結果

問題 No.74 貯金箱の退屈
コンテスト
ユーザー minami
提出日時 2019-03-21 21:36:39
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 5,000 ms
コード長 4,586 bytes
コンパイル時間 1,899 ms
コンパイル使用メモリ 180,392 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-09-19 02:01:41
合計ジャッジ時間 2,964 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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ファイルパターン 結果
sample AC * 3
other AC * 30
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;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = 1'000'000'007;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }


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

	ModInt &operator+=(ModInt that) { if ((x += that.x) >= kMod) x -= kMod; return *this; }
	ModInt &operator-=(ModInt that) { if ((x += kMod - that.x) >= kMod) x -= kMod; return *this; }
	ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % kMod; 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 = kMod, u = 1, v = 0;
		while (b) {
			signed t = a / b;
			a -= t * b; swap(a, b);
			u -= t * v; swap(u, v);
		}
		if (u < 0) u += kMod;
		ModInt res; res.x = (unsigned)u;
		return res;
	}
};
template<int MOD>
ostream &operator << (ostream &os, const ModInt<MOD> &m) { return os << m.x; }
template<int MOD>
istream &operator >> (istream &is, ModInt<MOD> &m) { signed long long s; is >> s; m = ModInt<MOD>(s); return is; };

using mint = ModInt<2>;

mint pow(mint a, unsigned long long k) {
	mint r = 1;
	while (k) {
		if (k & 1) r *= a;
		a *= a;
		k >>= 1;
	}
	return r;
}

vector<mint> fact, factinv;
void precomputeFactorial(int N) {
	N = min(N, mint::kMod - 1);
	fact.resize(N + 1); factinv.resize(N + 1);
	fact[0] = 1;
	rep(i, 1, N + 1) fact[i] = fact[i - 1] * i;
	factinv[N] = fact[N].inverse();
	for (int i = N; i >= 1; i--) factinv[i - 1] = factinv[i] * i;
}
mint combi(int n, int r) {
	if (n >= mint::kMod)
		return combi(n % mint::kMod, r % mint::kMod) * combi(n / mint::kMod, r / mint::kMod);
	return r > n ? 0 : fact[n] * factinv[n - r] * factinv[r];
}


// ガウスの消去法（Gauss elimination）
// O(n^3)
// 
// Verified:
//   http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=3437138
//   http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=3437187

using Num = mint;
using Vec = vector<Num>;
using Mat = vector<Vec>;
tuple<bool, int, Vec> gaussianElimination(Mat A, Vec b) {
	const int n = A.size(), m = A[0].size();
	int rank = 0, cj = 0;
	while (rank < n && cj < m) {
		// A[rank][cj] が最大になるように
		for (int i = rank + 1; i < n; i++) {
			if (A[i][cj].get() > A[rank][cj].get()) {
				A[i].swap(A[rank]);
				swap(b[i], b[rank]);
			}
		}
		if (A[rank][cj].get() > 0) {
			// 係数を 1 に
			Num d = A[rank][cj];
			for (int j = 0; j < m; j++)
				A[rank][j] /= d;
			b[rank] /= d;
			// 前進消去（forward elimination）
			for (int i = rank + 1; i < n; i++) {
				Num k = A[i][cj];
				for (int j = 0; j < m; j++)
					A[i][j] -= k * A[rank][j];
				b[i] -= k * b[rank];
			}
			rank++;
		}
		cj++;
	}
	// 0 != b[i] だったら不能
	for (int i = rank; i < n; i++)
		if (b[i].get())
			return make_tuple(false, rank, Vec());
	// 不定
	// rank != m
	// cj < m => n < m
	if (rank < m || cj < m)
		return make_tuple(true, rank, Vec());
	// 後退代入（back substitution）
	for (int j = m - 1; j >= 0; j--)
		for (int i = 0; i < j; i++)
			b[i] -= b[j] * A[i][j];
	return make_tuple(true, rank, b);
}

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N; cin >> N;
	vector<int> D(N); rep(i, 0, N) {
		cin >> D[i];
	}
	vector<int> W(N); rep(i, 0, N) {
		cin >> W[i];
	}

	Mat A(N, Vec(N));
	Vec b(N, 1);

	rep(i, 0, N) {
		int ai = (i + D[i] % N) % N;
		int bi = (i - D[i] % N + N) % N;
		if (ai == bi) {
			A[ai][i] = 1;
		}
		else {
			A[ai][i] = 1;
			A[bi][i] = 1;
		}
		if (W[i]) {
			b[i] = 0;
		}
	}

	auto res = gaussianElimination(A, b);

	if (get<0>(res)) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}

	return 0;
}
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