E(HDu3037Lucas比较综合的一道数论题目)

Saving Beans

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1666    Accepted Submission(s): 592

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

 

Recommend

gaojie

 

一道很强悍的数论题目,推都推不出来,但是好在,有别人的思路,明天再推推,看看能不能推出来,

但是有很多的知识点,比如说,费马小定理,比如说,lucas还比如说求c(n,m)

这些都是很基础的算法,但是自己以前都不知道,这一道题目都让我全部用到了,自己果然是菜鸟中的菜鸟,不过学到了知识,自己还是很开心的.

拿来别人的解题报告看看.

然后贴出自己的代码

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

下面简单介绍一下Lucas定理：

Lucas定理是用来求 C(n,m) mod p的值,p是素数（从n取m组合，模上p）。
描述为:
Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p)
Lucas(x,0,p)=1;

简单的理解就是：

以求解n! % p 为例，把n分段，每p个一段，每一段求得结果是一样的。但是需要单独处理每一段的末尾p,2p,...,把p提取出来，会发现剩下的数正好又是(n/p)! ，相当于

划归了一个子问题，这样递归求解即可。

这个是单独处理n！的情况，当然C(n,m)就是n！/(m! *（ｎ－ｍ）！），每一个阶乘都用上面的方法处理的话，就是Lucas定理了

Lucas最大的数据处理能力是p在10^5左右。

而C(a,b) =a! / ( b! * (a-b)! ) mod p

其实就是求 ( a! / (a-b)!)  * ( b! )^(p-2) mod p

  (上面这一步变换是根据费马小定理：假如p是质数，且a,p互质，那么a的（p-1）次方除以p的余数恒为1,

那么a和a^(p-2)互为乘法逆元，则（b / a) = (b * a^(p-2) ) mod p)

用下面的Lucas定理程序实现就能得出结果，实现过程中要注意乘法时的强制转换

#include 
#include 
#include 
#include 

using namespace std;

int N, M, P;

int fac[100011];

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

} 

int quick_pow(int a, int n)
{
	int ans = 1;

	while (n)
	{
		if (n & 1)
		{
			ans = (long long int)ans * a % P;
		}
		a = (long long int) a * a % P;
		n >>= 1;
	}
	return ans;
}

		

int C(int n, int m)
{
	if (n < m)
	{
		return 0;
	}
	return (long long int)fac[n] * (quick_pow((long long int)fac[m] * fac[n - m] % P, P - 2)) % P;
}
		
	

int lucas(long long int n, long long int m)
{
	if (m == 0)
	{
		return 1;
	}
	return (long long int)C(n % P, m % P) * lucas(n / P, m / P) % P;
}

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d%d", &N, &M, &P);
		factorial();
		int ans = lucas(N + M, M);
		printf("%d\n", ans);
	}
//	system("pause");
	return 0;
}