题意
Sol
李超线段树板子题。具体原理就不讲了。
一开始自己yy着写差点写自闭都快把叉积搬出来了。。。
后来看了下litble的写法才发现原来可以写的这么清晰简洁Orz
#include<bits/stdc++.h>
#define pdd pair<double, double>
#define MP make_pair
#define fi first
#define se second
using namespace std;
const int MAXN = 1e6 + 10, Lim = 1e9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
inline int read() {
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 = 39989, M;
int ls[MAXN], rs[MAXN], root, cnt, tot;
pdd mx[MAXN];
struct Line {
double k, b;
int id;
}s[MAXN];
pdd get(int x0, int y0, int x1, int y1) {
double k = (double) (y1 - y0) / (x1 - x0),
b = (double) y0 - k * x0;
return {k, b};
}
double calc(Line line, int x) {
return line.k * x + line.b;
}
double GetPoint(Line a, Line b) {
return (b.b - a.b) / (a.k - b.k);
}
pdd ret;
void Query(int k, int l, int r, int p) {//fi: val se: id
if(chmax(ret.fi, calc(s[k], p))) ret.se = s[k].id;
if(l == r) return ;
int mid = l + r >> 1;
if(p <= mid) Query(ls[k], l, mid, p);
else Query(rs[k], mid + 1, r, p);
}
void Modify(int &k, int l, int r, int ql, int qr, Line seg) {
if(!k) k = ++tot;
int mid = l + r >> 1;
if(ql <= l && r <= qr) {
if(!s[k].id) {s[k] = seg; return ;}
int p = GetPoint(s[k], seg);
int pl = calc(s[k], l), pr = calc(s[k], r), nl = calc(seg, l), nr = calc(seg, r);
if(pl > nl && pr > nr) return ;
if(pl < nl && pr < nr) {s[k] = seg; return ;}
if(pl < nl) {
if(p > mid) Modify(rs[k], mid + 1, r, mid + 1, r, s[k]), s[k] = seg;
else Modify(ls[k], l, mid, l, mid, seg);
} else {
if(p > mid) Modify(rs[k], mid + 1, r, mid + 1, r, seg);
else Modify(ls[k], l, mid, l, mid, s[k]), s[k] = seg;
}
return ;
}
if(l == r) return ;
if(ql <= mid) Modify(ls[k], l, mid, ql, qr, seg);
if(qr > mid) Modify(rs[k], mid + 1, r, ql, qr, seg);
}
signed main() {
M = read();
for(int i = 1, lastans = 0; i <= M; i++) {
int opt = read();
if(!opt) {
int k = read(), x = (k + lastans - 1) % 39989 + 1;
ret.fi = 0; ret.se = 0;
Query(root, 1, N, x);
printf("%d\n", lastans = (mx[x].fi > ret.fi ? mx[x].se : ret.se));
} else {
int x0 = (read() + lastans - 1) % 39989 + 1, y0 = (read() + lastans - 1) % Lim + 1,
x1 = (read() + lastans - 1) % 39989 + 1, y1 = (read() + lastans - 1) % Lim + 1;
if(x0 > x1) swap(x0, x1), swap(y0, y1);
if(x0 == x1 && chmax(mx[x0].fi, max(y0, y1))) mx[x0].se = i;
pdd li = get(x0, y0, x1, y1);
Modify(root, 1, N, x0, x1, {li.fi, li.se, ++cnt});
}
}
return 0;
}