Codeforces Round #396 (Div. 2) B. Mahmoud and a Triangle 贪心

B. Mahmoud and a Triangle

题目连接：

http://codeforces.com/contest/766/problem/B

Description

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.

Sample Input

5

1 5 3 2 4

Sample Output

YES

Hint

题意

问你能否从n个数字中抽出来三个，使得可以构成不退化的三角形。

题解：

一开始觉得好神呀……

然后发现只要排个序，然后判断a[i],a[i-1],a[i+1]能否组成三角形就好了，这样贪心肯定是对的。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int n;
long long a[maxn];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        cin>>a[i];
    sort(a+1,a+1+n);
    int flag = 0;
    for(int i=2;i<=n-1;i++){
        if(a[i]+a[i-1]>a[i+1])
            flag = 1;
    }
    if(flag)cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}