I just saw this question and have no idea how to solve it. can you please provide me with algorithms , C++ codes or ideas?
This is a very simple problem. Given the value of N and K, you need to tell us the value of the binomial coefficient C(N,K). You may rest assured that K <= N and the maximum value of N is 1,000,000,000,000,000. Since the value may be very large, you need to compute the result modulo 1009.
Input
The first line of the input contains the number of test cases T, at most 1000. Each of the next T lines consists of two space separated integers N and K, where 0 <= K <= N and 1 <= N <= 1,000,000,000,000,000.
Output
For each test case, print on a new line, the value of the binomial coefficient C(N,K) modulo 1009.
Example
Input:
3
3 1
5 2
10 3
Output:
3
10
120
Notice that 1009 is a prime.
Now you can use Lucas' Theorem.
Which states:
Let p be a prime.
If n = a1a2...ar when written in base p and
if k = b1b2...br when written in base p
(pad with zeroes if required)
Then
(n choose k) modulo p = (a1 choose b1) * (a2 choose b2) * ... * (ar choose br) modulo p.
i.e. remainder of n choose k when divided by p is same as the remainder of
the product (a1 choose b1) * .... * (ar choose br) when divided by p.
Note: if bi > ai then ai choose bi is 0.
Thus your problem is reduced to finding the product modulo 1009 of at most log N/log 1009 numbers (number of digits of N in base 1009) of the form a choose b where a <= 1009 and b <= 1009.
This should make it easier even when N is close to 10^15.
Note:
For N=10^15, N choose N/2 is more than
2^(100000000000000) which is way
beyond an unsigned long long.
Also, the algorithm suggested by
Lucas' theorem is O(log N) which is
exponentially faster than trying to
compute the binomial coefficient
directly (even if you did a mod 1009
to take care of the overflow issue).
Here is some code for Binomial I had written long back, all you need to do is to modify it to do the operations modulo 1009 (there might be bugs and not necessarily recommended coding style):
class Binomial
{
public:
Binomial(int Max)
{
max = Max+1;
table = new unsigned int * [max]();
for (int i=0; i < max; i++)
{
table[i] = new unsigned int[max]();
for (int j = 0; j < max; j++)
{
table[i][j] = 0;
}
}
}
~Binomial()
{
for (int i =0; i < max; i++)
{
delete table[i];
}
delete table;
}
unsigned int Choose(unsigned int n, unsigned int k);
private:
bool Contains(unsigned int n, unsigned int k);
int max;
unsigned int **table;
};
unsigned int Binomial::Choose(unsigned int n, unsigned int k)
{
if (n < k) return 0;
if (k == 0 || n==1 ) return 1;
if (n==2 && k==1) return 2;
if (n==2 && k==2) return 1;
if (n==k) return 1;
if (Contains(n,k))
{
return table[n][k];
}
table[n][k] = Choose(n-1,k) + Choose(n-1,k-1);
return table[n][k];
}
bool Binomial::Contains(unsigned int n, unsigned int k)
{
if (table[n][k] == 0)
{
return false;
}
return true;
}
Binomial coefficient is one factorial divided by two others, although the k! term on the bottom cancels in an obvious way.
Observe that if 1009, (including multiples of it), appears more times in the numerator than the denominator, then the answer mod 1009 is 0. It can't appear more times in the denominator than the numerator (since binomial coefficients are integers), hence the only cases where you have to do anything are when it appears the same number of times in both. Don't forget to count multiples of (1009)^2 as two, and so on.
After that, I think you're just mopping up small cases (meaning small numbers of values to multiply/divide), although I'm not sure without a few tests. On the plus side 1009 is prime, so arithmetic modulo 1009 takes place in a field, which means that after casting out multiples of 1009 from both top and bottom, you can do the rest of the multiplication and division mod 1009 in any order.
Where there are non-small cases left, they will still involve multiplying together long runs of consecutive integers. This can be simplified by knowing 1008! (mod 1009). It's -1 (1008 if you prefer), since 1 ... 1008 are the p-1 non-zero elements of the prime field over p. Therefore they consist of 1, -1, and then (p-3)/2 pairs of multiplicative inverses.
So for example consider the case of C((1009^3), 200).
Imagine that the number of 1009s are equal (don't know if they are, because I haven't coded a formula to find out), so that this is a case requiring work.
On the top we have 201 ... 1008, which we'll have to calculate or look up in a precomputed table, then 1009, then 1010 ... 2017, 2018, 2019 ... 3026, 3027, etc. The ... ranges are all -1, so we just need to know how many such ranges there are.
