题面
题解
李超线段树
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Code
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
const int N = 100005;
const int mod = 39989;
#define val(id, x) (k[id] * x + b[id])
using namespace std;
int Q, cnt, ans;
struct node
{
int id;
double k, b;
} t[N << 2];
double k[N], b[N];
template < typename T >
inline T read()
{
T x = 0, w = 1; char c = getchar();
while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * w;
}
void pushtag(int p, int l, int r, int id)
{
if(!t[p].id) { t[p] = (node) { id, k[id], b[id] }; return; }
int now = t[p].id;
double l1 = val(now, l), r1 = val(now, r), l2 = val(id, l), r2 = val(id, r);
if(l2 <= l1 && r2 <= r1) return;
if(l1 < l2 && r1 < r2) { t[p] = (node) { id, k[id], b[id] }; return; }
int mid = (l + r) >> 1; double x = (b[id] - b[now]) / (k[now] - k[id]);
if(x <= mid)
{
if(l1 <= l2) pushtag(p << 1, l, mid, id);
else pushtag(p << 1, l, mid, now), t[p] = (node) { id, k[id], b[id] };
}
else
{
if(l1 <= l2) pushtag(p << 1 | 1, mid + 1, r, now), t[p] = (node) { id, k[id], b[id] };
else pushtag(p << 1 | 1, mid + 1, r, id);
}
}
void modify(int p, int l, int r, int ql, int qr, int id)
{
if(ql <= l && r <= qr) { pushtag(p, l, r, id); return; }
int mid = (l + r) >> 1;
if(ql <= mid) modify(p << 1, l, mid, ql, qr, id);
if(mid < qr) modify(p << 1 | 1, mid + 1, r, ql, qr, id);
}
void check(int &a, int c, int x)
{
double y0 = val(a, x), y1 = val(c, x);
if(y1 > y0 || (fabs(y1 - y0) < 1e-7 && a > c)) a = c;
}
int query(int p, int l, int r, int x)
{
if(l == r) return t[p].id;
int mid = (l + r) >> 1, res = t[p].id;
if(x <= mid) check(res, query(p << 1, l, mid, x), x);
else check(res, query(p << 1 | 1, mid + 1, r, x), x);
return res;
}
int main()
{
/* freopen("cpp.in", "r", stdin);
freopen("cpp.out", "w", stdout);
*/ Q = read <int> ();
int opt, x0, x1, y0, y1;
while(Q--)
{
opt = read <int> ();
if(!opt)
{
// printf("%d
", x0);
x0 = read <int> (), x0 = ((x0 + ans - 1) % mod + 1);
printf("%d
", ans = query(1, 1, mod, x0));
}
else
{
x0 = read <int> (), y0 = read <int> (), x1 = read <int> (), y1 = read <int> ();
x0 = ((x0 + ans - 1) % mod + 1), y0 = ((y0 + ans - 1) % 1000000000 + 1);
x1 = ((x1 + ans - 1) % mod + 1), y1 = ((y1 + ans - 1) % 1000000000 + 1);
if(x0 > x1) swap(x0, x1), swap(y0, y1);
// printf("%d %d
", x0, x1);
k[++cnt] = 1.0 * (y1 - y0) / (x1 - x0), b[cnt] = 1.0 * y0 - k[cnt] * x0;
modify(1, 1, mod, x0, x1, cnt);
}
}
return 0;
}
/*
3
1 3 5 7 7
1 3 9 7 3
0 5
*/