oj.uz

Submission #58379

# Submission time Handle Problem Language Result Execution time Memory
58379 2018-07-17T15:45:52 Z Benq Energetic turtle (IZhO11_turtle) C++14
100 / 100
 970 ms 75640 KB 
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

namespace NT {
    vpi fac(int x) {
        vpi pri;
        
        for (int i = 2; i*i <= x; ++i) if (x % i == 0) {
            int t = 0;
            while (x % i == 0) x /= i, t ++;
            pri.pb({i,t});
        }
        
        if (x > 1) pri.pb({x,1});
        return pri;
    }

    int phi(int x) {
        for (auto a: fac(x)) x /= a.f, x *= a.f-1; 
        return x;
    }

    ll inv(ll a, ll b) { // 0 < a < b, gcd(a,b) = 1
        a %= b;
        if (a <= 1) return a;
        ll i = inv(b%a,a);
        ll tmp = -((b/a)*i+((b%a)*i)/a) % b;
        if (tmp < 0) tmp += b;
        return tmp;
    }
 
    pl CRT(pl a, pl b) { // Chinese Remainder Theorem, Verified by Kattis generalchineseremainder
        ll g = __gcd(a.s,b.s), l = a.s*b.s/g;
        if ((b.f-a.f) % g != 0) return {-1,-1};
        ll A = a.s/g, B = b.s/g;
        ll mul = (b.f-a.f)/g*inv(A%B,B) % B;
        return {((mul*a.s+a.f)%l+l)%l,l};
    }
};

template<int SZ> struct ComboPower {
    pl fac[SZ+1], ifac[SZ+1], mod;
    ll MOD = 1;
    
    void init(pl _mod) { // prime, power
        mod = _mod; F0R(i,mod.s) MOD *= mod.f;
        
        fac[0] = ifac[0] = {1,0};
        FOR(i,1,SZ+1) {
            fac[i] = fac[i-1];
            int I = i, z = 0;
            while (I % mod.f == 0) I /= mod.f, z++;
            fac[i].f = fac[i].f*I%MOD; fac[i].s += z;
            ifac[i] = {NT::inv(fac[i].f,MOD),fac[i].s};
        }
    }

    ll comb(ll a, ll b) {
        if (a < b || b < 0) return 0;
        ll tmp = (fac[a].f*ifac[b].f%MOD)*ifac[a-b].f % MOD;
        ll z = fac[a].s-fac[b].s-fac[a-b].s;
        if (z >= mod.s) return 0;
        F0R(i,z) tmp = tmp*mod.f % MOD;
        return tmp;
    }
};

template<int SZ> struct ComboGeneral {
    vector<ComboPower<SZ>> v;
    ll MOD;
    
    void init(ll _MOD) {
        MOD = _MOD;
        for (auto a: NT::fac(MOD)) {
            v.pb(ComboPower<SZ>());
            v.back().init(a);
        }
    }
    
    ll comb(ll a, ll b) {
        pl cur = {0,1};
        for (auto& x: v) cur = NT::CRT({x.comb(a,b),x.MOD},cur);
        return cur.f;
    }
};


ComboGeneral<600001> C;

int get(pi a, pi b) {
    return C.comb(b.f-a.f+b.s-a.s,b.s-a.s);
}

int N,M,K,T,Z, ways[22][22], dp[22][22];
vpi p;

ll ad(ll a, ll b) { return (a+b)%Z; }
ll sub(ll a, ll b) { return (a-b+Z)%Z; }
ll mul(ll a, ll b) { return a*b%Z; }

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M >> K >> T >> Z;
    C.init(Z);
    F0R(i,K) {
        int X,Y; cin >> X >> Y;
        p.pb({X,Y});
    }
    p.pb({0,0});
    sort(all(p));
    p.pb({N,M});
    F0R(i,sz(p)) F0Rd(j,i) {
        ways[j][i] = get(p[j],p[i]);
        FOR(k,j+1,i) ways[j][i] = sub(ways[j][i],mul(ways[j][k],get(p[k],p[i])));
        // cout << j << " " << i << " " << ways[j][i] << "\n";
    }
    dp[0][0] = 1;
    FOR(i,1,sz(p)) F0R(j,i) FOR(k,1,T+2) 
        dp[i][k] = ad(dp[i][k],mul(dp[j][k-1],ways[j][i]));
    
    int ans = 0;
    FOR(i,1,T+2) ans = ad(ans,dp[sz(p)-1][i]);
    cout << ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

Subtask #1
100.0 / 100.0

# Verdict Execution time Memory Grader output
1 Correct 274 ms 75640 KB Output is correct
2 Correct 917 ms 75468 KB Output is correct
3 Correct 414 ms 37880 KB Output is correct
4 Correct 536 ms 37880 KB Output is correct
5 Correct 418 ms 37880 KB Output is correct
6 Correct 529 ms 37880 KB Output is correct
7 Correct 529 ms 37892 KB Output is correct
8 Correct 531 ms 37852 KB Output is correct
9 Correct 529 ms 37880 KB Output is correct
10 Correct 558 ms 37880 KB Output is correct
11 Correct 512 ms 37884 KB Output is correct
12 Correct 533 ms 37880 KB Output is correct
13 Correct 922 ms 75460 KB Output is correct
14 Correct 920 ms 75472 KB Output is correct
15 Correct 936 ms 75512 KB Output is correct
16 Correct 938 ms 75476 KB Output is correct
17 Correct 941 ms 75512 KB Output is correct
18 Correct 970 ms 75512 KB Output is correct
19 Correct 920 ms 75512 KB Output is correct
20 Correct 943 ms 75512 KB Output is correct