Description
standard input/output
Announcement
- Statements
There is a set of n segments with the lengths li. 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 li separated by spaces (1 ≤ li ≤ 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
23 4
Output
YES2
Input
33 4 8
Output
YES6
Input
33 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>
2 using namespace std;
3 int n;
4 int a[200005];
5 int ans1,ans2;
6 int main()
7 {
8 scanf("%d",&n);
9 for(int i=0;i<n;i++)
10 scanf("%d",&a[i]);
11 sort(a,a+n);
12 ans1=a[n-1]-a[0];
13 ans2=a[0]+a[1];
14 if(ans1+1<ans2)
15 {
16 cout<<"YES"<<endl;
17 cout<<ans1+1<<endl;
18 }
19 else
20 cout<<"NO"<<endl;
21 return 0;
22 }