【链接】 我是链接,点我呀:)
【题意】
让你从一个集合中找出来一个子集
使得这个子集中任意两个数相减的绝对值是2^的整数次幂
且集合的大小最大
【题解】
考虑子集的个数为4个或4个以上
那么我们找到最小的4个a[1],a[2],a[3],a[4]
显然
dis(1,2)=2^a
dis(2,3)=2^b
dis(1,3)=dis(1,2)+dis(2,3) = 2^c
因为2^a+2^b=2^c
所以可以推出来a=b
也即dis(1,2)=dis(2,3)
同理对于a[2],a[3],a[4]
用同样的方法可以得到
dis(2,3)=dis(3,4)
那么dis(1,4)=dis(1,2)3=2^x3
显然不是2的整数幂
因此不存在大小大于等于4的集合满足题意。
所以只要考虑集合大小为3以及为2的了
大小为3的话,只要枚举中间那个数字,根据上面的推论dis(1,2)=dis(2,3)
则枚举2^j
看看x-2^j和x+2^j是否存在就好
大小为2就更简单啦
大小为1就随便输出一个数字就好
【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = (int)2e5;
static class Task{
int n;
int x[] = new int[N+10];
TreeMap<Integer,Integer> dic = new TreeMap<Integer,Integer>();
public void solve(InputReader in,PrintWriter out) {
n = in.nextInt();
for (int i = 1;i <= n;i++) {
x[i] = in.nextInt();
dic.put(x[i], 1);
}
for (int x:dic.keySet()) {
int cur = 1;
for (int j = 0;j <= 30;j++) {
int xl = x - cur,xr = x + cur;
if (dic.containsKey(xl) && dic.containsKey(xr)) {
out.println(3);
out.print(xl+" "+x+" "+xr);
return;
}
cur = cur * 2;
}
}
for (int x:dic.keySet()) {
int cur = 1;
for (int j = 0;j <= 30;j++) {
int xr = x + cur;
if (dic.containsKey(xr)) {
out.println(2);
out.print(x+" "+xr);
return;
}
cur = cur * 2;
}
}
out.println(1);
out.println(x[1]);
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
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());
}
}
}
【Codeforces 988D】Points and Powers of Two
原文:https://www.cnblogs.com/AWCXV/p/10461515.html