binomial-coefficients | 易学教程

Finding binomial coefficient for large n and k modulo m

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问题 I want to compute nCk mod m with following constraints: n<=10^18 k<=10^5 m=10^9+7 I have read this article: Calculating Binomial Coefficient (nCk) for large n & k But here value of m is 1009. Hence using Lucas theorem, we need only to calculate 1009*1009 different values of aCb where a,b<=1009 How to do it with above constraints. I cannot make a array of O(m*k) space complexity with given constraints. Help! 回答1: Just use the fact that (n, k) = n! / k! / (n - k)! = n*(n-1)*...*(n-k+1)/[k*(k-1)

Calculate Nth multiset combination (with repetition) based only on index

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问题 How can i calculate the Nth combo based only on it's index. There should be (n+k-1)!/(k!(n-1)!) combinations with repetitions. with n=2, k=5 you get: 0|{0,0,0,0,0} 1|{0,0,0,0,1} 2|{0,0,0,1,1} 3|{0,0,1,1,1} 4|{0,1,1,1,1} 5|{1,1,1,1,1} So black_magic_function(3) should produce {0,0,1,1,1}. This will be going into a GPU shader, so i want each work-group/thread to be able to figure out their subset of permutations without having to store the sequence globally. with n=3, k=5 you get: i=0, {0,0,0,0

Fast n choose k mod p for large n?

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问题 What I mean by "large n" is something in the millions. p is prime. I've tried http://apps.topcoder.com/wiki/display/tc/SRM+467 But the function seems to be incorrect (I tested it with 144 choose 6 mod 5 and it gives me 0 when it should give me 2) I've tried http://online-judge.uva.es/board/viewtopic.php?f=22&t=42690 But I don't understand it fully I've also made a memoized recursive function that uses the logic (combinations(n-1, k-1, p)%p + combinations(n-1, k, p)%p) but it gives me stack

Fast n choose k mod p for large n?

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binomial coefficient

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问题 I have a code which calculate binomial coefficient, but when number is bigger then 20, it start to calculate wrong, where is the problem? Thanks #include <iostream> using namespace std; long int bin(long int x) { if(x==0) return 1; long int r = x; for(int i = r-1;i>0;i--) { r = r*i; } return r; } int main() { cout << "Write n and k: " << endl; long int n=0; long int k=0; cin >> n; cin >> k; long int result = 0; long int fn = bin(n); long int fk = bin(k); long int fnk = bin(n-k); result = fn/

ncr in c (combinations)

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问题 I m trying to calculate ncr(combinations) in c using dp. But it is failing with n=70. Can anyone help? unsigned long long ncr( int n , int r) { unsigned long long c[1001]; int i=1; c[0]=1; for(i=1; i<=r; i++) c[i]= ((unsigned long long) (c[i-1]) * (unsigned long long)( n-i+1))%(unsigned long long) (1000000007)/ (unsigned long long)(i); return c[r]; } basic idea is ncr = ((n-r+1)/r)* nc(r-1) 回答1: The intermediate product (unsigned long long) (c[i-1]) * (unsigned long long)( n-i+1) is a very

Sage polynomial coefficients including zeros

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问题 If we have a multivariate polynomial in SAGE for instance f=3*x^3*y^2+x*y+3 how can i display the full list of coefficients including the zero ones from missing terms between maximum dregree term and constant. P.<x,y> = PolynomialRing(ZZ, 2, order='lex') f=3*x^2*y^2+x*y+3 f.coefficients() gives me the list [3, 1, 3] but i'd like the "full" list to put into a a matrix. In the above example it should be [3, ,0 , 0, 1, 0, 0, 0, 0, 3] corresponding to terms: x^2*y^2, x^2*y, x*y^2, x*y, x^2, y^2,

Efficient algorithm to get the combinations of all items in object

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问题 Given an array or object with n keys, I need to find all combinations with length x . Given X is variable. binomial_coefficient(n,x) . Currently I'm using this: function combine(items) { var result = []; var f = function(prefix, items) { for (var i = 0; i < items.length; i++) { result.push(prefix + items[i]); f(prefix + items[i], items.slice(i + 1)); } } f('', items); return result; } var combinations = combine(["a", "b", "c", "d"]); The output is: ["a", "ab", "abc", "abcd", "abd", "ac", "acd

%mod compatible ways of generating Binomial Coefficients

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问题 I would like to optimize a part of my program where I'm calculating the sum of Binomial Coefficients up to K. i.e. C(N,0) + C(N,1) + ... + C(N,K) Since the values go beyond the data type (long long) can support, I'm to calculate values mod M and was looking for procedures to do that. Currently, I've done it with Pascal's Triangle but it seems to be taking a bit of load. so, I was wondering if there's any other efficient way to do this. I've considered Lucas' Theorem, although M I have is

%mod compatible ways of generating Binomial Coefficients

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I would like to optimize a part of my program where I'm calculating the sum of Binomial Coefficients up to K. i.e. C(N,0) + C(N,1) + ... + C(N,K) Since the values go beyond the data type (long long) can support, I'm to calculate values mod M and was looking for procedures to do that. Currently, I've done it with Pascal's Triangle but it seems to be taking a bit of load. so, I was wondering if there's any other efficient way to do this. I've considered Lucas' Theorem, although M I have is already large enough so that C(N,k) goes out of hand! Any pointers as how can I do this differently, maybe