Codeforces Round #219 (Div. 2 )-----贪心--思维
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Description
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 Input
825769842
5
891626583
5
就是给n各袋鼠的袋子大小,一个袋鼠袋子是另一个的两倍以上是就可以把小的装进去,装进去之后从外边就看不到了,每个袋鼠只能装一个,问最少能看见多少个,升序排列,从中间开始,前半段与后半段比较就可以了
#include<cstdio>#include<algorithm>using namespace std;#define maxn 500005int s[maxn];int main(){int n;scanf("%d", &n);for(int i = 0; i < n; i++){scanf("%d", &s[i]);}sort(s, s+n);for(int i = n/2 - 1; i >= 0; i--){if(s[i]*2 <= s[n-1]){n--;}}printf("%d\n", n);return 0;}
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