結果
| 問題 | No.142 単なる配列の操作に関する実装問題 |
| コンテスト | |
| ユーザー |
kei
|
| 提出日時 | 2018-08-17 23:07:47 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.89.0) |
| 結果 |
AC
|
| 実行時間 | 2,476 ms / 5,000 ms |
| コード長 | 2,730 bytes |
| 記録 | |
| コンパイル時間 | 1,587 ms |
| コンパイル使用メモリ | 171,280 KB |
| 実行使用メモリ | 23,300 KB |
| 最終ジャッジ日時 | 2024-10-11 00:11:50 |
| 合計ジャッジ時間 | 10,884 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 5 |
ソースコード
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
template<class S,class T> ostream& operator << (ostream& out,const pair<S,T>& o){ out << "(" << o.first << "," << o.second << ")"; return out; }
template<class T> ostream& operator << (ostream& out,const vector<T> V){ for(int i = 0; i < V.size(); i++){ out << V[i]; if(i!=V.size()-1) out << " ";} return out; }
template<class T> ostream& operator << (ostream& out,const vector<vector<T> > Mat){ for(int i = 0; i < Mat.size(); i++) { if(i != 0) out << endl; out << Mat[i];} return out; }
template<class S,class T> ostream& operator << (ostream& out,const map<S,T> mp){ out << "{ "; for(auto it = mp.begin(); it != mp.end(); it++){ out << it->first << ":" << it->second; if(mp.size()-1 != distance(mp.begin(),it)) out << ", "; } out << " }"; return out; }
/*
<url:https://yukicoder.me/problems/no/142>
問題文============================================================
=================================================================
解説=============================================================
================================================================
*/
const ll BatchSize = 50;
ll Comp[2000000/50*2];
ll Sec[100000/50*2];
string solve(){
string res;
ll N,Seed,X,Y,Z; cin >> N >> Seed >> X >> Y >> Z;
vector<ll> A(N);
A[0] = Seed;
for(int i = 1; i < N;i++) A[i] = (X*A[i-1]+Y)%Z;
for_each(A.begin(), A.end(), [](ll& a){ a%=2;});
for(int i = 0; i < N;i++) Comp[i/BatchSize] |= A[i]<<(i%BatchSize);
ll Q; cin >> Q;
while(Q--){
ll S,T,U,V; cin >> S >> T >> U >> V;
ll len = T-S+1;
S--; U--;
ll b = S%BatchSize;
for(int i = 0; i < len/BatchSize+2;i++){
Sec[i+1] = Comp[S/BatchSize+i];
}
for(int i = 0; i < len/BatchSize+2;i++){
Sec[i+1] = (Sec[i+1]>>b) | (((Sec[i+2]&((1LL<<b)-1)))<<(BatchSize-b));
}
for(int i = 0; i < len/BatchSize+4;i++){
if((i+1)*BatchSize<=len) continue;
else if((i+1)*BatchSize-len<=BatchSize) Sec[i+1] &=(1LL<<(len%BatchSize))-1;
else Sec[i+1] = 0;
}
b = U%BatchSize;
for(ll i = len/BatchSize+2; i >= 0; i--) Sec[i+1] = ((Sec[i+1]<<b) | (Sec[i]>>(BatchSize-b))) & ((1LL<<BatchSize)-1);
for(int i = 0; i < len/BatchSize+2;i++){
Comp[i+U/BatchSize] ^= Sec[i+1];
}
}
for(int i = 0; i < N;i++){
res += "EO"[(Comp[i/BatchSize]>>(i%BatchSize))&1];
}
return res;
}
int main(void) {
cin.tie(0); ios_base::sync_with_stdio(false);
cout << solve() << endl;
return 0;
}
kei