Submission #799677

#	Time	Username	Problem	Language	Result	Execution time	Memory
799677		josiftepe	Exam (eJOI20_exam)	C++14	77 / 100	257 ms	99476 KiB

exam

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <set>
#include <stack>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
int n;
int a[maxn], b[maxn];
int L[maxn], R[maxn];
void solve_subtask2() {
    int x = b[0];
    int res = 0;
    for(int i = 0; i < n; i++) {
        if(a[i] == x) {
            res++;
            int j = i - 1;
            while(j >= 0 and a[j] <= x) {
                j--;
                res++;
            }
            j = i + 1;
            while(j < n and a[j] <= x) {
                res++;
                j++;
            }
            i = j - 1;
        }
    }
    cout << res << endl;
}
int dp[5005][5005];
int rec(int i, int j) {
    if(i < 0 or j < 0) {
        return 0;
    }
    if(dp[i][j] != -1) {
        return dp[i][j];
    }
    int res = 0;
    if(a[i] == b[j] and L[i] < j and R[i] > j) {
        res = max(res, rec(i, j - 1) + 1);
    }
    res = max(res, rec(i - 1, j));
    res = max(res, rec(i, j - 1));
    return dp[i][j] = res;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin >> n;
    set<int> st;
    for(int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for(int i = 0; i < n; i++) {
        cin >> b[i];
        st.insert(b[i]);
    }
    
    if((int) st.size() == 1) {
        solve_subtask2();
        return 0;
    }
    else if(n <= 5005){
        for(int i = 0; i < n; i++) {
            int idx1 = -1;
            for(int j = 0; j < i; j++) {
                if(a[i] < a[j]) {
                    idx1 = j;
                }
            }
            int idx2 = n;
            for(int j = i + 1; j < n; j++) {
                if(a[i] < a[j]) {
                    idx2 = j;
                    break;
                }
            }
            L[i] = idx1;
            R[i] = idx2;
        }
        memset(dp, -1, sizeof dp);
        cout << rec(n - 1, n - 1) << endl;
    }
    
    return 0;
}

• Subtask 1: 14 / 14
• Subtask 2: 12 / 12
• Subtask 3: 13 / 13
• Subtask 4: 0 / 23
• Subtask 5: 16 / 16
• Subtask 6: 22 / 22