HDU3037 Saving Beans (Lucas theorem + multiplicative inverse)

The topic probably asks that the items that are less than or equal to M are placed in n places in several ways.

That is, the number of nonnegative integer solutions of the N-ary equation $x_1+x_2+x_3+\dots+x_n=y$, which is 0<=y<=m.

The number of nonnegative integer solutions of this equation is a classical problem, and the number of solutions of the positive integers of +1 can be solved by inserting plate method: $C _{y+n-1}^{n-1}=c_{y+n-1}^y$.

And 0<=y<=m, the final result is--

$$\sum_{i=0}^m c_{i+n-1}^i$$

$ $C _{n-1}^0+c_{n}^1+c_{n+1}^2+\dots+c_{n-1+m}^m$$

$ $C _{n}^0+c_{n}^1+c_{n+1}^2+\dots+c_{n-1+m}^m$$

$ $C _{n+1}^1+c_{n+1}^2+\dots+c_{n-1+m}^m\ \tag{$C _{n+1}^1=c_{n}^0+c_{n}^1$}$$

$ $C _{n+2}^2+\dots+c_{n-1+m}^m\ \tag{$C _{n+2}^2=c_{n+1}^1+c_{n+1}^2$}$$

$$\vdots$$

$ $C _{n+m}^m$$

So the result is $c_{n+m}^m$. So is the calculation $c_{n+m}^m\ mod \ p$, where 1<=n,m<=1000000000,1<p<100000 and p are prime numbers.

This is then used to calculate the modulus of this large combination of Lucas theorem: $Lucas (n,m) \equiv c_{n\%p}^{m\%p}\times Lucas (n/p,m/p) \pmod p$.

In addition, the multiplication inverse of modulo p is used to $C _n^m\equiv\frac {n!} When calculating the combined number. {(n-m)!m!} \equiv N!\times ((n-m)!m!) ^{-1} \pmod p$

And it is not necessary to calculate the inverse of the Euclidean algorithm, because P is a prime number, the use of the Ma Xiao theorem can be introduced n in modulo p multiplication inverse of the $n^{p-2}\ mod\ p$.

1#include <cstdio>2#include <cstring>3 using namespacestd;4 Long LongIneLong LongNLong Longp) {5     Long Longres=1, m=p-2;6      while(m) {7         if(m&1) res=res*n%p;8n=n*n%p;9m>>=1;Ten     } One     returnRes; A } - Long Longfact[100000]={1}; - Long LongLucasLong LongNLong LongMLong Longp) {   the     Long Longres=1;  -      while(n&&m) {   -         Long Longa=n%p,b=m%p;  -         if(A&LT;B)return 0;  +Res=res*fact[a]*ine (fact[b]*fact[a-b]%p,p)%p;  -N/=p; M/=p; +     } A     returnRes; at } - intMain () { -     Long Longn,m,p; -     intT; -scanf"%d",&t); -      while(t--){ inscanf"%lld%lld%lld",&n,&m,&p); -          for(intI=1; i<p; ++i) fact[i]=fact[i-1]*i%p; toprintf"%lld\n", Lucas (n+m,m,p)); +     } -     return 0; the}

HDU3037 Saving Beans (Lucas theorem + multiplicative inverse)