T1:

#include<iostream>
using namespace std;
int main(void)
{
   return 0;
}

pow(a, b);

brust-force:

$a ^b$ = a* a* ... *a

for: 1- b
long long tim *= a;// tim < 10^9 a < 10^9
if(>10^9) return -1;
(10^9)


return tim;

solve:

$ 7 = (111)_2 = 1 \times2^0 + 1 \times2^1 + 1\times2^2$

b 二进制

$a^b = a^7 = a^{1} \times a^2 \times a^{4}$

T2:

$$\begin{aligned} & 设x = p, y = q, 则有 \\ & x \times y = n;\\ & e \times d = (x - 1)(y - 1) + 1 = x\times y -(x + y) + 2= n - (x + y) + 2 \\ & 则 x + y = n - e \times d + 2 \\ & x \times (n - e \times d + 2 - x) = n\\ & 设 n - e \times d + 2 = m \\ & x \times (m - x) = n;\\ & x^2 - mx + n = 0;\\ \end {aligned} $$$$\begin{aligned} & ax^2 + bx + c = 0;\\ &x_1 = \frac {-b + \sqrt{b^2-4ac}}{2a},x_2 = \frac {-b - \sqrt{b^2-4ac}}{2a}\\ & x_1 + x_2 = \frac {-2b}{2a} = -\frac b a \\ & x_1 x_2 = \frac{(-b + \sqrt{\Delta})(-b - \sqrt\Delta)}{{(2a)}^2}\\ & =\frac{b^2 - \Delta}{4a^2} = \frac { b^2 - (b^2 - 4ac)}{4a^2} = \frac {c}{a} \\ & x_1 x_2 = \frac c a\\ \end{aligned} $$

T3:

逻辑运算：

and(&) 有一个为0，结果为0， 全1，结果是1， 1 & 0 = 0

or(|): 有一个为1， 结果是1，全0，结果是0

not(!): !1 = 0, !0 = 1

xor(^): 不进位的加法