結果

問題 No.563 超高速一人かるた large
ユーザー antaanta
提出日時 2017-08-25 23:08:21
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 62 ms / 3,000 ms
コード長 3,456 bytes
コンパイル時間 2,656 ms
コンパイル使用メモリ 182,120 KB
実行使用メモリ 4,380 KB
最終ジャッジ日時 2023-08-05 19:01:40
合計ジャッジ時間 3,718 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(β)

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 1 ms
4,380 KB
testcase_01 AC 1 ms
4,376 KB
testcase_02 AC 1 ms
4,380 KB
testcase_03 AC 2 ms
4,380 KB
testcase_04 AC 2 ms
4,380 KB
testcase_05 AC 3 ms
4,376 KB
testcase_06 AC 13 ms
4,376 KB
testcase_07 AC 17 ms
4,376 KB
testcase_08 AC 33 ms
4,380 KB
testcase_09 AC 61 ms
4,380 KB
testcase_10 AC 54 ms
4,380 KB
testcase_11 AC 61 ms
4,376 KB
testcase_12 AC 61 ms
4,376 KB
testcase_13 AC 60 ms
4,380 KB
testcase_14 AC 60 ms
4,380 KB
testcase_15 AC 61 ms
4,376 KB
testcase_16 AC 61 ms
4,380 KB
testcase_17 AC 9 ms
4,380 KB
testcase_18 AC 62 ms
4,376 KB
testcase_19 AC 2 ms
4,376 KB
testcase_20 AC 36 ms
4,380 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];
}

int main() {
	int N;
	while (~scanf("%d", &N)) {
		vector<string> words(N);
		rep(i, N) {
			char S[200001];
			scanf("%s", S);
			words[i] = S;
		}
		sort(words.begin(), words.end());
		vector<int> counts(N + 1);
		{
			vector<pair<int, int>> q, nq;
			q.emplace_back(0, N);
			for (int lv = 0; !q.empty(); ++ lv) {
				nq.clear();
				for (auto p : q) {
					if(lv != 0)
						++ counts[p.second - p.first];
					for (int i = p.first, j = i; i != p.second; i = j) {
						for (++ j; j != p.second && words[j][lv] == words[i][lv]; ++ j);
						if (j - i > 1)
							nq.emplace_back(i, j);
					}
				}
				q.swap(nq);
			}
		}
		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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