Submission #225000

#	Time	Username	Problem	Language	Result	Execution time	Memory
225000		ne4eHbKa	Mobitel (COCI19_mobitel)	C++17	104 / 130	6054 ms	20004 KiB

mobitel

//{ <defines>
#include <bits/stdc++.h>
using namespace std;

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,-O3")

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int((x).size())
#define ip2(x) (1 << (x))
#define lp2(x) (1ll << (x))
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset((x), (y), sizeof (x))

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}

int md = 1e9 + 7;

inline int m_add(int&a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_sum(int a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_mul(int&a, int b) {re a = 1ll * a * b % md;}
inline int m_prod(int a, int b) {re 1ll * a * b % md;}

int m_bpow(ll A, ll b) {
    int a = A % md;
    ll ans = 1;
    for(ll p = lp2(63 - __builtin_clzll(b)); p; p >>= 1) {
        (ans *= ans) %= md;
        if(p & b)
            (ans *= a) %= md;
    }
    re ans;
}

//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int oo = 2e9;
const ll OO = 4e18;
//} </defines>
const int N = 305;

int r, s, n;

typedef list<pii> f;

void merge(const f &a, const f &b, f &ans) {
    auto it = a.begin(), e = a.end();
    #define pa (*it)
    for(const pii &s : b) {
        for(; it != e && pa.ft < s.ft; ++it) ans.push_back(pa);
        ans.push_back(s);
        for(; it != e && pa.ft == s.ft; ++it) (ans.back().sd += pa.sd) %= md;
    }
    for(; it != e; ++it) ans.push_back(pa);
    #undef pa
}

void divide(f &a, int k) {
    if(k == 1) re;
    for(pii &u : a) (u.ft += k - 1) /= k;
    for(auto it = a.begin(); it != a.end(); ++it) {
        while(1) {
            auto nx = it;
            ++nx;
            if(nx == a.end() || (*nx).ft > (*it).ft) break;
            ((*it).sd += (*nx).sd) %= md;
            a.erase(nx);
        }
    }
}

void solve() {
    cin >> r >> s >> n;
    f dp[r][s];
    int a[r][s];
    fo(r) fr(j, s) cin >> a[i][j];
    for(int i = r - 1; ~i; --i) {
        for(int j = s - 1; ~j; --j) {
            if(i + 1 == r) {
                if(j + 1 == s) {
                    dp[i][j].push_back({n, 1});
                } else {
                    dp[i][j] = dp[i][j + 1];
                }
            } ef(j + 1 == s) {
                dp[i][j] = dp[i + 1][j];
            } else {
                merge(dp[i + 1][j], dp[i][j + 1], dp[i][j]);
                dp[i + 1][j].clear();
            }
            divide(dp[i][j], a[i][j]);
        }
    }
    int ans = 0;
    for(pii &t : dp[0][0]) if(t.ft == 1) ans = t.sd;
    cout << ans << endl;
}

int main() {
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    for(int test = 1; test <= tests; ++test) {
        cerr << test << " {\n";
        solve();
        cerr << "}\n\n";
    }
#else
    //freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
    solve();
#endif
    return 0;
}

• Subtask 1: 104 / 130