Submission #17696

#	Time	Username	Problem	Language	Result	Execution time	Memory
17696		chrome	Hyper-minimum (IZhO11_hyper)	C++98	90 / 100	2000 ms	34112 KiB

hyper

#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 / 100