Saving Beans

Saving Beans

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5761    Accepted Submission(s):
2310

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

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.

Source

2009
Multi-University Training Contest 13 - Host by HIT

/*
题目相当于求n个数的和不超过m的方案数。
如果和恰好等于m，那么就等价于方程x1+x2+...+xn = m的解的个数，利用插板法可以得到方案数为：
(m+1)*(m+2)...(m+n-1)  = C(m+n-1,n-1) = C(m+n-1,m)
现在就需要求不大于m的，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)
= C(n+m,m)
现在就是要求C(n+m,m) % p,其中p是素数。
然后利用Lucas定理的模板就可以轻松的求得C(n+m,m) % p的值
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 100007

using namespace std;
long long f[N];

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

long long C(long long n,long long m,long long p)
{
    if(m>n)return 0;
    return  f[n]*Mi(f[m]*f[n-m]%p,p-2,p)%p;
}

long long Lcs(long long n,long long m,long long p)
{
    if(m==0)return 1;
    return (C(n%p,m%p,p)*Lcs(n/p,m/p,p))%p;
}

int main()
{
    long long n,m,p;long long t;
    cin>>t;
    while(t--)
    {
        cin>>n>>m>>p;
        f[0]=1;
        for(long long i=1;i<=p;i++)
          f[i]=f[i-1]*i%p;
        printf("%lld\n",Lcs(n+m,m,p));
    }
    return 0;
}