Creative Snap

C. Creative Snap time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.

Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 22. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:

• if the current length is at least 22, divide the base into 22 equal halves and destroy them separately, or
• burn the current base. If it contains no avenger in it, it takes AA amount of power, otherwise it takes his B⋅na⋅lB⋅na⋅l amount of power, where nana is the number of avengers and ll is the length of the current base.

Output the minimum power needed by Thanos to destroy the avengers' base. Input

The first line contains four integers nn, kk, AA and BB (1≤n≤301≤n≤30, 1≤k≤1051≤k≤105, 1≤A,B≤1041≤A,B≤104), where 2n2n is the length of the base, kkis the number of avengers and AA and BB are the constants explained in the question.

The second line contains kk integers a1,a2,a3,…,aka1,a2,a3,…,ak (1≤ai≤2n1≤ai≤2n), where aiai represents the position of avenger in the base.

Output

Output one integer — the minimum power needed to destroy the avengers base.

这题刚一看觉得很水

，仔细一看数据范围……

其实也不难，主要是数据范围过大。要用一种类似离散化的做法。

对数组a排序。对于一段区间[l,r]的英雄数量等于a中第一个比r大的数的下标减1减a中第一个大于等于l的数的下标加1，这样就可以做了(k很小)。

不过要注意剪枝：若区间[l,r]的英雄数量等于0，就直接返回A。

上代码：

#include <bits/stdc++.h>
using namespace std;
long long n, k, a, b;
long long hero[1000001];
long long solve(long long l, long long r) {
    long long num = upper_bound(hero + 1, hero + 1 + k, r) - lower_bound(hero + 1, hero + 1 + k, l);
    if (num == 0) return a;
    long long ans = (r - l + 1) * b * num;
    if (l >= r) return ans;
    ans = min(ans, solve(l, (l + r) / 2) + solve((l + r) / 2 + 1, r));
    return ans;
}
int main() {
    cin >> n >> k >> a >> b;
    for (long long i = 1; i <= k; i++) cin >> hero[i];
    sort(hero + 1, hero + 1 + k);
    cout << solve(1, (1 << n));
}