CF1111C Creative Snap

思路：

分治，递归实现就可以。不一定非得用前缀和，直接用一个数组记录avengers的位置然后二分即可。写的有点复杂了。

实现：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n, k, A, B;
ll dfs(ll l, ll r, vector<ll>& K, vector<ll>& V, vector<ll>& S)
{
    if (l == r)
    {
        if (binary_search(K.begin(), K.end(), l))
        {
            int p = lower_bound(K.begin(), K.end(), l) - K.begin();
            return B * V[p];
        }
        return A;
    }
    int p1 = lower_bound(K.begin(), K.end(), l) - K.begin();
    int p2 = upper_bound(K.begin(), K.end(), r) - K.begin() - 1;
    if (p1 > p2) return A;
    ll minn = B * (S[p2] - S[p1 - 1]) * (r - l + 1);
    ll m = l + r >> 1;
    minn = min(minn, dfs(l, m, K, V, S) + dfs(m + 1, r, K, V, S));
    return minn;
}
int main()
{
    while (cin >> n >> k >> A >> B)
    {
        map<ll, ll> mp;
        int x;
        for (int i = 1; i <= k; i++)
        {
            cin >> x;
            if (!mp.count(x)) mp[x] = 0;
            mp[x]++;
        }
        vector<ll> K, V, S;
        K.push_back(0); V.push_back(0); S.push_back(0);
        for (auto it: mp) { K.push_back(it.first); V.push_back(it.second); }
        for (int i = 1; i < V.size(); i++) S.push_back(S.back() + V[i]);
        cout << dfs(1, 1 << n, K, V, S) << endl;        
    }    
    return 0;
}