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Re: 11479 - Is this the Easiest Problem?
Posted: Wed Nov 26, 2008 10:55 pm
by Jorge-ThELaW
Re: 11479 - Is this the Easiest Problem?
Posted: Thu Nov 27, 2008 9:05 am
by shamim
Change the data type from long to long long and you will get AC
This is due to large values of sides whose sum wont fit in 32 bit signed integer.
Re: 11479 - Is this the Easiest Problem?
Posted: Thu Jul 09, 2009 12:14 am
by lnr
Can anyone give some special input output?
input:
Code: Select all
20
3 6 -1
5 3 5
6 2 9
1 2 7
0 9 3
6 0 6
2 6 1
8 7 9
2 0 2
3 7 5
9 2 2
8 9 7
3 6 1
2 9 3
1 9 4
7 8 4
5 0 3
6 1 0
6 3 2
0 6 1
output:
Code: Select all
Case 1: Invalid
Case 2: Isosceles
Case 3: Invalid
Case 4: Invalid
Case 5: Invalid
Case 6: Invalid
Case 7: Invalid
Case 8: Scalene
Case 9: Invalid
Case 10: Scalene
Case 11: Invalid
Case 12: Scalene
Case 13: Invalid
Case 14: Invalid
Case 15: Invalid
Case 16: Scalene
Case 17: Invalid
Case 18: Invalid
Case 19: Invalid
Case 20: Invalid
Accepted.
11479 pls pls pls help
Posted: Sat Oct 24, 2009 8:13 pm
by shorojini
#include<iostream>
using namespace std;
int main() {
long long a, b, c, n,i;
cin>>n;
for(i=1;i<=n;i++)
{
cin>>a>>b>>c;
if((a<=0||b<=0||c<=0)||((b+c)<=a)||((a+c)<=b)||((a+b)<=c))
cout<<"Case "<<i<<": "<<"Invalid\n";
else
{
if((a==b==c))
cout<<"Case "<<i<<": "<<"Equilateral\n";
else if((a==b)||(b==c)||(c==a))
cout<<"Case "<<i<<": "<<"Isosceles\n";
else
cout<<"Case "<<i<<": "<<"Scalene\n";
}
}
return 0;
}
pls tell me where is the wrong? why it is wrong ans. pls pls pls
Re: 11479 pls pls pls help
Posted: Sat Oct 24, 2009 8:45 pm
by coze
Try the following input:
Input
Output
is not equal to
Hope it helps.
Re: 11479 pls pls pls help
Posted: Fri Oct 30, 2009 1:03 pm
by shorojini
thanks coz 4 ur great help. i got accepted. i have started acm. and i have no help. will u pls give me ur mail number. so i will be able to share my probs.it is my request.
now , will u just tell me why (a==b==c) is not right. it will be great help.
Re: 11479 - Is this the Easiest Problem?
Posted: Mon Dec 20, 2010 4:48 pm
by wawa
Code: Select all
#include <iostream>
using namespace std;
int main()
{
long long N, a, b, c;
cin >> N;
for (long long i = 0; i < N; i++)
{
cin >> a >> b >> c;
if ((a + b) <= c || (a + c) <= b || (b + c) <= a)
cout << "Case " << i + 1 << ": Invalid" << endl;
else if ((a == b) && (a == c))
cout << "Case " << i + 1 << ": Equilateral" << endl;
else if ((a == b) || (a == c) || (a == b))
cout << "Case " << i + 1 << ": Isosceles" << endl;
else
cout << "Case " << i + 1 << ": Scalene" << endl;
}
return 0;
}
Why is it WA??????
plz help...~!!!
Re: 11479 - Is this the Easiest Problem?
Posted: Wed Dec 22, 2010 9:30 am
by asif_iut
else if ((a == b) || (a == c) || (a == b))
what is the difference between the first and the last condition??
Re: 11479 - Is this the Easiest Problem?
Posted: Sat Feb 05, 2011 4:31 pm
by kissu parina
long data type is enough for this problem.
Re: 11479 - Is this the Easiest Problem?
Posted: Mon Feb 21, 2011 9:02 am
by hotovaga
Where is the problem in my code? getting WA..
help me pls..
Re: 11479 - Is this the Easiest Problem?
Posted: Mon May 30, 2011 1:00 pm
by du_loser
getting wa
Code: Select all
#include<stdio.h>
int main(){
long long int c,tc,a[3],i,j,t;
scanf("%lld",&tc);
for(c=1;c<=tc;c++){
scanf("%lld%lld%lld",&a[0],&a[1],&a[2]);
for(i=0;i<2;i++){
for(j=0;j<2-i;j++){
if(a[j]<a[j+1]){
t=a[j];
a[j]=a[j+1];
a[j+1]=t;
}
}
}
printf("Case %lld: ",c);
if((a[0]<=(a[1]+a[2]))&&a[0]>0&&a[1]>0&&a[2]>0){
if(a[0]==a[1]&&a[1]==a[2])
printf("Equilateral\n");
else if(a[0]==a[1]||a[1]==a[2]||a[2]==a[0])
printf("Isosceles\n");
else
printf("Scalene\n");
}
else printf("Invalid\n");
}
return 0;
}
Re: 11479 - Is this the Easiest Problem?
Posted: Wed Jun 01, 2011 5:32 pm
by helloneo
Try this case..
My output is..
Re: 11479 - Is this the Easiest Problem?
Posted: Wed Jul 27, 2011 11:50 am
by ez0545
can anyone please tell me whats wrong with this code??? I am getting wrong answer every time.
Re: 11479 - Is this the Easiest Problem?
Posted: Thu Dec 08, 2011 4:09 am
by Help Me
What is the problem??? Why it Wrong answer?
#include<stdio.h>
#include<stdlib.h>
int comp(const void *a, const void *b)
{
return (*(long int *)a)-(*(long int *)b);
}
int main()
{
long int t,p,x,y,z,ary[10000],k,temp,max,small,mid,l;
scanf("%ld",&t);
for(p=1;p<=t;p++)
{
scanf("%ld%ld%ld",&x,&y,&z);
ary[1]=x;
ary[2]=y;
ary[3]=z;
qsort(ary,4,sizeof(ary[1]),comp);
small=ary[1];
mid=ary[2];
max=ary[3];
if((small+mid)<=max)
printf("Case %ld: Invalid\n",p);
else if(x==y==z)
printf("Case %ld: Equilateral\n",p);
else if((x==y)||(x==z)||(y==z))
printf("Case %ld: Isosceles\n",p);
else
printf("Case %ld: Scalene\n",p);
}
return 0;
}
Re: 11479 - Is this the Easiest Problem?
Posted: Fri Jan 13, 2012 2:52 am
by brianfry713
You're sorting an array of size 4 that should only be of size 3. Think of what happens with negatives.