[从头学数学] 第211节 带着计算机去高考（三）

剧情提要：

[机器小伟]在[工程师阿伟]的陪同下进入了[九转金丹]之第八转的修炼。设想一个场景：

如果允许你带一台不连网的计算机去参加高考，你会放弃选择一个手拿计算器和草稿本吗

？阿伟决定和小伟来尝试一下用计算机算高考题会是怎样的感觉。

正剧开始：

星历2016年05月18日 14:57:54, 银河系厄尔斯星球中华帝国江南行省。

[工程师阿伟]正在和[机器小伟]一起做着2002年的江苏省数学高考题]。

<span style="font-size:18px;">#题1
>>>
0.5000110003630133

def tmp1():
    d = geo.plDistance2D([1,0], [[0,0], [1.732, 1]]);
    print(d);</span>

<span style="font-size:18px;">#题2
def tmp2():
    a = (0.5+0.866j)**3;
    print(a);

>>>
(-0.9999339999999999+3.8104000000038774e-05j)</span>

<span style="font-size:18px;">//题3
	if (1) {
        var r = 20;
        config.setSector(1,1,1,1);
        config.graphPaper2D(0, 0, r);
        config.axis2D(0, 0,180);              

        //坐标轴设定
        var scaleX = 2*r, scaleY = 2*r;
        var spaceX = 1, spaceY = 1;
        var xS = -10, xE = 10;
        var yS = -10, yE = 10;
        config.axisSpacing(xS, xE, spaceX, scaleX, 'X');
        config.axisSpacing(yS, yE, spaceY, scaleY, 'Y');            

        var transform = new Transform();
        //存放函数图像上的点
        var a = [], b = [], c = [], d = [];          

        //需要显示的函数说明
                    //希腊字母表(存此用于Ctrl C/V
            //ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ
            //αβγδεζηθικλμνξοπρστυφχψω
        var f1 = '(1+x)(1-|x|)', f2 = '', f3 = '', f4 = '';        

		//(1-(x+2)^2)^0.5
		for (var x = xS; x <= xE; x+=0.2) {
			a.push([x, taskFun(x)]);

        }  

        //存放临时数组
        var tmp = [];        

        //显示变换
        if (a.length > 0) {
            a = transform.scale(transform.translate(a, 0, 0), scaleX/spaceX, scaleY/spaceY);
            //函数1
            tmp = [].concat(a);
            shape.pointDraw(tmp, 'red');
            tmp = [].concat(a);
            shape.multiLineDraw(tmp, 'pink');          

            plot.setFillStyle('red');
            plot.fillText(f1, 100, -90, 200);
        }  

    } 

function taskFun(x) {
	return (1+x)*(1-Math.abs(x));
}</span>

<span style="font-size:18px;">//题4
	if (1) {
        var r = 20;
        config.setSector(1,1,1,1);
        config.graphPaper2D(0, 0, r);
        config.axis2D(0, 0,180);              

        //坐标轴设定
        var scaleX = 2*r, scaleY = 2*r;
        var spaceX = 1, spaceY = 1;
        var xS = -10, xE = 10;
        var yS = -10, yE = 10;
        config.axisSpacing(xS, xE, spaceX, scaleX, 'X');
        config.axisSpacing(yS, yE, spaceY, scaleY, 'Y');            

        var transform = new Transform();
        //存放函数图像上的点
        var a = [], b = [], c = [], d = [];          

        //需要显示的函数说明
                    //希腊字母表(存此用于Ctrl C/V
            //ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ
            //αβγδεζηθικλμνξοπρστυφχψω
        var f1 = 'sinx-cosx', f2 = '', f3 = '', f4 = '';        

		//(1-(x+2)^2)^0.5
		for (var x = 0; x <= Math.PI*2; x+=0.2) {
			a.push([x, taskFun(x)]);

        }  

        //存放临时数组
        var tmp = [];        

