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Submission #58378

# Submission time Handle Problem Language Result Execution time Memory
58378 2018-07-17T15:40:16 Z Benq Energetic turtle (IZhO11_turtle) C++14
65 / 100
 1008 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;
    }
};

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

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;

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])));
    }
    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
65.0 / 100.0

# Verdict Execution time Memory Grader output
1 Correct 255 ms 75512 KB Output is correct
2 Correct 911 ms 75476 KB Output is correct
3 Correct 419 ms 37880 KB Output is correct
4 Correct 544 ms 37872 KB Output is correct
5 Correct 415 ms 37880 KB Output is correct
6 Incorrect 541 ms 37880 KB Output isn't correct
7 Correct 530 ms 37888 KB Output is correct
8 Correct 528 ms 37880 KB Output is correct
9 Correct 514 ms 37880 KB Output is correct
10 Correct 531 ms 37880 KB Output is correct
11 Correct 530 ms 37880 KB Output is correct
12 Correct 537 ms 37880 KB Output is correct
13 Incorrect 946 ms 75640 KB Output isn't correct
14 Incorrect 943 ms 75480 KB Output isn't correct
15 Correct 952 ms 75512 KB Output is correct
16 Incorrect 921 ms 75480 KB Output isn't correct
17 Incorrect 1008 ms 75500 KB Output isn't correct
18 Incorrect 992 ms 75512 KB Output isn't correct
19 Incorrect 939 ms 75512 KB Output isn't correct
20 Correct 935 ms 75472 KB Output is correct