結果
問題 | No.563 超高速一人かるた large |
ユーザー | anta |
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 | 63 ms / 3,000 ms |
コード長 | 3,456 bytes |
コンパイル時間 | 2,112 ms |
コンパイル使用メモリ | 186,968 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-10-15 16:09:26 |
合計ジャッジ時間 | 3,526 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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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
5,248 KB |
testcase_01 | AC | 2 ms
5,248 KB |
testcase_02 | AC | 2 ms
5,248 KB |
testcase_03 | AC | 2 ms
5,248 KB |
testcase_04 | AC | 2 ms
5,248 KB |
testcase_05 | AC | 3 ms
5,248 KB |
testcase_06 | AC | 13 ms
5,248 KB |
testcase_07 | AC | 17 ms
5,248 KB |
testcase_08 | AC | 34 ms
5,248 KB |
testcase_09 | AC | 62 ms
5,248 KB |
testcase_10 | AC | 55 ms
5,248 KB |
testcase_11 | AC | 62 ms
5,248 KB |
testcase_12 | AC | 62 ms
5,248 KB |
testcase_13 | AC | 62 ms
5,248 KB |
testcase_14 | AC | 61 ms
5,248 KB |
testcase_15 | AC | 62 ms
5,248 KB |
testcase_16 | AC | 63 ms
5,248 KB |
testcase_17 | AC | 9 ms
5,248 KB |
testcase_18 | AC | 63 ms
5,248 KB |
testcase_19 | AC | 3 ms
5,248 KB |
testcase_20 | AC | 36 ms
5,248 KB |
ソースコード
#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; }