思路（贪心）：

　　排序后分奇偶，小的放在奇数位，大的放在偶数位，最后for一遍查找满足条件的解即可（蒟蒻啥都不会，只能借鉴他人思路）

链接：(10条消息) D2. Sage‘s Birthday (hard version)(构造+贪心)_C_Dreamy的博客-CSDN博客

代码：

#include 
#define ll long long

using namespace std;

typedef pair PII;

const int N = 1e5 + 10, mod = 1e9 + 7;

int n;
int a[N], ans[N];

void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i ++)
    {
        cin >> a[i];
    }

    sort(a + 1, a + 1 + n);
    int num = 0;
    int p = 1;
    for (int i = 2; i <= n; i += 2)
    {
        ans[i] = a[p];
        p ++;
    }

    for (int i = 1; i <= n; i += 2)
    {
        ans[i] = a[p];
        p ++;
    }

    for (int i = 2; i <= n; i += 2)
    {
        if(ans[i] < ans[i - 1] && ans[i] < ans[i + 1]) num ++;
    }

    cout << num << endl;
    for (int i = 1; i <= n; i ++)
        cout << ans[i] << " ";
    cout << endl;
}

int main()
{
    ios::sync_with_stdio(0);
    //int T;
    //cin >> T;
    //while(T --)
    //{
        solve();
    //}
    return 0;
}