codeforces 626E. Simple Skewness 三分

 1 #include <iostream>
 2 #include <vector>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <cmath>
 7 #include <map>
 8 #include <set>
 9 #include <string>
10 #include <queue>
11 #include <stack>
12 #include <bitset>
13 using namespace std;
14 #define pb(x) push_back(x)
15 #define ll long long
16 #define mk(x, y) make_pair(x, y)
17 #define lson l, m, rt<<1
18 #define mem(a) memset(a, 0, sizeof(a))
19 #define rson m+1, r, rt<<1|1
20 #define mem1(a) memset(a, -1, sizeof(a))
21 #define mem2(a) memset(a, 0x3f, sizeof(a))
22 #define rep(i, n, a) for(int i = a; i<n; i++)
23 #define fi first
24 #define se second
25 typedef pair<int, int> pll;
26 const double PI = acos(-1.0);
27 const double eps = 1e-8;
28 const int mod = 1e9+7;
29 const int inf = 1061109567;
30 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
31 const int maxn = 2e5+5;
32 ll a[maxn], sum[maxn];
33 int n;
34 ll cal(int len, int pos) {
35     ll num = sum[pos]-sum[pos-len-1]+sum[n]-sum[n-len];
36     return num;
37 }
38 int main()
39 {
40     cin>>n;
41     for(int i = 1; i<=n; i++) {
42         scanf("%I64d", &a[i]);
43     }
44     if(n == 1 || n == 2) {
45         cout<<1<<endl;
46         cout<<a[1]<<endl;
47         return 0;
48     }
49     sort(a+1, a+1+n);
50     ll maxx = 0;
51     int len = 0, mid = 1;
52     for(int i = 1; i<=n; i++)
53         sum[i] = sum[i-1]+a[i];
54     for(int i = 2; i<=n-1; i++) {
55         int l = 0, r = min(i-1, n-i);
56         while(l<r) {
57             int lmid = (2*l+r)/3;
58             int rmid = (l+2*r+2)/3;
59             ll ans1 = cal(lmid, i);
60             ll ans2 = cal(rmid, i);
61             if(ans1*(2*rmid+1)<(2*lmid+1)*ans2) {
62                 l = lmid+1;
63             } else {
64                 r = rmid-1;
65             }
66         }
67         ll tmp = (sum[i]-sum[i-l-1]+sum[n]-sum[n-l])-(2*l+1)*a[i];
68         if(tmp*(2*len+1)>maxx*(2*l+1)) {
69             maxx = tmp;
70             len = l;
71             mid = i;
72         }
73     }
74     cout<<2*len+1<<endl;
75     for(int i = mid-len; i<=mid; i++)
76         printf("%d ", a[i]);
77     for(int i = n-len+1; i<=n; i++)
78         printf("%d ", a[i]);
79     return 0;
80 }