2018-BNUZ-ACM-GDCPC选拔赛 CodeForces 766B Mahmoud and a Triangle【补】

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

Copy

5
1 5 3 2 4

output

Copy

YES

input

Copy

3
4 1 2

output

Copy

NO

Note

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

题意： 输入n个数  只要其中任意三个数能组成一个三角形输出YES，否则输出NO

题解：首先组成三角形的条件是 两边之和大于第三边。 这道题如果直接暴力去找的话很明显会TLE。

所以我们要 先对 数组进行排序， 因为排好序的数组 从中 连续地选择3个数出来时，这三个数一定是差距最小的， 也就是最有可能组成三角形的。

AC代码：

#include <bits/stdc++.h>

#define ll long long
#define mem(a,b) memset(a,b,sieof(a))
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;
//-------------------------------------------------
int main() {
	int n, a[100050];
	cin >> n;
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	sort(a,a+n);
	int tmp;
	int flag = 0;
	for (int i = 0; i < n-2; i++) {
		int aa = a[i], bb = a[i+1], cc = a[i+2];
		if (aa+bb>cc && aa+cc>bb && bb+cc>aa) {
			flag = 1;
			goto gg;
		}
	}
gg:
	if (flag)
		printf("YES\n");
	else {
		puts("NO");
	}
	return 0;
}