oj.uz

Submission #17696

# Submission time Handle Problem Language Result Execution time Memory
17696 2016-01-12T08:48:58 Z chrome Hyper-minimum (IZhO11_hyper) C++
90 / 100
 2000 ms 34112 KB 
#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define foreach(it, S) for (__typeof (S.begin()) it = S.begin(); it != S.end(); it++)
#define all(x) x.begin(), x.end()
#define endl '\n'
#define _ ios_base :: sync_with_stdio(false); cin.tie(NULL);

#ifdef inputf
	#define fname ""
#else
	#define fname "" // <- Here
#endif

const double eps = 1e-9;
const int MaxN = int(2e5) + 256;
const int MOD = int(1e9) + 7;

template <typename T> inline T gcd(T a, T b) {
	return b ? gcd (b, a % b) : a;
}

inline bool Palindrome(const string& s) {
	return equal(s.begin(), s.end(), s.rbegin());
}

int x[45][45][45][45];
int y[45][45][45][45];
int z[45][45][45];
int mn = INT_MAX;

#define S(x, val) x = min(x, val)

int main() { // _
	#ifdef lcl
		freopen(fname".in", "r", stdin);
		freopen(fname".out", "w", stdout);
	#endif

	int n, m; scanf("%d%d", &n, &m);

	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j)
			for (int k = 1; k <= n; ++k)
				for (int l = 1; l <= n; ++l)
					z[i][j][k] = y[i][j][k][l] = INT_MAX;
	
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j)
			for (int k = 1; k <= n; ++k) {
				for (int l = 1; l <= n; ++l) {
					scanf("%d", &x[i][j][k][l]);
					int val = x[i][j][k][l];
					mn = min(mn, val);
					for (int p = 0, q = l; p < m && q > 0; ++p, --q) {
						S(y[i][j][k][q], val);
					}
					/* for (int r = 0, w = k; r < m && w > 0; ++r, --w) {
						S(z[i][j][w], y[i][j][k][l]);
					} */
				}
			}

	
	for (int i1 = 1; i1 <= n - m + 1; ++i1)
		for (int i2 = 1; i2 <= n - m + 1; ++i2)
			for (int i3 = 1; i3 <= n - m + 1; ++i3)
				for (int i4 = 1; i4 <= n - m + 1; ++i4) {
					int val = INT_MAX;
					for (int j1 = i1; val != mn && j1 <= i1 + m - 1; ++j1)
						for (int j2 = i2; val != mn && j2 <= i2 + m - 1; ++j2)
							for (int j3 = i3; val != mn && j3 <= i3 + m - 1; ++j3)
								S(val, y[j1][j2][j3][i4]);
									/* for (int j4 = i4; j4 <= i4 + m - 1; ++j4) {
									val = min(val, x[j1][j2][j3][j4]);
								} */
					printf("%d ", val);
				}
	
	
	return 0;
}

Subtask #1
90.0 / 100.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 34112 KB Output is correct
2 Correct 0 ms 34112 KB Output is correct
3 Correct 0 ms 34112 KB Output is correct
4 Correct 0 ms 34112 KB Output is correct
5 Correct 0 ms 34112 KB Output is correct
6 Correct 18 ms 34112 KB Output is correct
7 Correct 9 ms 34112 KB Output is correct
8 Correct 12 ms 34112 KB Output is correct
9 Correct 46 ms 34112 KB Output is correct
10 Correct 67 ms 34112 KB Output is correct
11 Correct 372 ms 34112 KB Output is correct
12 Correct 1250 ms 34112 KB Output is correct
13 Correct 184 ms 34112 KB Output is correct
14 Correct 379 ms 34112 KB Output is correct
15 Correct 372 ms 34112 KB Output is correct
16 Correct 313 ms 34112 KB Output is correct
17 Execution timed out 2000 ms 34112 KB Program timed out
18 Correct 573 ms 34112 KB Output is correct
19 Execution timed out 2000 ms 34112 KB Program timed out
20 Correct 396 ms 34112 KB Output is correct