That leaves 1009, 2018, 3027, which once we've cancelled them with 1009's from the bottom will just be 1, 2, 3, ... 1008, 1010, ..., plus some multiples of 1009^2, which again we'll cancel and leave ourselves with consecutive integers to multiply.
We can do something very similar with the bottom to compute the product mod 1009 of "1 ... 1009^3 - 200 with all the powers of 1009 divided out". That leaves us with a division in a prime field. IIRC that's tricky in principle, but 1009 is a small enough number that we can manage 1000 of them (the upper limit on the number of test cases).
Of course with k=200, there's an enormous overlap which could be cancelled more directly. That's what I meant by small cases and non-small cases: I've treated it like a non-small case, when in fact we could get away with just "brute-forcing" this one, by calculating ((1009^3-199) * ... * 1009^3) / 200!
I don't think you want to calculate C(n,k) and then reduce mod 1009. The biggest one, C(1e15,5e14) will require something like 1e16 bits ~ 1000 terabytes
Moreover executing the loop in snakiles answer 1e15 times seems like it might take a while.
What you might use is, if
n = n0 + n1*p + n2*p^2 ... + nd*p^d
m = m0 + m1*p + m2*p^2 ... + md*p^d
(where 0<=mi,ni < p)
then
C(n,m) = C(n0,m0) * C(n1,m1) *... * C(nd, nd) mod p
see, eg http://www.cecm.sfu.ca/organics/papers/granville/paper/binomial/html/binomial.html
One way would be to use pascal's triangle to build a table of all C(m,n) for 0<=m<=n<=1009.
psudo code for calculating nCk:
result = 1
for i=1 to min{K,N-K}:
result *= N-i+1
result /= i
return result
Time Complexity: O(min{K,N-K})
The loop goes from i=1 to min{K,N-K} instead of from i=1 to K, and that's ok because
C(k,n) = C(k, n-k)
And you can calculate the thing even more efficiently if you use the GammaLn function.
nCk = exp(GammaLn(n+1)-GammaLn(k+1)-GammaLn(n-k+1))
The GammaLn function is the natural logarithm of the Gamma function. I know there's an efficient algorithm to calculate the GammaLn function but that algorithm isn't trivial at all.
The following code shows how to obtain all the binomial coefficients for a given size 'n'. You could easily modify it to stop at a given k in order to determine nCk. It is computationally very efficient, it's simple to code, and works for very large n and k.
binomial_coefficient = 1
output(binomial_coefficient)
col = 0
n = 5
do while col < n
binomial_coefficient = binomial_coefficient * (n + 1 - (col + 1)) / (col + 1)
output(binomial_coefficient)
col = col + 1
loop
The output of binomial coefficients is therefore:
1
1 * (5 + 1 - (0 + 1)) / (0 + 1) = 5
5 * (5 + 1 - (1 + 1)) / (1 + 1) = 15
15 * (5 + 1 - (2 + 1)) / (2 + 1) = 15
15 * (5 + 1 - (3 + 1)) / (3 + 1) = 5
5 * (5 + 1 - (4 + 1)) / (4 + 1) = 1
I had found the formula once upon a time on Wikipedia but for some reason it's no longer there :(
Related
Can anyone explain how this code for computing of e works? Looks very easy for such complicated task, but I can't even understand the process. It has been created by Xavier Gourdon in 1999.
int main() {
int N = 9009, a[9009], x = 0;
for (int n = N - 1; n > 0; --n) {
a[n] = 1;
}
a[1] = 2, a[0] = 0;
while (N > 9) {
int n = N--;
while (--n) {
a[n] = x % n;
x = 10 * a[n-1] + x/n;
}
printf("%d", x);
}
return 0;
}
I traced the algorithm back to a 1995 paper by Stanley Rabinowitz and Stan Wagon. It's quite interesting.
A bit of background first. Start with the ordinary decimal representation of e:
e = 2.718281828...
This can be expressed as an infinite sum as follows:
e = 2 + 1⁄10(7 + 1⁄10(1 + 1⁄10(8 + 1⁄10(2 + 1⁄10(8 + 1⁄10(1 ...
Obviously this isn't a particularly useful representation; we just have the same digits of e wrapped up inside a complicated expression.
