Given three numbers n, r and p, compute value of nCr mod p. Here p is a prime number greater than n. Here nCr is Binomial Coefficient.
Input: n = 10, r = 2, p = 13 Output: 6 Explanation: 10C2 is 45 and 45 % 13 is 6. Input: n = 6, r = 2, p = 13 Output: 2
We have discussed following methods in previous posts.
Compute nCr % p | Set 1 (Introduction and Dynamic Programming Solution)
Compute nCr % p | Set 2 (Lucas Theorem)
In this post, Fermat Theorem based solution is discussed.
Fermat’s little theorem and modular inverse
Fermat’s little theorem states that if p is a prime number, then for any integer a, the number ap – a is an integer multiple of p. In the notation of modular arithmetic, this is expressed as:
ap = a (mod p)
For example, if a = 2 and p = 7, 27 = 128, and 128 – 2 = 7 × 18 is an integer multiple of 7.
If a is not divisible by p, Fermat’s little theorem is equivalent to the statement a p – 1 – 1 is an integer multiple of p, i.e
ap-1 = 1 (mod p)
If we multiply both sides by a-1, we get.
ap-2 = a-1 (mod p)
So we can find modular inverse as p-2.
We know the formula for nCr nCr = fact(n) / (fact(r) x fact(n-r)) Here fact() means factorial. nCr % p = (fac[n]* modIverse(fac[r]) % p * modIverse(fac[n-r]) % p) % p; Here modIverse() means modular inverse under modulo p.
Following is the implementation of the above algorithm. In the following implementation, an array fac is used to store all the computed factorial values.
Value of nCr % p is 6
In competitive programming, we can pre-compute fac for given upper limit so that we don’t have to compute it for every test case. We also can use unsigned long long int everywhere to avoid overflows.
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- Fermat's little theorem
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- Primality Test | Set 2 (Fermat Method)
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- Dilworth's Theorem
- Nicomachu's Theorem
- Wilson's Theorem
- Rosser's Theorem
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- How to compute mod of a big number?
- Compute n! under modulo p
- Hardy-Ramanujan Theorem
- Euclid Euler Theorem
- Extended Midy's theorem