Hi Luke,

​
> ​
> I disagree with this diagnosis. I think it faults with the large data set
> and gets the answer right if you make your array big enough.​


​You're right. My bad. I made a mistake while testing with both kinds of
input. The code indeed seg faults with only the large data set. ​

Thank you for clearing things up and apologies for any confusion.

Vinit

- - - - -
Check input and
​accepted ​
output for over
*​6​,​00​0* problems on uDebug <http://www.udebug.com/>!

​Now including problems from* ACM-ICPC Live Archive* and *Google Code Jam*!​


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On Sun, Sep 6, 2015 at 2:26 PM, Luke Pebody <[email protected]> wrote:

> ​​
> I disagree with this diagnosis. I think it faults with the large data set
> and gets the answer right if you make your array big enough.
>
> A quick note: the 'break' breaks you out of the 'for (k = j + 1; k < p;
> k++)' loop, but not the 'for (j = 0; j < p - 1; j++)' loop.
>
>
>
> On Sat, Sep 5, 2015 at 7:47 AM, uDebug <[email protected]> wrote:
>
>> Hi,
>>
>> Your code seg faults with both kinds of input.
>>
>> Thanks,
>>
>> Vinit
>>
>> - - - - -
>> Check input and
>> ​accepted ​
>> output for over
>> ​6​
>> *,​00​0* problems on uDebug <http://www.udebug.com/>!
>>
>> ​Now including problems from* ACM-ICPC Live Archive* and *Google Code
>> Jam*!​
>>
>>
>> Find us on Facebook
>> <https://www.facebook.com/pages/Udebug/1458086424467115>. Follow us on
>> Twitter <https://twitter.com/uDebug>.
>>
>> On Sat, Sep 5, 2015 at 12:04 PM, yatingarg12 <[email protected]>
>> wrote:
>>
>>> # include <iostream>
>>> using namespace std;
>>> int main()
>>> {int n,c,p,a[100],x,y;
>>> cin>>n;
>>> for(int i=0;i<n;i++)
>>> {cin>>c>>p;
>>> for(int j=0;j<p;j++)
>>> {cin>>a[j];}
>>> for(int j=0;j<p-1;j++)
>>> for(int k=j+1;k<p;k++)
>>> {if(a[j]+a[k]==c){x=j+1;y=k+1; break;}}
>>> cout<<"Case #"<<i+1<<": "<<x<<" "<<y<<endl;}
>>> return 0;
>>> }
>>>
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