codeforces 6A. Triangle
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 易做图 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 易做图 of a positive area, but it is possible to construct a degenerate 易做图. It can be so, that it is impossible to construct a degenerate 易做图 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 易做图. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate 易做图. Output IMPOSSIBLE if it is impossible to construct any 易做图. 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
[cpp]
/1.能组成三角形2面积为0的三角形，即一条边等于另两条之和3
：1，2都不满足 //1.能组成三角形2面积为0的三角形，即一条边等于另两条之和3：
1，2都不满足[cpp] view plaincopyprint?#include<iostream>
#include<cstdio>
#include<cstring>
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]]<a[b[2]]&&a[b[2]]-a[b[1]]<a[b[0]]&&a[b[0]]-a[b[2]]<a[b[1]])
return 1;
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]])
return 2;
return 0; }
int main(){ int res,i;
while(~scanf("%d%d%d%d",&a[0],&a[1],&a[2],&a[3])){
for(res=i=0;i<4;i++){
int k=fun(d[i]);
if(k==1){res=-1;printf("TRIANGLE\n");break;}
else if(k==2)res=k;
} if(res==2)printf("SEGMENT\n");
else if(res==0)printf("IMPOSSIBLE\n");
} return 0;
补充：软件开发 , C++ ,
