結果

問題 No.1596 Distance Sum in 2D Plane
ユーザー firiexpfiriexp
提出日時 2021-07-09 21:36:06
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 71 ms / 2,000 ms
コード長 3,462 bytes
コンパイル時間 930 ms
コンパイル使用メモリ 104,124 KB
実行使用メモリ 6,656 KB
最終ジャッジ日時 2024-07-01 15:25:48
合計ジャッジ時間 2,848 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 8 ms
6,400 KB
testcase_01 AC 8 ms
6,656 KB
testcase_02 AC 68 ms
6,656 KB
testcase_03 AC 67 ms
6,528 KB
testcase_04 AC 67 ms
6,656 KB
testcase_05 AC 67 ms
6,528 KB
testcase_06 AC 71 ms
6,528 KB
testcase_07 AC 67 ms
6,656 KB
testcase_08 AC 67 ms
6,528 KB
testcase_09 AC 66 ms
6,528 KB
testcase_10 AC 68 ms
6,656 KB
testcase_11 AC 57 ms
6,528 KB
testcase_12 AC 57 ms
6,656 KB
testcase_13 AC 57 ms
6,528 KB
testcase_14 AC 2 ms
5,376 KB
testcase_15 AC 2 ms
5,376 KB
testcase_16 AC 2 ms
5,376 KB
testcase_17 AC 2 ms
5,376 KB
testcase_18 AC 2 ms
5,376 KB
testcase_19 AC 2 ms
5,376 KB
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ソースコード

diff #

#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using uint = unsigned;
using ull = unsigned long long;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;

template <uint M>
struct modint {
    uint val;
public:
    static modint raw(int v) { modint x; x.val = v; return x; }
    modint() : val(0) {}
    template <class T>
    modint(T v) { ll x = (ll)(v%(ll)(M)); if (x < 0) x += M; val = uint(x); }
    modint(bool v) { val = ((unsigned int)(v) % M); }
    modint& operator++() { val++; if (val == M) val = 0; return *this; }
    modint& operator--() { if (val == 0) val = M; val--; return *this; }
    modint operator++(int) { modint result = *this; ++*this; return result; }
    modint operator--(int) { modint result = *this; --*this; return result; }
    modint& operator+=(const modint& b) { val += b.val; if (val >= M) val -= M; return *this; }
    modint& operator-=(const modint& b) { val -= b.val; if (val >= M) val += M; return *this; }
    modint& operator*=(const modint& b) { ull z = val; z *= b.val; val = (uint)(z % M); return *this; }
    modint& operator/=(const modint& b) { return *this = *this * b.inv(); }
    modint operator+() const { return *this; }
    modint operator-() const { return modint() - *this; }
    modint pow(long long n) const { modint x = *this, r = 1; while (n) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
    modint inv() const { return pow(M-2); }
    friend modint operator+(const modint& a, const modint& b) { return modint(a) += b; }
    friend modint operator-(const modint& a, const modint& b) { return modint(a) -= b; }
    friend modint operator*(const modint& a, const modint& b) { return modint(a) *= b; }
    friend modint operator/(const modint& a, const modint& b) { return modint(a) /= b; }
    friend bool operator==(const modint& a, const modint& b) { return a.val == b.val; }
    friend bool operator!=(const modint& a, const modint& b) { return a.val != b.val; }
};
using mint = modint<MOD>;
class Factorial {
    vector<mint> facts, factinv;
public:
    explicit Factorial(int n) : facts(n+1), factinv(n+1) {
        facts[0] = 1;
        for (int i = 1; i < n+1; ++i) facts[i] = facts[i-1] * mint(i);
        factinv[n] = facts[n].inv();
        for (int i = n-1; i >= 0; --i) factinv[i] = factinv[i+1] * mint(i+1);
    }
    mint fact(int k) const {
        if(k >= 0) return facts[k]; else return factinv[-k];
    }
    mint operator[](const int &k) const {
        if(k >= 0) return facts[k]; else return factinv[-k];
    }
    mint C(int p, int q) const {
        if(q < 0 || p < q) return 0;
        return facts[p] * factinv[q] * factinv[p-q];
    }
    mint P(int p, int q) const {
        if(q < 0 || p < q) return 0;
        return facts[p] * factinv[p-q];
    }
    mint H(int p, int q) const {
        if(p < 0 || q < 0) return 0;
        return q == 0 ? 1 : C(p+q-1, q);
    }
};


int main() {
    int n, m;
    cin >> n >> m;
    Factorial f(2*(n+1));
    mint ans = f.C(2*n, n)*(2*n);
    for (int i = 0; i < m; ++i) {
        int t, x, y;
        scanf("%d %d %d", &t, &x, &y);
        if(t == 1) ans -= f.C(x+y, x)*f.C((n-x-1)+(n-y), (n-x-1));
        else ans -= f.C(x+y, x)*f.C((n-x)+(n-y-1), (n-x));
    }
    cout << ans.val << "\n";
    return 0;
}
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