Submission #254845

#	Time	Username	Problem	Language	Result	Execution time	Memory
254845		MarcoMeijer	Fibonacci representations (CEOI18_fib)	C++14	50 / 100	4065 ms	10292 KiB

fib

#include <bits/stdc++.h>
using namespace std;

// macros
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> lll;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<lll> vlll;
#define REP(a,b,c) for(int a=int(b); a<int(c); a++)
#define RE(a,c) REP(a,0,c)
#define RE1(a,c) REP(a,1,c+1)
#define REI(a,b,c) REP(a,b,c+1)
#define REV(a,b,c) for(int a=int(c-1); a>=int(b); a--)
#define FOR(a,b) for(auto& a : b)
#define all(a) a.begin(), a.end()
#define INF 1e9
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
#define sz size()

// input
template<class T> void IN(T& x) {cin >> x;}
template<class H, class... T> void IN(H& h, T&... t) {IN(h); IN(t...); }

// output
template<class T1, class T2> void OUT(const pair<T1,T2>& x);
template<class T> void OUT(const T& x) {cout << x;}
template<class H, class... T> void OUT(const H& h, const T&... t) {OUT(h); OUT(t...); }
template<class T1, class T2> void OUT(const pair<T1,T2>& x) {OUT(x.fi," ",x.se);}
template<class... T> void OUTL(const T&... t) {OUT(t..., "\n"); }

//===================//
//  Added libraries  //
//===================//

// mod library
ll MOD=1e9+7;

inline ll mod(ll x_) {
    return (x_)%MOD;
}
ll modpow(ll x_, ll N_) {
    if(N_ == 0) return 1;
    ll a = modpow(x_,N_/2);
    a = (a*a)%MOD;
    if(N_%2) a = (a*x_)%MOD;
    return a;
}
ll inv(ll x_) {
    return modpow(x_, MOD-2);
}
class mi {
public:
    mi(ll v=0) {value = v;}
    mi  operator+ (ll rs) {return mod(value+rs);}
    mi  operator- (ll rs) {return mod(value-rs+MOD);}
    mi  operator* (ll rs) {return mod(value*rs);}
    mi  operator/ (ll rs) {return mod(value*inv(rs));}
    mi& operator+=(ll rs) {*this = (*this)+rs; return *this;}
    mi& operator-=(ll rs) {*this = (*this)-rs; return *this;}
    mi& operator*=(ll rs) {*this = (*this)*rs; return *this;}
    mi& operator/=(ll rs) {*this = (*this)/rs; return *this;}
    operator ll&() {return value;}

    ll value;
};
typedef vector<mi> vmi;

//===================//
//end added libraries//
//===================//

void program();
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    program();
}


//===================//
//   begin program   //
//===================//

const int MX = 5e5;
int n, a[MX];
map<int, int> mp;
set<int> inq;
queue<int> q;
int p[MX], m;
mi dp[MX][2];

void tryAdd(int x) {
    if(inq.count(x)) return;
    if(mp[x] >= 2) {
        q.push(x);
        inq.insert(x);
    }
}
void fix(int x) {
    int am = mp[x];
    int times = am/2;
    inq.erase(x);
    if(am%2) mp[x] = 1;
    else mp.erase(x);
    if(x >= 3) {
        mp[x-2] += times;
        mp[x+1] += times;
        tryAdd(x-2);
        tryAdd(x+1);
    }
    if(x == 2) {
        mp[1] += times;
        mp[3] += times;
        tryAdd(1);
        tryAdd(3);
    }
    if(x == 1) {
        mp[2] += times;
        tryAdd(2);
    }
}

mi cAns;
set<int> numbers;

void program() {
    IN(n);
    RE(i,n) IN(a[i]);

    RE(i,n) {
        mp[a[i]]++;
        tryAdd(a[i]);
        while(!q.empty()) fix(q.front()), q.pop();
        for(auto it=--mp.end(); it!=mp.begin(); it--) {
            auto nxt = it;
            nxt--;
            if(nxt->fi == it->fi-1) {
                int nx = it->fi+1;
                mp.erase(nxt);
                mp.erase(it);
                it = mp.insert({nx,1}).fi;
                it++;
                if(it != mp.end()) it++;
            }
        }
        int m=0;
        p[m++] = 0;
        for(auto it=mp.begin(); it!=mp.end(); it++)
            p[m++] = it->fi;
        reverse(p,p+m);

        dp[0][0] = 0;
        dp[0][1] = 1;
        REP(j,1,m) {
            ll dist = p[j-1]-p[j];
            dp[j][0] = ((dist%2)==0 ? dp[j-1][0]+dp[j-1][1] : (mi)0);
            dp[j][1] = (dp[j-1][0] + dp[j-1][1]) * mi((dist-1)/2) + dp[j-1][1];
        }
        OUTL(ll(dp[m-1][1]));
    }
}

• Subtask 1: 5 / 5
• Subtask 2: 20 / 20
• Subtask 3: 15 / 15
• Subtask 4: 10 / 10
• Subtask 5: 0 / 15
• Subtask 6: 0 / 35