結果
問題 | No.243 出席番号(2) |
ユーザー | anta |
提出日時 | 2015-07-10 22:31:04 |
言語 | C++11 (gcc 11.4.0) |
結果 |
MLE
|
実行時間 | - |
コード長 | 2,915 bytes |
コンパイル時間 | 727 ms |
コンパイル使用メモリ | 90,444 KB |
実行使用メモリ | 101,504 KB |
最終ジャッジ日時 | 2024-07-08 02:00:29 |
合計ジャッジ時間 | 3,235 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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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
6,816 KB |
testcase_01 | AC | 1 ms
6,944 KB |
testcase_02 | AC | 2 ms
6,944 KB |
testcase_03 | AC | 7 ms
7,552 KB |
testcase_04 | AC | 6 ms
7,424 KB |
testcase_05 | AC | 6 ms
7,552 KB |
testcase_06 | AC | 6 ms
7,680 KB |
testcase_07 | AC | 7 ms
7,424 KB |
testcase_08 | AC | 25 ms
19,072 KB |
testcase_09 | AC | 23 ms
19,072 KB |
testcase_10 | AC | 25 ms
18,944 KB |
testcase_11 | AC | 22 ms
19,072 KB |
testcase_12 | AC | 24 ms
19,200 KB |
testcase_13 | AC | 54 ms
38,656 KB |
testcase_14 | AC | 51 ms
38,784 KB |
testcase_15 | AC | 51 ms
38,656 KB |
testcase_16 | AC | 50 ms
38,784 KB |
testcase_17 | AC | 57 ms
38,656 KB |
testcase_18 | MLE | - |
testcase_19 | MLE | - |
testcase_20 | MLE | - |
testcase_21 | MLE | - |
testcase_22 | MLE | - |
testcase_23 | MLE | - |
testcase_24 | MLE | - |
testcase_25 | MLE | - |
testcase_26 | MLE | - |
testcase_27 | MLE | - |
testcase_28 | AC | 1 ms
6,940 KB |
testcase_29 | AC | 2 ms
6,944 KB |
testcase_30 | AC | 1 ms
6,944 KB |
testcase_31 | AC | 1 ms
6,944 KB |
testcase_32 | AC | 2 ms
6,944 KB |
ソースコード
#include <string> #include <vector> #include <algorithm> #include <numeric> #include <set> #include <map> #include <queue> #include <iostream> #include <sstream> #include <cstdio> #include <cmath> #include <ctime> #include <cstring> #include <cctype> #include <cassert> #include <limits> #include <functional> #include <bitset> #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)) #if defined(_MSC_VER) || __cplusplus > 199711L #define aut(r,v) auto r = (v) #else #define aut(r,v) __typeof(v) r = (v) #endif #define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it) #define all(o) (o).begin(), (o).end() #define pb(x) push_back(x) #define mp(x,y) make_pair((x),(y)) #define mset(m,v) memset(m,v,sizeof(m)) #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3fLL using namespace std; typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll; template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; } template<typename T, typename U> inline 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; } }; typedef ModInt<1000000007> mint; int main() { int N; while(cin >> N) { vi A(N); rep(i, N) cin >> A[i]; vi cnt(N, 0); rep(i, N) if(A[i] < N) ++ cnt[A[i]]; vector<vector<mint> > dp(N+1, vector<mint>(N+2)); dp[0][0] = 1; rep(i, N) rer(j, 0, N) { mint x = dp[i][j]; if(x.get() == 0) continue; dp[i+1][j] += x; dp[i+1][j+1] += x * cnt[i]; } mint ans; rer(j, 0, N) { mint x = dp[N][j]; rep(k, N - j) x *= N - j - k; if(j % 2 == 0) ans += x; else ans -= x; } printf("%d\n", ans.get()); } return 0; }