Submission #1010183

#	Time	Username	Problem	Language	Result	Execution time	Memory
1010183		dpsaveslives	Global Warming (CEOI18_glo)	C++17	0 / 100	40 ms	5516 KiB

glo

#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int N,X; cin >> N >> X;
    vector<int> arr(N);
    for(int i = 0;i<N;++i){
        cin >> arr[i];
    }
    vector<int> dp(N+1,INF), L(N,1);
    dp[0] = -INF; int ans = 1;
    for(int i = 0;i<N;++i){
        int ind = lower_bound(dp.begin(),dp.end(),arr[i])-dp.begin(); //first that is >= arr[i]
        dp[ind] = arr[i];
        L[i] = ind+1;
        ans = max(ans,L[i]);
    }
    dp = vector<int>(N+1,INF);
    dp[0] = -INF;
    for(int i = N-1;i>=0;--i){ //1 to i have been decreased by X and we want to find the LDS from N-1 to i -> this is equivalent to increasing i by X
        int ind = lower_bound(dp.begin(),dp.end(),-(arr[i]+X))-dp.begin(); //we want to find the negative version so that we can still lower bound
        ans = max(ans,L[i]+ind); //no -1 because in ind, i was not counted
        ind = lower_bound(dp.begin(),dp.end(),-arr[i])-dp.begin();
        dp[ind] = -arr[i];
    }
    cout << ans << "\n";
    return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 10
• Subtask 3: 0 / 13
• Subtask 4: 0 / 10
• Subtask 5: 0 / 20
• Subtask 6: 0 / 17
• Subtask 7: 0 / 25