sorry, dont bother, I got it.

On Fri, Mar 11, 2011 at 5:39 PM, Harshal <[email protected]> wrote:

> any test case where this fails by using brute force approach? like this one
> if c[l]<=c[r] then take c[l]*a and l++; else choose c[r]*a and r--; this is
> o(n) solution.
> any test case where it fails??
>
> On Thu, Mar 10, 2011 at 11:40 AM, Amol Sharma <[email protected]>wrote:
>
>> Bhaukali Launde !!
>>
>> Maa ka ladla bigad gya !!
>>
>>
>> On Thu, Mar 10, 2011 at 11:36 AM, UTKARSH SRIVASTAV <
>> [email protected]> wrote:
>>
>>> thnx ashutosh ..................tum i love u
>>>
>>>
>>> On Thu, Mar 10, 2011 at 11:11 AM, AnAnDiT AsHu <[email protected]>wrote:
>>>
>>>> launda pelu nikla.....
>>>>
>>>> On Thu, Mar 10, 2011 at 10:22 AM, UTKARSH SRIVASTAV <
>>>> [email protected]> wrote:
>>>>
>>>>> WELL I HAVE DONE THIS PROBLEM .HERE IS THE CODE
>>>>> #include<stdio.h>
>>>>> #include<algorithm>
>>>>> using namespace std;
>>>>> main()
>>>>> {
>>>>>     long long int t[2][2010],price[2010],r,c,i,j,n;
>>>>>     scanf("%lld",&n);
>>>>>     for(i=0;i<n;i++)
>>>>>     {
>>>>>         scanf("%lld",&price[i]);
>>>>>     }
>>>>>     for(r=n-1,c=0;r>=0&&c<=n-1;r--,c++)
>>>>>     {
>>>>>         for(i=r,j=n-1;i>=0&&j>=c;j--,i--)
>>>>>
>>>>> t[i&1][j]=max(price[i]*(n+i-j)+t[(i+1)&1][j],price[j]*(n+i-j)+t[i&1][j-1]);
>>>>>     }
>>>>>     printf("%lld\n",t[0][n-1]);
>>>>>     return 0;
>>>>>
>>>>> }
>>>>>
>>>>>
>>>>>
>>>>> On Wed, Mar 9, 2011 at 5:10 PM, Algoose chase <[email protected]>wrote:
>>>>>
>>>>>> Hi,
>>>>>>
>>>>>> Any solution other than brute force(exponential growth) for this
>>>>>> problem ?
>>>>>>
>>>>>>
>>>>>> On Sun, Mar 6, 2011 at 6:42 PM, UTKARSH SRIVASTAV <
>>>>>> [email protected]> wrote:
>>>>>>
>>>>>>> can anyone please tell me why i am getting wrong answer for
>>>>>>> problem.....https://www.spoj.pl/problems/TRT/
>>>>>>> .
>>>>>>> .
>>>>>>> .
>>>>>>> MY CODE IS THIS AND TO BE TESTED IN gcc COMPILER
>>>>>>>
>>>>>>>
>>>>>>> #include<stdio.h>
>>>>>>> double a[2100];
>>>>>>> double fun(long long int  m,long long int n,double count)
>>>>>>> {
>>>>>>>        double k,l;
>>>>>>>        count++;
>>>>>>>        if(m==n)
>>>>>>>        {
>>>>>>>
>>>>>>>                return count*a[m];
>>>>>>>        }
>>>>>>>        if((k=(fun(m+1,n,count)))>(l=(fun(m,n-1,count))))
>>>>>>>        {
>>>>>>>
>>>>>>>                return (count*a[m]+k);
>>>>>>>        }
>>>>>>>        else
>>>>>>>        {
>>>>>>>
>>>>>>>
>>>>>>>                return (count*a[n]+l);
>>>>>>>        }
>>>>>>> }
>>>>>>> int main()
>>>>>>> {
>>>>>>>        long long int i,m,n;
>>>>>>>        double ans,c=0;
>>>>>>>        scanf("%lld",&n);
>>>>>>>        for(i=1;i<=n;i++)
>>>>>>>        {
>>>>>>>                scanf("%lf",&a[i]);
>>>>>>>        }
>>>>>>>        m=1;
>>>>>>>        ans=fun(m,n,c);
>>>>>>>        printf("%.0lf\n",ans);
>>>>>>>        return 0;
>>>>>>> }
>>>>>>>
>>>>>>> --
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>>>>>>>
>>>>>>>
>>>>>>  --
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>>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> UTKARSH SRIVATAV
>>>>> CSE-3
>>>>> B-TECH 2nd YEAR
>>>>> MNNIT ALLAHABAD
>>>>>
>>>>>  --
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>>>>>
>>>>
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>>>
>>>
>>>
>>> --
>>> UTKARSH SRIVATAV
>>> CSE-3
>>> B-TECH 2nd YEAR
>>> MNNIT ALLAHABAD
>>>
>>>  --
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>>
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>
>
>
> --
> Harshal Choudhary,
> III Year B.Tech Undergraduate,
> Computer Science and Engineering,
> National Institute of Technology Surathkal, Karnataka
> India.
>
>


-- 
Harshal Choudhary,
III Year B.Tech Undergraduate,
Computer Science and Engineering,
National Institute of Technology Surathkal, Karnataka
India.

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