codeforces6A--Triangle
发布日期:2022-02-02 02:58:14 浏览次数:17 分类:技术文章

本文共 2006 字,大约阅读时间需要 6 分钟。

A. Triangle
time limit per test
2 seconds
memory limit per test
64 megabytes
input
standard input
output
standard output

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 test(s)
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE

这道题从一开始我就想错啦,它是在4个数中存在3个数满足下面三个条件:1.能构成三角形,输出“TRIANGLE”。2.如果两边之和等于第三边,输出“SEGMENT”。3.不满足1,2两个条件的输出“IMPOSSIBLE”。

代码实现:

#include
#include
#include
using namespace std;int d[][3]={
{0,1,2},{0,1,3},{0,2,3},{1,2,3}};int a[4];int fun(int *b){ if(a[b[0]]+a[b[1]]>a[b[2]]&&a[b[2]]+a[b[1]]>a[b[0]]&&a[b[0]]+a[b[2]]>a[b[1]]) if(a[b[0]]-a[b[1]]

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