UVA 11479 – Is this the easiest problem?

Accepted code:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int testcases,cases;
    long a,b,c;

    scanf("%d",&testcases);
    if(testcases>0 && testcases<20)
    {
        for(cases=1; cases<=testcases; cases++)
        {
            scanf("%ld %ld %ld",&a,&b,&c);
            if(a+b>c && b+c>a && c+a>b)
            {
                if(a==b && b==c)
                {
                    printf("Case %d: Equilateral\n",cases);
                }
                else if(a==b||b==c||a==c)
                {
                    printf("Case %d: Isosceles\n",cases);
                }
                else if(a!=b||b!=c||c!=a)
                {
                    printf("Case %d: Scalene\n",cases);
                }
            }
            else
            {
                printf("Case %d: Invalid\n",cases);
            }
        }

    }
    return 0;
}

I have tested with some critical input after implementing the first logic and also read the condition for testcases .It is very important to make it accepted.

Critical input:
4
100 200 300
500 501 502
1000000000 1000000001 1000000002
963 852 741

Critical output:
Case 1: Invalid
Case 2: Scalene
Case 3: Scalene
Case 4: Scalene