codeforces - 766B【三角形判断】

最新推荐文章于 2025-09-06 22:33:46 发布

转载最新推荐文章于 2025-09-06 22:33:46 发布 · 158 阅读

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原文链接：http://www.cnblogs.com/Jstyle-continue/p/6379673.html

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#数据结构与算法

本文介绍了一个关于使用三条线段构造非退化三角形的问题。通过给出的n个线段长度，文章提供了一种有效算法来判断是否能从中选出三个线段构成一个具有正面积的三角形，并附带了具体的实现代码。

Mahmoud has n line segments, the i-th of them has length a_i. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.

Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.

Input

The first line contains single integer n (3 ≤ n ≤ 10⁵) — the number of line segments Mahmoud has.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the lengths of line segments Mahmoud has.

Output

In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.

Examples

Input

5
1 5 3 2 4

Output

YES

Input

3
4 1 2

Output

NO

题意：给定n个数，问能否找到三个数可以构成三角形

题解：从大到小判断相邻三个即可

代码：

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <bitset>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <cmath>
10 #include <list>
11 #include <set>
12 #include <map>
13 #define rep(i,a,b) for(int i = a;i <= b;++ i)
14 #define per(i,a,b) for(int i = a;i >= b;-- i)
15 #define mem(a,b) memset((a),(b),sizeof((a)))
16 #define FIN freopen("in.txt","r",stdin)
17 #define FOUT freopen("out.txt","w",stdout)
18 #define IO ios_base::sync_with_stdio(0),cin.tie(0)
19 #define mid ((l+r)>>1)
20 #define ls (id<<1)
21 #define rs ((id<<1)|1)
22 #define N 100005
23 #define INF 0x3f3f3f3f
24 #define INFF ((1LL<<62)-1)
25 using namespace std;
26 typedef long long LL;
27 typedef pair<int, int> PIR;
28 const double eps = 1e-8;
29 
30 int n, a[N];
31 int main()
32 {IO;
33     while(cin >> n){
34         rep(i, 1, n)    cin >> a[i];
35         sort(a+1, a+n+1);
36         int ok = 0;
37         per(i, n, 3){
38             int x = a[i], y = a[i-1], z = a[i-2];
39             if(x+y > z && x+z > y && y+z > x){
40                 ok = 1;
41                 break;
42             }
43         }
44         cout << (ok ? "YES" : "NO") << endl;
45     }
46     return 0;
47 }