Submission #320159

#	Time	Username	Problem	Language	Result	Execution time	Memory
320159		Karen124	MP3 Player (CEOI10_mp3player)	C++14	10 / 100	1069 ms	2780 KiB

mp3player

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define ll long long int 
#define F first
#define S second
#define pb push_back
const ll N = 1e5 + 5;
const ll LOG = 30;
const ll MOD = 1e9 + 7;
const ll INF = 2e9 + 10;
int n, Vmax, V2, a[N], t[N], ansT = -1, ansV1;
vector <int> dif;
int R_Do_Operations(int e, int T){
    for (int i = n; i >= 2; i--){
        if (t[i] - t[i - 1] <= T){
            e -= a[i];
        }
        e = max(e, 0);
        e = min(e, Vmax);
    }
    return e;
}
int Do_Operations(int s, int T){
    for (int i = 2; i <= n; i++){
        if (t[i] - t[i - 1] <= T){
            s += a[i];
        }
        s = max(s, 0);
        s = min(s, Vmax);
    }  
    return s;
}
int ok(int T){
    int s = R_Do_Operations(V2, T);
    int e = Do_Operations(s, T);
    return (e == V2);
}
int main (){
    assert(n <= 4000);
    cin >> n >> Vmax >> V2;
    for (int i = 1; i <= n; i++){
        char c;
        cin >> c >> t[i];
        a[i] = ((c == '+') ? 1 : -1);
        if (i != 1) dif.pb(t[i] - t[i - 1]);
    }
    if (ok(INF)) return cout << "infinity\n", 0;
    dif.pb(0);
    sort(dif.begin(), dif.end());
    for (int x : dif){
        if (ok(x) && ansT <= x){
            ansT = x; ansV1 = R_Do_Operations(V2, x);
        }
    }
    for (int i = min(Vmax - ansV1, n); i >= 0; i--){
        if (Do_Operations(ansV1 + i, ansT) == V2){
            cout << ansT << ' ' << ansV1 + i << '\n';
            return 0;
        }
    }

    /*if (ok(INF, 0)) return cout << "infinity\n", 0;
    int l = 0, r = INF;
    while (l + 1 < r){
        int mid = (l + r) / 2;
        if (ok(mid, 0)) l = mid;
        else r = mid;
    }
    cout << l << ' ' << ok(l, 1) << '\n';*/
    return 0;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 10
• Subtask 3: 10 / 10
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10