Codeforces Round #219 (Div. 1) A. Counting Kangaroos is Fun 【二分】

A. Counting Kangaroos is Fun

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

8
2
5
7
6
9
8
4
2

output

5

input

8
9
1
6
2
6
5
8
3

output

5

比较喜欢Codeforces ，可以查看测试数据和别人代码，长见识。这道题二分可以不二分直接一次循环也可以做出来

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n, k[5000005];
bool judge(int x) {
    for (int i = x - 1, j = n - 1; i >= 0; i--, j--) {
        if (2*k[i] > k[j]) return false;
    }
    return true;
}
int main() {
    while (scanf("%d", &n) != EOF) {
        for (int i = 0; i < n; i++) {
            scanf("%d", &k[i]);
        }
        sort(k, k + n);
        int lb = 0, ub = n/2;
        int ans = 0;
        while (ub - lb >= 0) {
            int mid = (ub + lb)/2;
            if (judge(mid)) {
                ans = mid;
                lb = mid + 1;
            }
            else {
                ub = mid - 1;
            }
        }
        printf("%d
", n - ans);
    }
    return 0;
}

没有二分，直接进行查找

<pre code_snippet_id="1786486" snippet_file_name="blog_20160727_2_3662693" name="code" class="cpp" style="font-size: 14px;">sort(a,a+n);
int j=n-1;
for(i=n/2-1;i>=0;i--) {
<span>	</span>if(2*a[i]<=a[j])
	j--;
}
printf("%d
",j+1);