結果
問題 | No.1141 田グリッド |
ユーザー | hamath |
提出日時 | 2020-07-31 23:13:17 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 5,886 bytes |
コンパイル時間 | 2,448 ms |
コンパイル使用メモリ | 228,124 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-07-06 21:10:43 |
合計ジャッジ時間 | 5,097 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 1 ms
6,812 KB |
testcase_01 | AC | 1 ms
6,816 KB |
testcase_02 | AC | 2 ms
6,944 KB |
testcase_03 | AC | 1 ms
6,940 KB |
testcase_04 | AC | 1 ms
6,944 KB |
testcase_05 | AC | 1 ms
6,940 KB |
testcase_06 | AC | 2 ms
6,940 KB |
testcase_07 | AC | 1 ms
6,944 KB |
testcase_08 | AC | 2 ms
6,940 KB |
testcase_09 | AC | 2 ms
6,944 KB |
testcase_10 | AC | 2 ms
6,940 KB |
testcase_11 | AC | 2 ms
6,940 KB |
testcase_12 | AC | 2 ms
6,940 KB |
testcase_13 | AC | 93 ms
6,940 KB |
testcase_14 | AC | 98 ms
6,940 KB |
testcase_15 | AC | 93 ms
6,940 KB |
testcase_16 | AC | 91 ms
6,940 KB |
testcase_17 | AC | 95 ms
6,940 KB |
testcase_18 | WA | - |
testcase_19 | WA | - |
testcase_20 | WA | - |
testcase_21 | AC | 95 ms
6,944 KB |
testcase_22 | AC | 95 ms
6,940 KB |
testcase_23 | AC | 95 ms
6,940 KB |
testcase_24 | WA | - |
testcase_25 | WA | - |
testcase_26 | WA | - |
testcase_27 | WA | - |
testcase_28 | WA | - |
testcase_29 | WA | - |
testcase_30 | WA | - |
testcase_31 | WA | - |
testcase_32 | WA | - |
testcase_33 | WA | - |
ソースコード
#ifdef LOCAL //#define _GLIBCXX_DEBUG #endif //#pragma GCC target("avx512f,avx512dq,avx512cd,avx512bw,avx512vl") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #include <bits/stdc++.h> using namespace std; typedef long long ll; typedef pair<ll, ll> P; typedef pair<int, int> Pi; typedef vector<ll> Vec; typedef vector<int> Vi; typedef vector<string> Vs; typedef vector<char> Vc; typedef vector<P> VP; typedef vector<vector<ll>> VV; typedef vector<vector<int>> VVi; typedef vector<vector<char>> VVc; typedef vector<vector<vector<ll>>> VVV; typedef vector<vector<vector<vector<ll>>>> VVVV; #define REP(i, a, b) for(ll i=(a); i<(b); i++) #define PER(i, a, b) for(ll i=(a); i>=(b); i--) #define rep(i, n) REP(i, 0, n) #define per(i, n) PER(i, n, 0) const ll INF=1e18+18; const ll MOD=1000000007; #define Yes(n) cout << ((n) ? "Yes" : "No") << endl; #define YES(n) cout << ((n) ? "YES" : "NO") << endl; #define ALL(v) v.begin(), v.end() #define rALL(v) v.rbegin(), v.rend() #define pb(x) push_back(x) #define mp(a, b) make_pair(a,b) #define Each(a,b) for(auto &a :b) #define rEach(i, mp) for (auto i = mp.rbegin(); i != mp.rend(); ++i) #ifdef LOCAL #define dbg(x_) cerr << #x_ << ":" << x_ << endl; #define dbgmap(mp) cerr << #mp << ":"<<endl; for (auto i = mp.begin(); i != mp.end(); ++i) { cerr << i->first <<":"<<i->second << endl;} #define dbgset(st) cerr << #st << ":"<<endl; for (auto i = st.begin(); i != st.end(); ++i) { cerr << *i <<" ";}cerr<<endl; #define dbgarr(n,m,arr) rep(i,n){rep(j,m){cerr<<arr[i][j]<<" ";}cerr<<endl;} #define dbgdp(n,arr) rep(i,n){cerr<<arr[i]<<" ";}cerr<<endl; #else #define dbg(...) #define dbgmap(...) #define dbgset(...) #define dbgarr(...) #define dbgdp(...) #endif #define out(a) cout<<a<<endl #define