結果
| 問題 | No.1596 Distance Sum in 2D Plane | 
| コンテスト | |
| ユーザー |  nizi_24a | 
| 提出日時 | 2021-07-10 17:47:52 | 
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) | 
| 結果 | 
                                AC
                                 
                             | 
| 実行時間 | 207 ms / 2,000 ms | 
| コード長 | 4,334 bytes | 
| コンパイル時間 | 2,463 ms | 
| コンパイル使用メモリ | 202,212 KB | 
| 最終ジャッジ日時 | 2025-01-23 00:26:18 | 
| ジャッジサーバーID (参考情報) | judge2 / judge2 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 3 | 
| other | AC * 17 | 
ソースコード
#include <bits/stdc++.h>
using namespace std;
using i64 = int64_t;
using ll = long long;
using lint = long long;
typedef vector<long long> vint;
typedef pair<long long, long long> pint;
#define INF INT32_MAX / 2
#define INF64 INT64_MAX / 2
#define EPS 0.001
#define EPS14  1.0E-14
#define REP(i, n) for (ll i = 0; i < n; i++)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define ALL(f,c,...) (([&](decltype((c)) cccc) { return (f)(std::begin(cccc), std::end(cccc), ## __VA_ARGS__); })(c))
#define c(n) cout<<n<<endl;
#define cf(n) cout<<fixed<<setprecision(15)<<n<<endl;
template <class T>inline bool chmin(T&a,T b) {if(a>b){a=b;return true;}return false;}
template <class T>inline bool chmax(T&a,T b) {if(a<b){a=b;return true;}return false;}
template<class T>inline T sum(T n){return n*(n+1)/2;}
map<ll,ll> prime_fac(ll A) {map<ll,ll>mp;for(ll i=2;i*i<=A;i++){while(A%i== 0){mp[i]++;A/=i;}}if(A!=1){mp[A]=1;}return mp;}
bool is_prime(ll N){if(N<=1)return false;for(ll i=2;i*i<=N;i++){if(N%i==0) return false;}return true;}
template<class T>inline T myceil(T a,T b){return (a+(b-1))/b;}
ll pw(ll x, ll n){ll ret=1;while(n>0){if(n&1){ret*=x;}x *= x;n >>= 1;}return ret;}
bool is_product_overflow(long long a,long long b) {long prod=a*b;return (prod/b!=a);}
// modint: mod 計算を int を扱うように扱える構造体
template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};
#define MD 1000000007
using mint = Fp<MD>;
// 二項係数
const int MAX = 510000; // 問題ごとに変更する
const int MOD = 1000000007; // 問題ごとに変更する
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int main() {
    ll N, M;
    cin >> N >> M;
    vector<ll> t(M), x(M), y(M);
    REP (i, M) cin >> t[i] >> x[i] >> y[i];
    
    COMinit();
    mint ans = COM(N*2, N);
    ans *= N*2;
    REP (i, M) {
        if (t[i] == 1) ans -= (COM(x[i] + y[i], x[i]) * COM((N-x[i]-1) + (N-y[i]), N-y[i]));
        else ans -= (COM(x[i] + y[i], x[i]) * COM((N-x[i]) + (N-y[i]-1), N-x[i]));
    }
    c(ans)
}
            
            
            
        