Submission #642074

# Submission time Handle Problem Language Result Execution time Memory
642074 2022-09-18T11:56:30 Z vovamr Mobitel (COCI19_mobitel) C++17
130 / 130
3215 ms 9640 KB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define fi first
#define se second
#define ll long long
#define ld long double
#define sz(x) ((int)(x).size())
#define all(x) 	(x).begin(), (x).end()
#define pb push_back
#define mpp make_pair
#define ve vector
using namespace std;
using namespace __gnu_pbds;
template<class T> using oset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
const ll inf = 1e18; const int iinf = 1e9;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T> inline bool chmin(T& a, T b) { return (a > b ? a = b, 1 : 0); }
template <typename T> inline bool chmax(T& a, T b) { return (a < b ? a = b, 1 : 0); }

constexpr int md = 1e9 + 7; constexpr int iv2 = (md + 1) / 2;
inline int add(int a, int b) { if((a+=b)>=md)a-=md; return a; }
inline int sub(int a, int b) { if((a-=b)<0)a+=md; return a; }
inline int mul(int a, int b) { return 1ll*a*b%md; }
inline int bp(int a, int n) { int res = 1;
	for (; n; n >>= 1, a = mul(a, a)) {
		if (n & 1) res = mul(res, a);
	} return res;
}
inline int inv(int a) { return bp(a, md - 2); }

inline void solve() {
	int n, m, k;
	cin >> n >> m >> k;

	ve<ve<int>> a(n, ve<int> (m));
	for (auto &i : a) for (auto &j : i) cin >> j, chmin(j, k);

	int Z = k + 10;

	ve<int> al, pos(Z);
	for (int i = 1; i < Z; ++i) {
		int v = (k + i - 1) / i;
		if (al.empty() || al.back() != v) {
			pos[v] = sz(al);
			al.pb(v);
		}
	}

	int s = sz(al);

	ve<ve<int>> dp(m, ve<int>(s));
	dp[0][pos[(k + a[0][0] - 1) / a[0][0]]] = 1;

	for (int i = 0; i < n; ++i) {

		ve<ve<int>> f(m, ve<int>(s));

		for (int j = 0; j < m; ++j) {
			for (int z = 0; z < s; ++z) {
				if (!dp[j][z]) continue;

				if (i + 1 < n) {
					int x = a[i + 1][j];
					int d = pos[(al[z] + x - 1) / x];
					f[j][d] = add(f[j][d], dp[j][z]);
				}
				if (j + 1 < m) {
					int x = a[i][j + 1];
					int d = pos[(al[z] + x - 1) / x];
					dp[j + 1][d] = add(dp[j + 1][d], dp[j][z]);
				}
			}
		} if (i + 1 < n) dp.swap(f);
	}

	cout << dp[m - 1][pos[1]];
}

signed main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int q = 1; // cin >> q;
	while (q--) solve();
	cerr << fixed << setprecision(3) << "Time execution: " << (double)clock() / CLOCKS_PER_SEC << endl;
}
# Verdict Execution time Memory Grader output
1 Correct 52 ms 756 KB Output is correct
2 Correct 54 ms 768 KB Output is correct
3 Correct 113 ms 5424 KB Output is correct
4 Correct 118 ms 5900 KB Output is correct
5 Correct 263 ms 5900 KB Output is correct
6 Correct 234 ms 6068 KB Output is correct
7 Correct 104 ms 4992 KB Output is correct
8 Correct 1589 ms 8212 KB Output is correct
9 Correct 3215 ms 9640 KB Output is correct
10 Correct 3188 ms 9556 KB Output is correct