CF 6 A - Triangle

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本文介绍了一个编程挑战，任务是通过给定的四条线段长度判断是否可以构成非退化三角形、退化三角形或无法构成三角形，并提供了解决方案的示例代码。

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Description

Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.

The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.

Input

The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.

Output

Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.

Sample Input

Input

4 2 1 3

Output

TRIANGLE

Input

7 2 2 4

Output

SEGMENT

Input

3 5 9 1

Output

IMPOSSIBLE

题意：给出四个数，为能否用其中的三条组成三角形，如果能的话输出“TRIANGLE”，否则判断是不是退化三角形（就是有一个角是180度，一条线），是输出“SEGMENT",否则就输出”IMPOSSIBLE“。

此题主要是知道什么事退化三角形就OK了。

#include<iostream>
#include<stdio.h>
#include<cmath>
using namespace std;
int main()
{
    int m,n,p,q;
    while(scanf("%d%d%d%d",&m,&n,&p,&q)!=EOF)
    {
        if((m+n>p&&m+p>n&&n+p>m)||(m+n>q&&m+q>n&&n+q>m)||(m+p>q&&m+q>p&&p+q>m)||(p+n>q&&n+q>p&&p+q>n))
            cout<<"TRIANGLE"<<endl;
        else if((m+n==p)||(m+n==q)||(m+p==q)||(m+p==n)||(m+q==p)||(m+q==n)||(n+p==m)||(n+p==q)||(n+q==p)
                ||(n+q==m)||(p+q==m)||(p+q==n))
            cout<<"SEGMENT"<<endl;
        else
            cout<<"IMPOSSIBLE"<<endl;

    }
    return 0;
}