P3807 【模板】卢卡斯定理

卢斯卡定理

$\because C_{p}^{i}=\frac{p!}{i!(p-i)!} = \frac{p}{i}\frac{(p-1)!}{(i-1)!(p-1-(i-1))!} = \frac{p}{i}C_{i-1}^{p-1}$

$\therefore C_{p}^{i}\equiv \frac{p}{i}C_{p-1}^{i-1}\equiv 0\ (mod\ p)\ (1\leq i\leq p-1)$

$\therefore (1+x)^p\equiv C_{p}^{0}x^01^p+C_{p}^{1}x^11^{p-1}+C_{p}^{2}x^21^{p-2}+\cdot \cdot \cdot \ +C_{p}^{p}x^p1^0\equiv 1+x^p\ (mod\ p)$

 

令 $n=sp+q,m=lp+r\ (0\leq q,r\leq p-1)$

$\because n=sp+q$

$\therefore (1+x)^{n}=(1+x)^{sp}(1+x)^{q}$

$\because (1+x)^{p}\equiv 1+x^p\ (mod\ p)$

$\therefore (1+x)^{sp}\equiv ((1+x)^p)^s\equiv (1+x^p)^s\ (mod\ p)$

$\therefore (1+x)^n\equiv (1+x^p)^s(1+x)^q\ (mod\ p)$

$\therefore C_{n}^{m}x^m1^{n-m}\equiv C_{s}^{l}(x^p)^l1^{s-l}\cdot C_{q}^{r}x^r1^{q-r}\ (mod\ p)$

$\therefore C_{n}^{m}x^m\equiv C_{s}^{l}\cdot C_{q}^{r}x^{lp+r}\equiv C_{s}^{l}\cdot C_{q}^{r}x^m\ (mod\ p)$

$\therefore C_{n}^{m}x^m\equiv C_{s}^{l}\cdot C_{q}^{r}\equiv C_{[\frac{n}{p}]}^{[\frac{m}{p}]}\cdot C_{n\ mod\ p}^{m\ mod\ p}\ (mod\ p)$

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 1e5+5;
 5 ll a[maxn];
 6 ll qmul(ll a, ll b, ll p) {
 7     ll ans = 0;
 8     while (b) {
 9         if (b&1) ans = (ans+a)%p;
10         a = (a+a)%p;
11         b >>= 1;
12     }
13     return ans%p;
14 }
15 ll qpow(ll a, ll b, ll p) {
16     ll ans = 1;
17     while (b) {
18         if (b&1) ans = qmul(ans,a,p);
19         a = qmul(a,a,p);
20         b >>= 1;
21     }
22     return ans%p;
23 }
24 ll comb(ll n, ll m, ll p) {
25     if (m > n) return 0;
26     return ((a[n]*qpow(a[m],p-2,p))%p * qpow(a[n-m],p-2,p)%p)%p;
27 }
28 ll lucas(ll n, ll m, ll p) {
29     if (m == 0) return 1;
30     return comb(n%p,m%p,p)*lucas(n/p,m/p,p)%p;
31 }
32 void init(ll p) {
33     a[0] = 1;
34     for (int i = 1; i <= p; i++) {
35         a[i] = a[i-1]*i%p;
36     }
37 }
38 int main() {
39     int t; scanf("%d",&t);
40     while (t--) {
41         ll n, m, p; scanf("%lld%lld%lld",&n,&m,&p);
42         init(p);
43         printf("%lld\n",lucas(n+m,m,p));
44     }
45     return 0;
46 }

 

posted @ 2019-10-22 20:26  麻辣猪仔  阅读(135)  评论(0编辑  收藏  举报