CodeForces - 705C (16/600)

博客讲述了关于CodeForces中的一道题目,题目要求模拟手机应用通知的读取情况,包括新通知生成、特定应用的通知全部读取以及读取前t个通知的场景。博主分享了对题意的理解和解决过程。

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Thor is getting used to the Earth. As a gift Loki gave him a smartphone. There are n applications on this phone. Thor is fascinated by this phone. He has only one minor issue: he can’t count the number of unread notifications generated by those applications (maybe Loki put a curse on it so he can’t).

q events are about to happen (in chronological order). They are of three types:

Application x generates a notification (this new notification is unread).
Thor reads all notifications generated so far by application x (he may re-read some notifications).
Thor reads the first t notifications generated by phone applications (notifications generated in first t events of the first type). It’s guaranteed that there were at least t events of the first type before this event. Please note that he doesn’t read first t unread notifications, he just reads the very first t notifications generated on his phone and he may re-read some of them in this operation.
Please help Thor and tell him the number of unread notifications after each event. You may assume that initially there are no notifications in the phone.

Input
The first line of input contains two integers n and q (1 ≤ n, q ≤ 300 000) — the number of applications and the number of events to happen.

The next q lines contain the events. The i-th of these lines starts with an integer typei — type of the i-th event. If typei = 1 or typei = 2 then it is followed by an integer xi. Otherwise it is followed by an integer ti (1 ≤ typei ≤ 3, 1 ≤ xi ≤ n, 1 ≤ ti ≤ q).

Output
Print the number of unread notifications after each event.

Example
Input
3 4
1 3
1 1
1 2
2 3
Output
1
2
3
2
Input
4 6
1 2
1 4
1 2
3 3
1 3
1 3
Output
1
2
3
0
1
2
Note
In the first sample:

Application 3 generates a notification (there is 1 unread notification).
Application 1 generates a notification (there are 2 unread notifications).
Application 2 generates a notification (there are 3 unread notifications).
Thor reads the notification generated by application 3, there are 2 unread notifications left.
In the second sample test:

Application 2 generates a notification (there is 1 unread notification).
Application 4 generates a notification (there are 2 unread notifications).
Application 2 generates a notification (there are 3 unread notifications).
Thor reads first three notifications and since there are only three of them so far, there will be no unread notification left.
Application 3 generates a notification (there is 1 unread notification).
Application 3 generates a notification (there are 2 unread notifications).

就是模拟了一下。。
题意有点傻逼
他过去的也算

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
#include<set>
using namespace std;
int n, m, q, w, bj[300001], bz[300001];
set<int>tt[300001];
int main()
{
    cin >> n >> m;
    int tou = 1, wei = 0;
    int qb = 0;
    for (int a = 1; a <= m; a++)
    {
        scanf("%d%d", &q, &w);
        if (q == 1)
        {
            bj[++wei] = w;
            tt[w].insert(wei);
            qb++;
        }
        else if (q == 2)
        {
            set<int, int>::iterator it;
            for (it = tt[w].begin(); it != tt[w].end(); it++)
            {
                int sd = *it;
                if (bz[sd])continue;
                bz[sd] = 1;
                qb--;
            }
            tt[w].clear();
        }
        else
        {
            for (; tou<= w; )
            {
                if (tou > wei)break;
                if (bz[tou])
                {
                    tou++;
                    continue;
                }
                qb--;
                bz[tou] = 1;
                tou++;
            }
        }
        printf("%d\n", qb);
    }
}
### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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