結果
問題 | No.74 貯金箱の退屈 |
ユーザー |
|
提出日時 | 2022-08-27 16:54:50 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 69 ms / 5,000 ms |
コード長 | 5,404 bytes |
コンパイル時間 | 2,098 ms |
コンパイル使用メモリ | 205,772 KB |
最終ジャッジ日時 | 2025-01-31 06:26:59 |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 30 |
ソースコード
#include <bits/stdc++.h>#pragma GCC optimize("Ofast")#pragma GCC optimize("unroll-loops")#pragma GCC target("sse,sse2,sse3,ssse3,sse4,fma,abm,mmx,avx,avx2")#define rep(i, n) for (int i = 0; i < (int)(n); i++)#define rrep(i, n) for (int i = (int)(n - 1); i >= 0; i--)#define all(x) (x).begin(), (x).end()#define sz(x) int(x.size())#define yn(joken) cout<<((joken) ? "Yes" : "No")<<"\n"#define YN(joken) cout<<((joken) ? "YES" : "NO")<<"\n"using namespace std;using ll = long long;using vi = vector<int>;using vl = vector<ll>;using vs = vector<string>;using vc = vector<char>;using vd = vector<double>;using vld = vector<long double>;using vvi = vector<vector<int>>;using vvl = vector<vector<ll>>;using vvs = vector<vector<string>>;using vvc = vector<vector<char>>;using vvd = vector<vector<double>>;using vvld = vector<vector<long double>>;using vvvi = vector<vector<vector<int>>>;using vvvl = vector<vector<vector<ll>>>;using vvvvi = vector<vector<vector<vector<int>>>>;using vvvvl = vector<vector<vector<vector<ll>>>>;using pii = pair<int,int>;using pll = pair<ll,ll>;const int INF = 1e9;const ll LINF = 2e18;template <class T>bool chmax(T& a, const T& b) {if (a < b) {a = b;return 1;}return 0;}template <class T>bool chmin(T& a, const T& b) {if (b < a) {a = b;return 1;}return 0;}bool ispow2(int i) { return i && (i & -i) == i; }bool ispow2(ll i) { return i && (i & -i) == i; }template <class T>vector<T> make_vec(size_t a) {return vector<T>(a);}template <class T, class... Ts>auto make_vec(size_t a, Ts... ts) {return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));}template <typename T>istream& operator>>(istream& is, vector<T>& v) {for (int i = 0; i < int(v.size()); i++) {is >> v[i];}return is;}template <typename T>ostream& operator<<(ostream& os, const vector<T>& v) {for (int i = 0; i < int(v.size()); i++) {os << v[i];if (i < int(v.size()) - 1) os << ' ';}return os;}static uint32_t RandXor(){static uint32_t x=123456789;static uint32_t y=362436069;static uint32_t z=521288629;static uint32_t w=88675123;uint32_t t;t=x^(x<<11);x=y; y=z; z=w;return w=(w^(w>>19))^(t^(t>>8));}static double Rand01(){return (RandXor()+0.5)*(1.0/UINT_MAX);}// MAX_ROW,MAX_COLを適切に設定した上で, BitMatrix M(H,W); のようにする// int Gauss_Jordan(BitMatrix &A,bool is_extended=false):// -> 行列Aを掃き出してrankを返す,拡大係数行列のときはis_extended=trueにする// int linear_equation(BitMatrix A,vector<T> b,vector<T> &res):// -> 係数を表す行列Aと右辺を表すベクトルB,および空の配列resを与えるとrankを返した上でresに解を格納する// -> 解なしのときは-1を返すconst int MAX_ROW = 100; // to be set appropriatelyconst int MAX_COL = 100; // to be set appropriatelystruct BitMatrix {int H, W;bitset<MAX_COL> val[MAX_ROW];BitMatrix(int m = 1, int n = 1) : H(m), W(n) {}inline bitset<MAX_COL>& operator [] (int i) {return val[i];}};ostream& operator << (ostream& s, BitMatrix A) {s << endl;for (int i = 0; i < A.H; ++i) {for (int j = 0; j < A.W; ++j) {s << A[i][j] << ", ";}s << endl;}return s;}inline BitMatrix operator * (BitMatrix A, BitMatrix B) {BitMatrix R(A.H, B.W);BitMatrix tB(B.W, B.H);for (int i = 0; i < tB.H; ++i) for (int j = 0; j < tB.W; ++j) tB[i][j] = B[j][i];for (int i = 0; i < R.H; ++i) for (int j = 0; j < R.W; ++j) R[i][j] = ((A[i] & tB[j]).count() & 1);return R;}inline BitMatrix pow(BitMatrix A, long long n) {BitMatrix R(A.H, A.H);for (int i = 0; i < A.H; ++i) R[i][i] = 1;while (n > 0) {if (n & 1) R = R * A;A = A * A;n >>= 1;}return R;}int GaussJordan(BitMatrix &A, bool is_extended = false) {int rank = 0;for (int col = 0; col < A.W; ++col) {if (is_extended && col == A.W - 1) break;int pivot = -1;for (int row = rank; row < A.H; ++row) {if (A[row][col]) {pivot = row;break;}}if (pivot == -1) continue;swap(A[pivot], A[rank]);for (int row = 0; row < A.H; ++row) {if (row != rank && A[row][col]) A[row] ^= A[rank];}++rank;}return rank;}int linear_equation(BitMatrix A, vector<int> b, vector<int> &res) {int m = A.H, n = A.W;BitMatrix M(m, n + 1);for (int i = 0; i < m; ++i) {for (int j = 0; j < n; ++j) M[i][j] = A[i][j];M[i][n] = b[i];}int rank = GaussJordan(M, true);// check if it has no solutionfor (int row = rank; row < m; ++row) if (M[row][n]) return -1;// answerres.assign(n, 0);for (int i = 0; i < rank; ++i) res[i] = M[i][n];return rank;}void solve(){int N;cin>>N;vi D(N),W(N);cin>>D>>W;rep(i,N) W[i]^=1;BitMatrix M(N,N);rep(i,N){int a=(i+D[i])%N,b=(i-D[i]+N*1000)%N;M[a][i]=1;M[b][i]=1;}vi res;auto ret=linear_equation(M,W,res);yn(ret!=-1);}int main(){cin.tie(nullptr);ios::sync_with_stdio(false);solve();}