結果

問題 No.122 傾向と対策：門松列（その３）
コンテスト
ユーザー fumiphys
提出日時 2019-09-12 02:09:40
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 14 ms / 5,000 ms
コード長 4,584 bytes
コンパイル時間 1,618 ms
コンパイル使用メモリ 168,820 KB
実行使用メモリ 11,008 KB
最終ジャッジ日時 2024-07-02 16:57:15
合計ジャッジ時間 2,126 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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ファイルパターン 結果
other AC * 8
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 # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

ll dpu1[1<<4][20010], dpu2[1<<3][20010];
ll dpd1[1<<4][20010], dpd2[1<<3][20010];

int main(int argc, char const* argv[])
{
  ll amin[7], amax[7];
  rep(i, 7)cin >> amin[i] >> amax[i];
  dpd1[0][0] = 1;
  for(int i = 1; i <= 20000; i++){
    for(int j = 0; j < bit(4); j++){
      dpd1[j][i] = dpd1[j][i-1];
      for(int k = 0; k < 4; k++){
        if((j >> k) & 1 && amin[2*k] <= i && i <= amax[2*k]){
          (dpd1[j][i] += dpd1[j - bit(k)][i-1]) %= mod;
        }
      }
    }
  }
  dpd2[0][0] = 1;
  for(int i = 1; i <= 20000; i++){
    for(int j = 0; j < bit(3); j++){
      dpd2[j][i] = dpd2[j][i-1];
      for(int k = 0; k < 3; k++){
        if((j >> k) & 1 && amin[2*k+1] <= i && i <= amax[2*k+1]){
          (dpd2[j][i] += dpd2[j - bit(k)][i-1]) %= mod;
        }
      }
    }
  }
  dpu1[0][20001] = 1;
  for(int i = 20000; i > 0; i--){
    for(int j = 0; j < bit(4); j++){
      dpu1[j][i] = dpu1[j][i+1];
      for(int k = 0; k < 4; k++){
        if((j >> k) & 1 && amin[2*k] <= i && i <= amax[2*k]){
          (dpu1[j][i] += dpu1[j - bit(k)][i+1]) %= mod;
        }
      }
    }
  }
  dpu2[0][20001] = 1;
  for(int i = 20000; i > 0; i--){
    for(int j = 0; j < bit(3); j++){
      dpu2[j][i] = dpu2[j][i+1];
      for(int k = 0; k < 3; k++){
        if((j >> k) & 1 && amin[2*k+1] <= i && i <= amax[2*k+1]){
          (dpu2[j][i] += dpu2[j - bit(k)][i+1]) %= mod;
        }
      }
    }
  }
  ll res = 0;
  for(int x = 1; x <= 20000; x++){
    ll tmp = (dpu1[15][x] - dpu1[15][x+1]) * dpd2[7][x-1] % mod;
    res = (res + tmp) % mod;
    tmp = (dpd1[15][x] - dpd1[15][x-1]) * dpu2[7][x+1] % mod;
    res = (res + tmp) % mod;
  }
  if(res < 0)res += mod;
  cout << res << endl;
  return 0;
}
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