【CodeCraft-19 and Codeforces Round #537 (Div. 2) C】Creative Snap

横坐标1..2^n对应着2^n个复仇者的基地,上面有k个复仇者(位置依次给出)。你是灭霸你要用以下方法消灭这k个复仇者: 一开始你获取整个区间[1..2^n] 假设你当前获取的区间为[l,r] mid = (l+r)/2 那么你每次有两种选择 1.将整个区间全都毁掉,如果这个区间里没有复仇者,那么花费为A,否则花费为B*复仇者个数*区间长度 2.将区间分为[l,mid]和[mid+1,r]分开毁掉(即分别获取[l,mid]和[mid+1,r]这两个区间,然后累加递归处理后的花费) 问你最小花费是多少

【题解】

如果2^n比较大的话,k个复仇者其实所占据的区间会非常少也就是说有很多区间都是空着的。。而空着的区间其实就没有必要用第二种选择了，因为显然直接返回A是最好的所以我们只需要模拟题目说的方法就好。只需要用两个二分查找获取当前区间里面有多少个复仇者就好如果没有复仇者了就直接返回A，否则模拟题目描述就ok (同一个基地可能有多个复仇者).

【代码】

import java.io.*;
import java.util.*;

public class Main {
	
	final static int N = (int)1e5;
	static InputReader in;
	static PrintWriter out;
	static int n,k,A,B,a[],r;
	
	static int leftmorethan(int x) {
		int l = 1,r = k,temp=-1;
		while (l<=r) {
			int mid = (l+r)>>1;
			if (x<=a[mid]) {
				temp = mid;
				r = mid - 1;
			}else {
				l = mid + 1;
			}
		}
		return temp;
	}
	
	static int rightlessthan(int x) {
		int l = 1,r = k,temp = -1;
		while (l<=r) {
			int mid = (l+r)>>1;
			if (x>=a[mid]) {
				temp = mid;
				l = mid + 1;
			}else {
				r = mid-1;
			}
		}
		return temp;
	}
	
	
	static long dfs(int l,int r) {
		int L = leftmorethan(l);
		int R = rightlessthan(r);
		boolean ok = true;
		if (L==-1 || R==-1) ok = false;
		if (L>R) ok = false;
		if (!ok) return 1l*A;
		
		if (l==r) return 1l*B*(R-L+1);
		
		int mid = (l+r)>>1;
		long ans;
		//select all
		ans = 1l*B*(R-L+1)*(r-l+1);
		
		//divede
		ans = Math.min(ans, dfs(l,mid)+dfs(mid+1,r));
		return ans;
	}
		
	public static void main(String[] args) throws IOException{
		in = new InputReader();
		out = new PrintWriter(System.out);
		
		//code start from here
		a = new int[N+10];
		n = in.nextInt();k = in.nextInt();A = in.nextInt();B = in.nextInt();
		for (int i = 1; i <= k;i++) a[i] = in.nextInt();
		r = 1;
		for (int i = 1;i <= n;i++) r = r*2;
		Arrays.sort(a, 1,k+1);
		out.print(dfs(1,r));
		out.close();
	}

	static class InputReader{
		public BufferedReader br;
		public StringTokenizer tokenizer;
		
		public InputReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			tokenizer = null;
		}
		
		public String next(){
			while (tokenizer==null || !tokenizer.hasMoreTokens()) {
				try {
				tokenizer = new StringTokenizer(br.readLine());
				}catch(IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}