题解

#include <iostream>
using namespace std;
int n,m,k,op,x;
int h[1010],vis[1010];
void c1(int x) {
	vis[x] = k; 
}
void c2(int x) {
	vis[x] = 0; 
}
void c3(int x) {
	cout << h[x] << endl; 
}

int main() {
	cin >> n >> m >> k;
	for (int i = 0;i < m;i ++) {
		cin >> op;
		cin >> x;
		switch (op) {
			case 1: c1(x); break;
			case 2: c2(x); break;
			case 3: c3(x); break;
		}
			for (int i = 1;i <= n;i ++){
				if (vis[i] == 0){
					h[i] += 1;
				} else {
					h[i] += 2;
					vis[i] -- ;
				}
			}
	}
	return 0;
}