CareerCup

Following will be my strategy,
a) if difference between N and current score is greater than 8. I will choose 4
b) if difference between N and current score is less than 9 but greater than 4. I will (choose N - current_Score -5 )
c) if difference between N and current score is less than 5, I will choose the number it self.

If this strategy is used by both then person from whom we started will win.

I am not sure abt this .
keep choosing four and choose n-sum for the final value to be selected ?
Thats the fasted route to reach n.

actually unable to get the question completely.
be specific and try to provide complete data required.
Question above is incomplete...

The trick is to see that who ever gets to N-5 wins. So the strategy is reduces to getting to N-5. But then whoever gets to N-10 wins, etc. It, then is easy to see that whoever can get to N-5k where k is the largest number s.t. N-5k is positive, will win. That could be either player depending on the number.

By choosing 4 continuously and then 1 no. which is left
we can reach N at no. of step N/4 ( if N%4==0) otherwise N%4+1. Now, i don't think for any N it's possible to get lesser step then above. So this strategy is best.

the trick is to define the loosing and winning positions from 0 to N.
for example take N = 10;
and follow the algo:
0,1,2,3,4 are winning positions [base case]
for every i
if we can all winning positions from i-{1,2,3,4} then
i is a loosing positions
else
i is a winning position.

the strategy says that to be in a winning situation you must put the opponent in a
position where the remaining points to reach the score value N is divisible by 5(considering that we are allowed to choose only 1,2,3,4)and he/she has to choose.
Suppose, the remaining points to reach the score is x.

Then, if (x%5)=4 on your turn , choose 4
if (x%5)=3 on your turn , choose 3
if (x%5)=2 on your turn , choose 2
if (x%5)=1 on your turn , choose 1
if (x%5)=0 on your turn , you may not win if the opponent is smart enough
and follows the above strategy

u see all the coins and their positions

find sum of all coins at even pos and sum all coin at odd pos

which ever is greater. make sure that u pick that index coin when your chance is there, u can either win or hv a tie

if the question says the no one player will chose can not be chosen by other immediately for e.g. if player A has chosen 4 then player B cant choose 4 right at this moment then the no to be choose is = no/4;no = no - no/4;

else the question is trivial...the first person will win always..

let say A chooses number x
So, b should always choose a number 5- x so that B wins.

Not sure, how will A win if he itself is starting the series and b is already aware of strategy.