Mahmoud and a Triangle 【codeforces】【水题】

最新推荐文章于 2022-09-19 10:59:09 发布

原创最新推荐文章于 2022-09-19 10:59:09 发布 · 368 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces

水题同时被 2 个专栏收录

33 篇文章

订阅专栏

好题

7 篇文章

订阅专栏

本文介绍了一个编程挑战，即从给定长度的线段中选择三个能够组成非退化三角形的线段。通过排序和遍历算法实现解决方案，并提供完整的代码示例。

Mahmoud has n line segments, the i-th of them has length ai. 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, ..., an (1 ≤ ai ≤ 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

Note

For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.

题解：先排序，再遍历求和判定；

代码如下：

#include<cstdio>
#include<algorithm>
using namespace std;
int a[100005];
int main()
{
	int n;
	while(~scanf("%d",&n)){
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		sort(a,a+n);
		int flag=0;
		for(int i=0;i<n-2;i++){
			if(a[i]+a[i+1]>a[i+2]){
				
				flag=1;
				break;
			}
		}
		
		if(flag) printf("YES\n");
		else printf("NO\n");
	}
	
	return 0;
}