((x), (y), sizeof (x))

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}

int md = 998244353;

inline int m_add(int&a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_sum(int a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_mul(int&a, int b) {re a = 1ll * a * b % md;}
inline int m_prod(int a, int b) {re 1ll * a * b % md;}

int m_bpow(ll A, ll b) {
    int a = A % md;
    ll ans = 1;
    for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
        (ans *= ans) %= md;
        if(p & b)
            (ans *= a) %= md;
    }
    re ans;
}

//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int oo = 2e9;
const ll OO = 4e18;
//} </defines>
const int N = 1e5 + 5;

int n, k, m, a[N];
ll mx;

struct seg {
    int z {-1}, len {1};
    ll o {-1};
    list< pair<int, ll> > p, s;
    seg(int i = -1) { if(~i) init(i); }
    void init(int);
};

void seg::init(int f) {
    o = 1ll << a[f] - 1;
    p.clear();
    p.pb({1, o});
    s.clear();
    s.pb({1, o});
    z = k == 1 ? 1 : -1;
}

seg operator+(const seg &fi, const seg &se) {
    seg f;
    f.len = fi.len + se.len;
    f.o = fi.o | se.o;
    f.z = fi.z;
    if(f.z == -1 || se.z != -1 && se.z < f.z) f.z = se.z;
    auto ir = se.p.begin();
    for(auto il = fi.s.rbegin(); il != fi.s.rend(); ++il) {
        for(; ir != se.p.end() && (il->sd | ir->sd) != mx; ++ir);
        if(ir == se.p.end()) break;
        int z = ir->ft + il->ft;
        if(f.z < 0 || f.z > z) f.z = z;
    }
//    cerr << endl; for(auto i : fi.s) cerr << i.ft _ i.sd << endl;
//    for(auto i : se.p) cerr _ i.ft _ i.sd << endl; cerr << f.z << endl;
    f.p = fi.p;
    for(const auto &i : se.p)
        if((i.sd | fi.o) != f.p.back().sd)
            f.p.pb({fi.len + i.ft, i.sd | fi.o});
    f.s = se.s;
    for(const auto &i : fi.s)
        if((i.sd | se.o) != f.s.back().sd)
            f.s.pb({se.len + i.ft, i.sd | se.o});
    return f;
}

struct tree {
    tree *l, *r;
    seg v;
    tree(int tl = 0, int tr = n - 1) {
        if(tl < tr) {
            int tm = tl + tr >> 1;
            l = new tree(tl, tm);
            r = new tree(tm + 1, tr);
            v = l->v + r->v;
        } else {
            v = *new seg(tl);
        }
    }
    void upd(int i, int tl = 0, int tr = n - 1) {
        if(tl == tr) return v.init(i);
        int tm = tl + tr >> 1;
        i <= tm
            ? l->upd(i, tl, tm)
            : r->upd(i, tm + 1, tr);
        v = l->v + r->v;
    }
};

void solve() {
    cin >> n >> k >> m;
    mx = (1ll << k) - 1;
    fo(n) cin >> a[i];
    tree *t = new tree();
    fo(m) {
        char tp; cin >> tp;
        if(tp == '1') {
            int p, v; cin >> p >> v;
            a[--p] = v; t->upd(p);
        } else {
            cout << t->v.z << '\n';
        }
    }
}

int main() {
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    for(int test = 1; test <= tests; ++test) {
		cerr << "case #" << test << endl;
        solve();
        cerr << endl;
    }
#else
//    freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
    solve();
#endif
    return 0;
}

Compilation message (stderr)

nekameleoni.cpp: In function 'int m_bpow(ll, ll)':
nekameleoni.cpp:51:26: warning: suggest parentheses around '-' inside '<<' [-Wparentheses]
     for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
                       ~~~^~~~~~~~~~~~~~~~~~~~
nekameleoni.cpp: In member function 'void seg::init(int)':
nekameleoni.cpp:78:21: warning: suggest parentheses around '-' inside '<<' [-Wparentheses]
     o = 1ll << a[f] - 1;
                ~~~~~^~~
nekameleoni.cpp: In function 'seg operator+(const seg&, const seg&)':
nekameleoni.cpp:91:32: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
     if(f.z == -1 || se.z != -1 && se.z < f.z) f.z = se.z;
                     ~~~~~~~~~~~^~~~~~~~~~~~~
nekameleoni.cpp: In constructor 'tree::tree(int, int)':
nekameleoni.cpp:117:25: warning: suggest parentheses around '+' inside '>>' [-Wparentheses]
             int tm = tl + tr >> 1;
                      ~~~^~~~
nekameleoni.cpp: In member function 'void tree::upd(int, int, int)':
nekameleoni.cpp:127:21: warning: suggest parentheses around '+' inside '>>' [-Wparentheses]
         int tm = tl + tr >> 1;
                  ~~~^~~~

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 14 / 14
• Subtask 4: 0 / 14
• Subtask 5: 0 / 14
• Subtask 6: 0 / 14
• Subtask 7: 0 / 14
• Subtask 8: 0 / 14
• Subtask 9: 0 / 14
• Subtask 10: 0 / 14