#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include "treasure.h"
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define debug(x) cerr << #x << ": " << x << endl
#define debug2(x, y) debug(x), debug(y)
#define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i, a) for(int i = 0; i < (int)(a); i++)
#define all(v) v.begin(), v.end()
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define sq(x) ((x) * (x))
const int mxN = 105;
template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); }
int n, tot;
int l[mxN], r[mxN], u[mxN], d[mxN]; //all of these are inclusive
int LU[mxN][mxN], LD[mxN][mxN], RU[mxN][mxN], RD[mxN][mxN];
int qry(int r1, int c1, int r2, int c2){
int ret = countTreasure(r1, c1, r2, c2);
return ret;
}
bool check(int i, int j){
int cur = 0;
if(i > 1) cur += u[i - 1];
if(i < n) cur += d[i + 1];
if(j > 1) cur += l[j - 1];
if(j < n) cur += r[j + 1];
//now we do the intersections
if(i > 1 && j > 1) cur -= LU[i - 1][j - 1];
if(i < n && j > 1) cur -= LD[i + 1][j - 1];
if(i > 1 && j < n) cur -= RU[i - 1][j + 1];
if(i < n && j < n) cur -= RD[i + 1][j + 1];
//assert(cur == tot || cur == (tot - 1));
return cur != tot;
}
void solve1(int i, int j){
if(~RU[i][j]) RU[i][j] = qry(1, j, i, n);
if(j > 1) LU[i][j - 1] = u[i] - RU[i][j];
if(i < n){
RD[i + 1][j] = r[j] - RU[i][j];
if(j > 1) LD[i + 1][j - 1] = l[j - 1] - LU[i][j - 1];
}
}
void solve2(int i, int j){
if(~LU[i][j]) LU[i][j] = qry(1, 1, i, j);
if(i < n) LD[i + 1][j] = l[j] - LU[i][j];
if(j < n){
RU[i][j + 1] = u[i] - LU[i][j];
if(i < n) RD[i + 1][j + 1] = r[j + 1] - RU[i][j + 1];
}
}
void solve3(int i, int j){
if(~RD[i][j]) RD[i][j] = qry(i, j, n, n);
if(i > 1) RU[i - 1][j] = r[j] - RD[i][j];
if(j > 1){
LD[i][j - 1] = d[i] - RD[i][j];
if(i > 1) LU[i - 1][j - 1] = l[j - 1] - LD[i][j - 1];
}
}
void solve4(int i, int j){
if(~LD[i][j]) LD[i][j] = qry(i, 1, n, j);
if(i > 1) LU[i - 1][j] = l[j] - LD[i][j];
if(j < n){
RD[i][j + 1] = d[i] - LD[i][j];
if(i > 1) RU[i - 1][j + 1] = r[j + 1] - RD[i][j + 1];
}
}
void findTreasure(int N){
n = N;
tot = qry(1, 1, n, n);
memset(LD, -1, sizeof(LD));
memset(LU, -1, sizeof(LU));
memset(RD, -1, sizeof(RD));
memset(RU, -1, sizeof(RU));
//l and r
l[n] = r[1] = tot;
repn(i, 1, n){
if(i > (n - i)) l[i] = qry(1, 1, n, i);
else l[i] = tot - qry(1, i + 1, n, n);
}
repn(i, 2, n + 1) r[i] = tot - l[i - 1];
//u and d
u[n] = d[1] = tot;
repn(i, 1, n){
if(i > (n - i)) u[i] = qry(1, 1, i, n);
else u[i] = tot - qry(i + 1, 1, n, n);
}
repn(i, 2, n + 1) d[i] = tot - u[i - 1];
//corners
repn(i, 1, n + 1){
repn(j, 1, n + 1){
if(i <= (n - i)){
if(j <= (n - j)) solve3(i, j);
else solve4(i, j);
}
else{
if(j <= (n - j)) solve1(i, j);
else solve2(i, j);
}
}
}
repn(i, 1, n + 1){
repn(j, 1, n + 1){
assert(~LU[i][j]);
if(check(i, j)) Report(i, j);
}
}
}
/*
Things to look out for:
- Integer overflows
- Array bounds
- Special cases
Be careful!
*/
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Runtime error |
1 ms |
896 KB |
Execution killed with signal 11 (could be triggered by violating memory limits) |
| 2 |
Runtime error |
1 ms |
896 KB |
Execution killed with signal 11 (could be triggered by violating memory limits) |
| 3 |
Incorrect |
0 ms |
512 KB |
Error - no treasure at (r, c) : r = 1, c = 1 |
| 4 |
Incorrect |
0 ms |
512 KB |
Error - no treasure at (r, c) : r = 1, c = 1 |
| 5 |
Incorrect |
1 ms |
512 KB |
Error - no treasure at (r, c) : r = 1, c = 3 |
| 6 |
Incorrect |
1 ms |
512 KB |
Error - no treasure at (r, c) : r = 1, c = 1 |
| 7 |
Incorrect |
1 ms |
512 KB |
Error - no treasure at (r, c) : r = 1, c = 1 |
| 8 |
Runtime error |
2 ms |
896 KB |
Execution killed with signal 11 (could be triggered by violating memory limits) |
| 9 |
Incorrect |
1 ms |
512 KB |
Error - no treasure at (r, c) : r = 1, c = 2 |
| 10 |
Incorrect |
1 ms |
512 KB |
Error - no treasure at (r, c) : r = 1, c = 3 |
