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  • Lucas定理

    引理1:(x^p equiv x (mod p))(p)是素数(费马小定理)

    证明略

    引理2:((a + b)^n = sumlimits_{i = 0}^n inom{n}{i} a^i b^{n - i})

    证明略

    定理1:((x + 1)^p equiv x + 1 equiv x^p + 1 (mod p))

    由引理1可得

    定理2:(inom{n}{m} equiv inom{lfloor frac{n}{p} floor}{lfloor frac{m}{p} floor} inom{n mod p}{m mod p} (mod p))

    证明:

    ((x + 1)^n equiv (x + 1)^{lfloor frac{n}{p} floor imes p} imes (x + 1)^{n mod p} equiv (x^p + 1)^{lfloor frac{n}{p} floor imes p} imes (x + 1)^{n mod p} (mod p))

    两边用二项式定理展开,

    (sumlimits_{i = 0}^n inom{n}{i} x^i equiv (sumlimits_{i = 0}^{lfloor frac{n}{p} floor} inom{lfloor frac{n}{p} floor}{i} x^{ip}) imes (sumlimits_{i = 0}^{n mod p} inom{n mod p}{i}) x^i equiv (sumlimits_{i = 0}^{lfloor frac{n}{p} floor} inom{lfloor frac{n}{p} floor}{i} x^{ip}) imes (sumlimits_{i = 0}^{p - 1} inom{n mod p}{i}) x^i (mod p))

    所以(inom{n}{i} x^i equiv inom{lfloor frac{n}{p} floor}{lfloor frac{i}{p} floor} x^{lfloor frac{i}{p} floor imes p} imes inom{n mod p}{i mod p} x^{i mod p})

    证毕

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  • 原文地址:https://www.cnblogs.com/tkandi/p/10416771.html
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