But look what happens when we replace these 1⁄10 factors with the reciprocals of the natural numbers:
e = 2 + 1⁄2(1 + 1⁄3(1 + 1⁄4(1 + 1⁄5(1 + 1⁄6(1 + 1⁄7(1 ...
This so-called mixed-radix representation gives us a sequence consisting of the digit 2 followed by a repeating sequence of 1's. It's easy to see why this works. When we expand the brackets, we end up with the well-known Taylor series for e:
e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + ...
So how does this algorithm work? Well, we start by filling an array with the mixed-radix number (0; 2; 1; 1; 1; 1; 1; ...). To generate each successive digit, we simply multiply this number by 10 and spit out the leftmost digit.*
But since the number is represented in mixed-radix form, we have to work in a different base at each digit. To do this, we work from right to left, multiplying the nth digit by 10 and replacing it with the resulting value modulo n. If the result was greater than or equal to n, we carry the value x/n to the next digit to the left. (Dividing by n changes the base from 1/n! to 1/(n-1)!, which is what we want). This is effectively what the inner loop does:
while (--n) {
a[n] = x % n;
x = 10 * a[n-1] + x/n;
}
Here, x is initialized to zero at the start of the program, and the initial 0 at the start of the array ensures that it is reset to zero every time the inner loop finishes. As a result, the array will gradually fill with zeroes from the right as the program runs. This is why n can be initialized with the decreasing value N-- at each step of the outer loop.
The additional 9 digits at the end of the array are presumably included to safeguard against rounding errors. When this code is run, x reaches a maximum value of 89671, which means the quotients will be carried across multiple digits.
Notes:
This is a type of spigot algorithm, because it outputs successive digits of e using simple integer arithmetic.
As noted by Rabinowitz and Wagon in their paper, this algorithm was actually invented 50 years ago by A.H.J. Sale
* Except at the first iteration where it outputs two digits ("27")
We have to find the minimum number of digits required to make a given number, for example: 14 => 95 (9 + 5 = 14) is two digits which is the minimum to form 14.
int moves(int n) {
int m = 0; // Minimum count
while (n-9 >= 0) { // To place maximum number of 9's
n -= 9;
m++;
}
if (n == 0) { // If only nines made up the number
return m;
}
else {
m++;
return m;
}
}
I am getting a TLE (runtime time limit exceeded) by an online judge. How can I improve it or is there a better approach?
Your code starts by looking at how many times 9 fits into that number. This can be done way more easily:
int m = n/9;
This suffices since we do an integer division, in which the remainder is thrown away. Note that if n would be float or another floating type, this would not work.
The question left is if it is divisible by 9 or not. If not, we have one additional digit. This can be done by the modulo operator (made it verbose for ease of understanding):
bool divisible_by_nine = (n % 9 == 0);
Assuming that you might not know the modulo operator, it returns the remainder of an integer division, 47 % 9 = 2 since 47 / 9 = 5 remainder 2.
Without it, you would go with
int remainder = n - 9*m;
bool divisible = (remainder == 0);
Combined:
int required_digits(int number)
{
bool divisible = (number % 9 == 0);
return number/9 + (divisible ? 0 : 1);
}
Or in a single line, depending on how verbose you want it to be:
int required_digits(int number)
{
return number/9 + (number % 9 == 0 ? 0 : 1);
}
Since there isn't any loop, this is in Θ(1) and thus should work in your required time limit.
(Technically, the processor might as well handle the division somewhat like you did internally, but it is very efficient at that. To be absolutely correct, I'd have to add "assuming that division is a constant time operation".)
Your solution works fine. You can try the shorter:
return (n%9==0)? n/9 : n/9 +1 ;
Shorter, but less easy to read...
Or a compromise:
if (n%9==0) // n can be divided by 9
return n/9;
else
return n/9+1;
Explanation
We know that every number a can be represented as
(a_n * 10 ^ n) + ... + (a_2 * 10 ^ 2) + (a_1 * 10) + (a_0)
where a_k are digits
and 10^n = 11...11 * 9 + 1 (n digits 1).
Meaning that number 10^n can be represented as the sum of 11...11 + 1 digits.
Now we can write a as (a_n * 11..11 * 9 + a_n) + ...
After grouping by 9 (help, I don't know English term for this. Factoring?)
(a_n * 11..11 + a_n-1 * 11..11 + ... a_1) * 9 + (a_n + a_n-1 + ... + a_1 + a_0)
Which I'll write as b_9 * 9 + b_1.
