Submission #76231

#TimeUsernameProblemLanguageResultExecution timeMemory
76231born2ruleMechanical Doll (IOI18_doll)C++14
10 / 100
1 ms204 KiB
#include <bits/stdc++.h> #include "doll.h" #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; #define ll long long #define db long double #define ii pair<int,int> #define vi vector<int> #define fi first #define se second #define sz(a) (int)(a).size() #define all(a) (a).begin(),(a).end() #define pb push_back #define mp make_pair #define FN(i, n) for (int i = 0; i < (int)(n); ++i) #define FEN(i,n) for (int i = 1;i <= (int)(n); ++i) #define rep(i,a,b) for(int i=a;i<b;i++) #define repv(i,a,b) for(int i=b-1;i>=a;i--) #define SET(A, val) memset(A, val, sizeof(A)) typedef tree<int ,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set ; // order_of_key (val): returns the no. of values less than val // find_by_order (k): returns the kth largest element.(0-based) #define TRACE #ifdef TRACE #define trace(...) __f(#__VA_ARGS__, __VA_ARGS__) template <typename Arg1> void __f(const char* name, Arg1&& arg1){ cerr << name << " : " << arg1 << std::endl; } template <typename Arg1, typename... Args> void __f(const char* names, Arg1&& arg1, Args&&... args){ const char* comma = strchr(names + 1, ','); cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...); } #else #define trace(...) #endif const int N=200005; vi X,Y; int cnt=0,p=1,n; bool state[N*10]; int build(int low,int high) { if(low>=high) return 0; //if(high<p-n) return -1; int mid=(low+high)>>1; int ind=++cnt; X.pb(0); Y.pb(0); int x=build(low,mid),y=build(mid+1,high); X[ind-1]=x; Y[ind-1]=y; return -ind; } void solve(int ind,int nxt) { int &curr=((state[-ind])?Y[-ind-1]:X[-ind-1]); state[-ind]=!state[-ind]; if(curr==0) curr=nxt; else solve(curr,nxt); } void create_circuit(int m,vi A) { n=sz(A); if(n==1) { vi C(m+1,0); C[0]=A[0]; return answer(C,X,Y); } while(p<n) p<<=1; build(0,p-1); rep(i,1,n) solve(-1,A[i]); if(n&1) solve(-1,-1); solve(-1,0); vi C(m+1,-1); C[0]=A[0]; answer(C,X,Y); }
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