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ID提交记录ID题目HackerOwner结果提交时间测评时间
#1114#697914#9527. A Brand New Geometric ProblempropanepropaneFailed.2024-11-04 09:32:172024-11-04 17:10:52

詳細信息

the score gained by the hacked submission is not 100.
ID题目提交者结果用时内存语言文件大小提交时间测评时间
#697914#9527. A Brand New Geometric ProblempropaneTL 16ms5028kbC++203.8kb2024-11-01 16:22:512024-11-06 13:24:00

answer

#include<iostream>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
using LL = long long;

int cnt;

int c[10005];
int id1[100005], id2[100005];

int get(LL x){
    if (x <= 100000) return id1[x];
    else return id2[10'000'000'000 / x];
}

int main(){

#ifdef LOCAL
    freopen("data.in", "r", stdin);
    freopen("data.out", "w", stdout);
#endif

    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    int n; LL S, M;
    cin >> n >> S >> M;

    vector<LL> factor;
    for(LL i = 1; i * i <= M; i++){
        if (M % i == 0){
            factor.push_back(i);
            if (i * i != M){
                factor.push_back(M / i);
            }
        }
    }

    sort(factor.begin(), factor.end(), greater<>());
    factor.pop_back();
    reverse(factor.begin(), factor.end());
    while(!factor.empty() and factor.back() > S) factor.pop_back();
    reverse(factor.begin(), factor.end());
    for(int i = 0; i < factor.size(); i++){
        LL x = factor[i];
        if (x <= 100000) id1[x] = i;
        else id2[10'000'000'000 / x] = i;
    }

    int cnt1 = 0;

    for(int i = 0; i < n; i++){
        LL x;
        cin >> x;
        if (M % x != 0) continue;
        if (x == 1) cnt1 += 1;
        else c[get(x)] += 1;
    }

    if (cnt1 > S){
        LL keep = 0;
        LL t = M, ss = 0, tot = 0;
        for(LL i = 2; i * i <= t; i++){
            if (t % i == 0){
                int v = 0;
                while(t % i == 0){
                    ss += i;
                    t /= i;
                    v += 1;
                }
                if (i <= S){
                    if (i != 2){
                        keep += min(v, c[get(i)]);
                        tot += v;
                    }
                    else{
                        int cnt2 = c[get(2)];
                        int cnt4 = (v >= 2 ? c[get(4)] : 0);
                        if (cnt2 >= v){
                            tot += v;
                            keep += v;
                        }
                        else{
                            int t = cnt2 + min(cnt4, (v - cnt2) / 2);
                            keep += t;
                            tot += t;
                            int remain = v - t;
                            tot += (remain + 1) / 2;                            
                        }
                    }
                } 
            }
        }
        if (t > 1){
            ss += t;
            tot += 1;
            if (t <= S) keep += min(1, c[get(S)]);
        }
        if (ss > S){
            cout << -1 << '\n';
            return 0;
        }
        tot -= keep;
        keep += S - ss;
        cout << n - keep + tot << '\n';
        return 0;
    }

    const LL INF = 1e18;
    LL ans = INF;

    auto dfs = [&](auto &&dfs, int u, LL val, LL sum, int num){
        if (S - sum - M / val - 35 > ans) return;
        if (val == M){
            ans = min(ans, abs(S - sum - cnt1) + num);
            return;
        }
        if (u == factor.size()) return;
        vector<LL> p1{val}, p2{sum};
        {
            while(1){
                if (M / p1.back() % factor[u] == 0 and p2.back() + factor[u] <= S){
                    p1.push_back(p1.back() * factor[u]);
                    p2.push_back(p2.back() + factor[u]);
                }
                else break;
            }
        }
        for(int i = p1.size() - 1; i >= 0; i--){
            if (i <= c[u]) dfs(dfs, u + 1, p1[i], p2[i], num - i);
            else dfs(dfs, u + 1, p1[i], p2[i], num - c[u] + (i - c[u]));
        }
    };

    dfs(dfs, 0, 1, 0, n - cnt1);
    if (ans > INF / 2) cout << -1 << '\n';
    else cout << ans << '\n';

}