This means that number a can be represented as the sum of b_9 digits 9 + how much is needed for b_1 (this is recursive by the way)
To recapitulate:
Let's call function f
If -10 < digit < 10, the result is 1.
Two counters are needed, c1 and c2.
Iterate over digits
For every ith digit, multiply by i digit number 11..11 and add the result to c1
Add the ith digit to c2
The result is c_1 + f(c_2)
And for practice, implement this in a non-recursive way.
As you guess, you need to iterate on a lower number to a bigger one, like 111119 is fine, but we want the lowest one... Your answer is wrong. The lowest would be 59!
You can brute force and it will work, but for a bigger number you will struggle, so you need to guess first: How many minimum digits do I need to find my solution?
For instance, if you want to find 42, just add as much 9 you need to overflow the result!
9 + 9 + 9 + 9 + 9 = 45. When you find the overflow, you know that the answer is lower than 99999.
Now how much do I need to decrease the value to get the correct answer, 3 as expected?
So 99996, 99969, etc... will be valid! But you want to lower, so you have to decrease the greatest unit (the left one of course!).
The answer would be 69999 = 42!
int n = 14;
int r = 0;
for (int i = i; i < 10 /*if you play with long or long long*/; i++)
if (i * 9 >= n)
{
for (int j = 0; j < i; j++)
r = r * 10 + 9;
while (is_correct(r, n) == false)
{
// Code it yourself!!
}
return (r);
}
Now it correctly returns true or false. You can make it return the number that r is actually a decrease what you need to decrease! It's not the fastest way possible, and there is always a faster way, with a binary shift, but this algorithm would work just fine!
I have this function
public int numberOfPossiblePairs(int n)
{
int k=2;
if (k>n-k) { k=n-k;}
int b=1;
for (int i=1, m=n; i<=k; i++, m--)
b=b*m/i;
return b;
}
Which gets the number of pairs you can make from given number of items, so for example, if you have an array of 1000 items, you can make 499,500 pairs. But what I actually need is the opposite. In other words a function that would take 499500 as the parameter, and return 1000 as the original number of unique items that could produce that many pairs. (Would be a bonus if it could also handle imperfect numbers, like 499501, of which there is no number of unique items that makes exactly that many unique pairs, but it would still return 1000 as the closest since it produces 499500 pairs.)
I realize I could just incrementally loop numberOfPossiblePairs() until I see the number I am looking for, but seems like there should be an algorithmic way of doing this rather than brute forcing it like that.
Your problem boils down to a little algebra and can be solved in O(1) time. We first note that your function does not give the number of permutations of pairs, but rather the number of combinations of pairs. At any rate the logic that follows can be easily altered to accommodate permutations.
We start off by writing the formula for number of combinations choose k.
Setting n = 1000 and r = 2 gives:
1000! / (2!(998)!) = 1000 * 999 / 2 = 499500
Just as numberOfPossiblePairs(1000) does.
Moving on in our exercise, for our example we have r = 2, thus:
total = n! / ((n - 2)! * 2!)
We now simplify:
total = (n * (n - 1)) / 2
total * 2 = n^2 - n
n^2 - n - 2 * total = 0
Now we can apply the quadratic formula to solve for n.
Here we have x = n, a = 1, b = -1, and c = -2 * total which give:
n = (-(-1) +/- sqrt(1^2 - 4 * 1 * (-2 * total))) / 2
Since we are only interested in positive numbers we exclude the negative solution. In code we have something like (Note, it looks like the OP is using Java and I am not an expert here... the following is C++):
int originalNumber(int total) {
int result;
result = (1 + std::sqrt(1 - 4 * 1 * (-2 * total))) / 2;
return result;
}
As for the bonus question of returning the closest value if the result isn't a whole number, we could simply round the result before coercing to an integer:
int originalNumber(int total) {
int result;
double temp;
temp = (1 + std::sqrt(1 - 4 * 1 * (-2 * total))) / 2;
result = (int) std::round(temp);
return result;
}
Now when values like 500050 are passed, the actual result is 1000.55, and the above would return 1001, whereas the first solution would return 1000.
Sieve of Eratosthenes on the segment:
Sometimes you need to find all the primes that are in the range
[L...R] and not in [1...N], where R is a large number.
Conditions:
You are allowed to create an array of integers with size
(R−L+1).
