Submission #225101

#	Time	Username	Problem	Language	Result	Execution time	Memory
225101		kartel	Mobitel (COCI19_mobitel)	C++14	0 / 130	144 ms	6400 KiB

mobitel

#include <bits/stdc++.h>
#define in(x) freopen(x,"r",stdin)
#define out(x) freopen(x,"w",stdout)
#define F first
#define S second
#define pb push_back
#define M ll(1e9 + 7)
#define inf ll(2e9+7)
#define N ll(305)
#define sz(x) x.size()
//#pragma GCC optimize("unroll-loops")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("Ofast")
#define el '\n'
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree <ll, null_type, less_equal <ll>, rb_tree_tag, tree_order_statistics_node_update> os;

int n, m, x, i, j, f[2][N][1050][2], a[N][N], k, sq, ans, fc[N];

int sm(int x, int y) {return (0ll + x + y) % M;}

int pw(int x, int y)
{
    if (y == 0) return 1ll;
    if (y % 2) return pw(x, y - 1) * x % M;
    int a = pw(x, y / 2ll);
    return a * a % M;
}

int C(int n, int k)
{
    return (fc[n] * pw((fc[n - k] * fc[k]) % M, M - 2)) % M;
}

int main()
{
//    in("input.txt");
//    out("output.txt");

    ios_base::sync_with_stdio(false);
    iostream::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> m >> x;
    x--;

    sq = int(sqrt(x));

    for (i = 1; i <= n; i++)
        for (j = 1; j <= m; j++) cin >> a[i][j];

    fc[1] = 1;
    for (i = 2; i <= n + m; i++) fc[i] = (fc[i - 1] * i) % M;

    ans = C(n + m - 2, n - 1);

    if (sq == 1) f[0][1][0][1] = 1;
            else f[0][1][1][0] = 1;

    for (i = 1; i <= n; i++)
      {
         for (j = 1; j <= m; j++)
            for (k = 1; k <= max(x / sq, sq); k++) f[i % 2][j][k][0] = f[i % 2][j][k][1] = 0;
         for (j = 1; j <= m; j++)
            for (k = 1; k <= max(x / sq, sq); k++)
         {
             ll nk = k * a[i][j];
             if (nk < sq)
             {
                 f[i % 2][j][nk][0] = sm(f[i % 2][j][nk][0], f[(i + 1) % 2][j][k][0]);
                 f[i % 2][j][nk][0] = sm(f[i % 2][j][nk][0], f[i % 2][j - 1][k][0]);
             }
             else
             {
                 f[i % 2][j][x / nk][1] = sm(f[i % 2][j][x / nk][1], f[(i + 1) % 2][j][k][0]);
                 f[i % 2][j][x / nk][1] = sm(f[i % 2][j][x / nk][1], f[i % 2][j - 1][k][0]);
             }
             nk = k / a[i][j];
             f[i % 2][j][nk][1] = sm(f[i % 2][j][nk][1], f[(i + 1) % 2][j][k][1]);
             f[i % 2][j][nk][1] = sm(f[i % 2][j][nk][1], f[i % 2][j - 1][k][1]);
         }
      }
    for (i = 1; i <= max(x / sq, sq); i++)
    {
        ans = sm(ans, M - f[n % 2][m][i][1]);
        ans = sm(ans, M - f[n % 2][m][i][0]);
    }
    cout << ans;
}

• Subtask 1: 0 / 130