Just A Triangle
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Problem Description
This is an easy problem, just for you to warm up.
Give you three edges of a triangle. Can you tell me which kind of triangle it stands for?
If it’s a right triangle(直角三角形), please output “good”. If it’s a isosceles triangle(等腰三角形), please output “perfect”. Otherwise, please output “just a triangle”. You may suppose the input is legal.
Give you three edges of a triangle. Can you tell me which kind of triangle it stands for?
If it’s a right triangle(直角三角形), please output “good”. If it’s a isosceles triangle(等腰三角形), please output “perfect”. Otherwise, please output “just a triangle”. You may suppose the input is legal.
Input
The first line contains an integer t means the number of test cases.
The each case contains three integers a, b, c in a line which stands for the length of the three edges.
(0 <a, b, c < 300).
The each case contains three integers a, b, c in a line which stands for the length of the three edges.
(0 <a, b, c < 300).
Output
For each case, output the answer in one line.
Sample Input
43 4 52 2 3 1 4 44 6 3
Sample Output
goodperfectperfectjust a triangle
Source
HDU 2009-11 Programming Contest
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题目分析
判断三角形的类型,并按给定要求输出,直角三角形输出good,等腰三角形输出perfect,其他三角形输出just a triangle
解题思路
水题,判断一下就行
源代码
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
//#include<bits\stdc++.h>
using namespace std;
void solve()
{
vector<int>vec(3);
cin>>vec[0]>>vec[1]>>vec[2];
sort(vec.begin(),vec.end());
if(vec[0]*vec[0]+vec[1]*vec[1]==vec[2]*vec[2])
cout<<"good"<<endl;
else
if(vec[0]==vec[1]||vec[1]==vec[2])
cout<<"perfect"<<endl;
else
cout<<"just a triangle"<<endl;
}
int main()
{
int t;
cin>>t;
while(t--)
{
solve();
}
}
#include<string>
#include<algorithm>
#include<vector>
//#include<bits\stdc++.h>
using namespace std;
void solve()
{
vector<int>vec(3);
cin>>vec[0]>>vec[1]>>vec[2];
sort(vec.begin(),vec.end());
if(vec[0]*vec[0]+vec[1]*vec[1]==vec[2]*vec[2])
cout<<"good"<<endl;
else
if(vec[0]==vec[1]||vec[1]==vec[2])
cout<<"perfect"<<endl;
else
cout<<"just a triangle"<<endl;
}
int main()
{
int t;
cin>>t;
while(t--)
{
solve();
}
}
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