Codeforces Round #572 (Div. 2) Number Circle

You are given n numbers a1,a2,…,an. Is it possible to arrange them in a circle in such a way that every number is strictly less than the sum of its neighbors?

For example, for the array [1,4,5,6,7,8], the arrangement on the left is valid, while arrangement on the right is not, as 5≥4+1 and 8>1+6.

Input
The first line contains a single integer n (3≤n≤105) — the number of numbers.

The second line contains n integers a1,a2,…,an (1≤ai≤109) — the numbers. The given numbers are not necessarily distinct (i.e. duplicates are allowed).

Output
If there is no solution, output “NO” in the first line.

If there is a solution, output “YES” in the first line. In the second line output n numbers — elements of the array in the order they will stay in the circle. The first and the last element you output are considered neighbors in the circle. If there are multiple solutions, output any of them. You can print the circle starting with any element.

Examples
inputCopy
3
2 4 3
outputCopy
YES
4 2 3
inputCopy
5
1 2 3 4 4
outputCopy
YES
4 4 2 1 3
inputCopy
3
13 8 5
outputCopy
NO
inputCopy
4
1 10 100 1000
outputCopy
NO
Note
One of the possible arrangements is shown in the first example:

4<2+3;

2<4+3;

3<4+2.

One of the possible arrangements is shown in the second example.

No matter how we arrange 13,8,5 in a circle in the third example, 13 will have 8 and 5 as neighbors, but 13≥8+5.

There is no solution in the fourth example.

题意理解：
大意：
给你n个数，让你排成一个圆圈，要满足a[i-1]+a[i+1]>a[i]，输出最后序列
对数组排序。 首先，如果a[n]≥a[n-1] + a[n-2]则答案是 NO（因为否则an不小于邻居的总和。
其他情况下，答案是 YES。
同样答案不唯一，其中一个可行解为：a[n] a[n-2] … a[n-1]
注意环状满足条件即可

#include
#include
using namespace std;
const int maxn=100005;
int num[maxn];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>num[i];
    }
    sort(num,num+n);
    if(num[n-1]>=num[n-2]+num[n-3]) cout<<"NO";
    else
    {
        cout<<"YES"<<endl;
        cout<<num[n-1]<<" "<<num[n-2]<<" ";
        for(int i=n-4;i>=0;i--) cout<<num[i]<<" ";
        cout<<num[n-3]<<endl;
    }
}