codeforces 766 B

B. Mahmoud and a Triangle
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 24 and 5 to form a non-degenerate triangle.


問題就是給出邊長,問是否能構成三角形。


兩種方法:第一種就是排序啊a[n]+a[n+1]>a[n+2]就輸出YES。(nlogn)
第二種:可以觀察到斐波那契數列是符合題意的最長數列,而到fib(50)時已經超過題目範圍的邊長10^9。因此當n>50時一定有解。n<50時則暴力n^3或採用第一種方法算出解。
這裏的代碼是第一種的方法。
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=100000;
int n,a[maxn+10];
int main(){
	scanf("%d",&n);
	for(int i = 1; i <= n; i++){
		scanf("%d",&a[i]);
	}
	sort(a+1,a+n+1);
	for(int i = 1; i <= n-2; i++){
		if(a[i] + a[i+1] > a[i+2]){
			printf("YES");
			return 0;
		}
	}
	printf("NO");
	return 0;
}


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