HDU 3037:卢卡斯定理

Description
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
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.

分析： 题目意思为，不超过m个不同的豆豆要放到n颗树上，可知豆豆数目可能为0，1，……，m。由排列组合中的隔板法我们知道，如果当前有m个豆豆，将豆豆和树排成一排，中间有n+m-1个间隔，我们要向其中插入n-1块隔板将其分成n部分，即求C(n+m-1,n-1)，也即C(n+m-1,m)。最终结果需要计算C(n-1,0)+C(n,1)+C(n+1,2)+……+C(n+m-1,m)=C(n+m,m) (由公式C(n,m)=C(n-1,m)+C(n-1,m-1)得到)。

AC代码

#include
#include
using namespace std;
long long f[100010];
long long pow(long long base,long long exponent,long long p)
{
    long long ans=1;
    while(exponent)
    {
        if(exponent&1)
            ans=(long long)ans*base%p;
        base=base*base%p;
        exponent>>=1;
    }
    return ans;
}
void init(long long p)
{//f[n]=n!
    f[0]=1;
    for(int i=1;i<=p;i++)
        f[i]=(f[i-1]*i)%p;
}
long long Lucas(long long n,long long m,long long p)
{
    long long ans=1;
    while(n&&m)
    {
        if(n%preturn 0;
        ans=(ans*f[n%p]*pow(f[m%p]*f[n%p-m%p]%p,p-2,p))%p;
        n/=p;
        m/=p;
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        long long n,m,p;
        scanf("%lld%lld%lld",&n,&m,&p);
        init(p);
        cout<return 0;
}