Question is to find the number of coins received on Nth day and not the
total number of coins received after N iterations.

On Wed, Feb 16, 2011 at 11:48 PM, Praveen <[email protected]> wrote:

> Hi All
>
>    total number of coins = 1+(2+2)+(3+3+3)+(4+4+4+4)+........(N+N+.....N
> times)
>                                   = 1+(2*2)+(3*3)+4*4+.......+(N*N)
>                                   = (N*(N+1)*(2N+1))/6
>
>
>     Please do correct me if i am wrong
>
> Regards
> Praveen
>
> On Thu, Feb 17, 2011 at 6:26 AM, Rel Guzman Apaza <[email protected]>wrote:
>
>> I did it.
>>
>> #include <iostream>
>> #include <cmath>
>> using namespace std;
>>
>> int main(){
>>     int n,ac,k,sum;
>>     while(cin>>n && n){
>>         ac=0;
>>         k=ceil((sqrt(1+8*n)-1)/2)-1;
>>         ac+=k*(k+1)*(2*k+1)/6;
>>         sum=(k+1)*(k+2)/2;
>>         ac+=(k+1)*((k+1)-(sum-n));
>>         cout<<n<<" "<<ac<<endl;
>>     }
>> }
>>
>> 2011/2/16 nphard nphard <[email protected]>
>>
>> Let f(n) = n(n+1)/2
>>> We have to find n1 and n2 such that f(n1) < N <= f(n2) and n2 = n1 + 1.
>>> Solution is n2.
>>>
>>> Can be done in O(1) as follows:
>>>
>>> Solve N = n(n+1)/2 for unknown n.
>>> Requires us to solve quadratic equation: n^2 + n - 2N = 0
>>> Find positive root of the equation which could be a real number. n2 =
>>> ceil(n).
>>>
>>>  On Wed, Feb 16, 2011 at 5:14 PM, Pedro Rezende <[email protected]>wrote:
>>>
>>>> It seems to be a very easy problem, but I'm not finding an *equation *that
>>>> solves it... could someone help me with the steps?
>>>>
>>>> Brief:
>>>> A king pays 1 gold coin to a knight on the first day. 2 gold coins for
>>>> the next 2 days, 3 gold coins for the next 3 days, and so on...
>>>> Given a day N, how much gold coins the knight must receive?
>>>>
>>>> Link:
>>>> http://acm.uva.es/archive/nuevoportal/data/problem.php?p=3045
>>>>
>>>> Thank you all! :-)
>>>>
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>>>
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>
>
>
> --
> B. Praveen
>
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