        //显示变换
        if (a.length > 0) {
            a = transform.scale(transform.translate(a, 0, 0), scaleX/spaceX, scaleY/spaceY);
            //函数1
            tmp = [].concat(a);
            shape.pointDraw(tmp, 'red');
            tmp = [].concat(a);
            shape.multiLineDraw(tmp, 'pink');          

            plot.setFillStyle('red');
            plot.fillText(f1, 100, -90, 200);
        }  

    } 

}  

function taskFun(x) {
	return Math.sin(x) - Math.cos(x);
}</span>

<span style="font-size:18px;">#题5
def tmp5():
    M = [];
    N = [];

    for i in range(-10, 10, 1):
        M.append(i/2+1/4);
        N.append(i/4+1/2);

    print('M=', M);
    print('N=', N);

    setM = set(M);
    setN = set(N);

    print(setM.intersection(setN));

>>>
M= [-4.75, -4.25, -3.75, -3.25, -2.75, -2.25, -1.75, -1.25, -0.75, -0.25, 0.25, 0.75, 1.25, 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75]
N= [-2.0, -1.75, -1.5, -1.25, -1.0, -0.75, -0.5, -0.25, 0.0, 0.25, 0.5, 0.75, 1.0, 1.25, 1.5, 1.75, 2.0, 2.25, 2.5, 2.75]
{-0.75, -0.25, 2.75, 1.75, 0.25, 1.25, 2.25, -1.25, 0.75, -1.75}
</span>

<span style="font-size:18px;">//题6
	if (1) {
        var r = 20;
        config.setSector(1,1,1,1);
        config.graphPaper2D(0, 0, r);
        config.axis2D(0, 0,180);              

        //坐标轴设定
        var scaleX = 2*r, scaleY = 2*r;
        var spaceX = 2, spaceY = 2;
        var xS = -10, xE = 10;
        var yS = -10, yE = 10;
        config.axisSpacing(xS, xE, spaceX, scaleX, 'X');
        config.axisSpacing(yS, yE, spaceY, scaleY, 'Y');            

        var transform = new Transform();
        //存放函数图像上的点
        var a = [], b = [], c = [], d = [];          

        //需要显示的函数说明
                    //希腊字母表(存此用于Ctrl C/V
            //ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ
            //αβγδεζηθικλμνξοπρστυφχψω
        var f1 = 'x=t^[2], y=2t', f2 = '', f3 = '', f4 = '';        

		//(1-(x+2)^2)^0.5
		for (var x = xS; x <= xE; x+=0.2) {
			a.push([x*x, 2*x]);

        }  

        //存放临时数组
        var tmp = [];        

        //显示变换
        if (a.length > 0) {
            a = transform.scale(transform.translate(a, 0, 0), scaleX/spaceX, scaleY/spaceY);
            //函数1
            tmp = [].concat(a);
            shape.pointDraw(tmp, 'red');
            tmp = [].concat(a);
            shape.multiLineDraw(tmp, 'pink');          

            plot.setFillStyle('red');
            plot.fillText(f1, 100, -180, 200);
        }  

    } </span>

<span style="font-size:18px;">#题7
def tmp7():
    s = ['2/3pr^[3]','1/3pr^[2]h'];
    #h = 2r
    thita = 2*math.atan(0.5);
    print(math.cos(thita));

>>>
0.6000000000000001</span>

<span style="font-size:18px;">//题10
	if (1) {
        var r = 20;
        config.setSector(1,1,1,1);
        config.graphPaper2D(0, 0, r);
        config.axis2D(0, 0,180);              

        //坐标轴设定
        var scaleX = 2*r, scaleY = 2*r;
        var spaceX = 2, spaceY = 2;
        var xS = -10, xE = 10;
        var yS = -10, yE = 10;
        config.axisSpacing(xS, xE, spaceX, scaleX, 'X');
        config.axisSpacing(yS, yE, spaceY, scaleY, 'Y');            

        var transform = new Transform();
        //存放函数图像上的点
        var a = [], b = [], c = [], d = [];          

