I'm wondering if there is a trick with number theory to compute this remainder without need to implement a BigInt division algorithm.
Haha, it's easy!
I can iterate over all digits, adding each parcel...
Using the properties:
1) (a+b) mod c = (a mod c + b mod c) mod c
2) (a*b) mod c = (a mod c * b mod c) mod c
The power of ten can be increased mod 1500 each step.
Its simple, just check these three things:
Divisibility by 1500
it has to be divisible by 100 (last two digits must be 00)
it has to be divisible by 5 (third digit from right has to be 0 or 5)
it has to be divisible by 3 (iterate over all digits, sum them, and the result has to be divisible by 3)
And if you want to know the remainder, its again simple:
Check for divisible by 5 and get remainder
get remainder from last 4 digits after division by 500, it will be from 0 to 499.
Check for divisible by 3 and get remainder
iterate over all digits, sum them, and get remainder from that after division by 3, it will be from 0 to 2.
and depending from this remainder increase the remainder from 1st step by this remainder multiplied by 500.
Example 1
1234567890 % 1500 = 390
7890 % 500 = 390
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 = 45 and 45 % 3 = 0, so nothing has to be added to 390 and the result is then 390.
Example 2
12345678901 % 1500 = 901
8901 % 500 = 401
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 + 1 = 46 and 46 % 3 = 1, so we have to add 1 * 500 to the result from 1st step, so 401 + 1 * 500 = 901.
Example 3
1357913579 % 1500 = 1079
3579 % 500 = 79
1 + 3 + 5 + 7 + 9 + 1 + 3 + 5 + 7 + 9 = 50 and 50 % 3 = 2, so we have to add 2 * 500 to the result from 1st step, so 79 + 2 * 500 = 1079.
Hope this helps you.
Related
I need help with the following problem.
Given an integer m, I need to find the number of positive integers n and the integers, such that the factorial of n ends with exactly m zeroes.
I wrote this code it works fine and i get the right output, but it take way too much time as the numbers increase.
a = input()
while a:
x = []
m, n, fact, c, j = input(), 0, 1, 0, 0
z = 10*m
t = 10**m
while z - 1:
fact = 1
n = n + 1
for i in range(1, n + 1):
fact = fact * i
if fact % t == 0 and ((fact / t) % 10) != 0:
x.append(int(n))
c = c + 1
z = z - 1
for p in range(c):
print x[p],
a -= 1
print c
Could someone suggest me a more efficient way to do this. Presently, it takes 30 seconds for a test case asking for numbers with 250 trailing zeros in its factorial.
Thanks
To get number of trailing zeroes of n! efficiently you can put
def zeroes(value):
result = 0;
d = 5;
while (d <= value):
result += value // d; # integer division
d *= 5;
return result;
...
# 305: 1234! has exactly 305 trailing zeroes
print zeroes(1234)
In order to solve the problem (what numbers have n trailing zeroes in n!) you can use these facts:
number of zeroes is a monotonous function: f(x + a) >= f(x) if a >= 0.
if f(x) = y then x <= y * 5 (we count only 5 factors).
if f(x) = y then x >= y * 4 (let me leave this for you to prove)
Then implement binary search (on monotonous function).
E.g. in case of 250 zeroes we have the initial range to test [4*250..5*250] == [1000..1250]. Binary search narrows the range down into [1005..1009].
1005, 1006, 1007, 1008, 1009 are all numbers such that they have exactly 250 trainling zeroes in factorial
Edit I hope I don't spoil the fun if I (after 2 years) prove the last conjecture (see comments below):
Each 5**n within facrtorial when multiplied by 2**n produces 10**n and thus n zeroes; that's why f(x) is
f(x) = [x / 5] + [x / 25] + [x / 125] + ... + [x / 5**n] + ...
where [...] stands for floor or integer part (e.g. [3.1415926] == 3). Let's perform easy manipulations:
f(x) = [x / 5] + [x / 25] + [x / 125] + ... + [x / 5**n] + ... <= # removing [...]
x / 5 + x / 25 + x / 125 + ... + x / 5**n + ... =
x * (1/5 + 1/25 + 1/125 + ... + 1/5**n + ...) =
x * (1/5 * 1/(1 - 1/5)) =
x * 1/5 * 5/4 =
x / 4
So far so good
f(x) <= x / 4
Or if y = f(x) then x >= 4 * y Q.E.D.
