답안 #1073097

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
1073097 2024-08-24T09:28:53 Z coolboy19521 힘 센 거북 (IZhO11_turtle) C++17
45 / 100
311 ms 476 KB
// Thanks Benq

// #pragma GCC optimize("Ofast")
#include "bits/stdc++.h"
#define ll long long
#define int ll
 
using namespace std;
 
const int sz = 6e5 + 2;
const int sm = 30;
 
vector<pair<int,int>> vp;
vector<int> ps;
ll wz[sm][sm];
ll dp[sm][sm];
bool sv[sz];
 
ll bpw(ll x, ll b, ll z) {
    ll r = 1;
    while (b) {
        if (b & 1)
            (r *= x) %= z;
        (x *= x) %= z;
        b >>= 1;
    }
    return r;
}

vector<pair<ll, ll>> factorize(ll z) {
    vector<pair<ll, ll>> factors;
    for (ll i = 2; i * i <= z; ++i) {
        if (z % i == 0) {
            ll count = 0;
            while (z % i == 0) {
                z /= i;
                ++count;
            }
            factors.push_back({i, count});
        }
    }
    if (z > 1) {
        factors.push_back({z, 1});
    }
    return factors;
}
ll chinese_remainder(vector<ll>& remainders, vector<ll>& moduli) {
    ll result = 0, prod = 1;
    for (auto m : moduli) prod *= m;
    
    for (int i = 0; i < remainders.size(); i++) {
        ll pp = prod / moduli[i];
        result += remainders[i] * bpw(pp, moduli[i] - 2, moduli[i]) * pp;
        result %= prod;
    }
    
    return result;
}

ll factorial_mod(ll n, ll p, ll pe) {
    ll result = 1;
    for (ll i = 1; i <= n; ++i) {
        if (i % p != 0) {
            result = (result * i) % pe;
        }
    }
    return result;
}

ll nck_mod(ll n, ll k, ll p, ll e) {
    if (k > n) return 0;
    if (k == 0 || n == k) return 1;
    
    ll pe = pow(p, e);
    
    ll numerator = 1, denominator = 1;
    
    while (n > 0 || k > 0) {
        ll n_mod_p = n % p;
        ll k_mod_p = k % p;
        
        if (k_mod_p > n_mod_p) return 0;
        
        numerator = (numerator * factorial_mod(n_mod_p, p, pe)) % pe;
        denominator = (denominator * factorial_mod(k_mod_p, p, pe)) % pe;
        denominator = (denominator * factorial_mod(n_mod_p - k_mod_p, p, pe)) % pe;
        
        n /= p;
        k /= p;
    }
    
    return (numerator * bpw(denominator, pe - pe/p - 1, pe)) % pe;
}


 
ll nck(ll n, ll k, ll z) {
    if (n < k || n < 0) return 0;
    if (k == 0 || n == k) return 1;
    if (1 == k || n-1 == k) return n;

    vector<pair<ll, ll>> factors = factorize(z);
    vector<ll> remainders, moduli;

    for (auto& factor : factors) {
        ll p = factor.first;
        ll e = factor.second;
        ll pe = pow(p, e); // Compute p^e
        moduli.push_back(pe);
        remainders.push_back(nck_mod(n, k, p, e));
    }

    return chinese_remainder(remainders, moduli);
}

 
ll gt(pair<int,int>& f, pair<int,int>& s, int z) {
    int xds = s.first - f.first;
    int yds = s.second - f.second;
    return nck(xds + yds, yds, z);
}
 
ll sb(ll a, ll b, ll z) {
    return (a + z - b) % z;
}
 
ll ml(ll a, ll b, ll z) {
    return a * b % z;
}
 
ll ad(ll a, ll b, ll z) {
    return (a + b) % z;
}
 
signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);
 
    ll n, m, k, t, z;
    cin >> n >> m >> k >> t >> z;
 
    for (int i = 0; i < k; i ++) {
        int x, y;
        cin >> x >> y;
        vp.push_back(make_pair(x, y));
    }
 
    vp.push_back(make_pair(0, 0));
    vp.push_back(make_pair(n, m));
 
    sort(begin(vp), end(vp));
    k = vp.size();
 
    for (int i = 1; i < k; i ++) {
        for (int j = i - 1; 0 <= j; j --) {
            wz[j][i] = gt(vp[j], vp[i], z);
            // cout << j << ' ' << i << ' ' << wz[j][i] << '\n';
            for (int l = j + 1; l < i; l ++) {
                wz[j][i] = sb(wz[j][i], ml(wz[j][l], gt(vp[l], vp[i], z), z), z);
                // cout << "hi: " << l << ' ' << i << ' ' << gt(vp[l], vp[i], z) << '\n';
            }
            // cout << j << ' ' << i << ' ' << wz[j][i] << '\n';
        }
    }
 
    dp[0][1] = 1;
 
    for (int i = 1; i < k; i ++)
        for (int j = 0; j < i; j ++)
            for (int l = 2; l <= t + 2; l ++)
                dp[i][l] = ad(dp[i][l], ml(dp[j][l - 1], wz[j][i], z), z);
 
    ll r = 0;
 
    for (int i = 1; i <= t + 2; i ++)
        r = ad(r, dp[k - 1][i], z);

    // cout << ps.size() << '\n';
    cout << r << '\n';
}

Compilation message

turtle.cpp: In function 'long long int chinese_remainder(std::vector<long long int>&, std::vector<long long int>&)':
turtle.cpp:51:23: warning: comparison of integer expressions of different signedness: 'long long int' and 'std::vector<long long int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   51 |     for (int i = 0; i < remainders.size(); i++) {
      |                     ~~^~~~~~~~~~~~~~~~~~~
# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 348 KB Output is correct
2 Incorrect 0 ms 348 KB Output isn't correct
3 Correct 0 ms 348 KB Output is correct
4 Correct 1 ms 468 KB Output is correct
5 Correct 2 ms 348 KB Output is correct
6 Correct 2 ms 348 KB Output is correct
7 Correct 4 ms 348 KB Output is correct
8 Incorrect 2 ms 348 KB Output isn't correct
9 Incorrect 25 ms 472 KB Output isn't correct
10 Correct 52 ms 444 KB Output is correct
11 Correct 311 ms 440 KB Output is correct
12 Correct 185 ms 452 KB Output is correct
13 Incorrect 1 ms 348 KB Output isn't correct
14 Incorrect 1 ms 472 KB Output isn't correct
15 Incorrect 1 ms 348 KB Output isn't correct
16 Incorrect 2 ms 348 KB Output isn't correct
17 Incorrect 1 ms 476 KB Output isn't correct
18 Incorrect 1 ms 348 KB Output isn't correct
19 Incorrect 1 ms 344 KB Output isn't correct
20 Incorrect 1 ms 348 KB Output isn't correct