大组合数取模

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1119

考虑从(1,1)->(n,m)必定会向下走n-1步,向右走m-1步,那么总的走法是C(n-1+m-1,m-1)。

关于组合数取模:大神博客：http://blog.csdn.net/acdreamers/article/details/8037918

 1 #include <iostream>
 2 #include <string.h>
 3 #include <stdio.h>
 4 
 5 using namespace std;
 6 typedef long long LL;
 7 const LL p = 1e9+7;
 8 LL n,m;
 9 
10 LL quick_mod(LL a, LL b)
11 {
12     LL ans = 1;
13     a %= p;
14     while(b)
15     {
16         if(b & 1)
17         {
18             ans = ans * a % p;
19             b--;
20         }
21         b >>= 1;
22         a = a * a % p;
23     }
24     return ans;
25 }
26 
27 LL C(LL n, LL m)
28 {
29     if(m > n) return 0;
30     LL ans = 1;
31     for(int i=1; i<=m; i++)
32     {
33         LL a = (n + i - m) % p;
34         LL b = i % p;
35         ans = ans * (a * quick_mod(b, p-2) % p) % p;
36     }
37     return ans;
38 }
39 
40 LL Lucas(LL n, LL m)
41 {
42     if(m == 0) return 1;
43     //printf("%I64d
",C(n,m));
44     return C(n % p, m % p) * Lucas(n / p, m / p) % p;
45 }
46 
47 int main()
48 {
49     scanf("%I64d%I64d", &n, &m);
50     printf("%I64d
", Lucas(n+m-2,m-1));
51     return 0;
52 }