out2(a,b) cout<<a<<" "<<b<<endl #define vout(v) rep(i,v.size()){cout<<v[i]<<" ";}cout<<endl #define Uniq(v) v.erase(unique(v.begin(), v.end()), v.end()) #define fi first #define se second 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<typename T1, typename T2> ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; } template<typename T>istream& operator>>(istream&i,vector<T>&v) {rep(j,v.size())i>>v[j];return i;} // vector template<typename T> ostream &operator<<(ostream &s, const vector<T> &v) { int len=v.size(); for(int i=0; i<len; ++i) { s<<v[i]; if(i<len-1) s<<" "; } return s; } // 2 dimentional vector template<typename T> ostream &operator<<(ostream &s, const vector<vector<T> > &vv) { int len=vv.size(); for(int i=0; i<len; ++i) { s<<vv[i]<<endl; } return s; } //mint struct mint { ll x; // typedef long long ll; mint(ll x=0):x((x%MOD+MOD)%MOD){} mint operator-() const { return mint(-x);} mint& operator+=(const mint a) { if ((x += a.x) >= MOD) x -= MOD; return *this; } mint& operator-=(const mint a) { if ((x += MOD-a.x) >= MOD) x -= MOD; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= MOD; return *this;} mint operator+(const mint a) const { return mint(*this) += a;} mint operator-(const mint a) const { return mint(*this) -= a;} mint operator*(const mint a) const { return mint(*this) *= a;} mint pow(ll t) const { if (!t) return 1; mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } // for prime MOD mint inv() const { return pow(MOD-2);} mint& operator/=(const mint a) { return *this *= a.inv();} mint operator/(const mint a) const { return mint(*this) /= a;} }; istream& operator>>(istream& is, const mint& a) { return is >> a.x;} ostream& operator<<(ostream& os, const mint& a) { return os << a.x;} typedef vector<mint> Vmint; typedef vector<vector<mint>> VVmint; typedef vector<vector<vector<mint>>> VVVmint; struct combination { vector<mint> fact, ifact; combination(int n):fact(n+1),ifact(n+1) { assert(n < MOD); fact[0] = 1; for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i; ifact[n] = fact[n].inv(); for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i; } mint operator()(int n, int k) { if (k < 0 || k > n) return 0; return fact[n]*ifact[k]*ifact[n-k]; } }; int solve(){ ll h,w; cin>>h>>w; VV g(h,Vec(w)); mint sum = 1; Vec rzero(h); Vec czero(w); bool sumzero = false; rep(i,h){ rep(j,w){ cin>>g[i][j]; if(g[i][j]!=0){ sum *= g[i][j]; }else{ sumzero = true; rzero[i] = 1; czero[j] = 1; } } } dbg(g); ll rz = 0; ll cz = 0; rep(i,h){ rz += rzero[i]; } rep(j,w){ cz += czero[j]; } if(rz>=2 || cz >= 2){ out(0); return 0; } Vmint r(h); Vmint c(w); rep(i,h){ mint tmp = 1; rep(j,w){ tmp *= g[i][j]; } r[i] = tmp; } rep(j,w){ mint tmp = 1; rep(i,h){ tmp *= g[i][j]; } c[j] = tmp; } ll Q; cin>>Q; while(Q--){ ll ri,ci; cin>>ri>>ci; ri--; ci--; mint ans = sum; ans /= r[ri]; ans /= c[ci]; ans *= g[ri][ci]; if(rz==1 && !rzero[ri]){ out(0); continue; } if(cz==1 && !czero[ci]){ out(0); continue; } out(ans); } return 0; } int main() { cin.tie(0); ios::sync_with_stdio(false); cout<<std::setprecision(10); // ll T; // cin>>T; // while(T--) solve(); }