With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked to carefully design the cheapest route to go.
若每项比一个收敛级数小，收敛；若每项比一个发散级数大，发散。
两个不同级数的项\(u_n\)、\(v_n\)：
\[\lim_{n\to\infty}\frac{u_n}{v_n}=l\] 若比值极限为有限非零常数：同收敛发散；
若为0（v大很多）：若\(v_n\)收敛则\(u_n\)收敛；
若为\(\infty\)（v小很多）：若\(v_n\)发散则\(u_n\)发散。
\[\lim_{n\to\infty}\frac{a_{n+1} }{a_n}=\rho\begin{cases} \rho\lt 1 收敛 \\ \rho\gt 1 发散 \\ \rho=1 不确定 \end{cases}\]
\[\lim_{n\to\infty}\sqrt[n]{u_n}=\rho \begin{cases} \rho \lt 1 收敛 \\ \rho \gt 1 发散 \\ \rho = 1 不确定 \end{cases}\]
\[\sum_{n=1}^{\infty}(-1)^{n-1}u_n,(u_n\gt 0)\] 若\(u_n\)递减且极限为0：收敛。
\[e^x=\sum_{n=0}^\infty \frac{x^n}{n!}\] \[\sin x=\sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1} }{(2n+1)!}\] \[\cos x=\sum_{n=0}^\infty \frac{(-1)^{n}x^{2n} }{(2n)!}\] \[\frac{1}{1-x}=\sum_{n=0}^\infty x^n\] \[\frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n\] \[\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}\] \[-\ln(1-x)=\sum_{n=1}^\infty \frac{x^n}{n}\]
分离，化简，凑成基本形式
凑不出？可以求它的导数
假设想求\(f(x)\)的展开式：
这就是利用逐项可积性，所以先求导再积分并不是无用功。
Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.
Given two increasing sequences of integers, you are asked to find their median.
Excel can sort records according to any column. Now you are supposed to imitate this function.
A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For any pair of players, if there are some tables open when they arrive, they will be assigned to the available table with the smallest number. If all the tables are occupied, they will have to wait in a queue. It is assumed that every pair of players can play for at most 2 hours.
Your job is to count for everyone in queue their waiting time, and for each table the number of players it has served for the day.
Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.
对弧长的线积分（第一类）
1.判断奇偶性和对称性
\(\begin{cases} 若轴对称，则偶为2倍奇为0 \\ 若关于y=x对称，则可交换xy \end{cases}\)
2.计算
参数方程
\[\begin{cases} x=x(t) \\ y=y(t) \\ \alpha\le t\le\beta \\ \int_\alpha^\beta f(x(t),y(t))\sqrt{x'^2(t)+y'^2(t)}dt \end{cases} \]
直角坐标计算
\[\begin{cases} L:y=y(x),a\le x\le b \\ \int_a^bf(x,y(x))\sqrt{1+y'^2(x)}dx\end{cases}\]
极坐标计算
\[\begin{cases} \rho=\rho(\theta) \\ \alpha\le x\le\beta \\ \int_{\alpha}^{\beta} f( \rho ( \theta ) \cos \theta, \rho ( \theta ) \sin \theta ) \sqrt { \rho^2 + \rho'^2} d \theta \end{cases} \]
一奇偶对称，二替换变量，三乘根号；
参数方程导数平方和开根号，直角坐标函数导数平方加一开根号，极坐标函数和导数平方和开根号
对坐标的线积分（第二类）
\(\int_LPdx+Qdy=\int_L(P\cos\alpha+Q\cos\beta)ds\)
\(\int_LPdx+Qdy\)
\(\begin{cases} 直角坐标： \int_a^bP(x,\phi(x))dx+Q(x,\phi x)\phi'(x)dx\\ 参数方程：\int_\alpha^\beta P(\phi(t),\Phi(t))\phi'(t)dt+Q(\phi(t),\Phi(t))\Phi'(t)dt \end{cases}\)
直接替换求积分，记得\(d(x)\)里面的\(x\)替换后要求导。
