제출 #235203

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
235203		ne4eHbKa	Garaža (COCI17_garaza)	C++17	160 / 160	2724 ms	209508 KiB

garaza

//{ <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 = 3e5 + 5;

int n, q;
int a[N];

vi sieve() {
    vi primes;
    bool f[N] {};
    for(int i = 2; i * i <= 1e9; ++i) {
        if(f[i]) continue;
        primes.pb(i);
        for(int j = i * i; j < N; j += i) f[j] = true;
    }
    re primes;
}

vi primes;

inline void factorize(int v, list<pii> &f) {
    ifn(v) return;
    for(int i : primes) {
        if(i * i > v) break;
        if(!(v % i)) {
            f.pb({i, 1});
            do v /= i; while(!(v % i));
        }
    }
    if(v - 1) f.pb({v, 1});
}

struct seg {
    list<pii> p, s;
    int gcd, len;
    ll ans;
    seg(): len(0) {}
    seg(int v) {
        list<int> f;
        factorize(v, p);
        s = p;
        gcd = v;
        len = 1;
        ans = v > 1;
    }
};

typedef list<pii>::const_iterator lit;

#define F (*fi)
#define S (*se)

seg operator+ (const seg &a, const seg &b) {
    if(!a.len) re b;
    if(!b.len) re a;
    seg n {};
    n.gcd = __gcd(a.gcd, b.gcd);
    n.ans = a.ans + b.ans;
    n.len = a.len + b.len;
    lit fi = b.p.begin();
    for(const pii &z : a.p) {
        n.p.pb(z);
        if(z.sd == a.len) {
            if(b.gcd % z.ft) {
                for(; fi != b.p.end() && F.ft < z.ft; ++fi);
                if(fi != b.p.end() && F.ft == z.ft)
                    n.p.back().sd += F.sd;
            } else {
                n.p.back().sd += b.len;
            }
        }
    }
    lit se = a.s.begin();
    for(const pii &z : b.s) {
        n.s.pb(z);
        if(z.sd == b.len) {
            if(a.gcd % z.ft) {
                for(; se != a.s.end() && S.ft < z.ft; ++se);
                if(se != a.s.end() && S.ft == z.ft)
                    n.s.back().sd += S.sd;
            } else {
                n.s.back().sd += a.len;
            }
        }
    }
    fi = a.s.begin(), se = b.p.begin();
    map< int, int, greater<int> > f;
//cerr << "initial: " << n.ans << endl;
//for(pii i : a.s) cerr << i.ft _ i.sd << endl;
//cerr << endl;
//for(pii i : b.p) cerr << i.ft _ i.sd << endl;
//cerr << endl;
    while(fi != a.s.end() && se != b.p.end()) {
        if(F.ft == S.ft) {
//cerr << F.ft _ F.sd _ S.ft _ S.sd << endl;
            umax(f[F.sd], S.sd);
            ++fi, ++se;
        } else F.ft < S.ft ? fi++ : se++;
    }
//cerr << endl;
    int pr = -1, mx = 0;
    for(const pii &t : f) {
        if(~pr) n.ans += 1ll * mx * (pr - t.ft);
        pr = t.ft, umax(mx, t.sd);
    }
    n.ans += 1ll * mx * pr;
//cerr << "final: " << n.ans << endl;
//cerr << endl;
    re n;
}

struct tree {
    tree *l, *r;
    seg v;
    tree(int tl = 0, int tr = n - 1): l(r = 0), v() {
        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(a[tl]);
        }
    }
    void upd(int p, int x, int tl = 0, int tr = n - 1) {
        if(tl == tr) re void(v = *new seg(x));
        int tm = (tl + tr) >> 1;
        p <= tm
            ? l->upd(p, x, tl, tm)
            : r->upd(p, x, tm + 1, tr);
        v = l->v + r->v;
    }
    seg req(int ql, int qr, int tl = 0, int tr = n - 1) {
        if(ql > qr) re *new seg();
        if(tl == ql && tr == qr) re v;
        int tm = (tl + tr) >> 1;
        re l->req(ql, min(qr, tm), tl, tm) +
           r->req(max(tm + 1, ql), qr, tm + 1, tr);
    }
};

void solve() {
    cin >> n >> q;
    fo(n) cin >> a[i];
    primes = sieve();
    tree t {};
    while(q--) {
        int tp; cin >> tp;
        if(tp == 1) {
            int x, v;
            cin >> x >> v;
            t.upd(--x, v);
        } else {
            int l, r;
            cin >> l >> r;
            cout << t.req(--l, --r).ans << '\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;
}

컴파일 시 표준 에러 (stderr) 메시지

garaza.cpp: In function 'int m_bpow(ll, ll)':
garaza.cpp:50:26: warning: suggest parentheses around '-' inside '<<' [-Wparentheses]
     for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
                       ~~~^~~~~~~~~~~~~~~~~~~~

• Subtask 1: 40 / 40
• Subtask 2: 40 / 40
• Subtask 3: 40 / 40
• Subtask 4: 40 / 40