判断三角形

Triangle

                                                                             Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                     Total Submission(s): 1038    Accepted Submission(s): 648

Problem Description

You are given the side lengths of a triangle, determine is it an acute triangle, right triangle or obtuse triangle.

 

Input

The first line contains an integer T indicating the number of test cases.
For each test case, there is a single line contains three integer A, B and C indicating the side lengths of the triangle. (You can assume it is a valid triangle)

Technical Specification

1. 1 <= T <= 50
2. 1 <= A,B,C <= 100

 

Output

For each test case, output the case number first, then output "Acute triangle", "Right triangle" or "Obtuse triangle".

 

Sample Input

32 2 24 8 53 4 5

 

Sample Output

Case 1: Acute triangleCase 2: Obtuse triangleCase 3: Right triangle

 

Author

hanshuai

 

Source

The 6th Central China Invitational Programming Contest and 9th Wuhan University Programming Contest Preliminary

 

代码如下：

#include <iostream>#include <cstdio>using namespace std;int cases = 1;int main(){    int a , b , c,t;    cin >> t;    while(t --)    {        cin >>a >> b >> c;        int sum = a+b+c;        c = max(max(a , b), c);        a = min(min(a , b), c);        b = sum - a - c ;        if(a*a + b*b > c*c) printf("Case %d: Acute triangle\n",cases ++);        else if(a*a+b*b==c*c) printf("Case %d: Right triangle\n",cases ++);        else if(a*a+b*b < c*c)printf("Case %d: Obtuse triangle\n",cases ++);    }    return 0;}

判断一个三角形的形状，先选出两条最小的边，

如果最小的两条边的平方和大于第三边的平方，是锐角三角形。

如果最小的两条边的平方和等于第三边的平方，是直角三角形。

如果最小的两条边的平方和小于第三边的平方，是钝角三角形。

判断给定三边能不能构成三角，选出两条最小的边，

如果两边之和大于第三条边，则是三角形。

a >0 , b >0 , c >0;

a < c , b < c ;

所以a-b <c , b-a < c;

int is_trangle(int a , int b , int c){    int sum = a+b+c;    int t= max(max(a , b), c);    if(sum -t > t) return 1;    return 0;}