Focus on the number of 2s and 5s that makes up a number. e.g. 150 is made up of 2*3*5*5, there 1 pair of 2&5 so there's one trailing zero. Each time you increase the tested number, try figuring out how much 2 and 5s are in the number. From that, adding up previous results you can easily know how much zeros its factorial contains.
For example, 15!=15*...*5*4*3*2*1, starting from 2:
Number 2s 5s trailing zeros of factorial
2 1 0 0
3 1 0 0
4 2 0 0
5 2 1 1
6 3 1 1
...
10 5 2 2
...
15 7 3 3
..
24 12 6 6
25 12 8 8 <- 25 counts for two 5-s: 25 == 5 * 5 == 5**2
26 13 8 8
..
Refer to Peter de Rivaz's and Dmitry Bychenko's comments, they have got some good advices.
I was recently solving a problem when I encountered this one: APAC Round E Q2
Basically the question asks to find the smallest base (>1) in which if the number (input) is written then the number would only consist of 1s. Like 3 if represented in base 2 would become 1 (consisting of only 1s).
Now, I tried to solve this the brute force way trying out all bases from 2 till the number to find such a base. But the constraints required a more efficient one.
Can anyone provide some help on how to approach this?
Here is one suggestion: A number x that can be represented as all 1s in a base b can be written as x = b^n + b^(n-1) + b^(n-2) + ... + b^1 + 1
If you subtract 1 from this number you end up with a number divisble by b:
b^n + b^(n-1) + b^(n-2) + ... + b^1 which has the representation 111...110. Dividing by b means shifting it right once so the resulting number is now b^(n-1) + b^(n-2) + ... + b^1 or 111...111 with one digit less than before. Now you can repeat the process until you reach 0.
For example 13 which is 111 in base 3:
13 - 1 = 12 --> 110
12 / 3 = 4 --> 11
4 - 1 = 3 --> 10
3 / 3 = 1 --> 1
1 - 1 = 0 --> 0
Done => 13 can be represented as all 1s in base 3
So in order to check if a given number can be written with all 1s in a base b you can check if that number is divisble by b after subtracting 1. If not you can immediately start with the next base.
This is also pretty brute-forcey but it doesn't do any base conversions, only one subtraction, one divisions and one mod operation per iteration.
We can solve this in O( (log2 n)^2 ) complexity by recognizing that the highest power attainable in the sequence would correspond with the smallest base, 2, and using the formula for geometric sum:
1 + r + r^2 + r^3 ... + r^(n-1) = (1 - r^n) / (1 - r)
Renaming the variables, we get:
n = (1 - base^power) / (1 - base)
Now we only need to check power's from (floor(log2 n) + 1) down to 2, and for each given power, use a binary search for the base. For example:
n = 13:
p = floor(log2 13) + 1 = 4:
Binary search for base:
(1 - 13^4) / (1 - 13) = 2380
...
No match for power = 4.
Try power = 3:
(1 - 13^3) / (1 - 13) = 183
(1 - 6^3) / (1 - 6) = 43
(1 - 3^3) / (1 - 3) = 13 # match
For n around 10^18 we may need up to (floor(log2 (10^18)) + 1)^2 = 3600 iterations.
I'd like to create a function where for an arbitrary integer input value (let's say unsigned 32 bit) and a given number of d digits the return value will be a d digit B base number, B being the smallest base that can be used to represent the given input on d digits.
Here is a sample input - output of what I have in mind for 3 digits:
Input Output
0 0 0 0
1 0 0 1
2 0 1 0
3 0 1 1
4 1 0 0
5 1 0 1
6 1 1 0
7 1 1 1
8 0 0 2
9 0 1 2
10 1 0 2
11 1 1 2
12 0 2 0
13 0 2 1
14 1 2 0
15 1 2 1
16 2 0 0
17 2 0 1
18 2 1 0
19 2 1 1
20 0 2 2
21 1 2 2
22 2 0 2
23 2 1 2
24 2 2 0
25 2 2 1
26 2 2 2
27 0 0 3
28 0 1 3
29 1 0 3
30 1 1 3
.. .....
