Submission #939723

#	Time	Username	Problem	Language	Result	Execution time	Memory
939723		sysia	Boat (APIO16_boat)	C++17	36 / 100	2069 ms	30024 KiB

boat

//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;

void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

#define int long long
typedef pair<int, int> T;
const int oo = 1e18, oo2 = 1e9+7;
const int mod = 1e9+7;

int mul(int a, int b){
    return (a*b)%mod;
}

void add(int &a, int b){
    a += b;
    if (a >= mod) a-=mod;
}

int power(int a, int b){
    if (!b) return 1ll;
    int k = power(a, b/2);
    k = mul(k, k);
    if (b&1) k = mul(k, a);
    return k;
}

void solve(){
    int n; cin >> n;
    vector<int>a(n+1), b(n+1);
    vector<int>s = {0, 1};
    for (int i = 1; i<=n; i++) {
        cin >> a[i] >> b[i];
        s.emplace_back(a[i]);
        s.emplace_back(b[i]);
    }
    sort(s.begin(), s.end());
    s.erase(unique(s.begin(), s.end()), s.end());
    debug(s);
    vector<T>seg;
    for (int i = 0; i<(int)s.size()-1; i++){
        seg.emplace_back(s[i], s[i]);
        if (s[i]+1 <= s[i+1]-1) seg.emplace_back(s[i]+1, s[i+1]-1);
    }
    seg.emplace_back(s.back(), s.back());
    debug(seg);
    vector<int>inv(n+1, 1);
    for (int i = 1; i<=n; i++){
        inv[i] = mul(inv[i-1], power(i, mod-2));
    }
    // auto nck = [&](int N, int K){
    //     if (N < 0 || K < 0 || N < K) return 0ll;
    //     return mul(f[N], mul(inv[K], inv[N-K]));
    // };
    int m = (int)seg.size();
    //len choose k moze byc bardzo dlugie, ale na szczescie dlugosci len jest O(n), wartosci k jest O(n)
    //i mozna uzyc lepszego wzorku by to spreprocessowac
    vector nck(m, vector<int>(n+1, 1)), I(m, vector<int>(n+1));
    for (int j = 0; j<m; j++){
        int len = seg[j].second - seg[j].first+1;
        for (int k = 0; k<=n; k++){
            for (int rep = 0, val = len; rep < k; rep++, val--){
                nck[j][k] = mul(nck[j][k], val);
            }
            nck[j][k] = mul(nck[j][k], inv[k]);
            I[j][k] = power(nck[j][k], mod-2);
        }
    }
    //dp[i][j][k] = na ile sposobow da sie wybrac dzialajace rozwiazanie na prefiksie i, gdzie 
    //ostatnie k elementow nalezy do przedzialu seg[j]
    vector dp(m, vector<int>(n+1, 0));
    dp[0][1] = 1; //zerowy przedzial
    for (int i = 1; i<=n; i++){
        auto new_dp = dp;
        vector<int>pref(m);
        for (int s = 0; s < m; s++){
            if (s) pref[s] += pref[s-1];
            for (int k = 0; k <= n; k++) add(pref[s], dp[s][k]);
        }
        // dp[i] = dp[i-1]; //nic nie dokladamy
        for (int j = 1; j<m; j++){
            if (seg[j].second < a[i] || seg[j].first > b[i]) continue;
            int len = seg[j].second - seg[j].first + 1;
            // dp[i][j][1] = suma po dp[i-1][<j][cokolwiek]
            add(new_dp[j][1], mul(pref[j-1], len)); //dokladamy pierwszy element tego przedzialu
            // for (int s = 0; s < j; s++){
            //     for (int k = 0; k <= n; k++){
            //         if (dp[i-1][s][k]){
            //             debug(dp[i-1][s][k], len);
            //         }
            //         add(dp[i][j][1], mul(dp[i-1][s][k], len));
            //     }
            // } 
            for (int k = 2; k <= min(i, len); k++){
                //dokladamy juz nie pierwszy element tego przedzialu
                // add(new_dp[j][k], mul(dp[j][k-1], mul(power(nck(len, k-1), mod-2), nck(len, k))));
                add(new_dp[j][k], mul(dp[j][k-1], mul(I[j][k-1], nck[j][k])));
            }
        }
        dp.swap(new_dp);
    }
    int ans = mod-1;
    for (int i = 0; i<m; i++){
        for (int j = 0; j<=n; j++){
            add(ans, dp[i][j]);
        }
    }
    cout << ans << "\n";
}

int32_t main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    //cin >> t;
    while (t--) solve();

    return 0;
}

• Subtask 1: 9 / 9
• Subtask 2: 0 / 22
• Subtask 3: 27 / 27
• Subtask 4: 0 / 42