推薦博客：http://blog.csdn.net/u010582475/article/details/47707739

排列組合公式不多說了。。。

盧卡斯定理：

C（n, m）%p == C(n%p, m%p)*C(n/p, m/p),,當m == 0， 遞歸結束返回1

特例：C(n,m)%2 == 1當且僅當（n&m == m）{C(a,b)是奇數當且僅當把a,b二進制表達後b中1的位置是a中1的位置的子集}

範德蒙恒等式：

關於範德蒙推薦博客：http://blog.csdn.net/acdreamers/article/details/31032763

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http://acm.hdu.edu.cn/showproblem.php?pid=3037

盧卡斯定理的應用

經過化解得到求C（n+m， m）；;直接跑lks

 1
 
 #include <iostream>
 2 using namespace std;
 3 #define ll long long
 4 ll n, m, p;
 5 ll a[100009];
 6 void init(){
 7     a[0] = 1LL;
 8     for(int i = 1; i <= p; ++i){
 9         a[i] = a[i-1]*i%p;
10     }
11 }
12 ll q_m(ll x, ll y){
13     ll xx = 1LL;
14     while(y){
15         if
 
(y&1) xx = xx*x%p;
16         x = x*x%p;
17         y >>= 1;
18     }
19     return xx;
20 }
21 ll C(ll x, ll y){
22     return x >= y ? a[x]*q_m(a[y]*a[x-y]%p, p-2LL)%p : 0;
23 }
24 ll lks(ll x, ll y){
25     return y ? C(x%p, y%p)*lks(x/p, y/p)%p : 1LL;
26 }
27 void la(){
28     init();
29     printf("%lld\n", lks(n+m, m));
30 }
31 int main(){
32     int t;
33     scanf("%d", &t);
34     while(t--){
35         scanf("%lld%lld%lld", &n, &m, &p);
36         la();
37     }
38     return 0;
39 }

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範德蒙恒等式的應用：

裸的

http://codeforces.com/contest/785/problem/D

 1 #include <iostream>
 2 #include <string.h>
 3 using namespace std;
 4 #define ll long long
 5 const int maxn = 200009;
 6 const ll Mod = 1e9+7;
 7 char str[maxn];
 8 ll p[maxn], sum = 0, x, num;
 9 void init(){
10     p[0] = p[1] = 1;
11     for(int i = 2; i < maxn; ++i){
12         p[i] = p[i-1]*i%Mod;
13     }
14 }
15 ll q_m(ll x, ll y){
16     ll sum = 1;
17     while(y){
18         if(y&1) sum = sum*x%Mod;
19         x = x*x%Mod;
20         y >>= 1;
21     }
22     return sum;
23 }
24 ll C(ll x, ll n){
25     return p[n]*q_m(p[x]*p[n-x]%Mod, Mod-2)%Mod;
26 }
27 int main(){
28     init();
29     scanf("%s", str);
30     int len = strlen(str);
31     for(int i = 0; str[i]; ++i){
32         x += (str[i] == ‘)‘);
33     }
34     for(int j = 0; j < len; ++j){
35         if(str[j] == ‘)‘) --x;
36         else{
37             sum = (sum + C(x-1, x+(num++)))%Mod;
38         }
39     }
40     printf("%lld\n", sum);
41     return 0;
42 }

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wenbao與組合數