Saving Beans

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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Total Submission(s): 2978    Accepted Submission(s): 1120



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

Hint

 


Source

​​2009 Multi-University Training Contest 13 - Host by HIT​​

 


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解析：n棵树上挂 x beans的方案数实际上就是：x1+x2+...+xn=x的方案数。

           设b[i]=x[i]+1，则有：b1+b2+.....bn=x+n

           将 x+n 看成一条线段，将此线段分为 n 段的方案数即为解的个数，即为：C(n+x-1,n-1)=C(n+x-1,x)；

          综上：n棵树上挂不超过 m beans 的方案数就为：

                     ans=ans1+ans2+....+ansm

                           =C(n-1,0)+C(n,1)+C(n+1,2)+....+C(n+m-1,m)

                           =C(n,0)+C(n,1)+C(n+1,2)+......+C(n+m-1,m)

                     然后根据：C(n+1,r)=C(n,r-1)+C(n,r)

                     ==>ans=C(n+m,m)

           原问题就转化为求:C(n+m,m)%p的值，用lucas定理即可解决。

代码：

#include<cstdio>
#define maxn 100000
using namespace std;

typedef long long LL;
LL fac[maxn+20];

LL pow_mod(LL x,LL y,LL p)
{
  LL ans=1;
  while(y>0)
    {
      if(y&1)ans=ans*x%p;
      x=x*x%p,y>>=1;
  }
  return ans;
}
  
LL lucas(LL n,LL m,LL p)
{
  LL ans=1,a,b,c;
  while(n && m)
    {
      a=n%p,b=m%p;
      if(a<b)return 0;
      c=fac[a]*pow_mod(fac[b]*fac[a-b]%p,p-2,p)%p;
      ans=ans*c%p,n/=p,m/=p;
  }
  return ans;
}

int main()
{
  LL n,m,p,t,i;
  scanf("%I64d",&t);
  while(t--)
    {
      scanf("%I64d%I64d%I64d",&n,&m,&p);
    fac[0]=fac[1]=1;
    for(i=2;i<=p;i++)fac[i]=fac[i-1]*i%p;
    printf("%I64d\n",lucas(n+m,m,p));
  }
  return 0;
}