제출 #738526

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
738526		GrindMachine	Brunhilda’s Birthday (BOI13_brunhilda)	C++17	100 / 100	972 ms	80060 KiB

brunhilda

// Om Namah Shivaya

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

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

ig i read the edi for this problem a long time ago (dont remember the approach in the edi tho)

greedy strategy
=> in each op, try to reduce the number as much as possible
because the #of ops is a montonic sequence

if we need k ops to make n1 = 0, then we would need <= k ops to make n2 = 0, n2 < n1
because we can apply the same k ops we applied to n1 to n2 and n2 would also become 0

i remember seeing this idea somewhere => maybe in the edi of this problem/some other problem

*/

const int MOD = 1e9 + 7;
const int N = 1e7 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    int n, q; cin >> n >> q;
    vector<int> a(n);
    rep(i, n) cin >> a[i];

    priority_queue<pii, vector<pii>, greater<pii>> pq;
    rep(i, n) pq.push({0, a[i]});

    vector<ll> dp(N, inf1);
    dp[0] = 0;

    rep1(i, N - 1) {
        while (!pq.empty()) {
            auto [val, p] = pq.top();
            pq.pop();

            int curr = (i / p) * p;

            if (curr != val) {
                pq.push({curr, p});
                conts;
            }

            pq.push({val, p});
            dp[i] = dp[val] + 1;
            break;
        }
    }

    while (q--) {
        int m; cin >> m;
        if (dp[m] < inf1) cout << dp[m] << endl;
        else cout << "oo" << endl;
    }
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 60 / 60