## nagabhushana.s

BAN USERIf 10units diameter sphere can fit in 10x10x10units cube............................

100x100x100units cube can fit in 10x10x10 number of 10*10*10 cubes..........

so, it is 10x10x10 number of spheres.

3!

- nagabhushana.s July 02, 2010ha ha ....6 = inverted 9 is the idea ...great guys!

- nagabhushana.s July 02, 2010sab desi hai bhai...

- nagabhushana.s July 02, 2010Since key-value pairs are not interrelated, we cant use topological sort. Linked list looks as only the option.

- nagabhushana.s July 02, 2010great!

- nagabhushana.s July 01, 2010Sexy dude.

- nagabhushana.s July 01, 2010Explanation:-

To understand this we have to compare 2 clocks.

At the end of day1, suppose Normal clock shows 12AM. Screwed up clock will be 11 PM 59Min 58Sec. Lost 2 days.

One Interesting fact is that Screwed up clock will reach 12AM only after 2 secs. :)

At the end of Day2, as usual screwed up clock will lose 2 more seconds.

Means, 2+2(make up for day1's lost seconds) +2 = 6secs!

sexy dude!

- nagabhushana.s July 01, 2010Great, patron!

- nagabhushana.s July 01, 2010Its not a clear answer problem....Its answer will be algorithmic complexity terms...I feel 'netappreject' is correct!

- nagabhushana.s July 01, 20101. Divide into groups of 27, 27 and 26

2. Weight 27 against 27, its either group with 27 or 26. For Ex, lets take 27

3. Divide into 9, 9 and 9

4. Weigh 9 against 9, we know group 9 balls which are heavier

5. Now divide them into 3,3,3

6. Similarly we will end up with group of 3

7. Divide into 1, 1, 1

7. Now compare 1 and 1...U know the answer

4 Steps! he he

for whole number case , it is (3,4)...great guys...I am enjoying

- nagabhushana.s July 01, 2010Guys....its either (x,y)=(0,0) or (x,y)=(2,2)...........

CodeMonkey, u r great!

My friend came up with a better solution. Its by defining intial condition. Both robots will start from each end of the line facing away from each other and in a perpendicular direction to the line. EX: Suppose line is drawn from north to south. Robot1 will be in north end of the line facing towards east whereas Robot2 is at the other end facing West. With this simple algorithm will be...

while(!AmIpoint())

moveright();

Next 24 hours is actually cake-walk!!!

- nagabhushana.s July 01, 2010Next 24 hours is actually cake-walk!!!

- nagabhushana.s July 01, 2010Bhavik's solution is correct but its best understood if its explained starting from all 5 slaves sharing 1/5 of same barrel.

1.Lets share 20% barrel1 to S1,S2,S3,S4,S5. [This is to achieve this:If all of them die at the end of 24th hour, its barrel1 which is poisoned]

2.Lets share (barrel2+barrel3) equally among S1,S2,S3,S4. If we continue like this, we have spent 2*10=20 barrels.

3.Lets share (barrel12+barrel13+barrel14+barrel15) equally among S1,S2,S3. If we continue like this, we have spent 4*10=40barrels.

4.Lets share (barrel62+....+barrel68) equally among S1,S2. If we continue like this, we have spent 8*10=80barrels.

5.Lets share (barrel143+...+barrel159) to S1. If we continue like this, we will have spent 16*5=80 barrels.

That leaves us with 240-1-20-40-80-80 = 29barrels for next 24 hours.

If all the slaves die, Barrel1 is poisonous.

If (S1+S2+S3+S4) die, then (Barrel2+barrel3) is poisonous. With S5, we can easily find out whether Barrel2 or Barrel3 is poisonous!

Like this, we can move ahead!

1. Group 9 balls into sets of 3,3 and 3

2. Compare group1 and group2, if both are equal, go to step 4(proceed with group3)

3. Else, Comapare group1 with group3, if they are equal proceed with group2, else with group1

4. Divide balls into 1, 1 and 1

5. Compare ball1 with ball2 once, then with ball3 once.

6. We know the lightest of heaviest ball by now!

Guatam is correct!

- nagabhushana.s July 01, 2010No. Here are all answers for all possible cases.

(Note : I have taken 'r' as side length of the cube.

Interior(If you cut through): srqoot(3)*r

Exterior(Along Edges):3*r

Exterior(shortest along the surface):sqroot(5)*r

Heretic and gouti are correct!

sorry...shoonya mohit has already done that. great job, man!

- nagabhushana.s June 30, 2010I think this is the best algorithm....

1.Sort ascendancy, remove duplicates also in the process

2.Remove numbers which are >= k

3.Do steps 4 through 8 till i != j

4.Start with i = 0

5.Start with j = size - 1

6.If the sum is == k, display a[i]+a[j] = k, i++, j--

7.If the sum is < k, i++, go to step 5

8.If the sum is > k, j--, go to step 5

Can somebody calculate the complexity for this please? I don't know how to do...

Lets take case of one robot. Here is the algorithm

1. Check whether current point is the desired point by using AmIPoint()

2. If yes, program ends, Robot found the point, Exit. Else, go to step 3.

3. MoveLeft()

4. MoveRight()

5. MoveRight(). Robot is now on the next point on the line. Go to step1.

Repeat the above steps for Robot 2.

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488! sexy!

- nagabhushana.s July 03, 2010