条件
区域为闭曲线，方向逆时针（正向）
\(P(x,y)\)与\(Q(x,y)\)处处连续有一阶偏导数（路径无关）
公式
\(\oint_L Pdx+Qdy=\iint_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy\)
注意\(P\rightarrow dx\rightarrow y\)偏导，在后面；\(Q\rightarrow dy\rightarrow x\)偏导，在前面。
例 曲线为\(x^2+y^2+z^2=1\)与\(x+y+z=0\)的交线，则\(\oint_L (x+2y)^2ds\)为__________
解 直接算不好算，因此把曲线方程凑积分方程，尽量化简成常数。
1. 首先展开：\((x+2y)^2=x^2+4y^2+4y\)
2. 根据变量对称性：\(\oint x^2ds=\oint y^2ds=\oint z^2ds\)，\(\oint xyds=\oint yzds=\oint xzds\)
3. 原式=\(\oint \frac{5}{3}(x^2+y^2+z^2)+\frac{2}{3}[(x+y+z)^2-(x^2+y^2+z^2)]ds\)
4. =\(2Pi\)
若出现\(x^2+y^2\)，通常将直角坐标转换为极坐标。
若原式为\(\iint f(x,y)dxdy\)： 令\(\begin{cases} x=r\cos\theta \\ y=r\sin\theta \end{cases}\)，写成\(\int d\theta \int f(r\cos\theta, r\sin\theta)rdr\)的形式，然后考虑积分范围。 \(\theta\)的范围为积分区域所占的范围，\(r\)的上界是离原点远的边界函数，下界是离原点近的边界函数。
例如给定\(x^2+y^2=-2ax\)，易转换为\((x-a)^2+y^2=a^2\)，此时圆的边界为\(2a\cos\theta\)，写成\(\int^{2a\cos\theta}_{0}frdr\)。为什么下界是0呢？因为圆与y轴相切，那么从原点出发与圆相交的线只有两个交点，原点与相对的圆上的点。
例如给定\(x^2+y^2=-2ay\)，易转换为\(x^2+(y-a)^2=a^2\)，此时圆的边界为\(2a\sin\theta\)，写成\(\int^{2a\cos\theta}_{0}frdr\)。
例如给定\(x^2+y^2=a^2\)，圆的边界为\(a\cos\theta\)。
令积分部分等于A，对等式两边同时积分
A的积分为常数（考虑乘以积分范围）
计算式子中剩余的积分，即可算出A
最后把A带入最初给的式子中，即可得到函数f(x)
两边积分：\(A=\int \phi(x)dx+\int Adx\)
两边积分：\(A=\iint \phi(x,y)dxdx+\iint Adxdy\)
例如： \[\int_{-\frac{\pi}{4} }^{\frac{\pi}{2} }d\theta\int_0^{2\cos \theta}f(\rho\cos\theta,\rho\sin\theta)\rho d\rho\]
转换成 \[\int_0^{\sqrt{2} }\rho d\rho\int_{-\frac{\pi}{4} }^{\arccos\frac{\rho}{2} }f(\rho\cos\theta,\rho\sin\theta)d\theta+\int_{\sqrt{2} }^2\rho d\rho\int_{-\arccos\frac{\rho}{2} }^{\arccos\frac{\rho}{2} }f(\rho\cos\theta,\rho\sin\theta)d\theta+\]
黑色线为\(\rho=\sqrt 2\)的曲线，左侧为第一个积分的积分区域，右侧为第二个积分的积分区域。因为要先按半径积分，而圆不是一个完整的圆，被直线切掉了一块，因此以线段画弧先积分，再加上弧右侧的积分即可。
注意，积分\(\theta\)的角度不是显而易见的常数，而是关于\(\rho\)的函数，而这个关于\(\rho\)的函数随着\(\rho\)的轨迹便是圆的边缘线。
这个函数怎么得出的？\(\rho=2\cos\theta\to\theta=\arccos{\frac{\rho}{2} }\)
\[\Gamma(n)=\int^{+\infty}_0x^{n-1}e^{-x}dx\]
\[\Gamma(n)=(n-1)!\]
\[\Gamma(\frac{1}{2})=\sqrt{\pi}\]
\[\int_0^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi} }{2}\]
\[\int xe^xdx=\int (xe^x+e^x-e^x)dx=(x-1)e^x+C\]
\[\int \frac{dx}{\sin x}=\int \frac{dx}{2\sin \frac{x}{2}\cos \frac{x}{2} }=\int \frac{\sec^2 \frac{x}{2} d \frac{x}{2} }{\tan \frac{x}{2} }=\int \frac{d\tan \frac{x}{2} }{\tan \frac{x}{2} }=\ln (\tan \frac{x}{2})+C\]
\[\int \frac{1}{1+e^x}dx=\int \frac{e^{-x} }{1+e^{-x} }dx=-\int \frac{1}{1+e^{-x} }d(1+e^{-x})=-\ln (1+e^{-x})+C\]
\[\int \sqrt{\frac{1-t}{1+t} }dt\to令\frac{1-t}{1+t}=u^2\to t^2=\frac{1-u}{1+u}\]
\[dt=\frac{-4u}{(1+u^2)^2}du\to \int \frac{-4u^2}{(1+u^2)^2}du=-4\int(\frac{1}{1+u^2}-\frac{1}{(1+u^2)^2})du\]
后面部分令\(u=\tan\theta\)，即可计算出。