結果

問題 No.562 超高速一人かるた small
ユーザー antaanta
提出日時 2017-08-25 23:00:45
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 24 ms / 3,000 ms
コード長 3,450 bytes
コンパイル時間 1,993 ms
コンパイル使用メモリ 178,708 KB
実行使用メモリ 47,488 KB
最終ジャッジ日時 2024-10-15 16:39:08
合計ジャッジ時間 3,043 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
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 23 ms
47,488 KB
testcase_01 AC 24 ms
47,488 KB
testcase_02 AC 24 ms
47,360 KB
testcase_03 AC 24 ms
47,360 KB
testcase_04 AC 23 ms
47,488 KB
testcase_05 AC 24 ms
47,488 KB
testcase_06 AC 24 ms
47,488 KB
testcase_07 AC 24 ms
47,488 KB
testcase_08 AC 24 ms
47,488 KB
testcase_09 AC 23 ms
47,488 KB
testcase_10 AC 22 ms
47,360 KB
testcase_11 AC 24 ms
47,488 KB
testcase_12 AC 22 ms
47,360 KB
testcase_13 AC 23 ms
47,488 KB
testcase_14 AC 22 ms
47,480 KB
testcase_15 AC 22 ms
47,488 KB
testcase_16 AC 24 ms
47,488 KB
testcase_17 AC 24 ms
47,360 KB
testcase_18 AC 22 ms
47,360 KB
testcase_19 AC 23 ms
47,488 KB
testcase_20 AC 23 ms
47,488 KB
testcase_21 AC 22 ms
47,360 KB
testcase_22 AC 22 ms
47,488 KB
testcase_23 AC 23 ms
47,488 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;
	}
};
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
	ModInt<MOD> r = 1;
	while (k) {
		if (k & 1) r *= a;
		a *= a;
		k >>= 1;
	}
	return r;
}
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];
}

struct Node {
	Node *next[27];
	int num;
	Node() : next{}, num(0) {}
};

int main() {
	int N;
	while (~scanf("%d", &N)) {
		vector<Node> nodes(202001);
		int nNodes = 0;
		Node *root = new(&nodes[nNodes ++]) Node;
		const int A = 26;
		vector<Node*> path;
		rep(i, N) {
			char S[200001];
			scanf("%s", S);
			path = { root };
			Node *t = root;
			for (const char *p = S; *p; ++ p) {
				auto &x = t->next[*p - 'a'];
				if (!x) x = new(&nodes[nNodes ++]) Node;
				t = x;
				path.push_back(t);
			}
			for (auto p : path)
				++ p->num;
		}
		vector<int> counts(N + 1);
		reu(i, 1, nNodes)
			++ counts[nodes[i].num];
		nCr_computeFactinv(N);
		vector<mint> ans(N + 1);
		rer(n, 2, N) {
			rer(K, 1, N) {
				mint sum = nCr(N - 1, K - 1) * n;
				//k = n
				if(n <= K)
					sum -= nCr(N - n, K - n);
				sum *= fact[K];
				ans[K] += sum * counts[n];
			}
		}
		rer(K, 1, N) {
			mint total = fact[N] * factinv[N - K];
			ans[K] += total * K;
			printf("%d\n", ans[K].get());
		}
	}
	return 0;
}
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