Gym 100971C 水&愚&三角形

Description

standard input/output
Announcement

 

• Statements

  There is a set of n segments with the lengths l_i. Find a segment with an integer length so that it could form a non-degenerate triangle with any two segments from the set, or tell that such segment doesn't exist.

Input

The first line contains a single integer n(2 ≤ n ≤ 200000) — the number of segments in the set.

The second line contains n integers l_i separated by spaces (1 ≤ l_i ≤ 10⁹) — the lengths of the segments in the set.

Output

If the required segment exists, in the first line output «YES» (without quotes). In this case in the second line output a single integer x — the length of the needed segment. If there are many such segments, output any of them.

If the required segment doesn't exist, output «NO» (without quotes).

Sample Input

 

Input

2
3 4

Output

YES
2

Input

3
3 4 8

Output

YES
6

Input

3
3 4 9

Output

NO

题意：给你一些边 判断是否存在一条边 使得和其中的任意两条边组合都能形成三角形 

题解：将边排序之后 所要添加的边存在上下界
     ans1=a[n-1]-a[0];      
     ans2=a[0]+a[1];
     范围内 任意输出

 1 #include<bits/stdc++.h>
 using namespace std;
 int n;
 int a[];
 int ans1,ans2;
 int main()
 {
     scanf("%d",&n);
     for(int i=;i<n;i++)
         scanf("%d",&a[i]);
     sort(a,a+n);
     ans1=a[n-]-a[];
     ans2=a[]+a[];
     if(ans1+<ans2)
     {
         cout<<"YES"<<endl;
         cout<<ans1+<<endl;
     }
     else
         cout<<"NO"<<endl;
     return ;
 }