結果

問題 No.681 Fractal Gravity Glue
コンテスト
ユーザー anta
提出日時 2018-04-27 23:25:45
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 2,437 bytes
コンパイル時間 1,528 ms
コンパイル使用メモリ 167,516 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-27 22:18:59
合計ジャッジ時間 2,232 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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ファイルパターン 結果
other AC * 20
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 namespace std;


template<int MOD>
struct ModInt {
	static const int Mod = MOD;
	unsigned x;
	ModInt() : x(0) { }
	ModInt(signed sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	ModInt(signed long long sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	int get() const { return (int)x; }

	ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
	ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }

	ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
	ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }

	ModInt inverse() const {
		signed a = x, b = MOD, u = 1, v = 0;
		while (b) {
			signed t = a / b;
			a -= t * b; std::swap(a, b);
			u -= t * v; std::swap(u, v);
		}
		if (u < 0) u += Mod;
		ModInt res; res.x = (unsigned)u;
		return res;
	}

	bool operator==(ModInt that) const { return x == that.x; }
	bool operator!=(ModInt that) const { return x != that.x; }
	ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
	ModInt<MOD> r = 1;
	while (k) {
		if (k & 1) r *= a;
		a *= a;
		k >>= 1;
	}
	return r;
}
typedef ModInt<1000000007> mint;

mint solve1(int b, int d) {
	return ((mint(d + 1) ^ b) + ((mint(d + 1) ^ b) - b - 1) * d - 1) / d;
}

int solve2(int b, int d) {
	static const int INF = 0x3f3f3f3f;
	long long sum = 0;
	long long p = 1;
	for (int i = b; i >= 1; -- i) {
		sum += p * d;
		if (sum > INF) return INF;
		p *= d + 1;
	}
	return (int)sum;
}

mint solve3(int n, int b, int d) {
	if (b == 1) return n;
	int x = solve2(b - 1, d);
	int q = min(n / (x + 1), d);
	int r = n - q * (x + 1);
	mint sum;
	sum += (solve1(b - 1, d) + b) * q;
	sum += solve3(min(r, x), b - 1, d);
	return sum;
}


int main() {
	int n; int b; int d;
	while (~scanf("%d%d%d", &n, &b, &d)) {
		//a(b,d) = a(b-1,d) * (d+1) + b*d
		//\sum_{i=1}^b i*(d+1)^(b-1)*d
		mint ans = solve1(b, d);
		ans -= solve3(n, min(b, 100), d);
		printf("%d\n", ans.get());
	}
}
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