Submission #66169

#	Time	Username	Problem	Language	Result	Execution time	Memory
66169		Benq	Sequence (BOI14_sequence)	C++11	42 / 100	1084 ms	2564 KiB

sequence

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const ll INF = 1e18;
const int MX = 100000;

ll ans = INF;
bitset<MX> bad[10], res[10], res2[10];
bitset<MX> BAD, RES, RES2;
int K, B[MX], digYes[MX], digNo[MX];

ll tri(int x, int y) {
    if (x == 0) return y;
    if (x == 1) x ^= 2;
    ll t = 0;
    FOR(i,1,10) if (x&(1<<i)) {
        t = i;
        x ^= 1<<i;
        break;
    }
    F0R(i,10) if (x&(1<<i)) {
        t = 10*t+i;
        x ^= 1<<i;
    }
    return t;
}

ll solve(pi a, pi b) { // what about leading zeroes??
    ll t = INF;
    F0R(i,9) {
        int A = min(a.f,a.f^(1<<i));
        int B = min(b.f,b.f^(1<<(i+1)));
        t = min(t,10*tri(A|B,(a.f&1) || (i == 0 && a.s))+i);
        
        A = min(A,A^(1<<9));
        B = min(B,B^(1<<0));
        t = min(t,100*tri(A|B,a.s)+10*i+9);
    }
    return MX*t;
}

pi get0(int st) {
    pi ret = {0,0};
    F0R(i,10) if (res[i][st] == 1) ret.f ^= 1<<i;
    if (RES[st] == 1) ret.s = 1;
    return ret;
}

pi get1(int st) {
    pi ret = {0,0};
    F0R(i,10) if (res2[i][st] == 1) ret.f ^= 1<<i;
    if (RES2[st] == 1) ret.s = 1;
    return ret;
}

ll test(int st) {
    pi a = get0(st), b = get1(st);
    // if (st == 0) cout << a.f << " " << a.s << "\n";
    ll ret = st+solve(a,b);
    return ret;
}

void gen() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    FOR(i,1,MX) digNo[i] = digNo[i/10]|(1<<(i%10));
    F0R(i,MX) {
        digYes[i] = digNo[i];
        if (i < MX/10) digYes[i] |= 1;
    }
    
    F0R(i,MX) F0R(j,10) {
        if (!(digYes[i]&(1<<j))) bad[j][i] = 1;
        if (j == 0 && !(digNo[i]&(1<<j))) BAD[i] = 1;
    }
    
    cin >> K;
    F0R(i,K) cin >> B[i];
    F0R(i,K) {
        res[B[i]] |= (bad[B[i]]>>i);
        res2[B[i]] |= (bad[B[i]]<<(MX-i));
        if (B[i] == 0) {
            RES |= (BAD>>i);
            RES2 |= (BAD<<(MX-i));
        }
    }
}

int main() {
    gen();
    F0R(i,MX) ans = min(ans,test(i));
    cout << ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 9 / 9
• Subtask 2: 33 / 33
• Subtask 3: 0 / 25
• Subtask 4: 0 / 33