제출 #1193867

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1193867		Mkswll	Feast (NOI19_feast)	C++20	71 / 100	1002 ms	34488 KiB

feast

// I will become better someday.

#include <bits/stdc++.h>
using namespace std;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, int> pli;
typedef pair <int, ll> pil;
typedef pair <ll, ll> pll;
typedef pair <ld, ld> pdd;
#define debug(x) cout << '[' << #x << ": " << x << "] "
#define cio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define cases int _; cin >> _; while(_--)
#define pb push_back
#define eb emplace_back
#define space << " " <<
#define lb lower_bound
#define ub upper_bound
#define F first
#define S second
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define Unique(v) v.erase(unique(all(v)), v.end())
#define mset(x) memset(x, 0, sizeof(x))
#define sflush fflush(stdout)
#define cflush cout.flush()
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define lyes cout << "Yes\n"
#define lno cout << "No\n"
#define nl cout << "\n";
#define vt vector
#define ar array
#define uid uniform_int_distribution 

template <typename T> 
istream& operator >> (istream& in, vector<T>& a){
	for(auto &x : a) in >> x; 
	return in;
}

template <typename T> 
ostream& operator << (ostream& out, vector<T>& a){
	for(auto &x : a) out << x << ' '; 
	return out;
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());

ll read(){
    int w = 1, c;
    ll ret;
    while((c = getchar()) > '9' || c < '0'){
        w = (c == '-' ? -1 : 1);
    }
    ret = c - '0';
    while((c = getchar()) >= '0' && c <= '9'){
        ret = ret * 10 + c - '0';
    }
    return ret * w;
}

ll rd(){
	ll in;
	cin >> in;
	return in;
}
 
void write(ll x){
    if(x < 0){
        putchar('-');
        x = -x;
    }
    if(x > 9){
        write(x / 10);
    }
    putchar(x % 10 + '0');
}

const int MAXN = 3e5 + 5, MAXM = 2e5 + 5, INF = 1e9 + 5, MOD = 1e9 + 7;
const ll LMOD = (ll) 1e18 + 9;
const ll LINF = 1e18 + 5;
const ld ep = 1e-8, Pi = acos(-1.0);

int n, m, k, x; 
int a[MAXN];
string s;


pli solve(ll pen){
	vt dp(n + 1, vt <ll> (2, 0));
	vt cdp(n + 1, vt <int> (2, 0));
	dp[0][1] = -pen;
	cdp[0][1] = 1;
	for(int i = 1; i <= n; ++i){
		dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]);
		if(dp[i][0] == dp[i - 1][0]) cdp[i][0] = max(cdp[i][0], cdp[i - 1][0]);
		if(dp[i][0] == dp[i - 1][1]) cdp[i][0] = max(cdp[i][1], cdp[i - 1][1]);
		
		dp[i][1] = max(dp[i - 1][0] + a[i] - pen, dp[i - 1][1] + a[i]);
		if(dp[i][1] == dp[i - 1][0] + a[i] - pen) cdp[i][1] = max(cdp[i][1], cdp[i - 1][0] + 1);
		if(dp[i][1] == dp[i - 1][1] + a[i]) cdp[i][1] = max(cdp[i][1], cdp[i - 1][1]);
	}
	return max((pli) {dp[n][0], cdp[n][0]}, (pli) {dp[n][1], cdp[n][1]});
}

void clear(){
    
}

int main(){
    cio;
    cin >> n >> k;
    int cnt = 0;
    ll sm = 0;
    for(int i = 1; i <= n; ++i){
    	cin >> a[i];
    	// if(a[i] >= 0){
    		// ++cnt;
    		// sm += a[i];
    	// }
    }
    // if(cnt <= k){
    	// cout << sm << "\n";
    	// return 0;
    // }
    ll l = 0, r = 1e15, ans = 0;
    while(l <= r){
    	ll mid = (l + r) >> 1;
    	auto res = solve(mid);
    	if(res.S >= k){
    		ans = mid;
    		l = mid + 1;
    	}
    	else r = mid - 1;
    }
    auto res = solve(ans);
    cout << res.F + ans * k << "\n";
    
    return 0;
}   

• Subtask 1: 4 / 4
• Subtask 2: 8 / 8
• Subtask 3: 18 / 18
• Subtask 4: 10 / 10
• Subtask 5: 11 / 11
• Subtask 6: 20 / 20
• Subtask 7: 0 / 29