結果

問題 No.498 ワープクリスタル (給料日編)
ユーザー antaanta
提出日時 2017-03-24 22:32:32
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 18 ms / 2,000 ms
コード長 3,128 bytes
コンパイル時間 1,595 ms
コンパイル使用メモリ 170,532 KB
実行使用メモリ 4,380 KB
最終ジャッジ日時 2023-09-20 07:06:23
合計ジャッジ時間 3,228 ms
ジャッジサーバーID
(参考情報)
judge14 / judge15
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 2 ms
4,376 KB
testcase_01 AC 2 ms
4,376 KB
testcase_02 AC 2 ms
4,376 KB
testcase_03 AC 1 ms
4,380 KB
testcase_04 AC 18 ms
4,376 KB
testcase_05 AC 2 ms
4,380 KB
testcase_06 AC 18 ms
4,380 KB
testcase_07 AC 18 ms
4,380 KB
testcase_08 AC 18 ms
4,376 KB
testcase_09 AC 18 ms
4,380 KB
testcase_10 AC 2 ms
4,380 KB
testcase_11 AC 9 ms
4,380 KB
testcase_12 AC 1 ms
4,380 KB
testcase_13 AC 8 ms
4,380 KB
testcase_14 AC 2 ms
4,380 KB
testcase_15 AC 2 ms
4,376 KB
testcase_16 AC 2 ms
4,380 KB
testcase_17 AC 2 ms
4,376 KB
testcase_18 AC 18 ms
4,376 KB
testcase_19 AC 18 ms
4,380 KB
testcase_20 AC 18 ms
4,376 KB
testcase_21 AC 18 ms
4,380 KB
testcase_22 AC 18 ms
4,380 KB
testcase_23 AC 18 ms
4,376 KB
testcase_24 AC 18 ms
4,376 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#include "bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> static void amin(T &x, U y) { if (y < x) x = y; }
template<typename T, typename U> static void amax(T &x, U y) { if (x < y) x = y; }


template<int MOD>
struct ModInt {
	static const int Mod = MOD;
	unsigned x;
	ModInt() : x(0) { }
	ModInt(signed sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	ModInt(signed long long sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	int get() const { return (int)x; }

	ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
	ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }

	ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
	ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }

	ModInt inverse() const {
		signed a = x, b = MOD, u = 1, v = 0;
		while (b) {
			signed t = a / b;
			a -= t * b; std::swap(a, b);
			u -= t * v; std::swap(u, v);
		}
		if (u < 0) u += Mod;
		ModInt res; res.x = (unsigned)u;
		return res;
	}

	bool operator==(ModInt that) const { return x == that.x; }
	bool operator!=(ModInt that) const { return x != that.x; }
	ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
typedef ModInt<1000000007> mint;

vector<mint> fact, factinv;
void nCr_computeFactinv(int N) {
	N = min(N, mint::Mod - 1);
	fact.resize(N + 1); factinv.resize(N + 1);
	fact[0] = 1;
	rer(i, 1, N) fact[i] = fact[i - 1] * i;
	factinv[N] = fact[N].inverse();
	for (int i = N; i >= 1; i --) factinv[i - 1] = factinv[i] * i;
}
mint nCr(int n, int r) {
	if (n >= mint::Mod)
		return nCr(n % mint::Mod, r % mint::Mod) * nCr(n / mint::Mod, r / mint::Mod);
	return r > n ? 0 : fact[n] * factinv[n - r] * factinv[r];
}

int main() {
	int Gx; int Gy; int K;
	while (~scanf("%d%d%d", &Gx, &Gy, &K)) {
		nCr_computeFactinv(K * 15);
		vector<int> xs(K), ys(K), ns(K);
		rep(i, K) {
			int x; int y; int n;
			scanf("%d%d%d", &x, &y, &n);
			xs[i] = x, ys[i] = y, ns[i] = n;
		}
		mint ans;
		rep(S, 1 << (K * 4)) {
			mint prod = 1;
			int p = 0;
			int X = 0, Y = 0;
			rep(i, K) {
				int n = S >> (i * 4) & 15;
				if (n > ns[i]) {
					prod = mint();
					break;
				}
				prod *= factinv[n];
				p += n;
				X += xs[i] * n;
				Y += ys[i] * n;
			}
			if (X == Gx && Y == Gy) {
				prod *= fact[p];
				ans += prod;
			}
		}
		printf("%d\n", ans.get());
	}
	return 0;
}
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