[ACM] hdu 3037 Saving Beans (Lucas定理，組合數取模）
Saving Beans
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
題意為：將不超過m個豆子放在n棵不同的樹上，一共有多少種方法。
以下轉載於：
http://blog.csdn.net/tju_virus/article/details/7843248
題目相當於求n個數的和不超過m的方案數。
如果和恰好等於m，那麼就等價於方程x1+x2+...+xn = m的解的個數，利用插板法可以得到方案數為：
(m+1)*(m+2)...(m+n-1) = C(m+n-1,n-1) = C(m+n-1,m)
現在就需要求不大於m的，相當於對i = 0,1...,m對C(n+i-1,i)求和，根據公式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)
現在就是要求C(n+m,m) % p,其中p是素數。
然後利用Lucas定理的模板就可以輕松的求得C(n+m,m) % p的值
下面簡單介紹一下Lucas定理：
Lucas定理是用來求 C(n,m) mod p的值,p是素數（從n取m組合，模上p）。
描述為:
Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p)
Lucas(x,0,p)=1;
A、B是非負整數，p是質數。AB寫成p進制：A=a[n]a[n-1]...a[0]，B=b[n]b[n-1]...b[0]。
