이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define fast ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define all(v) v.begin(),v.end()
#define pb push_back
#define sz size()
#define ft first
#define sd second
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef unsigned long long ull;
const int N = 1e6 + 5;
const ll M = 1e8;
const ll inf = 1e18;
const ll mod = 1e9;
const double Pi = acos(-1);
ll binpow(ll x, ll ti) { ll res = 1;while (ti){if(ti & 1)res *= x;x *= x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll binmul(ll x, ll ti) { ll res = 0;while (ti){if(ti & 1)res += x;x += x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll nok(ll a, ll b) { return (a*b)/__gcd(abs(a),abs(b)) * (a*b > 0 ? 1 : -1); }
bool odd(ll n) { return (n % 2 == 1); }
bool even(ll n) { return (n % 2 == 0); }
struct dinamike {
ll cnt = 0, sum = 0;
} dp[N];
ll n, a[N], pr[N], cur;
multiset <pll> st[N];
const void solve(/*Armashka*/) {
cin >> n;
for (int i = 1; i <= n; ++ i) {
cin >> a[i];
pr[i] = pr[i - 1] + a[i];
dp[i] = {1, pr[i]};
}
ll cur = 1;
st[1].insert({dp[1].sum + pr[1], 1});
cur = 1;
for (int i = 2; i <= n; ++ i) {
auto it = st[cur].upper_bound({pr[i], inf});
if (it != st[cur].begin()) {
-- it;
//cout << it->sd << " ";
dp[i] = {++ cur, pr[i] - pr[it->second]};
} else {
dp[i] = {dp[i - 1].cnt, dp[i - 1].sum + a[i]};
}
st[cur].insert({dp[i].sum + pr[i], i});
}
cout << dp[n].cnt;
}
signed main() {
fast;
//freopen("divide.in", "r", stdin);
//freopen("divide.out", "w", stdout);
int tt = 1;
//cin >> tt;
while (tt --) {
solve();
}
}
/*
5 4 4
1 2
3 1
3 4
5 3
4 5 2 3
2 1 3 1
1 3 5
2 3 4 5
2 1 3 1
*/
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