最快，不服来战

显然，这是一道模拟。

Obviously, this is an imitate question.

可是，这道题的题面有些坑。

Fyxj, there are a lot of tricks in the question.

我们不需要找到最优解，

Finding the best solution isn't our work,

我们只需要贪心地有券用券。

Else than using the tickets greedily.

运行，提交，AC，我很快就完成了这道题。

Run, Submit, Accepted, I finished this question quickly.

现在给上我的第一版代码。

My first edition of code is given now.

// Execute in 2023-09-21 20:31:30.

#include <stdio.h>

struct tic{
    int t, m;
};
tic queue[114514];
int head=0, tail=0, tmptail=0;
int n, s=0, tmp;
int type, price, time__;

int main(){
	scanf("%d", &n);
	while(n--){
		scanf("%d%d%d", &type, &price, &time__);
		if(type==0){
			s += price;
			queue[tail].t = time__;
            queue[tail].m = price;
            tail++;
		}
		else{
			tmp = 0;
			tmptail = tail;
			while(head<tmptail){
				if(time__-queue[head].t<=45){
                    if(price<=queue[head].m){
						head++;
						tmp = 1;
						break;
					}
					queue[tail++] = queue[head];
				}
				head++;
			}
			if(!tmp) s += price;
            while(head<tmptail){
				queue[tail++] = queue[head++];
			}
		}
        if(head==tail) head=tail=0;
	}
	printf("%d", s);
	
	return 0;
}

可是，它运行了 28ms。

However, it had run for 28ms.

考虑增加优化。

Addition of optimization be thought.

$q$ 数组的长度太臭了，

Length of array $q$ is too awful,

最后我们把它缩短至 $100001$。

Finally, we shorten it into $100001$.

但是，这不能优化多少。

However, this can't optimize a lot.

嗯……对了，可以用快读！

Ummm... Yes, We can use "fast read"!

正常的快读是这样的：

Normal fast read is like this:

inline int read(){
	int x=0;bool w=0;
	char ch=gc();
	while(ch<'0'||ch>'9'){
		if(ch=='-') w=1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		x=(x<<3)+(x<<1)+(ch^48);
		ch=getchar();
	}
	return w?-x:x;
}

读数字是读整数中最耗时的一步。

Digit reading is the most time-consuming method in interger reading.

最近，我在 OI wiki 上看到了一种使用 fread() 的快读。

Recently, I saw a kind of fast reading which using fread().

把许多次读数字合并成一次是个好主意。

Unite a lot of digit readings into one time is a good idea.

当~这就是最终代码：（3ms, 2.51MB, $147$ 行）

Dang~This is the final code: (3ms, 2.51MB, $147$ lines)

#include <cstdio>
using namespace std;

#define ll long long
#define ull unsigned long long
namespace io{
const int size=(1<<20)+1;
char buf[size],*p1=buf,*p2=buf;
int op1=-1;
const int op2=size-1;
inline char readchar() {
	if(p1!=p2) {
		return *p1++;
	}
	return p1==(p2=(p1=buf)+fread(buf,1,size-1,stdin))?EOF:*p1++;
}
#ifndef ONLINE_JUDGE
#define gc getchar
#else
#define gc readchar
#endif
#define pc putchar
#define el pc(10)
#define sp pc(32)
inline int readi(){
	int x=0;bool w=0;
	char ch=gc();
	while(ch<'0'||ch>'9'){
		if(ch==EOF) return EOF;
		if(ch=='-') w=1;
		ch=gc();
	}
	while(ch>='0'&&ch<='9'){
		x=(x<<3)+(x<<1)+(ch^48);
		ch=gc();
	}
	return w?-x:x;
}

inline ll readl(){
	ll x=0;bool w=0;
	char ch=gc();
	while(ch<'0'||ch>'9'){
		if(ch=='-') w=1;
		ch=gc();
	}
	while(ch>='0'&&ch<='9'){
		x=(x<<3)+(x<<1)+(ch^48);
		ch=gc();
	}
	return w?-x:x;
}

inline ll readull(){
	ll x=0;
	char ch=gc();
	while(ch>=32){
		if(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48);
		ch=gc();
	}
	if(ch==-1) x=-x;
	return x;
}

inline void reads(char* s){
	char ch=gc();
	while(ch>32) *(s++)=ch,ch=gc();
	*s=0;
}

inline void readline(char* s){
	char ch=gc();
	while(ch>=32) *(s++)=ch,ch=gc();
	*s=0;
}

inline void clearline(){
	char ch=gc();
	while(ch>=32) ch=gc();
}

void write(int x){
	if(x<0) pc('-'),x=-x;
	if(x>9) write(x/10);
	pc((x%10)^48);
}

void write(ll x){
	if(x<0) pc('-'),x=-x;
	if(x>9) write(x/10);
	pc((x%10)^48);
}

void write(ull x){
	if(x>9) write(x/10);
	pc((x%10)^48);
}

void writes(const char* s){
	const char*pt=s;
	while(*pt!=0)pc(*pt++);
}
};
using namespace io;

struct tic{
    int t, m;
};
tic queue[100001];
int head=0, tail=0, tmptail=0;
int n, s=0, tmp;
int type, price, time__;

int main(){
	n=readi();
	while(n--){
		type=readi(),price=readi(),time__=readi();
		if(type==0){
			s += price;
			queue[tail].t = time__;
            queue[tail].m = price;
            tail++;
		}
		else{
			tmp = 0;
			tmptail = tail;
			while(head<tmptail){
				if(time__-queue[head].t<=45){
                    if(price<=queue[head].m){
						head++;
						tmp = 1;
						break;
					}
					queue[tail++] = queue[head];
				}
				head++;
			}
			if(!tmp) s += price;
            while(head<tmptail){
				queue[tail++] = queue[head++];
			}
		}
        if(head==tail) head=tail=0;
	}
	write(s);
	return 0;
}

! 一举拿下最优解。

! It got the best solution.

---

好了，你们猜我为什么会写英文？因为把所有加粗的字符（各段首字符）拼在一起，就是：
Offer me half hundred!
v 我 50！

FInished.