Submission #867069

#	Time	Username	Problem	Language	Result	Execution time	Memory
867069		Huseyn123	Crayfish scrivener (IOI12_scrivener)	C++17	5 / 100	97 ms	168252 KiB

scrivener

#include <bits/stdc++.h>
#define MAX 1000010
#define INF INT_MAX
#define MOD 1000000007
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define ins insert
#define ff first
#define ss second
#define gett(x,m) get<m>(x)
#define all(a) a.begin(),a.end()
#define lb(a,b) lower_bound(all(a),b)
#define ub(a,b) upper_bound(all(a),b)
#define sortv(a) sort(all(a))
#define sorta(a,sz) sort(a,a+sz)
#define inputar(a,b){\
    for(int i=0;i<b;i++){\
        cin >> a[i];\
    }\
}
#define inputvec(a,b){\
    for(int i=0;i<b;i++){\
        ll num;\
        cin >> num;\
        a.pb(num);\
    }\
}
#define outputar(a,b){\
    for(int i=0;i<b;i++){\
        cout << a[i] << " ";\
    }\
    cout << "\n";\
}
#define outputvec(a){\
    for(auto x:a){\
        cout << x << " ";\
    }\
    cout << "\n";\
}
#define reset(a,n,v){\
    for(int i=0;i<n;i++){\
        a[i]=v;\
    }\
}
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef tuple<ll,ll,ll> tll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef double db;
typedef long double ldb;
inline void USACO(string filename){
  freopen((filename+".in").c_str(),"r",stdin);
  freopen((filename+".out").c_str(),"w",stdout);
}
int cnt=1; 
int lastU=0;
int a[MAX],b[MAX],e[MAX];
vector<vector<int>> c;
vector<int> d; 
vector<char> d1;
void Init() {
    c.resize(21,vector<int>(MAX,INF)); 
    d.pb(0);
    d1.pb('0'); 
}

void TypeLetter(char L) {
    d1.pb(L);
    a[cnt]=a[cnt-1]+1; 
    cnt++;
}
void UndoCommands(int U) {
    d1.pb('0');
    auto it=lb(d,max(cnt-U,1)); 
    --it; 
    c[0][cnt]=*it; 
    int h=*it; 
    for(int i=1;i<=20;i++){
        if(h!=INF){
            h=c[i-1][h]; 
        }
        c[i][cnt]=h;
    }
    d.pb(cnt);
    a[cnt]=a[cnt-1];
    b[cnt]=a[max(cnt-U-1,0)]-a[*it]+b[*it];
    e[cnt]=U;
    lastU=cnt;
    cnt++; 
}

char GetLetter(int P) {
    a[cnt]=a[cnt-1]; 
    b[cnt]=a[cnt]-a[lastU]+b[lastU]; 
    e[cnt]=0;
    P+=1;
    int h=lastU;
    for(int i=20;i>=0;i--){
        if(c[i][h]!=INF && b[c[i][h]]>=P){
            h=c[i][h]; 
        }
    }
    if(b[h]<P){
        h=cnt;
    }
    return d1[h-e[h]-1-(b[h]-P)]; 
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 7
• Subtask 3: 0 / 22
• Subtask 4: 0 / 26
• Subtask 5: 0 / 40