766B 很水但是有点点启示

最新推荐文章于 2021-09-12 14:01:03 发布

原创最新推荐文章于 2021-09-12 14:01:03 发布 · 636 阅读

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博客内容讲述了如何通过排序和避免重复判断来优化三角形边长检查的算法，减少了三重循环的时间消耗，提高了效率。

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.

Example
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.

3
组成三角形嘛
就是输入
判断两边和（差）和第三边的关系
输出
但是要是利用一点点三角形的数学特性就能优化
就不用三重循环呀
少了很多很多的时间呢好棒棒鼓鼓掌~
嗯呢
就是先排序
在判断（少了重复的情况）
只判断最长边（或者最短边就好啦呀）
代码~

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
int n;
int a[100005];
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    sort(a+1,a+n+1);
    for(int i=3;i<=n;i++){
        if(a[i-2]>a[i]-a[i-1]){
            cout<<"YES"<<endl;
                return 0;
        }
    }
   cout<<"NO"<<endl;
    return 0;
}