        //需要显示的函数说明
                    //希腊字母表(存此用于Ctrl C/V
            //ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ
            //αβγδεζηθικλμνξοπρστυφχψω
        var f1 = 'y = 1 - 1/(x-1)', f2 = '', f3 = '', f4 = '';
        var  y = 0;
		//(1-(x+2)^2)^0.5
		for (var x = xS; x <= xE; x+=0.2) {  

			y = taskFun(x);

			if (y != Number.NaN) {
				a.push([x, y]);
			}

        }  

        //存放临时数组
        var tmp = [];        

        //显示变换
        if (a.length > 0) {
            a = transform.scale(transform.translate(a, 0, 0), scaleX/spaceX, scaleY/spaceY);
            //函数1
            tmp = [].concat(a);
            shape.pointDraw(tmp, 'red');
            tmp = [].concat(a);
            shape.multiLineDraw(tmp, 'pink');          

            plot.setFillStyle('red');
            plot.fillText(f1, 100, -180, 200);
        }  

    } 

}  

function taskFun(x) {
	return 1- 1/(x-1);
}
</span>

<span style="font-size:18px;">#题11
def tmp11():
    #从正方体中选两个相对的面
    step1 = 3;

    #每组对面有各个中间面
    step2 = 4;

    result = step1 * step2;
    print(result);

>>>
12</span>

<span style="font-size:18px;">#题12
def tmp12():
    print(95993*(1.073)**4);

>>>
127244.3338498723</span>

<span style="font-size:18px;">#题14
def tmp14():
    #5/k-1 = 4 =>k =1</span>

<span style="font-size:18px;">#题15
def tmp15():
	a = ['x^[2]', '1'];
    b = ['x', '-2'];

    a = strformat(a);
    b = strformat(b);

    #乘方7次
    c = strpow_n(b, 7);
    #整理
    d = strcombine(c);

    #多项式乘法一次
    e = strdot(a, d);
    #整理
    f = strcombine(e);
    print(f);

>>>
['(1)*x^[9]', '(-14)*x^[8]', '(85)*x^[7]', '(-294)*x^[6]', '(644)*x^[5]', '(-952)*x^[4]', '(1008)*x^[3]', '(-800)*x^[2]', '(448)*x^[1]', '(-128)']
</span>

<span style="font-size:18px;">#题16
def tmp16():
    a = [1, 2, 1/2, 3, 1/3, 4, 1/4];

    sum_ = 0;

    for i in range(len(a)):
        sum_ += f(a[i]);

    print(sum_);

def f(x):
    return x**2/(1+x**2);
</span>

<span style="font-size:18px;">#题20
def tmp20():
    #a_[1] = 30;
    #a_[n] = a_[n-1]*0.94+x
    q = alg.strformat(['0.94']);
    x = alg.strformat(['x']);

    a_1 = alg.strformat(['30']);
    a_2 = alg.stradd(alg.strdot(q, a_1), x);

    a_2 = alg.strcombine(a_2);

    for i in range(25):
        a_2 = alg.stradd(alg.strdot(q, a_2), x);
        a_2 = alg.strcombine(a_2);
        print(a_2);

    #13x+6<60 x<3

    #x 是以1为首项, 0.94为公比的等比数列的前N项和
    Sx = 1/0.06 = 16;
    #x < 60/16

#这是25年的数据
>>>
['(26.508)', '(1.94)*x^[1]']
['(24.91752)', '(2.8236)*x^[1]']
['(23.422469)', '(3.654184)*x^[1]']
['(22.017121)', '(4.434933)*x^[1]']
['(20.696094)', '(5.168837)*x^[1]']
['(19.454328)', '(5.858707)*x^[1]']
['(18.287068)', '(6.507185)*x^[1]']
['(17.189844)', '(7.116754)*x^[1]']
['(16.158453)', '(7.689749)*x^[1]']
['(15.188946)', '(8.228364)*x^[1]']
['(14.277609)', '(8.734662)*x^[1]']
['(13.420952)', '(9.210582)*x^[1]']
['(12.615695)', '(9.657947)*x^[1]']
['(11.858753)', '(10.07847)*x^[1]']
['(11.147228)', '(10.473762)*x^[1]']
['(10.478394)', '(10.845336)*x^[1]']
['(9.84969)', '(11.194616)*x^[1]']
['(9.258709)', '(11.522939)*x^[1]']
['(8.703186)', '(11.831563)*x^[1]']
['(8.180995)', '(12.121669)*x^[1]']
['(7.690135)', '(12.394369)*x^[1]']
['(7.228727)', '(12.650707)*x^[1]']
['(6.795003)', '(12.891665)*x^[1]']
['(6.387303)', '(13.118165)*x^[1]']
['(6.004065)', '(13.331075)*x^[1]']
</span>

