计蒜客ACM ICPC 2017 Warmup Contest 9--I题-Sticky Situation

本文链接：https://blog.youkuaiyun.com/zhengxiangmaomao/article/details/78308697

本文介绍了一个关于在森林中玩捉迷藏游戏时选择安全区的问题。为了明确划定安全区域，需要用三根棍子围成一个非退化的三角形。文章提供了一段程序代码，用于判断收集到的棍子是否能够组成这样的三角形。

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While on summer camp, you are playing a game of hide-and-seek in the forest. You need to designate a “safe zone”, where, if the players manage to sneak there without being detected,they beat the seeker. It is therefore of utmost importance that this zone is well-chosen.

You point towards a tree as a suggestion, but your fellow hide-and-seekers are not satisfied. After all, the tree has branches stretching far and wide, and it will be difficult to determine whether a player has reached the safe zone. They want a very specific demarcation for the safe zone. So, you tell them to go and find some sticks, of which you will use three to mark anon-degenerate triangle (i.e. with strictly positive area) next to the tree which will count as the safe zone. After a while they return with a variety of sticks, but you are unsure whether you can actually form a triangle with the available sticks.

Can you write a program that determines whether you can make a triangle with exactly three of the collected sticks?

Input

The first line contains a single integer N , with 3 ≤ N ≤ 20 000, the number of sticks collected. Then follows one line with Npositive integers, each less than $2^{60}$ , the lengths of the sticks which your fellow campers have collected.

Output

Output a single line containing a single word: possible if you can make a non-degenerate triangle with three sticks of the provided lengths, and impossible if you can not.

样例输入1

3
1 1 1

样例输出1

possible

样例输入2

5
3 1 10 5 15

样例输出2

impossible

#include<iostream>
#include<algorithm>
using namespace std;
int n;
long long a[20001];
long long sum;
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;i<n;i++) scanf("%lld",&a[i]);
		sort(a,a+n);
		bool flag=false;
		for(int i=0;i<n-2;i++)
		{
			sum=a[i]+a[i+1];
			if(sum<=a[i+2]) continue;
			if(sum>a[i+2])
			{
				flag=true;
				break;
			}
		}
		if(flag) printf("possible\n");
		else printf("impossible\n");
	}
	return 0;
}