#676288 (C++17) No.1596 Distance Sum in 2D Plane

• 問題: No.1596 Distance Sum in 2D Plane
• ユーザー: Example0911
• 提出日時: 2021-07-09 22:31:58
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 155 ms / 2,000 ms
• コード長: 2,034 bytes
• コンパイル時間: 2,315 ms
• コンパイル使用メモリ: 197,856 KB
• 最終ジャッジ日時: 2025-01-22 22:15:16
• ジャッジサーバーID （参考情報）: judge2 / judge3

ファイルパターン	結果
sample	AC * 3
other	AC * 17

ソースコード

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#include "bits/stdc++.h"
#define int long long
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const ll INF = (1LL << 61);
ll mod = 1000000007;
struct mint {
    ll x; // typedef long long ll;
    mint(ll x = 0) :x((x%mod + mod) % mod) {}
    mint operator-() const { return mint(-x); }
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }
    // for prime mod
    mint inv() const {
        return pow(mod - 2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res /= a;
    }
};
istream& operator>>(istream& is, mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
struct combination {
    vector<mint> fact, ifact;
    combination(int n) :fact(n + 1), ifact(n + 1) {
        //assert(n < mod);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
};
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    combination com(400010);
    int N, M; cin >> N >> M;
    mint ans = com(2 * N, N) * 2 * N;
    for (int i = 0; i < M; i++) {
        int t, x, y; cin >> t >> x >> y;
        int nx = x, ny = y;
        if (t == 1)nx = x + 1;
        else ny = y + 1;
        ans -= com((x + y), x) * com((N  - nx) + (N - ny), N - nx);
    }
    cout << ans << endl;
    return 0;
}
 
 
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