hdu3037 LUCAS定理运用。

最新推荐文章于 2021-09-23 19:38:10 发布

原创最新推荐文章于 2021-09-23 19:38:10 发布 · 262 阅读

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#数论

本文探讨了一道经典的组合数学问题：如何计算不超过m颗相同的豆子存储在n棵树中的不同方式数量，并给出了解决方案。该问题通过隔板法进行简化，并利用Lucas定理进行大数模运算。

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Saving Beans

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4440 Accepted Submission(s): 1738

Problem 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

题意可以转化成求 x1+x2+x3+...xn<=m的解;x1+x2+x3...xn=m的解用隔板法，将 n+m-1取m个的方法就是解的个数及 ans=sum(c(n+m-1,m)) m从0到m;

C(n,k) = C(n-1,k)+C(n-1,k-1) 从而得到 ans=c(n+m,m);然后就可以用lucas了 C(n,m)%p=C(n/p,m/p)*C(n%p,m%p)%p

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long int llg;

const int N =150000;

llg n, m, p, fac[N];

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

llg pow(llg a, llg b)
{
    llg tmp = a % p, ans =1;
    while(b)
    {
        if(b &1)  ans = ans * tmp % p;
        tmp = tmp*tmp % p;
        b >>=1;
    }
    return  ans;
}

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

llg Lucas(llg n, llg 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("%I64d%I64d%I64d",&n,&m,&p);
       init();
       printf("%I64d\n",Lucas(m+n,m));
   }
    return 0;
}