gcd,exgcd,crt

gcd证明

\[\begin{align*} &d=gcd(x,y)\\ &\therefore x|d,y|d\\ &x=ad,y=bd\\ &x\%y=ad\%bd=(a\%b)*d \\&\because(a\%b)\not\mid b \\&\therefore gcd(a,b)=gcd(b,a\%b)=1 \\&\therefore gcd(x,y)=gcd(y,x\%y) \end{align*} \]

exgcd

\[\begin{align}ax+by&=gcd(a,b)\nonumber\\bx'+(a\%b)y'&=gcd(b,a\%b)\nonumber\\bx'+(a-[\frac{a}{b}]*b)*y'&=gcd(b,a\%b)\nonumber\\ay'+b(x'-[\frac{a}{b}]*by')&=gcd(b,a\%b)\nonumber\end{align}\\\therefore ax+by=ay'+b(x'-[\frac{a}{b}]*by')\\\therefore x=y',y=x'-[\frac{a}{b}]*by' \]

方程\(ax+by=c\)当且仅当\(c\mid gcd(a,b)\)

\[ax'+by'=gcd(a,b)\\ a\frac{x'c}{gcd(a,b)}+b\frac{y'c}{gcd(a,b)}=c \]

crt

对于形如

\[\left\{ \begin{align} x&\equiv a_1(mod\ n_1)\nonumber\\ x&\equiv a_2(mod\ n_2)\nonumber\\ x&\equiv a_3(mod\ n_3)\nonumber\\ &~~\vdots\nonumber\\ x&\equiv a_k(mod\ n_k)\nonumber\\ \end{align} \right. \]

的线性同余方程组,设

\[n=\prod_{i=1}^k n_i\\m_i=\frac{n}{n_i}\\m_i^{-1}\equiv m_i(mod\ n_i)\\c_i=m_im_i^{-1}\nonumber \]

有

\[x\equiv\sum_{i=1}^{k}a_ic_i(mod\ n) \]

证明:我们需要证明x符合上面的所有方程式,易知当\(j\neq i\)时,该项\(mod\ n_i=0\)

\[\begin{align} x&\equiv\sum_{j=1}^{k}a_jc_j(mod\ n_i)\\ &\equiv a_ic_i(mod\ n_i) \\&\equiv a_i(mod\ n_i) \end{align} \]

证毕.