11479 - Is this the easiest problem?

All about problems in Volume 114. If there is a thread about your problem, please use it. If not, create one with its number in the subject.

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lighted
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Location: Kyrgyzstan, Bishkek

Re: 11479 - Is this the easiest problem?

Post by lighted » Mon Sep 15, 2014 3:20 pm

Try input in this thread first.
lnr wrote:Can anyone give some special input output?
input:

Code: Select all

1
9 2 2
output:

Code: Select all

Case 1: Invalid
Accepted.
sohel wrote:A triangle is invalid if any side_length <= 0 or (sum of smaller 2 sides <= largest side).
A triangle is scalene if it is valid && all the sides are distinct.
renatov wrote:To check if the sides form a triangle, you should performe 2 tests:

1. See if every input is greater than zero
2. | b - c | < a < b + c
I got accepted and I didn't check if input is negative. Maybe judge doesn't contain negative input
A person who sees the good in things has good thoughts. And he who has good thoughts receives pleasure from life... Bediuzzaman

LanceHAOH
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Posts: 9
Joined: Tue Aug 27, 2013 9:04 am

11479 - WA

Post by LanceHAOH » Sun Jan 11, 2015 11:05 am

I have researched this forums for possible hints to why my code is wrong. But I cannot find any. Could someone please help me?

Code: Select all

class Main {

    public static void main(String[] args) {
        java.util.Scanner in = new java.util.Scanner(System.in);
        long[] sides = new long[3];
        long cases = in.nextLong();
        String triangleType;

        for (int i = 1; i <= cases; i++) {

            sides[0] = in.nextLong();
            sides[1] = in.nextLong();
            sides[2] = in.nextLong();

            triangleType = "Invalid";

            /* Ensure sum of any two sides > third side */
            if (sides[0] > 0 && sides[1] > 0 && sides[2] > 0) {
                if (sides[0] + sides[1] > sides[2] && sides[1] + sides[2] > sides[0]
                        && sides[0] + sides[2] > sides[1]) {
                    if (sides[0] == sides[1] && sides[1] == sides[2]) {
                        /* Check for equilateral triangle */
                        triangleType = "Equilateral";
                    } else if (sides[0] != sides[1] && sides[0] != sides[2] && sides[1] != sides[2]) {
                        /* Check for scalene triangle */
                        triangleType = "Scalene";
                    } else {
                        /* Check for isoceles triangle */
                        triangleType = "Isosceles";
                    }
                }
                System.out.println("Case " + i + ": " + triangleType);
            }
        }
    }
}
Last edited by brianfry713 on Tue Jan 13, 2015 12:21 am, edited 1 time in total.
Reason: Added code blocks

brianfry713
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Posts: 5947
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Re: 11479 - Is this the easiest problem?

Post by brianfry713 » Tue Jan 13, 2015 12:21 am

Next time post in the existing thread and use code blocks.
It looks like you figured it out.
Check input and AC output for thousands of problems on uDebug!

Sabrina Haque
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Posts: 3
Joined: Fri Jan 23, 2015 9:33 am

Re: 11479 - Is this the easiest problem?

Post by Sabrina Haque » Sat Jan 24, 2015 10:24 am

got WA

Code: Select all

#include<stdio.h>

int main()
{
    int test_case,i;
    long long int a,b,c;

    scanf("%d",&test_case);

    for(i=1;i<=test_case;i++)
    {
        scanf("%llu %llu %llu",&a,&b,&c);

        if(a>=b+c || b>=a+c || c>=a+b || a<=0 || b<=0 || c<=0)
            printf("case %d: Invalid\n",i);
        else if(a==b && b==c)
            printf("case %d: Equilateral\n",i);
        else if(a==b || a==c || b==c || c==a)
            printf("case %d: Isosceles\n",i);
        else if(a!=b && b!=c)
             printf("case %d: Scalene\n",i);
    }

    return 0;
}
Last edited by brianfry713 on Tue Jan 27, 2015 1:03 am, edited 1 time in total.
Reason: Added code blocks

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