제출 #1268880

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1268880		MunkhErdene	Art Exhibition (JOI18_art)	C++17	100 / 100	625 ms	43716 KiB

art

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define ff first
#define ss second
#define _ << " " <<
#define yes cout<<"YES\n"
#define no cout<<"NO\n"
#define ull unsigned long long
#define lll __int128
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define BlueCrowner ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define FORD(i, a, b) for (ll i = (a); i >= (b); i--)
const ll mod = 1e9 + 7;
const ll mod1 = 998244353;
const ll naim = 1e9;
const ll max_bit = 60;
const ull tom = ULLONG_MAX;
const ll MAXN = 100005;
const ll LOG = 20;
const ll NAIM = 1e18;
const ll N = 2e6 + 5;
// ---------- GCD ----------
ll gcd(ll a, ll b) {
    while (b) {
        a %= b;
        swap(a, b);
    }
    return a;
}
// ---------- LCM ----------
ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}
// ---------- Modular Exponentiation ----------
ll modpow(ll a, ll b, ll m = mod) {
    ll c = 1;
    a %= m;
    while (b > 0) {
        if (b & 1) c = c * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return c;
}
// ---------- Modular Inverse (Fermat’s Little Theorem) ----------
ll modinv(ll a, ll m = mod) {
    return modpow(a, m - 2, m);
}
// ---------- Factorials and Inverse Factorials ----------
ll fact[N], invfact[N];
void pre_fact(ll n = N-1, ll m = mod) {
    fact[0] = 1;
    for (ll i = 1; i <= n; i++) fact[i] = fact[i-1] * i % m;
    invfact[n] = modinv(fact[n], m);
    for (ll i = n; i > 0; i--) invfact[i-1] = invfact[i] * i % m;
}
// ---------- nCr ----------
ll nCr(ll n, ll r, ll m = mod) {
    if (r < 0 || r > n) return 0;
    return fact[n] * invfact[r] % m * invfact[n-r] % m;
}
// ---------- Sieve of Eratosthenes ----------
vector<ll> primes;
bool is_prime[N];
void sieve(ll n = N-1) {
    fill(is_prime, is_prime + n + 1, true);
    is_prime[0] = is_prime[1] = false;
    for (ll i = 2; i * i <= n; i++) {
        if (is_prime[i]) {
            for (ll j = i * i; j <= n; j += i)
                is_prime[j] = false;
        }
    }
    for (ll i = 2; i <= n; i++)
        if (is_prime[i]) primes.pb(i);
}
void solve() {
    ll n; cin >> n;
    map<ll,ll> mp;
    FOR(i, 0, n) {
        ll x, y; cin >> x >> y;
        mp[x] += y;
    }
    ll m = mp.size();
    vector<ll> a;
    vector<ll> b;
    for (auto x : mp) { a.pb(x.first); b.pb(x.second); }
    vector<ll> p(m+1, 0);
    FOR(i, 0, m) p[i+1] = p[i] + b[i];
    ll ans = -NAIM;
    ll mn = NAIM; 
    FOR(i, 0, m) {
        mn = min(mn, p[i] - a[i]);
        ll cand = (p[i+1] - a[i]) - mn;
        ans = max(ans, cand);
    }
    cout << ans << '\n';
}
int main() {
    BlueCrowner;
    ll t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 50 / 50