Submission #407421

#	Time	Username	Problem	Language	Result	Execution time	Memory
407421		534351	Cigle (COI21_cigle)	C++17	48 / 100	1092 ms	10204 KiB

cigle

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()

const int MAXN = 5013;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N;
int arr[MAXN], pref[MAXN];
int dp[MAXN][MAXN];
int ans;

int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    cin >> N;
    FOR(i, 0, N)
    {
        cin >> arr[i];
        pref[i + 1] = pref[i] + arr[i];
    }
    dp[0][0] = 0;
    FOR(i, 1, N + 1)
    {
        FOR(j, i, N + 1)
        {
            FOR(k, 1, i)
            {
                ckmax(dp[i][j], dp[k][i - 1]);
            }
            int c = 2 * pref[i - 1] - pref[j];
            int idx = LB(pref, pref + N + 1, c) - pref;
            // cerr << "dp " << i << ' ' << j << " = " << dp[i][j] << endl;
            if (pref[idx] == c)
            {
                //c(x, i-1) += 1 for all x <= idx.
                // #warning check this
                FOR(k, 0, idx + 1)
                {
                    // cerr << "ADD " << k << ' ' << i - 1 << endl;
                    dp[k][i - 1]++;
                }
            }
            //i...j. that means you use i-1....j-1.
            //count how many s i-2...k = si-1...c
            //pref[i-1] - pref[k] = pref[j] - pref[i - 1].
            //pref[k] = 2 pref[i - 1] - pref[j]
            //find dp[i][j].
            //dp[i][j] = max(dp[i][j], dp[k][i - 1] + c(k, i, j))
            //c(k, i, j) is # of positions l such that s[k] +
            //there's always some poition
        }
    }
    FOR(i, 1, N + 1)
    {
        ckmax(ans, dp[i][N]);
    }
    cout << ans << '\n';
    return 0;
}

• Subtask 1: 9 / 9
• Subtask 2: 11 / 11
• Subtask 3: 13 / 13
• Subtask 4: 15 / 15
• Subtask 5: 0 / 52