The assignment should be 1:1, for each input value there should be exactly one, unique output value. Think of it as if the function should return the nth value from the list of strangely sorted B base numbers.
Actually this is the only approach I could come up so far with - given an input value, generate all the numbers in the smallest possible B base to represent the input on d digits, then apply a custom sorting to the results ('penalizing' the higher digit values and putting them further back in the sort), and return the nth value from the sorted array. This would work, but is a spectacularly inefficient implementation - I'd like to do this without generating all the numbers up to the input value.
What would be an efficient approach for implementing this function? Any language or pseudocode is fine.
MBo's answer shows how to find the smallest base that will represent an integer number with a given number of digits.
I'm not quite sure about the ordering in your example. My answer is based on a different ordering: Create all possible n-digit numbers up to base b (e.g. all numbers up to 999 for max. base 10 and 3 digits). Sort them according to their maximum digit first. Numbers are sorted normalls within a group with the same maximum digit. This retains the characteristic that all values from 8 to 26 must be base 3, but the internal ordering is different:
8 0 0 2
9 0 1 2
10 0 2 0
11 0 2 1
12 0 2 2
13 1 0 2
14 1 1 2
15 1 2 0
16 1 2 1
17 1 2 2
18 2 0 0
19 2 0 1
20 2 0 2
21 2 1 0
22 2 1 1
23 2 1 2
24 2 2 0
25 2 2 1
26 2 2 2
When your base is two, life is easy: Just generate the appropriate binary number.
For other bases, let's look at the first digit. In the example above, five numbers start with 0, five start with 1 and nine start with 2. When the first digit is 2, the maximum digit is assured to be 2. Therefore, we can combine 2 with a 9 2-digit numbers of base 3.
When the first digit is smaller than the maximum digit in the group, we can combine it with the 9 2-digit numbers of base 3, but we must not use the 4 2-digit numbers that are ambiguous with the 4 2-digit numbers of base 2. That gives us five possibilites for the digits 0 and 1. These possibilities – 02, 12, 20, 21 and 22 – can be described as the unique numbers with two digits according to the same scheme, but with an offset:
4 0 2
5 1 2
6 2 0
7 2 1
8 2 2
That leads to a recursive solution:
for one digit, just return the number itself;
for base two, return the straightforward representation in base 2;
if the first number is the maximum digit for the determined base, combine it with a straighforward representations in that base;
otherwise combine it with a recursively determined representation of the same algorithm with one fewer digit.
Here's an example in Python. The representation is returned as list of numbers, so that you can represent 2^32 − 1 as [307, 1290, 990].
import math
def repres(x, ndigit, base):
"""Straightforward representation of x in given base"""
s = []
while ndigit:
s += [x % base]
x /= base
ndigit -= 1
return s
def encode(x, ndigit):
"""Encode according to min-base, fixed-digit order"""
if ndigit <= 1:
return [x]
base = int(x ** (1.0 / ndigit)) + 1
if base <= 2:
return repres(x, ndigit, 2)
x0 = (base - 1) ** ndigit
nprev = (base - 1) ** (ndigit - 1)
ncurr = base ** (ndigit - 1)
ndiff = ncurr - nprev
area = (x - x0) / ndiff
if area < base - 1:
xx = x0 / (base - 1) + x - x0 - area * ndiff
return [area] + encode(xx, ndigit - 1)
xx0 = x0 + (base - 1) * ndiff
return [base - 1] + repres(x - xx0, ndigit - 1, base)
for x in range(32):
r = encode(x, 3)
print x, r
Assuming that all values are positive, let's make simple math:
d-digit B-based number can hold value N if
Bd > N
so
B > N1/d
So calculate N1/d value, round it up (increment if integer), and you will get the smallest base B.
