hdu 3037 Saving Beans 隔板法+lucas

Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel family thinks that they have to solve a problem. They suppose that they will save beans in n different trees. However, since the food is not sufficient nowadays, they will get no more than m beans. They want to know that how many ways there are to save no more than m beans (they are the same) in n trees.

Now they turn to you for help, you should give them the answer. The result may be extremely huge; you should output the result modulo p, because squirrels can’t recognize large numbers.
Input
The first line contains one integer T, means the number of cases.

Then followed T lines, each line contains three integers n, m, p, means that squirrels will save no more than m same beans in n different trees, 1 <= n, m <= 1000000000, 1 < p < 100000 and p is guaranteed to be a prime.
Output
You should output the answer modulo p.
Sample Input
2
1 2 5
2 1 5
Sample Output
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.

题意：从n棵树上得到 不超过 m颗豆子 的所有情况
豆子是一样的 树是不一样的 也就是 m颗豆子分给n颗树，其中树可以为空，那么就是 隔板法，那么得到公式 C(n+i-1,n-1) = C(n+i-1,i);
所以 求的是 ∑C(n+i-1,i)
转完以后 得到
相当于对i = 0,1…,m对C(n+i-1,i)求和，根据公式C(n,k) = C(n-1,k)+C(n-1,k-1)得
C(n-1,0)+C(n,1)+…+C(n+m-1,m)
= C(n,0)+C(n,1)+C(n+1,2)+…+C(n+m-1,m)

#include 
using namespace std;
typedef long long ll;

const int N = 150000;
ll n,m,p;
ll fac[N];
void init()
{
    fac[0]=1;
    for(int i=1;i<=p;i++)
    {
        fac[i]=(fac[i-1]*i)%p;
    }
}

ll qpow(ll a,ll b)
{
    ll res=1;
    while(b)
    {
        if(b&1) res=res*a,res%=p;
        a*=a;
        a%=p;
        b>>=1;
    }
    return res;
}



ll C(ll n, ll m)
{
    if(m>n) return 0;
    return fac[n]*qpow(fac[m],p-2)%p*qpow(fac[n-m],p-2)%p;
}


ll Lucas(ll n,ll m)
{
    if(m==0) return 1;
    else return (C(n%p,m%p)*Lucas(n/p,m/p))%p;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&n,&m,&p);
        init();
        printf("%lld\n",Lucas(n+m,m) );
    }
}