silent__killer1's blog

By silent__killer1, history, 6 weeks ago, In English,

Hello i am just trying to solve this series 1/12 + 11/12^2 + (11*11)/12^3+...... first i need to tell you what my approach to solve which prove me wrong what i got is this is a gp so sum of finite term of gp is :Sn=a(1-r^n)/(1-r) for this series we know 0<r<1 as r comes out to be r=11/12 so consideration of this series — values come out to be a=1/12 , r=11/12 Sn=((1/12)(1-(11/12)^n)) / (1-11/12)

to compute 11^n and 12^n it take so much time with brute force so i optimize and take time log(n) now after some solving what i get is Sn=(12^n-11^n)/12^n. now n could be very large so 12^n or 11^n could be very large and we out of bounds so we need to modulo him to to get 12^n and 11^n with in the range .

lets take an example where i take n=36 my fastpower function which take less time


ll power(ll x,ll y){

ll te=1;

if(y==0)return 1;


if(y%2==0)return (te*te)%mod;

return (te*te*x)%mod;

} int main(){

long long a=(power(12,36) - power(11,36)) / power(12,36);

} i need to find a as this is sum of finite series bu it comes out to be negative....

when i solve this i get 12^36 <11^36 and gives me finite sum of this series negative ans

Sn=(12^n-11^n)/12^n as 12^n<11^n Sn =-ve i have got i really didnt know how to solve after that i just need this series in p/q form and i need to take multiplicaive inverse of q which is later thing but first i need to solve this series and get the p/q form

someone help me out..... thankyou....

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By silent__killer1, history, 6 weeks ago, In English,

You are given an array A of n integers.The value of this array is defined as the sum of absolute value of differnce of consecutive elements in the array. Example |a2-a1|+|a3-a2|+.....|an-an-1|.you are allowed to select any subarray and reverse it.You can perform this operation only once.find the max value of the array achievable.

n=no of elements in an array. ex- n=5

2 3 1 5 4 ans-10


2 4 9 24 2 1 10 ans-68

question link--

i really didnt get how to do that please anyone ...

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