Description：

Solution：

Code：

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstdio>

#include<cmath>

#include<cctype>

#include<cstring>

using namespace std;

#define N 50000

int n;

struct line

{

	double k,b;

	line(int x = 0,double y = 0.0,double k_ = 0.0)

	{

		k = k_;b = y - k * x;

		return;

	}

	double f(int x)

	{

		return k * x + b;

	}

};

struct node

{

	int lc,rc;

	line l;

}t[N << 1];

int ptr = 0;

int newnode(){return ++ptr;}

int root;

#define mid ((l + r) >> 1)

void build(int &rt,int l,int r)

{

	rt = newnode();

	if(l == r)return;

	build(t[rt].lc,l,mid);

	build(t[rt].rc,mid + 1,r);

	return;

}

void add(int rt,line e,int l,int r)

{

	if(t[rt].l.f(l) <= e.f(l) && t[rt].l.f(r) <= e.f(r))

	{

		t[rt].l = e;

		return;

	}

	if(t[rt].l.f(l) >= e.f(l) && t[rt].l.f(r) >= e.f(r))return;

	if(t[rt].l.f(mid) < e.f(mid))swap(t[rt].l,e);

	if(t[rt].l.f(l) < e.f(l))

	{

		add(t[rt].lc,e,l,mid);

	}

	else

	{

		add(t[rt].rc,e,mid + 1,r);

	}

	return;

}

double query(int rt,int p,int l,int r)

{

	if(l == r)return t[rt].l.f(l);

	double res;

	if(p <= mid)res = query(t[rt].lc,p,l,mid);

	else res = query(t[rt].rc,p,mid + 1,r);

	return max(res,t[rt].l.f(p));

}

int main()

{

	scanf("%d",&n);

	build(root,1,N);

	int k;

	double x,y;

	char c[20];

	for(int i = 1;i <= n;++i)

	{

		scanf("%s",c);

		if(c[0] == 'Q')

		{

			scanf("%d",&k);

			printf("%d\n",(int)(query(root,k,1,N) / 100));

		}

		else

		{

			scanf("%lf%lf",&x,&y);

			add(root,line(1,x,y),1,N);

		}

	}

	return 0;

}