結果

問題 No.1086 桁和の桁和2
ユーザー Haar
提出日時 2020-06-19 22:57:26
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 289 ms / 3,000 ms
コード長 4,185 bytes
コンパイル時間 2,347 ms
コンパイル使用メモリ 196,220 KB
最終ジャッジ日時 2025-01-11 07:40:58
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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ファイルパターン 結果
sample AC * 5
other AC * 31
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 #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#ifdef DEBUG
#include <Mylib/Debug/debug.cpp>
#else
#define dump(...)
#endif
/**
 * @title Modint
 * @docs mint.md
 */
template <uint32_t M> class ModInt{
public:
  constexpr static uint32_t MOD = M;
  uint64_t val;
  
  constexpr ModInt(): val(0){}
  constexpr ModInt(int64_t n){
    if(n >= M) val = n % M;
    else if(n < 0) val = n % M + M;
    else val = n;
  }
  
  inline constexpr auto operator+(const ModInt &a) const {return ModInt(val + a.val);}
  inline constexpr auto operator-(const ModInt &a) const {return ModInt(val - a.val);}
  inline constexpr auto operator*(const ModInt &a) const {return ModInt(val * a.val);}
  inline constexpr auto operator/(const ModInt &a) const {return ModInt(val * a.inv().val);}
  
  inline constexpr auto& operator=(const ModInt &a){val = a.val; return *this;}
  inline constexpr auto& operator+=(const ModInt &a){if((val += a.val) >= M) val -= M; return *this;}
  inline constexpr auto& operator-=(const ModInt &a){if(val < a.val) val += M; val -= a.val; return *this;}
  inline constexpr auto& operator*=(const ModInt &a){(val *= a.val) %= M; return *this;}
  inline constexpr auto& operator/=(const ModInt &a){(val *= a.inv().val) %= M; return *this;}
  
  inline constexpr bool operator==(const ModInt &a) const {return val == a.val;}
  inline constexpr bool operator!=(const ModInt &a) const {return val != a.val;}
  
  inline constexpr auto& operator++(){*this += 1; return *this;}
  inline constexpr auto& operator--(){*this -= 1; return *this;}
  
  inline constexpr auto operator++(int){auto t = *this; *this += 1; return t;}
  inline constexpr auto operator--(int){auto t = *this; *this -= 1; return t;}
  
  inline constexpr static ModInt power(int64_t n, int64_t p){
    if(p < 0) return power(n, -p).inv();
    
    int64_t ret = 1, e = n % M;
    for(; p; (e *= e) %= M, p >>= 1) if(p & 1) (ret *= e) %= M;
    return ret;
  }
  
  inline constexpr static ModInt inv(int64_t a){
    int64_t b = M, u = 1, v = 0;
    
    while(b){
      int64_t t = a / b;
      a -= t * b; std::swap(a,b);
      u -= t * v; std::swap(u,v);
    }
    
    u %= M;
    if(u < 0) u += M;
    
    return u;
  }
  
  inline constexpr static auto frac(int64_t a, int64_t b){return ModInt(a) / ModInt(b);}
  
  inline constexpr auto power(int64_t p) const {return power(val, p);}
  inline constexpr auto inv() const {return inv(val);}
  
  friend inline constexpr auto operator-(const ModInt &a){return ModInt(-a.val);}
  
  friend inline constexpr auto operator+(int64_t a, const ModInt &b){return ModInt(a) + b;}
  friend inline constexpr auto operator-(int64_t a, const ModInt &b){return ModInt(a) - b;}
  friend inline constexpr auto operator*(int64_t a, const ModInt &b){return ModInt(a) * b;}
  friend inline constexpr auto operator/(int64_t a, const ModInt &b){return ModInt(a) / b;}
  
  friend std::istream& operator>>(std::istream &s, ModInt<M> &a){s >> a.val; return s;}
  friend std::ostream& operator<<(std::ostream &s, const ModInt<M> &a){s << a.val; return s;}
  template <int N>
  inline static auto div(){
    static auto value = inv(N);
    return value;
  }
  explicit operator int32_t() const noexcept {return val;}
  explicit operator int64_t() const noexcept {return val;}
};
using mint = ModInt<1000000007>;
mint g(int64_t N){
  return (mint::power(10, N) - 1) / 9;
}
mint f(int64_t L, int64_t R, int m){
  if(m == 0) return 1;
  return g(R) - g(L);
}
int main(){
  int N;
  while(std::cin >> N){
    std::vector<int64_t> L(N), R(N), D(N);
    for(int i = 0; i < N; ++i) std::cin >> L[i];
    for(int i = 0; i < N; ++i) std::cin >> R[i];
    for(int i = 0; i < N; ++i) std::cin >> D[i];
    mint ans = 1;
    for(int i = 0; i < N; ++i){
      if(i == 0){
        ans *= f(L[i], R[i], D[i]);
      }else{
        if(D[i-1] != 0 and D[i] == 0){
          ans = 0;
          break;
        }
        if(D[i] != 0){
          mint t = 0;
          for(int k = 0; k <= 9; ++k){
            if((D[i-1] + k) % 9 == D[i] % 9){
              t += f(L[i], R[i], k);
            }
          }
          ans *= t;
        }
      }      
    }
    
    std::cout << ans << "\n";
  }
  return 0;
}
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