Implementation:
bool isPrime[r - l + 1]; //filled by true
for (long long i = 2; i * i <= r; ++i) {
for (long long j = max(i * i, (l + (i - 1)) / i * i); j <= r; j += i) {
isPrime[j - l] = false;
}
}
for (long long i = max(l, 2); i <= r; ++i) {
if (isPrime[i - l]) {
//then i is prime
}
}
What is the logic behind setting the lower limit of 'j' in second for loop??
Thanks in advance!!
Think about what we want to find. Ignore the i*i part. We have only
(L + (i - 1)) / i * i) to consider. (I wrote the L capital since l and 1 look quite similar)
What should it be? Obviously it should be the smallest number within L..R that is divisible by i. That's when we want to start to sieve out.
The last part of the formula, / i * i finds the next lower number that is divisible by i by using the properties of integer division.
Example: 35 div 4 * 4 = 8 * 4 = 32, 32 is the highest number that is (equal or) lower than 35 which is divisible by 4.
The L is where we want to start, obviously, and the + (i-1) makes sure that we don't find the highest number equal or lower than but the smallest number equal or bigger than L that is divisible by i.
Example: (459 + (4-1)) div 4 * 4 = 462 div 4 * 4 = 115 * 4 = 460.
460 >= 459, 460 | 4, smallest number with that property
(the max( i*i, ...) is only so that i is not sieved out itself if it is within L..R, I think, although I wonder why it's not 2 * i)
For reasons of readability, I'd made this an inline function next_divisible(number, divisor) or the like. And I'd make it clear that integer division is used. If not, somebody clever might change it to regular division, with which it wouldn't work.
Also, I strongly recommend to wrap the array. It is not obvious to the outside that the property for a number X is stored at position X - L. Something like a class RangedArray that does that shift for you, allowing you a direct input of X instead of X - L, could easily take the responsibility. If you don't do that, at least make it a vector, outside of a innermost class, you shouldn't use raw arrays in C++.
Given an input n , find the sum of all the possible combinations of numbers 1 ... n.
For example, if n=3 , then all the possible combinations are
(1),(2),(3),(1,2),(1,3),(2,3),(1,2,3)
and their sum is
1 + 2 + 3 + (1+2) + (1+3) + (2+3) + (1+2+3) =24
I am able to solve this problem using recursion. How can I solve this problem using Dynamic Programming ?
#include<iostream>
using namespace std;
int sum=0,n;
int f(int pos,int s)
{
if(pos>n)
{
return 0;
}
else
{
for(int i=pos+1;i<=n;++i)
{
sum+=s+i;
f(i,s+i);
}
}
}
int main()
{
cin>>n;
sum=0;
f(0,0);
cout<<sum<<'\n';
}
}
EDIT
Though this problem can be solved in constant time using this series.
But I want to know how this can be done using Dynamic Programming as I am very weak at it.
You do not need to use dynamic programming; you can use simple arithmetic if you want.
The number of cases is 2 ^ n, since each number is either on or off for a given sum.
Each number from 1 to n is used in exactly half of the sums, so each number comes 2 ^ (n-1) times.
1 + 2 + ... + n = (n - 1) * n / 2.
So the sum is (n - 1) * n / 2 * 2 ^ (n-1).
For n = 3, it is (4*3/2) * 4 = 24.
EDIT: if you really want to use dynamic programming, here's one way.
Dynamic programming makes use of saving the results of sub-problems to make the super problem faster to solve. In this question, the sub-problem would be the sum of all combinations from 1 ... n-1.
So create a mapping from n -> (number of combinations, sum of combinations).
Initialize with 1 -> (2,1). Because there are two combinations {0,1} and the sum is 1. Including 0 just makes the math a bit easier.
Then your iteration step is to use the mapping.
Let's say (n-1) -> (k,s), meaning there are k sets that sum to s for 1 ... n-1.
Then the number of sets for n is k * 2 (each combination either has n or does not).
And the sum of all combinations is s + (s + k * n), since you have the previous sum (where n is missing) plus the sum of all the combinations with n (which should be k * n more than s because there are k new combinations with n in each).
So add n -> (2*k,2*s + k*n).
And your final answer is the s in n -> (k,s).
let dp[n] be the result, Therefore:
dp[1] = 1
dp[n] = 2 * dp[n-1] + 2^(n-1) * n
First, it is obvious that dp[1] = 1
Second, dp[n] is the sum which contains n and sum which didn't contains n
E.G: dp[3] = {(1) (2) (1,2)} + {(3), (1,3), (2,3), (1,2,3)}
We can find dp[n-1] appear twice and the number of n appear 2^(n-1) times
I think maybe it is what you want.