QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#745576 | #8780. Training, Round 2 | cyj888 | WA | 8ms | 127248kb | C++14 | 2.0kb | 2024-11-14 10:38:49 | 2024-11-14 10:38:53 |
Judging History
answer
#include <bits/stdc++.h>
//#define int long long
#define fi first
#define se second
#define pb push_back
#define ott(i, l, r) for (register int i = (l); i <= (r); i ++)
#define tto(i, l, r) for (register int i = (r); i >= (l); i --)
using namespace std;
typedef long long ll;
typedef long double ld;
int read () {
int x = 0; bool f = false; char c = getchar ();
while (!isdigit (c)) f |= (c == '-'), c = getchar ();
while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
return f ? -x : x;
}
const int N = 5500, mod = 1e9 + 7;
int n, s1, s2, res;
int vis[N][N]; bool ins[N][N];
vector <pair <int, int> > cac;
inline void add (const int &x, const int &y) {
if (!vis[x + 1][y]) cac.pb ({x + 1, y}), vis[x + 1][y] = true;
if (!vis[x][y + 1]) cac.pb ({x, y + 1}), vis[x][y + 1] = true;
return ;
}
struct Segment_tree {
int cnt, id; int siz[N];
#define ls p << 1
#define rs p << 1 | 1
void pup (int p) {
return void (siz[p] = siz[ls] + siz[rs]);
}
void ins (int x, int p = 1, int l = 0, int r = n) {
if (l >= r) return void (++ siz[p]);
int mid = l + r >> 1;
x <= mid ? ins (x, ls, l, mid) : ins (x, rs, mid + 1, r);
return pup (p);
}
void del (int x, int y, int p = 1, int l = 0, int r = n) {
if (!siz[p]) return ;
if (l >= r) {
add (id, l); return void (-- siz[p]);
}
int mid = l + r >> 1;
if (x <= mid) del (x, y, ls, l, mid); if (y > mid) del (x, y, rs, mid + 1, r);
return pup (p);
}
} has[N];
int main () {
n = read (), s1 = read (), s2 = read ();
ott (i, 0, n) has[i].id = i;
vis[0][0] = true, has[0].ins (0);
ott (i, 1, n) {
register int l1 = read () - s1, r1 = read () - s1, l2 = read () - s2, r2 = read () - s2;
if (l1 < 0) l1 = 0; if (l2 < 0) l2 = 0; if (r1 > n) r1 = n; if (r2 > n) r2 = n;
if (l2 <= r2) ott (x, l1, r1) has[x].del (l2, r2);
for (auto &[x, y] : cac) has[x].ins (y); cac.clear ();
}
ott (i, 0, n)
ott (j, 0, n - i)
if (vis[i][j])
res = max (res, i + j);
printf ("%d\n", res);
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 7944kb
input:
3 0 0 0 1 0 1 1 1 0 1 1 1 1 1
output:
3
result:
ok single line: '3'
Test #2:
score: 0
Accepted
time: 7ms
memory: 127248kb
input:
5000 801577551 932138594 801577551 801577551 932138594 932138594 801577552 801577552 932138594 932138594 801577552 801577552 932138595 932138595 801577552 801577552 932138596 932138596 801577553 801577553 932138596 932138596 801577553 801577553 932138597 932138597 801577553 801577553 932138598 93213...
output:
5000
result:
ok single line: '5000'
Test #3:
score: -100
Wrong Answer
time: 8ms
memory: 122444kb
input:
5000 932138594 801577551 932138594 932138594 801577551 801577551 932138594 932138594 801577552 801577552 932138595 932138595 801577552 801577552 932138596 932138596 801577552 801577552 932138596 932138596 801577553 801577553 932138597 932138597 801577553 801577553 932138598 932138598 801577553 80157...
output:
3267
result:
wrong answer 1st lines differ - expected: '5000', found: '3267'