Solution of 11479 - Is this the easiest problem

Problem Description
source:https://uva.onlinejudge.org/external/114/11479.html

A triangle is a geometric shape with three positive sides. However, any given three sides won’t necessarily form a triangle. The three sides must form a closed region. Triangles are categorized depending on the values of the sides of a valid triangle. In this problem you are required to determine the type of a triangle. 

Input 

The first line of input will contain a positive integer T < 20, where T denotes the number of test cases. Each of the next T lines will contain three 32 bit signed integer. 

Output 

For each case of input there will be one line of output. It will be formatted as: 
         Case x: triangle type. 
Where x denotes the case number being processed and triangle type is the type of the triangle. triangle type will be one of the following, depending on the values of the three sides: 
    • Invalid - The three sides can not form a triangle 
    • Equilateral - All three sides of valid triangle are equal 
    • Isosceles - Exactly two of the sides of a valid triangle are equal. 
    • Scalene - No pair of sides are equal in a valid triangle. 

Sample Input 

4 
1 2 5 
1 1 1 
4 4 2 
3 4 5 

Sample Output 

Case 1: Invalid 
Case 2: Equilateral 
Case 3: Isosceles 
Case 4: Scalene

Solution:

#include <algorithm>
#include <cstdio>
#include<stdio.h>
#include<stdlib.h>

using namespace std;

int main()
{
    long long int a,b,c,tem,i,t;
    while(scanf("%lld",&t)==1)
    {
        for(i=1;i<=t;i++)
        {
            scanf("%lld%lld%lld",&a,&b,&c);
            ((a>b)?((a>c)?(tem=a,a=c,c=tem):(tem=c)):((b>c)?(tem=b,b=c,c=tem):(tem=c)));
            if((a+b)>c)
            {
                if(a==b&&b==c)
                    printf("Case %lld: Equilateral\n",i);
                else if(a==b ||a==c || b==c)
                    printf("Case %lld: Isosceles\n",i);
                else
                    printf("Case %lld: Scalene\n",i);
            }
            else
                printf("Case %lld: Invalid\n",i);

        }
    }
    return 0;
}