卡片取模生成树(Sirni) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    sort(a.begin(), a.end());
    a.erase(unique(a.begin(), a.end()), a.end());
    int m = (int)a.size();
    if (m <= 1) {
        cout << 0 << "\n";
        return 0;
    }
    int maxV = a.back();
    vector<int> nxtIdx(maxV + 2, -1);
    for (int i = 0; i < m; ++i) nxtIdx[a[i]] = i;
    for (int v = maxV; v >= 1; --v) {
        if (nxtIdx[v] == -1) nxtIdx[v] = nxtIdx[v + 1];
    }
    vector<vector<pair<int, int>>> buckets(maxV + 1);
    for (int i = 0; i + 1 < m; ++i) {
        int u = a[i];
        int j = nxtIdx[u + 1];
        if (j != -1 && a[j] < 2 * u) {
            buckets[a[j] % u].push_back({i, j});
        }
        for (int mult = 2 * u; mult <= maxV; mult += u) {
            int k = nxtIdx[mult];
            if (k != -1 && (k == i + 1 || a[k] / u > a[k - 1] / u)) {
                buckets[a[k] % u].push_back({i, k});
            }
        }
    }
    vector<int> p(m);
    iota(p.begin(), p.end(), 0);
    auto find = [&](int x) {
        while (p[x] != x) x = p[x] = p[p[x]];
        return x;
    };
    long long ans = 0;
    int cnt = 0;
    for (int w = 0; w <= maxV && cnt < m - 1; ++w) {
        for (auto [u, v] : buckets[w]) {
            int ru = find(u), rv = find(v);
            if (ru != rv) {
                p[ru] = rv;
                ans += w;
                if (++cnt == m - 1) break;
            }
        }
    }
    cout << ans << "\n";
    return 0;
}