Higher Math
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 2219 Accepted
Submission(s): 1219
Problem Description
You are building a house. You’d prefer if all the
walls have a precise right angle relative to the ground, but you have no device
to measure angles. A friend says he has a great idea how you could ensure that
all walls are upright: All you need to do is step away a few feet from the wall,
measure how far away you are from the wall, measure the height of the wall, and
the distance from the upper edge of the wall to where you stand. You friend
tells you to do these measurements for all walls, then he’ll tell you how to
proceed. Sadly, just as you are done, a timber falls on your friend, and an
ambulance brings him to the hospital. This is too bad, because now you have to
figure out what to do with your measurements yourself.
Given the three
sides of a triangle, determine if the triangle is a right triangle, i.e. if one
of the triangle’s angles is 90 degrees.
Input
The inputs start with a line containing a single
integer n. Each of the n following lines contains one test case. Each test case
consists of three integers 1 <= a, b, c <= 40000 separated by a space. The
three integers are the lengths of the sides of a triangle.
Output
The output for every scenario begins with a line
containing “Scenario #i:”, where i is the number of the scenario counting from
1. After that, output a single line containing either the string “yes” or the
string “no”, depending on if the triangle in this test case has a right angle.
Terminate each test case with an empty line.
Sample Input
2
36 77 85
40 55 69
Sample Output
Scenario #1: yes
Scenario #2: no
Source
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题意：给你三个数，作为三角形的三边，判断这个数是不是直角三角形。
代码：
1 #include <iostream>
2
3 using namespace std;
4
5 int main()
6 {
7 int i,a,b,c,T;
8 cin>>T;
9 for(i=1;i<=T;i++){
10 cin>>a>>b>>c;
11 if(a*a+b*b==c*c || b*b+c*c==a*a || a*a+c*c==b*b)
12 cout<<"Scenario #"<<i<<‘:‘<<endl<<"yes"<<endl;
13 else
14 cout<<"Scenario #"<<i<<‘:‘<<endl<<"no"<<endl;
15 cout<<endl;
16 }
17 return 0;
18 }
Freecode : www.cnblogs.com/yym2013