結果

問題 No.1513 simple 門松列 problem
ユーザー maimai
提出日時 2021-06-24 08:11:36
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 95 ms / 3,000 ms
コード長 7,588 bytes
コンパイル時間 3,086 ms
コンパイル使用メモリ 190,436 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-25 07:27:59
合計ジャッジ時間 4,251 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 2 ms
5,248 KB
testcase_01 AC 2 ms
5,376 KB
testcase_02 AC 95 ms
5,376 KB
testcase_03 AC 2 ms
5,376 KB
testcase_04 AC 2 ms
5,376 KB
testcase_05 AC 2 ms
5,376 KB
testcase_06 AC 2 ms
5,376 KB
testcase_07 AC 2 ms
5,376 KB
testcase_08 AC 3 ms
5,376 KB
testcase_09 AC 2 ms
5,376 KB
testcase_10 AC 2 ms
5,376 KB
testcase_11 AC 2 ms
5,376 KB
testcase_12 AC 2 ms
5,376 KB
testcase_13 AC 2 ms
5,376 KB
testcase_14 AC 93 ms
5,376 KB
testcase_15 AC 94 ms
5,376 KB
testcase_16 AC 93 ms
5,376 KB
testcase_17 AC 60 ms
5,376 KB
testcase_18 AC 32 ms
5,376 KB
testcase_19 AC 3 ms
5,376 KB
testcase_20 AC 14 ms
5,376 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 # 

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include <bits/stdc++.h>

using namespace std;
using ll = long long int;

#define all(v) (v).begin(),(v).end()
#define repeat(cnt,l) for(typename remove_const<typename remove_reference<decltype(l)>::type>::type cnt={};(cnt)<(l);++(cnt))
#define rrepeat(cnt,l) for(auto cnt=(l)-1;0<=(cnt);--(cnt))
#define iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))
#define diterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);--(cnt))
const long long MD = 998244353; const long double PI = 3.1415926535897932384626433832795L;
template<typename T1, typename T2> inline ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << '(' << p.first << ':' << p.second << ')'; return o; }
template<typename T> inline T& chmax(T& to, const T& val) { return to = max(to, val); }
template<typename T> inline T& chmin(T& to, const T& val) { return to = min(to, val); }
void bye(string s, int code = 0) { cout << s << endl; exit(code); }
mt19937_64 randdev(8901016);
template<typename T, typename Random = decltype(randdev), typename enable_if<is_integral<T>::value>::type* = nullptr>
inline T rand(T l, T h, Random& rand = randdev) { return uniform_int_distribution<T>(l, h)(rand); }
template<typename T, typename Random = decltype(randdev), typename enable_if<is_floating_point<T>::value>::type* = nullptr>
inline T rand(T l, T h, Random& rand = randdev) { return uniform_real_distribution<T>(l, h)(rand); }template<typename T>
static ostream& operator<<(ostream& o, const std::vector<T>& v) {
  o << "[ "; for(const auto& e : v) o<<e<<' ';
  return o << ']';
}

template <typename I>
struct MyRangeFormat{ I b,e; MyRangeFormat(I _b, I _e):b(_b),e(_e){} };
template<typename I>
static ostream& operator<<(ostream& o, const MyRangeFormat<I>& f) {
  o << "[ "; iterate(i,f.b,f.e) o<<*i<<' ';
  return o << ']';
}
template <typename I>
struct MyMatrixFormat{
  const I& p; long long n, m;
  MyMatrixFormat(const I& _p, long long _n, long long _m):p(_p),n(_n),m(_m){}
};
template<typename I>
static ostream& operator<<(ostream& o, const MyMatrixFormat<I>& f) {
  o<<'\n';
  repeat(i,(f.n)) {
    repeat(j,f.m) o<<f.p[i][j]<<' ';
    o<<'\n';
  }
  return o;
}
struct LOG_t { ~LOG_t() { cout << endl; } };
#define LOG (LOG_t(),cout<<'L'<<__LINE__<<": ")
#define FMTA(m,w) (MyRangeFormat<decltype(m+0)>(m,m+w))
#define FMTR(b,e) (MyRangeFormat<decltype(e)>(b,e))
#define FMTV(v) FMTR(v.begin(),v.end())
#define FMTM(m,h,w) (MyMatrixFormat<decltype(m+0)>(m,h,w))

