ABC C にありそうな教育的問題!
問題概要
正の整数 が与えられる。 かつ を満たすすべての整数組 についての
の総和を 1000000007 で割ったあまりを求めよ。
制約
- (テストケース数)
考えたこと
結局、
を計算すればよい。それぞれ
となる。modint があれば実装も簡単。
#include <bits/stdc++.h> using namespace std; // modint template<int MOD> struct Fp { long long val; constexpr Fp(long long v = 0) noexcept : val(v % MOD) { if (val < 0) val += MOD; } constexpr int getmod() const { return MOD; } constexpr Fp operator - () const noexcept { return val ? MOD - val : 0; } constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; } constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; } constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; } constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; } constexpr Fp& operator += (const Fp& r) noexcept { val += r.val; if (val >= MOD) val -= MOD; return *this; } constexpr Fp& operator -= (const Fp& r) noexcept { val -= r.val; if (val < 0) val += MOD; return *this; } constexpr Fp& operator *= (const Fp& r) noexcept { val = val * r.val % MOD; return *this; } constexpr Fp& operator /= (const Fp& r) noexcept { long long a = r.val, b = MOD, u = 1, v = 0; while (b) { long long t = a / b; a -= t * b, swap(a, b); u -= t * v, swap(u, v); } val = val * u % MOD; if (val < 0) val += MOD; return *this; } constexpr bool operator == (const Fp& r) const noexcept { return this->val == r.val; } constexpr bool operator != (const Fp& r) const noexcept { return this->val != r.val; } friend constexpr istream& operator >> (istream& is, Fp<MOD>& x) noexcept { is >> x.val; x.val %= MOD; if (x.val < 0) x.val += MOD; return is; } friend constexpr ostream& operator << (ostream& os, const Fp<MOD>& x) noexcept { return os << x.val; } friend constexpr Fp<MOD> modpow(const Fp<MOD>& r, long long n) noexcept { if (n == 0) return 1; if (n < 0) return modpow(modinv(r), -n); auto t = modpow(r, n / 2); t = t * t; if (n & 1) t = t * r; return t; } friend constexpr Fp<MOD> modinv(const Fp<MOD>& r) noexcept { long long a = r.val, b = MOD, u = 1, v = 0; while (b) { long long t = a / b; a -= t * b, swap(a, b); u -= t * v, swap(u, v); } return Fp<MOD>(u); } }; // Binomial coefficient template<class T> struct BiCoef { vector<T> fact_, inv_, finv_; constexpr BiCoef() {} constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) { init(n); } constexpr void init(int n) noexcept { fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1); int MOD = fact_[0].getmod(); for(int i = 2; i < n; i++){ fact_[i] = fact_[i-1] * i; inv_[i] = -inv_[MOD%i] * (MOD/i); finv_[i] = finv_[i-1] * inv_[i]; } } constexpr T com(int n, int k) const noexcept { if (n < k || n < 0 || k < 0) return 0; return fact_[n] * finv_[k] * finv_[n-k]; } constexpr T fact(int n) const noexcept { if (n < 0) return 0; return fact_[n]; } constexpr T inv(int n) const noexcept { if (n < 0) return 0; return inv_[n]; } constexpr T finv(int n) const noexcept { if (n < 0) return 0; return finv_[n]; } }; const int MOD = 1000000007; using mint = Fp<MOD>; BiCoef<mint> bc; int main() { int N; while (cin >> N, N) { mint a, b, c, d; cin >> a >> b >> c >> d; mint x = (a + c) * (c - a + 1) / 2; mint y = (b + d) * (d - b + 1) / 2; cout << x * y << endl; } }