제출 #23333

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
23333		Nurlykhan	수열 (APIO14_sequence)	C++14	89 / 100	1396 ms	83668 KiB

sequence

#include <bits/stdc++.h>
 
#define pii pair<int, int>
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define ll long long 
#define ld long double
#define sz(v) int(v.size())
#define all(v) v.begin(), v.end()
 
using namespace std;
 
const int N = (int) 1e5 + 7;
const int M = (int) 2e6 + 7;
const int K = 202;
const ll LINF = (ll) 1e16;
const int INF = (int) 1e9 + 7;
const double EPS = (double) 1e-9;
ll dp[N][2], a[N];
int p[N][K];
int n, k;
 
ll sum(int l, int r) {
    if (l == 0) {
        while (1) {

        }
    }
    return a[r] - a[l - 1];
}
 
ld inter(int i, int x, int y) {
    if (a[y] == a[x]) return INF;
    return (dp[y][i % 2] - dp[x][i % 2]) * 1.0 / (a[y] - a[x]);
}
 
int st[N];
int l, r;
 
ll f(int i, int j, ll x) {
    return x *  1ll * a[i] + dp[i][j % 2];
}
 
int get(int j, int x) {
    int id = l;
    while (id + 1 <= r && f(st[id + 1], j, x) >= f(st[id], j, x)) id++;
    return st[id];
}

void solve() {
    for (int i = 1; i <= n; i++) {
        dp[i][1] = sum(1, i) * sum(i + 1, n);
    }
    for (int i = 2; i <= k; i++) {
        for (int j = i; j <= n; j++) {
            dp[j][i%2]=0;
            for (int t = 1; t < j; t++) {
                if (dp[j][i%2] < dp[t][1 - i % 2] + sum(j + 1, n) * sum(t + 1, j)) {
                    dp[j][i%2] = dp[t][1 - i % 2] + sum(j + 1, n) * sum(t + 1, j);
                    p[j][i] = t;
                }
            }
        }
    }
    int j = 1;
    for (int i = 1; i <= n; i++) {
        if (dp[j][k%2] < dp[i][k%2]) j = i;
    }
    cout << dp[j][k%2] << endl;
    vector<int> ans;
    while (j > 0 && k > 0) {
        ans.pb(j);
        j = p[j][k];
        k--;
    }
    reverse(all(ans));
    for (auto it : ans) cout << it << " ";
    exit(0);
}
 
int main() {
    #define fn "balls"
    #ifdef witch
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #else
//        freopen(fn".in", "r", stdin);
//        freopen(fn".out", "w", stdout);
    #endif
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[i] += a[i - 1];
    }
    if (n <= 200) {
        solve();
    }
    for (int i = 1; i <= n; i++) {
        dp[i][1] = sum(1, i) * sum(i + 1, n);
    }
    if (k == 1) {
        ll mx = 0;
        for (int i = 1; i <= n; i++) {
            mx = max(mx, dp[i][1]);
        }
        for (int i = 1; i <= n; i++) {
            if (mx == dp[i][1]) {
                cout << mx << "\n" << i;
                return 0;
            }
        }
    }
    for (int i = 2; i <= k; i++) {
        l = 1, r = 0;
        for (int j = 1; j <= n; j++) {
            while (l + 1 <= r && inter(i - 1, st[r], j) - inter(i - 1, st[r - 1], st[r]) > -EPS) {
                --r;
            }
            while (l + 1 <= r && f(st[l], i - 1, -sum(j + 1, n)) <= f(st[l + 1], i - 1, -sum(j + 1, n))) {
                ++l;
            }
            /*
            y[t] = (a[j] - a[t]) * (a[n] - a[j]) + dp[t][i - 1]
            a[j]*a[n]-a[j]*a[j]-a[t]*a[n]+a[t]*a[j]
            
            a[t]*(-a[n]+a[j])+dp[t][i-1]+a[j]*(a[n]-a[j])
 
            k=a[j]-a[t]
            */
            dp[j][i % 2] = 0;
            //if (l <= r) {
                for (int it = max(1, l - 2); it <= min(l + 2, r); it++) {
                    int t = st[it];
                    if (dp[j][i % 2] < dp[t][1 - i % 2] + sum(j + 1, n) * sum(t + 1, j)) {
                        dp[j][i % 2] = dp[t][1 - i % 2] + sum(j + 1, n) * sum(t + 1, j);
                        p[j][i] = t;
                    }
                }
            //}
            st[++r] = j;
        }
    }
    int j = 1;
    for (int i = 1; i <= n; i++) {
        if (dp[j][k % 2] < dp[i][k % 2]) j = i;
    }
    cout << dp[j][k % 2] << endl;
    vector<int> ans;
    while (j > 0 && k > 0) {
        ans.pb(j);
        j = p[j][k];
        k--;
    }
    reverse(all(ans));
    for (auto it : ans) cout << it << " ";
    return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 0 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 29 / 29