(note that numerical errors might occur)
Examples:
d=2, N=99 => 9.95 => B=10
d=2, N=100 => 10 => B=11
d=2, N=57 => 7.55 => B=8
d=2, N=33 => 5.74 => B=6
Delphi code
function GetInSmallestBase(N, d: UInt32): string;
const
Digits = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
var
Base, i: Byte;
begin
Base := Ceil(Power(N, 1/d) + 1.0E-12);
if Base > 36 then
Exit('Big number, few digits...');
SetLength(Result, d);
for i := d downto 1 do begin
Result[i] := Digits[1 + N mod Base]; //Delphi string is 1-based
N := N div Base;
end;
Result := Result + Format(' : base [%d]', [Base]);
end;
begin
Memo1.Lines.Add(GetInSmallestBase(99, 2));
Memo1.Lines.Add(GetInSmallestBase(100, 2));
Memo1.Lines.Add(GetInSmallestBase(987, 2));
Memo1.Lines.Add(GetInSmallestBase(1987, 2));
Memo1.Lines.Add(GetInSmallestBase(87654321, 6));
Memo1.Lines.Add(GetInSmallestBase(57, 2));
Memo1.Lines.Add(GetInSmallestBase(33, 2));
99 : base [10]
91 : base [11]
UR : base [32]
Big number, few digits...
H03LL7 : base [22]
71 : base [8]
53 : base [6]
This question already has answers here:
How Does Modulus Divison Work
(19 answers)
Closed 8 years ago.
What does the % symbol mean in Ruby? For example, I use the following code:
puts "Roosters #{100 - 25 * 3 % 4}"
And get the following output:
97
Where the deuce did the 97 come from? I've looked up what the modulo operator is and still have no idea what it does in this simple mathematics example.
Modulo operator.
It does division and returns the remainder. So, in your case, 75 / 4 is 18 with a remainder of 3.
25 * 3 = 75
75 % 4 = 3 (the remainder)
100 - 3 = 97
modulo - divide with remainder
divide, and take the remainder from the integer division.
10 / 3 = 3 (with remainder 1 that we discard with integer division)
10 % 3 = 1 (the part we normally discard is the part we are interested in with mod)
It is also used to create cycles. If we had a sequence of 1 to N, we could mod it by M and produce a cycle. Assume M = 3 again
for n in 0..10
m = n % 3
puts "#{n} mod 3 = #{m}"
end
0 mod 3 = 0
1 mod 3 = 1
2 mod 3 = 2
3 mod 3 = 0
4 mod 3 = 1
5 mod 3 = 2
6 mod 3 = 0
7 mod 3 = 1
8 mod 3 = 2
9 mod 3 = 0
10 mod 3 = 1
Let's say I have a number of base 3, 1211. How could I check this number is divisible by 2 without converting it back to base 10?
Update
The original problem is from TopCoder
The digits 3 and 9 share an interesting property. If you take any multiple of 3 and sum its digits, you get another multiple of 3. For example, 118*3 = 354 and 3+5+4 = 12, which is a multiple of 3. Similarly, if you take any multiple of 9 and sum its digits, you get another multiple of 9. For example, 75*9 = 675 and 6+7+5 = 18, which is a multiple of 9. Call any digit for which this property holds interesting, except for 0 and 1, for which the property holds trivially.
A digit that is interesting in one base is not necessarily interesting in another base. For example, 3 is interesting in base 10 but uninteresting in base 5. Given an int base, your task is to return all the interesting digits for that base in increasing order. To determine whether a particular digit is interesting or not, you need not consider all multiples of the digit. You can be certain that, if the property holds for all multiples of the digit with fewer than four digits, then it also holds for multiples with more digits. For example, in base 10, you would not need to consider any multiples greater than 999.
Notes
- When base is greater than 10, digits may have a numeric value greater than 9. Because integers are displayed in base 10 by default, do not be alarmed when such digits appear on your screen as more than one decimal digit. For example, one of the interesting digits in base 16 is 15.
Constraints
- base is between 3 and 30, inclusive.
This is my solution:
class InterestingDigits {
public:
vector<int> digits( int base ) {
vector<int> temp;
for( int i = 2; i <= base; ++i )
if( base % i == 1 )
temp.push_back( i );
return temp;
}
};
The trick was well explained here : https://math.stackexchange.com/questions/17242/how-does-base-of-a-number-relate-to-modulos-of-its-each-individual-digit
Thanks,
Chan
If your number k is in base three, then you can write it as
k = a0 3^n + a1 3^{n-1} + a2 3^{n-2} + ... + an 3^0
where a0, a1, ..., an are the digits in the base-three representation.
