codeforces 766B Mahmoud and a Triangle

最新推荐文章于 2019-12-23 16:41:35 发布

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本文探讨了一个关于使用三条线段构建非退化三角形的问题。通过给出特定条件下的输入输出示例，介绍了如何验证给定长度的线段能否形成具有正面积的三角形。文章还提供了一段C++代码实现，利用排序和遍历策略来解决该问题。

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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.

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e5;
int a[N + 5] = {0};
int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        scanf("%d", &a[i]);
    }
    sort(a, a + n);
    bool flag = false;
    int p, q, r;
    for(int i = 0; i < n - 2; i++)
    {
        p = a[i], q = a[i + 1], r = a[i + 2];

        if(p + q > r && p + r > q && q + r > p)
        {
            flag = true;
            break;
        }
    }
    if(flag)
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}