#if defined(_WIN32) || defined(_WIN64)
#define getc_x _getc_nolock
#define putc_x _putc_nolock
#elif defined(__GNUC__)
#define getc_x getc_unlocked
#define putc_x putc_unlocked
#else
#define getc_x getc
#define putc_x putc
#endif
class MaiScanner {
  FILE* fp_;
  constexpr bool isvisiblechar(char c) noexcept { return (0x21<=(c)&&(c)<=0x7E); }
public:
  inline MaiScanner(FILE* fp):fp_(fp){}
  template<typename T> void input_integer(T& var) noexcept {
    var = 0; T sign = 1;
    int cc = getc_x(fp_);
    for (; cc < '0' || '9' < cc; cc = getc_x(fp_))
      if (cc == '-') sign = -1;
    for (; '0' <= cc && cc <= '9'; cc = getc_x(fp_))
      var = (var << 3) + (var << 1) + cc - '0';
    var = var * sign;
  }
  inline int c() noexcept { return getc_x(fp_); }
  template<typename T, typename enable_if<is_integral<T>::value, nullptr_t>::type = nullptr>
  inline MaiScanner& operator>>(T& var) noexcept { input_integer<T>(var); return *this; }
  inline MaiScanner& operator>>(string& var) {
    int cc = getc_x(fp_);
    for (; !isvisiblechar(cc); cc = getc_x(fp_));
    for (; isvisiblechar(cc); cc = getc_x(fp_))
      var.push_back(cc);
    return *this;
  }
  template<typename IT> inline void in(IT begin, IT end) { for (auto it = begin; it != end; ++it) *this >> *it; }
};
class MaiPrinter {
  FILE* fp_;
public:
  inline MaiPrinter(FILE* fp):fp_(fp){}
  template<typename T>
  void output_integer(T var) noexcept {
    if (var == 0) { putc_x('0', fp_); return; }
    if (var < 0)
      putc_x('-', fp_),
      var = -var;
    char stack[32]; int stack_p = 0;
    while (var)
      stack[stack_p++] = '0' + (var % 10),
      var /= 10;
    while (stack_p)
      putc_x(stack[--stack_p], fp_);
  }
  inline MaiPrinter& operator<<(char c) noexcept { putc_x(c, fp_); return *this; }
  template<typename T, typename enable_if<is_integral<T>::value, nullptr_t>::type = nullptr>
  inline MaiPrinter& operator<<(T var) noexcept { output_integer<T>(var); return *this; }
  inline MaiPrinter& operator<<(char* str_p) noexcept { while (*str_p) putc_x(*(str_p++), fp_); return *this; }
  inline MaiPrinter& operator<<(const string& str) {
    const char* p = str.c_str();
    const char* l = p + str.size();
    while (p < l) putc_x(*p++, fp_);
    return *this;
  }
  template<typename IT> void join(IT begin, IT end, char sep = ' ') { for (bool b = 0; begin != end; ++begin, b = 1) b ? *this << sep << *begin : *this << *begin; }
};
MaiScanner scanner(stdin);
MaiPrinter printer(stdout);

bool kado(int x, int y, int z) {
  return (x!=z)&&((x<y&&y>z)||(x>y&&y<z));
}

//

int N, K;

//

int dp1[2][200][200];
int dp2[2][200][200];

int cjk[200][200];
// int jtotal[200];

int main() {
  scanner >> N >> K;
  
  repeat(i, K) {
    repeat(j, K) {
      dp1[0][i][j] = i != j;
    }
  }
  repeat(i, K) {
    repeat(j, K) {
      dp2[0][i][j] = i != j ? i + j : 0;
    }
  }
  
  repeat(j, K) {
    repeat(k, K) {
      repeat(i, K) {
        if (kado(i,j,k)) {
          cjk[j][k] += 1;
        }
      }
    }
  }
  
  repeat(c, N-2) {
    int p1 = c&1;
    int p2 = !p1;
    fill(dp1[p2][0], dp1[p2][K], 0);
    fill(dp2[p2][0], dp2[p2][K], 0);
    repeat(j, K) {
      ll t1kj = 0, t2kj = 0;
      for (int i = 0; i < j; ++i) {
        // if (i == k) continue;
        t1kj += dp1[p1][i][j];
        t2kj += dp2[p1][i][j];
      }
      t1kj %= MD;
      t2kj %= MD;
      ll t1jk = 0, t2jk = 0;
      for (int i = j+1; i < K; ++i) {
        // if (i == k) continue;
        t1jk += dp1[p1][i][j];
        t2jk += dp2[p1][i][j];
      }
      t1jk %= MD;
      t2jk %= MD;
      
      repeat(k, K) {
        if (j > k) {
          ll t1 = (t1kj - dp1[p1][k][j] + MD)%MD;  // reduce i == k
          ll t2 = (t2kj - dp2[p1][k][j] + MD)%MD;  // reduce i == k
          dp1[p2][j][k] += t1;
          dp1[p2][j][k] %= MD;
          dp2[p2][j][k] += (ll(t2) + ll(t1)*k)%MD;
          dp2[p2][j][k] %= MD;
          // for (int i = 0; i < j; ++i) {
          //   if (i == k) continue;
          //   dp1[p2][j][k] += dp1[p1][i][j];
          //   dp1[p2][j][k] %= MD;
          //   dp2[p2][j][k] += (ll(dp2[p1][i][j]) + ll(dp1[p1][i][j])*k)%MD;
          //   dp2[p2][j][k] %= MD;
          // }
        } else if (j < k) {
          ll t1 = (t1jk - dp1[p1][k][j] + MD)%MD;  // reduce i == k
          ll t2 = (t2jk - dp2[p1][k][j] + MD)%MD;  // reduce i == k
          dp1[p2][j][k] += t1;
          dp1[p2][j][k] %= MD;
          dp2[p2][j][k] += (ll(t2) + ll(t1)*k)%MD;
          dp2[p2][j][k] %= MD;
          // for (int i = j+1; i < K; ++i) {
          //   if (i == k) continue;
          //   dp1[p2][j][k] += dp1[p1][i][j];
          //   dp1[p2][j][k] %= MD;
          //   dp2[p2][j][k] += (ll(dp2[p1][i][j]) + ll(dp1[p1][i][j])*k)%MD;
          //   dp2[p2][j][k] %= MD;
          // }
        }
      }
    }
  }
  const int p = N&1;
  ll total1 = 0, total2 = 0;
  repeat(i, K) {
    repeat(j, K) {
      total1 += dp1[p][i][j];
    }
  }
  repeat(i, K) {
    repeat(j, K) {
      total2 += dp2[p][i][j];
    }
  }
  cout << (total1%MD) << ' ' << (total2%MD) << endl;
  
  return 0;
}
0