Sort 19 Discussions, By:
Please Login in order to post a comment
Python is love, python is life.
"The square root of two is 1.41421356237309504880..., and the digital sum of the first one hundred decimal digits is 475."
The explanation can be clearer, its not first one hundred decimal digits but first one hundred digits. Decimal digits starts after the decimal point. But the answer counts all digits including the one before decimal point.
Also, set precision to higher than P, then truncate, e.g. P = 100, set precision to 110, then truncate the last 10 digits.
Man, this one is tough to get in under the wire.
For C++/Java, it seems to require:
* BigInteger (or optimized custom class for c++)
* Newton's method (faster than digit-by-digit calculation)
* Strong initial guess using double-precision sqrt
* Reducing the number of actual roots computed the long way (the final speed-up 'trick')
Even so, my Java implementation barely comes in under the wire at 3.2 s for test case 9, and C++ implementation is hit-and-miss around 2 s.
I used all those tricks including Newton's method on the inverse square root in C++. Results: case 6 in 0.96 s, case 7 in 1.68 s and case 8 and 9 time-out. Should I just hand in the Python solution or do you have a final suggestion?
optimized sum of digits of google's bigint and passed testcase 9 in 0.67 seconds using C++ (see my other comment)
You are doing right. Just optimize your multiplication and division. I was able to pass all the cases with my custom bigint class.
Finally solved this problem after a lot of efforts. Approach:
With F# (so .NET), this approach solves all test cases but the last one. With java, this exact same approach solves all test cases (test case 9 takes 3 seconds).
I'm disappointed to see that all my efforts to solve this problem in F# have failed. I see that some people managed to solve this problem with C#, so it could be that another approach than mine would be fast enough also in F#.
You only need to compute the square roots of prime numbers, because the square root of a composite number c is the product of the square roots of its prime factors.
E.g. for c=12: sqrt(12) = sqrt(2*2*3) = sqrt(2) * sqrt(2) * sqrt(3)
Actually you don't need a full factorization. Choose any two factors a and b where ab=c: sqrt(12) = sqrt(3) * sqrt(4) (or sqrt(2) * sqrt(12))
You need to be a bit careful, though: a factor can be a perfect square (such as 4), therefore memoize all (!) square roots, even the trivial ones of perfect squares. And even more, you have to increase your precision by a few digits because you'll introduce some rounding errors when multiplying. Adding 15 extra digits is more than enough.
After I added this trick to my code it became about 4x faster for N=100 and P=10000.
I'm coding in python.
The sample test case works perfectly. But, all other test cases fail badly..I'm using precision of 10000 digits..
Can you give more test cases to check my code??
The precision is given as P.
This line from the explanation of the problem also confuses me a lot..
"For the first N natural numbers, find the total of the digital sums of the first P digits for all the irrational square roots x such that x≤N "
Should we perform our calculations only for the natural number given as input, or should we take it as an indicator for 'first N natural numbers?' @shashank21j
Only for a natural number N
@shashank21j We should calculate for all the number below N right?
"Only for a natural number N"
"Yes. We should calculate for all the numbers below N"
Both these statements are confusing me..!
@shashank21j Should we round the decimal to P digits scale with HALF_UP method or else take the first 100 digits without rounding?
first 100 digits without rounding.
What do constraints 1 and 2 mean?
Does it means 1<= x <= N?
So for N=2, x will be 1 and 2
when x=1, don't add to sum because it is not irrational number
It means either/or. If P is over 1000, N will be limited to 100. Otherwise, the limit for both is 1000.
hey guys, firstly python is magic!, secondly i wrote my python solution but i get the first simple case right and the rest is worng :D, any help.
from decimal import *
Sum = 0
N = int(raw_input())
P = int(raw_input())
for i in range(1, N+1):
if (math.sqrt(i) - int(math.sqrt(i))):
getcontext().prec = P
All the testcases are failing but is code is correct any one please help me.
Here is my submission
You code isn't correct if it fails the testcases. Verify your answers for different values :)
can this question be done in c++?