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  • 【Codeforces 988D】Points and Powers of Two

    【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^x*3 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());
            }
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/10461515.html
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