扩展中国剩余定理
前置知识
问题
已知\(\begin{cases}x\equiv c_1\pmod{m_1}\\x\equiv c_2\pmod{m_2}\\x\equiv c_3\pmod{m_3}\\...\\x\equiv c_n\pmod{m_n}\end{cases}\)，求未知数\(x\)
求解
若能将这两个式子\(\begin{cases}x\equiv c_1\pmod{m_1}\\x\equiv c_2\pmod{m_2}\end{cases}\)合并为一个，则可以递推合并整个式子，故问题转化为求解
\[\begin{cases}x\equiv c_1\pmod{m_1}\\x\equiv c_2\pmod{m_2}\end{cases}
\]
可以转化为$$c_1+m_1k_1=c_2+m_2k_2$$
两边同时除以\(\gcd(m_1,m_2)\)得
\[\frac{m_1k_1}{\gcd(m_1,m_2)}=\frac{c_2-c_1}{\gcd(m_1,m_2)}+\frac{m_2k_2}{\gcd(m_1,m_2)}
\]
得
\[\frac{m_1k_1}{\gcd(m_1,m_2)}\equiv \frac{c_2-c_1}{\gcd(m_1,m_2)}\pmod{\frac{m_2}{\gcd(m_1,m_2)}}
\]
两边同时除以\(\frac{m_1}{\gcd(m_1,m_2)}\)得
\[k_1=\frac{c_2-c_1}{\gcd(m_1,m_2)}*inv(\frac{m_1}{\gcd(m_1,m_2)})\pmod{\frac{m_2}{\gcd(m_1,m_2)}}
\]
故
\[k_1=\frac{c_2-c_1}{\gcd(m_1,m_2)}*inv(\frac{m_1}{\gcd(m_1,m_2)})+{\frac{m_2}{\gcd(m_1,m_2)}}*y
\]
代入\(x=k_1m_1+c_1\)得
\[x=\frac{c_2-c_1}{\gcd(m_1,m_2)}*inv(\frac{m_1}{\gcd(m_1,m_2)})*m_1+c_1+{\frac{m_1m_2}{\gcd(m_1,m_2)}}*y
\]
即
\[x\equiv{{\frac{c_2-c_1}{\gcd(m_1,m_2)}*inv(\frac{m_1}{\gcd(m_1,m_2)})*m_1+c_1}\pmod{{\frac{m_1m_2}{\gcd(m_1,m_2)}}}}
\]
故
\[c_1'={\frac{c_2-c_1}{\gcd(m_1,m_2)}*inv(\frac{m_1}{\gcd(m_1,m_2)})*m_1+c_1}
\]
\[m_1'={\frac{m_1m_2}{\gcd(m_1,m_2)}}
\]
不断递推直到只剩下一个式子\(x\equiv{c'\pmod{m’}}\)，得\(x=c'+km'\)
