# Zr × 317. [18 increased by 2] a (Computational Geometry dichotomy)

Posted by caaronbeasley on Thu, 02 Jan 2020 01:41:17 +0100

## Sol

It's a very good question. Fortunately, I didn't play this game. Otherwise, I guess I will die on this question qwq

It doesn't mean how difficult it is. The key is that there are too many details. wcz and I had a chat about my ideas. Then he wrote it all night and didn't call out the qwq

It's quite a routine to solve it. First put forward a \$x\$

And then maintain a bunch of convex shells corresponding to the straight lines

It can be divided into two parts on convex shell.

Because the \$x \$of this question is very small, just deal with the answer directly

```/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7, BB = 32323;
const double eps = 1e-9;
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N;
struct Node {
double a, b;
bool operator < (const Node &rhs) const {
return a == rhs.a ? b < rhs.b : a < rhs.a;
}
}P[MAXN], s1[MAXN], s2[MAXN];
int t1 = 0, t2 = 0, ans[MAXN];
double cross(Node x, Node y) {
//    printf("%lf\n", 1.0 * (y.b - x.b) / (x.a - x.b));
return 1.0 * (y.b - x.b) / (x.a - y.a);
}
void Get() {
sort(P + 1, P + N + 1);
s1[++t1] = P[1];
for(int i = 2; i <= N; i++) {
if(t1 && P[i].a == s1[t1].a) t1--;
while(t1 > 1 && cross(P[i], s1[t1]) <= cross(s1[t1], s1[t1 - 1])) t1--;
s1[++t1] = P[i];
}
for(int i = 1; i <= N; i++) P[i].b = -P[i].b;
sort(P + 1, P + N + 1);
s2[++t2] = P[1];
for(int i = 2; i <= N; i++) {
if(t1 && P[i].a == s2[t2].a) t2--;
while(t2 > 1 && cross(P[i], s2[t2]) <= cross(s2[t2], s2[t2 - 1])) t2--;
s2[++t2] = P[i];
}
}
int Query(Node p, int x) {
return p.a * x * x + p.b * x;
}
void MakeAns() {
for(int i = 1, c = 1; i <= BB; i++) {
//printf("%lf\n", cross(s1[c], s1[c + 1]));
while(c < t1 && cross(s1[c], s1[c + 1]) <= (double)i) c++;
ans[i + BB] = Query(s1[c], i);
}
for(int i = 1, c = 1; i <= BB; i++) {
while(c < t2 && cross(s2[c], s2[c + 1]) <= (double)i) c++;
ans[BB - i] = Query(s2[c], i);
}
}
main() {
//    freopen("a.in", "r", stdin);
for(int i = 1; i <= N; i++) P[i].a = read(), P[i].b = read();
Get();
MakeAns();
while(Q--) {