Saving Beans

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

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.

尼玛看了一天，推导了一天。。真不容易

题解传送门：http://blog.csdn.net/wukonwukon/article/details/7341270

组合数取模需要用的到，

快速幂取模传送门：http://blog.csdn.net/lsldd/article/details/5506933

插板法：http://blog.sina.com.cn/s/blog_7cc6f2770100red3.html

费马小定理：http://baike.baidu.com/view/263807.htm

其中除法取模还需要求逆元。

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#define SIZE 100010
using namespace std ;
long long f[SIZE] ;

long long modPow(long long a , long long n , long long p)
{
	long long ret = 1 , A=a ;
	while(n)
	{
		if(n & 1)
			ret = (ret*A)%p ;
		A = (A*A)%p ;
		n >>= 1 ;
	}
	return ret ;
}

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

long long lucas(long long a , long long b , long long p)
{
	long long re = 1 ;
	while(a && b)
	{
		long long aa = a%p , bb = b%p ;
		if(aa<bb)
			return 0 ;
		re = re*f[aa]*modPow(f[bb]*f[aa-bb]%p,p-2,p)%p ;
		a /= p ;
		b /= p ;
	}
	return re ;
}

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

与君共勉