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  • 数学相关结论整理(没有证明)

    Gcd 与 Lcm

    [ ext{lcm}(S)=prod_{Tsubseteq S}gcd(T)^{{(-1)}^{|T|+1}}\ f(n)=af(n-1)+bf(n-2),gcd(a,b)=1 \ gcd(f(x),f(y))=f(gcd(x,y))\ gcd(x^a-1,x^b-1)=x^{gcd(a,b)}-1 ]

    欧拉定理

    [a^{phi(p)}=1 ( ext{mod } p) (gcd(a,p=1)) ]

    扩展欧拉定理

    [a^b=a^{min (b\%phi(p)+phi(p),b)}( ext {mod }p) ]

    中国剩余定理

    [forall i<j le n gcd(b_i,b_j)=1\ egin{cases} x≡a_1mod b_1\ x≡a_2mod b_2\ ...\ x≡a_nmod b_n\ end{cases}\ x=sum_{i = 1}^na_i imes ans_i imesprod_{j=1}^n b_j imesfrac 1 {b_i} ]

    扩展中国剩余定理

    [∃ i<j le n gcd(b_i,b_j)>1\ egin{cases} x≡a_1mod b_1\ x≡a_2mod b_2\ end{cases}\ x=b_1x_1+a_1=b_2x_2+a_2\ b_1x_1+(-b_2)x_2=a_2-a_1\ x=b_1x_1+a_1 mod ext{lcm}(b_1,b_2) ]

    卢卡斯定理

    [pin m Prime \inom n m =inom {lfloorfrac n p floor}{lfloorfrac m p floor} imesinom {n ext{ mod }p}{m ext{ mod }p} ]

    扩展卢卡斯定理

    [P = prod_{i = 1}^np_i^{k_i} \ inom n m=frac{frac {n!}{p_i^{u}}}{frac {m!}{p_i^{v}} frac{{(n-m)!}}{p_i^w}} imes p_i^{u-v-w}mod p_i^{k_i}\ ext{Let } f(n)=max(x) ext{ makes }n!mod p_i^x=0\ g(n)=frac{n!}{p_i^{f(n)}}\ f(n)=f(lfloorfrac{n}{p_i} floor) imeslfloorfrac{n}{p_i} floor\ g(n)=g(lfloorfrac{n}{p_i} floor) imes frac{p_i^{k_i}!}{p_i^{u_1}} imes frac {(n ext{ mod }p_i^{k_i})!}{p_i^{u2}} \ ext{Finally just merge each ans for mod }p_i^{k_i} ext{ by CRT.} ]

    BSGS 算法

    [a^x=b ext{ (mod p)}\ x = pm-q (m=sqrt p)\ a^{pm-q}=b ext{ (mod p)}\ a^{pm}=b imes a^q ext{ (mod p)}\ ]

    扩展 BSGS 算法

    [a imes b ext{ mod } p=frac a d imes frac b d ext{ mod } frac p d\ ext{Let d = gcd(a, p)}\ a^{x-1} imesfrac a d = frac{b}{d}( ext{mod } frac p {d})\ ext{Specially When b = 1, x = 0.} ]

    组合数

    [inom n m =inom {n - 1} m+inom {n - 1}{m - 1}\ sum_{i = 0}^ninom n i=2^i \ sum_{i = B}^{n} inom i B=inom {n + 1} {B + 1}\ inom n {a + b}=sum_{i = 0}^n inom i a inom {n - i } b\ inom n m =sum_{j = 0}^m inom {n - m - 1 + j} {j} ]

    二项式定理

    [(a+b)^n=sum_{i = 0}^n inom n i a^ib_{n-i} ]

    二项式反演

    [f(n)=sum_{i = 0}^n inom n ig(i)→g(n)=sum_{i = 0}^n(-1)^{n-i}inom n i f(i)\ f(k)=sum_{i = k}^ninom i kg(i)→g(k)=sum_{i = k}^n(-1)^{i-k}inom i k f(i) ]

    上升幂与下降幂

    [x^{underline{k}}=prod_{i = 0}^{k-1}(x-i),x^{overline{k}}=prod_{i = 0}^{k - 1}(x+i)\x^{underline{k}}=inom x k imes k! ]

    斯特林数

    [genfrac{[}{]}{0pt}{}{n}{m}=(n-1)genfrac{[}{]}{0pt}{}{n-1}{m}+genfrac{[}{]}{0pt}{}{n-1}{m-1}\ genfrac{{}{}}{0pt}{}{n}{m}=mgenfrac{{}{}}{0pt}{}{n-1}{m}+genfrac{{}{}}{0pt}{}{n-1}{m-1}\ genfrac{{}{}}{0pt}{}{n}{m}=frac 1 {m!}sum_{k = 0}^m(-1)^kinom m k (m-k)^n\  x^{overline n}=sum_{i = 0}^n genfrac{[}{]}{0pt}{}{n}{i}x^i,x^n=sum_{i = 0}^ngenfrac{{}{}}{0pt}{}{n}{i}x^{underline{i}}\ x^{underline n}=(-1)^n(-x)^{overline n}\ x^{underline{n}}=sum_{i = 0}^n(-1)^{n-i}genfrac{[}{]}{0pt}{}{n}{i}x^i,x^n=sum_{i = 0}^n(-1)^{n-i}genfrac{{}{}}{0pt}{}{n}{i}x^{overline i}\ n!=sum_{i = 0}^ngenfrac{[}{]}{0pt}{}{n}{i} ]

    斯特林反演

    [f(n)=sum_{i = 0}^ngenfrac{{}{}}{0pt}{}{n}{i}g(i)→g(n)=sum_{i = 0}^n(-1)^{n-i}genfrac{[}{]}{0pt}{}{n}{i}f(i) ]

    常见积性函数

    [ ext{Let n = }prod_{i = 1}^mp_i^{k_i}\ e(n)=[n=1]\ I(n)=1\id(n)=(n)\ mu(n)=[max_{i = 1}^m(k_i)le1](-1)^m \ phi(n)=sum_{i = 1}^n[gcd(i,n)=1]]

    狄利克雷卷积

    [(f*g)(n) = sum_{d|n}f(d)g(frac n d)\ f*g=g*f\(f*g)*h=f*(g*h)\ (f+g)*h=f*h+g*h\f*e=f]

    积性函数相关套路

    [sum_{d|n}mu(d)=[n=1],e=mu*1\ sum_{d | n}phi(d)=n, phi*I=id\ sum_{d|n}frac {mu(d)}{d}=frac {phi(n)} n,phi=mu*id ]

    杜教筛

    [f*g=h, ext{How to get the sum of f.}\sum_{i = 1}^nh(i)=sum_{i = 1}^nsum_{d | i}f(d)g(frac i d) \ sum_{i = 1}^nh(i)=sum_{d = 1}^ng(d)sum_{k = 1}^{lfloorfrac n d floor}f(k) \ sum_{i = 1}^nh(i)-sum_{d = 2}^ng(d)sum_{k = 1}^{lfloorfrac n d floor}f(k) =g(1)sum_{i = 1}^nf(i) ]

    FFT 共轭优化

    [F(a+bi)=F(a)+iF(b)\ (overline{F(a+bi)})'=F(a)-iF(b)\ ]

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  • 原文地址:https://www.cnblogs.com/brunch/p/10638783.html
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