首页 > ACM题库 > HDU-杭电 > HDU 3037-Saving Beans-数论-[解题报告]HOJ
2014
02-27

HDU 3037-Saving Beans-数论-[解题报告]HOJ

Saving Beans

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

2
1 2 5
2 1 5

样例输出:

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.

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 <iostream>
#include <cstdio>

//typedef __int64 lld;
typedef long long lld;
lld N,M,P;

int Pow(lld a,lld n,lld p)
{
	lld x = a;
	lld res = 1;
	while(n)
	{
		if(n & 1)
		{
			res = ((lld)res * (lld)x) % p;
		}
		n >>= 1;
		x = ((lld)x*(lld)x) % p;
	}
	return res;
}

int Cm(lld n,lld m,lld p)
{
	lld a = 1,b = 1;
	if(m > n) return 0;
	//实现(a!/(a-b)!) * (b!)^(p-2)) mod p,由于n比较大,所以,此处不知道有什么好的优化
	while(m)
	{
		a = (a * n) % p;
		b = (b * m) % p;
		m--;
		n--;
	}
	return ((lld)a * (lld)Pow(b,p-2,p))%p;
}

int Lucas(lld n,lld m,lld p)
{
	if(m==0)
		return 1;
	return((lld)Cm(n%p,m%p,p)*(lld)Lucas(n/p,m/p,p))%p;
}

int main()
{
	int t;
	//freopen("in.txt","r",stdin);
	scanf("%d",&t);
	while(t--)
	{
		scanf("%I64d%I64d%I64d",&N,&M,&P);
		printf("%d\n",Lucas(N+M,M,P));
	}
	return 0;
}

参考:http://blog.csdn.net/tju_virus/article/details/7843248


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  2. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮

  3. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?

  4. 学算法中的数据结构学到一定程度会乐此不疲的,比如其中的2-3树,类似的红黑树,我甚至可以自己写个逻辑文件系统结构来。

  5. 一开始就规定不相邻节点颜色相同,可能得不到最优解。我想个类似的算法,也不确定是否总能得到最优解:先着一个点,随机挑一个相邻点,着第二色,继续随机选一个点,但必须至少有一个边和已着点相邻,着上不同色,当然尽量不增加新色,直到完成。我还找不到反例验证他的错误。。希望LZ也帮想想, 有想法欢迎来邮件。谢谢