codeforces 372A Counting Kangaroos is Fun

原创
2015/04/04 15:43
阅读数 302

There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.

Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.

The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.

Input

The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).

Output

Output a single integer — the optimal number of visible kangaroos.

Sample test(s)

input

8
2
5
7
6
9
8
4
2

output

5

input

8
9
1
6
2
6
5
8
3

output

5


技巧在于最小的K个袋鼠会被最大的K个所包含,二分找K就行

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class CF372A {

	private static int[] s;
	private static int n;

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		n = Integer.parseInt(in.readLine());
		s = new int[n];
		for (int i = 0; i < n; i++) {
			s[i] = Integer.parseInt(in.readLine());
		}
		Arrays.sort(s);
		int lo = 0, hi = n / 2;
		while (lo <= hi) {
			int mid = lo + (hi - lo) / 2;
			if (ok(mid)) {
				if (lo == n / 2) {
					break;
				}
				lo = mid + 1;
			} else {
				hi = mid - 1;
			}
		}
		System.out.println(n - lo);
	}

	private static boolean ok(int x) {
		for (int i = 0; i <= x; i++) {
			if (2 * s[x - i] > s[n - i - 1]) {
				return false;
			}
		}
		return true;
	}
}

提交的时候一直卡在了当可以被全部包含的时候,lo任然会递增+1,想了半天只能取巧在lo == n / 2的时候break掉

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