結果

問題 No.1253 雀見椪
コンテスト
ユーザー keijak
提出日時 2020-10-09 22:56:12
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
RE  
実行時間 -
コード長 5,965 bytes
コンパイル時間 2,009 ms
コンパイル使用メモリ 196,748 KB
最終ジャッジ日時 2025-01-15 05:34:50
ジャッジサーバーID
（参考情報） 		judge2 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 3 RE * 11
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 # 

#include <bits/stdc++.h>
using i64 = long long;
using u64 = unsigned long long;
#define REP(i, n) for (int i = 0, REP_N_ = int(n); i < REP_N_; ++i)
#define ALL(x) std::begin(x), std::end(x)
#define SIZE(a) (int)((a).size())

template <class T>
inline bool chmax(T &a, T b) {
  return a < b and ((a = std::move(b)), true);
}
template <class T>
inline bool chmin(T &a, T b) {
  return a > b and ((a = std::move(b)), true);
}

template <typename T>
using V = std::vector<T>;
template <typename T>
std::istream &operator>>(std::istream &is, std::vector<T> &a) {
  for (auto &x : a) is >> x;
  return is;
}
template <typename Container>
std::ostream &pprint(const Container &a, std::string_view sep = " ",
                     std::string_view ends = "\n", std::ostream *os = nullptr) {
  if (os == nullptr) os = &std::cout;
  auto b = std::begin(a), e = std::end(a);
  for (auto it = std::begin(a); it != e; ++it) {
    if (it != b) *os << sep;
    *os << *it;
  }
  return *os << ends;
}
template <typename T, typename = void>
struct is_iterable : std::false_type {};
template <typename T>
struct is_iterable<T, std::void_t<decltype(std::begin(std::declval<T>())),
                                  decltype(std::end(std::declval<T>()))>>
    : std::true_type {};

template <typename T,
          typename = std::enable_if_t<is_iterable<T>::value &&
                                      !std::is_same<T, std::string>::value>>
std::ostream &operator<<(std::ostream &os, const T &a) {
  return pprint(a, ", ", "", &(os << "{")) << "}";
}
template <typename T, typename U>
std::ostream &operator<<(std::ostream &os, const std::pair<T, U> &a) {
  return os << "(" << a.first << ", " << a.second << ")";
}

#ifdef ENABLE_DEBUG
template <typename T>
void pdebug(const T &value) {
  std::cerr << value;
}
template <typename T, typename... Ts>
void pdebug(const T &value, const Ts &... args) {
  pdebug(value);
  std::cerr << ", ";
  pdebug(args...);
}
#define DEBUG(...)                                   \
  do {                                               \
    std::cerr << " \033[33m (L" << __LINE__ << ") "; \
    std::cerr << #__VA_ARGS__ << ":\033[0m ";        \
    pdebug(__VA_ARGS__);                             \
    std::cerr << std::endl;                          \
  } while (0)
#else
#define pdebug(...)
#define DEBUG(...)
#endif

template <u64 M>
struct ModInt {
  constexpr ModInt(long long val = 0) : _v(0) {
    if (val < 0) {
      long long k = (abs(val) + M - 1) / M;
      val += k * M;
    }
    assert(val >= 0);
    _v = val % M;
  }

  static constexpr int mod() { return M; }
  static constexpr u64 umod() { return M; }
  inline u64 val() const { return _v; }

  ModInt &operator++() {
    _v++;
    if (_v == umod()) _v = 0;
    return *this;
  }
  ModInt &operator--() {
    if (_v == 0) _v = umod();
    _v--;
    return *this;
  }
  ModInt operator++(int) {
    auto result = *this;
    ++*this;
    return result;
  }
  ModInt operator--(int) {
    auto result = *this;
    --*this;
    return result;
  }

  constexpr ModInt operator-() const { return ModInt(-_v); }
  constexpr ModInt &operator+=(const ModInt &a) {
    if ((_v += a._v) >= M) _v -= M;
    return *this;
  }
  constexpr ModInt &operator-=(const ModInt &a) {
    if ((_v += M - a._v) >= M) _v -= M;
    return *this;
  }
  constexpr ModInt &operator*=(const ModInt &a) {
    _v = ((unsigned long long)(_v)*a._v) % M;
    return *this;
  }
  constexpr ModInt pow(unsigned long long t) const {
    ModInt base = *this;
    ModInt res = 1;
    while (t) {
      if (t & 1) res *= base;
      base *= base;
      t >>= 1;
    }
    return res;
  }

  constexpr ModInt inv() const {
    // Inverse by Extended Euclidean algorithm.
    // M doesn't need to be prime, but x and M must be coprime.
    assert(_v != 0);
    // auto [g, x, y] = ext_gcd(_v, M);
    // assert(g == 1LL);  // The GCD must be 1.
    // return x;

    // Inverse by Fermat's little theorem.
    // M must be prime. It's often faster.
    //
    return pow(M - 2);
  }
  constexpr ModInt &operator/=(const ModInt &a) { return *this *= a.inv(); }

  friend constexpr ModInt operator+(const ModInt &a, const ModInt &b) {
    return ModInt(a) += b;
  }
  friend constexpr ModInt operator-(const ModInt &a, const ModInt &b) {
    return ModInt(a) -= b;
  }
  friend constexpr ModInt operator*(const ModInt &a, const ModInt &b) {
    return ModInt(a) *= b;
  }
  friend constexpr ModInt operator/(const ModInt &a, const ModInt &b) {
    return ModInt(a) /= b;
  }
  friend constexpr bool operator==(const ModInt &a, const ModInt &b) {
    return a._v == b._v;
  }
  friend constexpr bool operator!=(const ModInt &a, const ModInt &b) {
    return a._v != b._v;
  }
  friend std::istream &operator>>(std::istream &is, ModInt &a) {
    return is >> a._v;
  }
  friend std::ostream &operator<<(std::ostream &os, const ModInt &a) {
    return os << a._v;
  }

 private:
  // Extended Euclidean algorithm
  // Returns (gcd(a,b), x, y) where `a*x + b*y == gcd(a,b)`.
  static std::tuple<int, int, int> ext_gcd(int a, int b) {
    int ax = 1, ay = 0, bx = 0, by = 1;
    for (;;) {
      if (b == 0) break;
      auto d = std::div(a, b);
      a = b;
      b = d.rem;
      ax -= bx * d.quot;
      std::swap(ax, bx);
      ay -= by * d.quot;
      std::swap(ay, by);
    }
    return {a, ax, ay};
  }

  u64 _v;  // raw value
};
const u64 MOD = 1'000'000'007;
using Mint = ModInt<MOD>;

using namespace std;

void solve() {
  int n;
  Mint ag, bg, ac, bc, ap, bp;
  cin >> n >> ag >> bg >> ac >> bc >> ap >> bp;
  Mint allg = (ag / bg).pow(n), allc = (ac / bc).pow(n),
       allp = (ap / bp).pow(n);
  Mint ans = 0;
  ans += ((bp - ap) / bp).pow(n);
  ans += ((bg - ag) / bg).pow(n);
  ans += ((bc - ac) / bc).pow(n);
  ans -= 2 * allg + 2 * allc + 2 * allp;
  ans = 1 - ans;
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  int T;
  cin >> T;
  REP(i, T) { solve(); }
}
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