【大组合数取模 Lucas定理模板】HDU - 3037 A - Saving Beans

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本文探讨了松鼠如何在不同树木中存豆的问题，通过数学建模将其转化为组合计数问题，并利用Lucas定理解决大规模组合数的计算。介绍了Lucas定理的应用场景和算法实现。

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A - Saving Beans HDU - 3037

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.

        
多测几组样例，会发现就是求C(n+m,m)%p ,因为p是小于10^5的素数所以用到了Lucas定理

Lucas(n,m,p)=c(n%p,m%p)*Lucas(n/p,m/p,p)

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll fac[100005];

void init(int p)
{
    fac[0]=1;
    for(int i=1;i<=p;i++)
        fac[i]=(fac[i-1]*i)%p;
}

ll inv(ll a,ll m)
{
    if(a==1) return 1;
    return inv(m%a,m)*(m-m/a)%m;
}

ll lucas(ll n,ll m,ll p)
{
    ll ans=1;
    while(n&&m)
    {
        ll a=n%p,b=m%p;
        if(a<b) return 0;
        ans=ans*fac[a]%p*inv(fac[b]*fac[a-b]%p,p)%p;
        n/=p,m/=p;
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll n,m,p;
        scanf("%lld%lld%lld",&n,&m,&p);
        init(p);
        printf("%lld\n",lucas(n+m,m,p));
    }
    return 0;
}