$\text{Question\ 1}$
$Description\ 1$
求解同于方程组:
\[\begin{cases} x \equiv2 \pmod7\\ x \equiv1 \pmod8\\ x \equiv3 \pmod{11}\\ \end{cases}\]$Ansewer\ 1$
运用中国剩余定理，有
\[\begin{align} &b_1=2, b_2=1, b_3=3 \\ &m_1=7, m_2=8, m_3=11 \\ &M_1=m_2m_3=88, M_2=m_1m_3=77, M_3=m_1m_2=56\\ &M_1^{-1}=2\pmod7,M_2^{-1}=5\pmod8,M_3^{-1}=1\pmod{11} \end{align}\]最终得到
\[\begin{align} x\equiv &\ b_1M_1M_1^{-1}+b_2M_2M_2^{-1}+b_3M_3M_3^{-1}\pmod{m_1m_2m_3} \\ x\equiv &\ 352+385+168 \pmod{616} \\ x\equiv &\ 289 \pmod{616} \end{align}\]$\text{Question\ 2}$
$Description\ 2$
求 $1112$ 和 $695$ 的最大公约数
$Ansewer\ 2$
\[\begin{align} 1112=& 1\times695+417 \\ 695 =& 1\times417+278 \\ 417 =& 1\times278+139 \\ 278 =& 12\times139+0 \end{align}\]由上得 $\gcd(1112,695)=139$
$\text{Question\ 3}$
$Description\ 3$
求 $GF_{(2)}[x]$ 上的多项式 $f(x)=x^6+x^5+x^4+x^2+1$ 与 $g(x)=x^3+1$ 的最大公因式，并求出多项式 $s(x),t(x)$，使 $s(x)f(x)+t(x)g(x)=\gcd(f(x),g(x))$，且 $\deg(s(x))<\deg(g(x))$，$\deg(t(x))<\deg(f(x))$
$Ansewer\ 3$
\[\begin{align} f(x)=& x^6+x^5+x^4+x^2+1,g(x)=x^3+1\\ f(x)=& q_0(x)g(x)+r_0(x) \ q_0(x)=x^3+x^2+x+1,\ r_0(x)=x+1 \\ g(x)=& q_1(x)r_0(x)+r_1(x) \ q_1(x)=x^2+x+1,\ r_1(x)=0 \\ \end{align}\]由上得
\[\begin{align} \gcd(f(x),g(x))=&r_0(x)=x+1 \\ r_0(x)=& f(x)+q_0(x)g(x) \\ s(x)f(x)+t(x)g(x)=&\gcd(f(x),g(x)) \\ f(x)+(x^3+x^2+x+1)g(x)=& x+1 \\ \end{align}\]$\text{Question\ 4}$
$Description\ 4$
计算勒让得符号 $\left(\cfrac{5}{439}\right)$
$Ansewer\ 4$
\[\begin{align} \left(\cfrac{5}{439}\right) =& (-1)^{\frac{5-1}{2}\cdot\frac{439-1}{2}} \left(\cfrac{439}{5}\right)\\ =&\left(\cfrac{4}{5}\right)\\ =&\left(\cfrac{2^2}{5}\right)\\ =&1 \end{align}\]$\text{Question\ 5}$
$Description\ 5$
用模重复平方算法计算 $2^{31} \pmod {37}$
$Ansewer\ 5$
有 $2^{31}_{10}=(11111)_2$
\[\begin{aligned} Init\ \ :ans=&1,base=2\\ step1: ans=& 1 \times 2 = 2\pmod {37},\ \ \ \ \ \ \ \ base=2\times 2 = 4 \pmod {37}\\ step2: ans=& 2 \times 4 = 8\pmod {37}, \ \ \ \ \ \ \ \ base=4\times 4 = 16\pmod {37}\\ step3: ans=& 8 \times 16 = 17\pmod {37}, \ \ \ \ base=16\times 16 = 34 \pmod {37}\\ step4: ans=& 17 \times 34 = 23\pmod {37}, \ \ base=34\times 34 = 9\pmod {37}\\ step5: ans=& 23 \times 9 = 22\pmod {37} \\ \end{aligned}\]最终得到 $2^{31}=22\pmod{37}$