Summary of several methods for calculating combinatorial number by [combinatorial number]

Summary of methods for C (n,m)%mod

1. When the n,m are very small, you can use the Yang Hui Triangle direct request.
C (n,m) =c (n-1,m) +c (n-1,m-1);

2. Use the multiplication inverse element.
Multiplication Inverse: (A/b)%mod=a* (b^ (mod-2)) mod is a prime number.
The inverse can be calculated using the extended Euclidean or Euler function:

 1).扩展欧几里德：b*x+p*y=1 有解，x就是所求 2).费马小定理：b^(p-1)=1(mod p),故b*b^(p-2)=1(mod p),所以x=b^(p-2)

1. n!/(m!* (n-m) =x%p, first to calculate the n! M! , (N-M)! The remainder of P modulus is converted to a/b=x%p; Since P is a prime number, it is equivalent to Bx+py=a, and then the solution of BX ' +py ' =1 is calculated with the extended Euclidean theorem, x=x ' *a, and the final x value, i.e. C (m,n)%p, is obtained.

2. Inverse element In fact, if the MoD is a prime number, the inverse of B is actually b^ (mod-2)
i.e. (m! ( N-M)!) The inverse is (m! ( N-M)!) P-2;

int inv (inta) {//return Fpow (A, MOD-2, MOD);     return a==1?1: (Long Long) (Mod-mod/a) * INV (MOD%a)% MOD; } ll C (ll n,ll m) {if(M <0)return 0;if(N < m)return 0;if(M > n-m) m = n-m; LL up =1, down =1; for(LL i =0; I < m;          i + +) {up = up * (n-i)% MOD; Down = down * (i+1)% MOD; }returnUp * INV (down)% MOD; }

3. When the N and M ratios are large, the mod is prime and relatively small (around 10^5), calculated by the Lucas theorem

Lucas theorem: A, B are non-negative integers, p is prime. A B is written in P-system: a=a[n]a[n-1]...a[0],b=b[n]b[n-1]...b[0].
Then the combined number C (A, B) and C (A[n],b[n])C (a[n-1],b[n-1]) ... *c (a[0],b[0]) mod p congruence
namely: Lucas (n,m,p) =c (n%p,m%p) *lucas (n/p,m/p,p)

#include <iostream>//#include <algorithm>using namespace STD;typedef Long LongllintQuick_power_mod (intAintBintm) {//pow (A, b)%m    intresult =1;intbase = A; while(b>0){if(B &1==1) {result = (result*base)% m;         } base = (base*base)%m; b>>=1; }returnResult;}//Calculate the combination number to take the modelLL Comp (ll A, ll B,intP) {//composite Num C (A, b)%p    if(A < b)return 0;if(A = = b)return 1;if(b > A) b = a A;intAns =1, CA =1, CB =1; for(LL i =0; I < b;        ++i) {CA = (CA * (a-i))%p;    CB = (CB * (b-i))%p; } ans = (Ca*quick_power_mod (CB, P-2, p))% p;returnAns;} ll Lucas (ll N, ll M, ll P) {ll ans =1; while(N&&m&&ans) {ans = (Ans*comp (n%p, m%p, p))% p;//also can recusiven/= p;     M/= p; }returnAns;}intMain () {ll m,n; while(Cin&GT;&GT;N&GT;&GT;M) {cout<<lucas (N,m,10007) <<endl; }return 0;}

C (n mod, m% mod)% MoD; If it is too large, use the inverse of the above to handle it.

Semi-pretreatment
Since the Lucas theorem guarantees that the number of factorial is less than p, all factorial preprocessing is possible, and the C (N,M) is optimized.
MoD requirements: p<10^6, and the prime number
Valid range: 1<=n,m<=10^9

//semi-pretreatmentConstll MAXN =100000; ll fac[maxn+ -];voidInitint MoD) {fac[0] =1; for(intI=1; i<MoD; i++) {Fac[i] = fac[i-1] * I%MoD; }  }//semi-pretreatment inverse element for C (n,m)%modll C (ll N, ll m) {if(m>n)return 0;returnFac[n] * (Getinverse (FAC[M]*FAC[N-M),MoD)) %MoD; }

4. There is also a decomposition of the quality factor, this is more troublesome.
Http://www.mamicode.com/info-detail-621758.html

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

Summary of several methods for calculating combinatorial number by [combinatorial number]