To see if the number is divisible by two, you're interested in whether the number, modulo 2, is equal to zero. Well, k mod 2 is given by
k mod 2 = (a0 3^n + a1 3^{n-1} + a2 3^{n-2} + ... + an 3^0) mod 2
= (a0 3^n) mod 2 + (a1 3^{n-1}) mod 2 + ... + an (3^0) mod 2
= (a0 mod 2) (3^n mod 2) + ... + (an mod 2) (3^0 mod 2)
The trick here is that 3^i = 1 (mod 2), so this expression is
k mod 2 = (a0 mod 2) + (a1 mod 2) + ... + (an mod 2)
In other words, if you sum up the digits of the ternary representation and get that this value is divisible by two, then the number itself must be divisible by two. To make this even cooler, since the only ternary digits are 0, 1, and 2, this is equivalent to asking whether the number of 1s in the ternary representation is even!
More generally, though, if you have a number in base m, then that number is divisible by m - 1 iff the sum of the digits is divisible by m. This is why you can check if a number in base 10 is divisible by 9 by summing the digits and seeing if that value is divisible by nine.
You can always build a finite automaton for any base and any divisor:
Normally to compute the value n of a string of digits in base b
you iterate over the digits and do
n = (n * b) + d
for each digit d.
Now if you are interested in divisibility you do this modulo m instead:
n = ((n * b) + d) % m
Here n can take at most m different values. Take these as states of a finite automaton, and compute the transitions depending on the digit d according to that formula. The accepting state is the one where the remainder is 0.
For your specific case we have
n == 0, d == 0: n = ((0 * 3) + 0) % 2 = 0
n == 0, d == 1: n = ((0 * 3) + 1) % 2 = 1
n == 0, d == 2: n = ((0 * 3) + 2) % 2 = 0
n == 1, d == 0: n = ((1 * 3) + 0) % 2 = 1
n == 1, d == 1: n = ((1 * 3) + 1) % 2 = 0
n == 1, d == 2: n = ((1 * 3) + 2) % 2 = 1
which shows that you can just sum the digits 1 modulo 2 and ignore any digits 0 or 2.
Add all the digits together (or even just count the ones) - if the answer is odd, the number is odd; if it's even, the nmber is even.
How does that work? Each digit from the number contributes 0, 1 or 2 times (1, 3, 9, 27, ...). A 0 or a 2 adds an even number, so no effect on the oddness/evenness (parity) of the number as a whole. A 1 adds one of the powers of 3, which is always odd, and so flips the parity). And we start from 0 (even). So by counting whether the number of flips is odd or even we can tell whether the number itself is.
I'm not sure on what CPU you have a number in base-3, but the normal way to do this is to perform a modulus/remainder operation.
if (n % 2 == 0) {
// divisible by 2, so even
} else {
// odd
}
How to implement the modulus operator is going to depend on how you're storing your base-3 number. The simplest to code will probably be to implement normal pencil-and-paper long division, and get the remainder from that.
0 2 2 0
_______
2 ⟌ 1 2 1 1
0
---
1 2
1 1
-----
1 1
1 1
-----
0 1 <--- remainder = 1 (so odd)
(This works regardless of base, there are "tricks" for base-3 as others have mentioned)
Same as in base 10, for your example:
1. Find the multiple of 2 that's <= 1211, that's 1210 (see below how to achieve it)
2. Substract 1210 from 1211, you get 1
3. 1 is < 10, thus 1211 isn't divisible by 2
how to achieve 1210:
1. starts with 2
2. 2 + 2 = 11
3. 11 + 2 = 20
4. 20 + 2 = 22
5. 22 + 2 = 101
6. 101 + 2 = 110
7. 110 + 2 = 112
8. 112 + 2 = 121
9. 121 + 2 = 200
10. 200 + 2 = 202
... // repeat until you get the biggest number <= 1211
it's basically the same as base 10 it's just the round up happens on 3 instead of 10.