题目链接：https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1420
由于最少的袋鼠数量也就是n/2取上整，那么可以排序，然后从[1,n/2] [n/2+1,n]两个部分中匹配。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 500500; 5 int n, s[maxn]; 6 7 int main() { 8 // freopen("in", "r", stdin); 9 while(~scanf("%d", &n)) { 10 for(int i = 1; i <= n; i++) scanf("%d", &s[i]); 11 sort(s+1, s+n+1); 12 int ret = 0; 13 int mid = n / 2; 14 for(int i = n; i >= mid; i--) { 15 while(s[i] / 2 < s[mid] && mid >= 1) mid--; 16 if(mid <= 0) break; 17 ret++; mid--; 18 } 19 printf("%d\n", n - ret); 20 } 21 return 0; 22 }