[ICPC2019西安M] value

[ICPC2019西安M] value

Description

给出N个A，B寻找一个集合使价值最大价值即为集合内A的和如果 i，j属于集合内并且 (i>1, j>1, i^k=j(k>1)) 则价值需要减去j的B值

Solution

对于每个非幂数，我们枚举它的所有幂，单独拎出来，因为不同部分的贡献是相互独立的

对于每个非幂数的所有次幂，暴力枚举一遍选谁不选谁

VP 时一开始居然想到 MST 上去了

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e5 + 5;

int n, a[N], b[N], v[N];

int solve(vector<pair<int, int>> p)
{
    int n = p.size();
    int ans = 0;
    for (int i = 0; i < (1 << n); i++)
    {
        bitset<20> s(i);
        int tmp = 0;
        for (int j = 0; j < n; j++)
            if (s[j])
                tmp += p[j].first;
        for (int j = 0; j < n; j++)
            for (int k = 0; k < n; k++)
                if (j > k && s[j] && s[k] && (j + 1) % (k + 1) == 0)
                    tmp -= p[j].second;
        ans = max(ans, tmp);
    }
    return ans;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
        cin >> b[i];
    int ans = a[1];
    for (int i = 2; i <= n; i++)
    {
        if (v[i])
            continue;
        vector<pair<int, int>> vec;
        for (int j = i; j <= n; j *= i)
        {
            v[j] = 1;
            vec.push_back({a[j], b[j]});
        }
        ans += solve(vec);
    }
    cout << ans << endl;
}