Submission #58978

#	Time	Username	Problem	Language	Result	Execution time	Memory
58978		Benq	Brunhilda’s Birthday (BOI13_brunhilda)	C++14	100 / 100	556 ms	83992 KiB

brunhilda

#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 int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 10000001;

int m,Q, mx = MOD, ans[MX], mxP[MX];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> m >> Q;
    F0R(i,m) {
        int p; cin >> p;
        for (int j = 0; j < MX; j += p) mxP[j] = max(mxP[j],p);
    }
    array<int,3> cur = {0,0,0};
    while (1) {
        int nex = cur[2];
        FOR(i,cur[1],cur[2]+1) nex = max(nex,i+mxP[i]-1);
        nex = min(nex,MX-1);
        if (nex == cur[2]) {
            mx = cur[2];
            break;
        }
        cur[0] ++;
        cur[1] = cur[2]+1; cur[2] = nex;
        FOR(i,cur[1],cur[2]+1) ans[i] = cur[0];
    }
    F0R(i,Q) {
        int x; cin >> x;
        if (x > mx) cout << "oo";
        else cout << ans[x];
        cout << "\n";
    }
}

/* 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: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 60 / 60