2014
02-27

# Saving Beans

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.

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.

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.

2
1 2 5
2 1 5

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.

(m+1)*(m+2)…(m+n-1)  = C(m+n-1,n-1) = C(m+n-1,m)

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)

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）。

Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p)
Lucas(x,0,p)=1;

Lucas最大的数据处理能力是p在10^5左右。

(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,

#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;
}

1. [email protected]

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