假设有n+1个树，第n+1个树埋不足m的种子，隔板法C【n+m】【m】

大组合数取mod用Lucas定理:

Lucas(n,m,p) = C[n%p][m%p] × Lucas(n/p,m/p,p) ;

Saving Beans

2
1 2 5
2 1 5

3
3
Hint
Hint

For sample 1, squirrels will put no more than 2 beans in one tree. Since trees are different, we can label them as 1, 2 … and so on. 
The 3 ways are: put no beans, put 1 bean in tree 1 and put 2 beans in tree 1. For sample 2, the 3 ways are:
 put no beans, put 1 bean in tree 1 and put 1 bean in tree 2.
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

LL n,m,p;

LL fact[100100];

LL QuickPow(LL x,LL t,LL m)
{
	if(t==0) return 1LL;
	LL e=x,ret=1LL;
	while(t)
	{
		if(t&1) ret=(ret*e)%m;
		e=(e*e)%m;
		t>>=1LL;
	}
	return ret%m;
}

void get_fact(LL p)
{
	fact[0]=1LL;
	for(int i=1;i<=p+10;i++)
		fact[i]=(fact[i-1]*i)%p;
}

LL Lucas(LL n,LL m,LL p)
{
	///lucas(n,m,p)=c[n%p][m%p]*lucas(n/p,m/p,p);
	LL ret=1LL;
	while(n&&m)
	{
		LL a=n%p,b=m%p;
		if(a<b) return 0;
		ret=(ret*fact[a]*QuickPow((fact[b]*fact[a-b])%p,p-2,p))%p;
		n/=p; m/=p;
	}
	return ret%p;
}

int main()
{
	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		LL n,m,p;
		cin>>n>>m>>p;
		get_fact(p);
		cout<<Lucas(n+m,m,p)<<endl;
	}
	return 0;
}

HDOJ 3037 Saving Beans,布布扣,bubuko.com

HDOJ 3037 Saving Beans

原文：http://blog.csdn.net/ck_boss/article/details/38471743