经典的splay维护凸壳，但是看了看zky学长的题解最后决定写线段树维护标记永久化。
Round1考到了这个之后一直没有理解标记永久化，CTSC也因为自己的缺陷丢掉了一些部分分，so sad
看来以后不懂的东西要及时学啊QwQ
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node {
double k, b; bool p;
node(int xa = 0, int ya = 0, int xb = 0, int yb = 0, bool num = 0) {
p = num;
if (xa == xb) {k = 0; b = max(ya, yb);}
else {k = (ya - yb) / (xa - xb); b = ya - xa * k;}
}
double get(int x) {return k * x + b;}
};
bool lessthan(node A, node B, double x) {
if (!A.p) return 1;
double na = A.get(x), nb = B.get(x);
return na == nb ? A.p < B.p : na < nb;
} int n = 50000;
node T[200003];
node Q(int rt, int l, int r, int pos) {
if (l == r) return T[rt];
node tmp; int mid = (l + r) >> 1;
tmp = (pos <= mid) ? Q(rt << 1, l, mid, pos) : Q(rt << 1 | 1, mid + 1, r, pos);
return lessthan(T[rt], tmp, pos) ? tmp : T[rt];
}
void ins2(int rt, int l, int r, node se) {
if (!T[rt].p) {T[rt] = se; return;}
if (lessthan(T[rt], se, l)) swap(T[rt], se);
if (l == r || T[rt].k == se.k) return; //!!!
double x = (T[rt].b - se.b) / (se.k - T[rt].k); int mid = (l + r) >> 1;
if (x < l || x > r) return;
if (x <= mid) ins2(rt << 1, l, mid, T[rt]), T[rt] = se; else ins2(rt << 1 | 1, mid + 1, r, se);
}
void ins1(int rt, int l, int r, int L, int R, node se) {
if (L <= l && r <= R) {ins2(rt, l, r, se); return;}
int mid = (l + r) >> 1;
if (L <= mid) ins1(rt << 1, l, mid, L, R, se);
if (R > mid) ins1(rt << 1 | 1, mid + 1, r, L, R, se);
} int main() {
double S, P; node tmp; int T, pos; scanf("%d", &T); char c[15];
while (~scanf("%s", c)) {
if (c[0] == 'Q') {
scanf("%d", &pos);
printf("%lld\n", (long long) (Q(1, 1, n, pos).get(pos) / 100 + 1e-8));
} else {
scanf("%lf%lf", &S, &P);
tmp.k = P; tmp.b = S - P; tmp.p = 1;
ins1(1, 1, n, 1, n, tmp);
}
}
return 0;
}
Round2加油吧ovo