Lucas定理

定义

对于任意质数p

$Huge C_m^nequiv C_{iggllfloorfrac{m}{p}iggr floor}^{iggllfloorfrac{n}{p}iggr floor}*C_{m mod p}^{n mod p} (MOD p)$

证明

对于任意质数p都有

$huge C_p^iequiv0 MOD p(i ot= 0&&i ot=p)$

通过二项式定理，我们可得

$huge (x+1)^p=sumlimits_{i=0}^pC_p^ix^i$

因为$huge C_p^iequiv0 MOD p(i ot= 0&&i ot=p)$

所以$huge (x+1)^pequiv x^p+1 (MOD p)$

于是我们对于任意正整数m有

$huge (x+1)^m=(x+1)^{iggllfloorfrac{m}{p}iggr floor*p}*(x+1)^{m mod p}equiv (x^p+1)^{iggllfloorfrac{m}{p}iggr floor}*(x+1)^{m mod p} (MOD p)$

用二项式定理展开就是

 $huge sumlimits_{i=0}^mC_m^ix^iequiv sumlimits_{i=0}^{iggllfloorfrac{m}{p}iggr floor}C_{iggllfloorfrac{m}{p}iggr floor}^ix^{i*p}*sumlimits_{i=0}^{m mod p}C_{m mod p}^ix^i (MOD p)$

当x指数为n的项就为,

$huge C_m^nx^nequiv C_{iggllfloorfrac{m}{p}iggr floor}^{iggllfloorfrac{n}{p}iggr floor}x^{iggllfloorfrac{m}{p}iggr floor*p}*C_{m mod p}^{n mod p}x^{n mod p} (MOD p)$

两边约去x的n次方就为

$huge C_m^nequiv C_{iggllfloorfrac{m}{p}iggr floor}^{iggllfloorfrac{n}{p}iggr floor}*C_{m mod p}^{n mod p} (MOD p)$