<span style="font-size:18px;">//题21
	if (1) {
        var r = 20;
        config.setSector(1,1,1,1);
        config.graphPaper2D(0, 0, r);
        config.axis2D(0, 0,180);              

        //坐标轴设定
        var scaleX = 2*r, scaleY = 2*r;
        var spaceX = 2, spaceY = 2;
        var xS = -10, xE = 10;
        var yS = -10, yE = 10;
        config.axisSpacing(xS, xE, spaceX, scaleX, 'X');
        config.axisSpacing(yS, yE, spaceY, scaleY, 'Y');            

        var transform = new Transform();
        //存放函数图像上的点
        var a = [], b = [], c = [], d = [];          

        //需要显示的函数说明
                    //希腊字母表(存此用于Ctrl C/V
            //ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ
            //αβγδεζηθικλμνξοπρστυφχψω
        var f1 = 'y = x^2+|x-2|+1', f2 = 'y = x^2+|x+2|+1', f3 = '', f4 = '';
        var  y1 = y2 = 0;
		//(1-(x+2)^2)^0.5
		for (var x = xS; x <= xE; x+=0.2) {  

			y1 = x*x+Math.abs(x-2)+1;
			y2 = x*x+Math.abs(x+2)+1;

			if (y1 != Number.NaN) {
				a.push([x, y1]);

			}

			if (y2 != Number.NaN) {
				b.push([x, y2]);
			}

        }  

        //存放临时数组
        var tmp = [];        

        //显示变换
        if (a.length > 0) {
            a = transform.scale(transform.translate(a, 0, 0), scaleX/spaceX, scaleY/spaceY);
            //函数1
            tmp = [].concat(a);
            shape.pointDraw(tmp, 'red');
            tmp = [].concat(a);
            shape.multiLineDraw(tmp, 'pink');          

            plot.setFillStyle('red');
            plot.fillText(f1, 100, -90, 200);
        }  

		if (b.length > 0) {
            b = transform.scale(transform.translate(b, 0, 0), scaleX/spaceX, scaleY/spaceY);
            //函数1
            tmp = [].concat(b);
            shape.pointDraw(tmp, 'blue');
            tmp = [].concat(b);
            shape.multiLineDraw(tmp, 'blue');          

            plot.setFillStyle('blue');
            plot.fillText(f2, 100, -120, 200);
        }  

    } </span>

好了，这次题就做到这里，这张试卷的难度比前两年的要低上2环。

先说明一下这个环的意思：

你找到一把钥匙，开了一个宝箱，然后拿走宝物，这叫一环难度；

这把钥匙开了一个宝箱，里面没有宝物，只有另一把钥匙，

你再找一个宝箱开了，宝物在里面，你拿到手，这叫二环难度；

以此类推啦，这张试卷的最难的题也就四环难度，前两年的达到六环。

一般两环难度以下是送分题，大概30到40分，三环难度的再给30分。

四环难度要是给个50分，这张卷子就已经比较厚道了。然后是五环难度，

一般出到六环难度就是伤天害理的感觉了。

前两年的卷子大致是：二环20分，三环20分，四环40分，五环40分，然后是六环，

然后陷阱还多，明明一个四环题，伪装成两环的，特别阴险，这就是江苏卷的特色。

这张是很温和的啦，运算量也不大。

这次[工程师阿伟]终于把早就想做的那个代数式运算工具搞出来了，这可是很好玩的一个东西。

<span style="font-size:18px;">###
# @usage   代数式字符串的运算
# @author  mw
# @date    2016年05月17日  星期二  16:48:56
# @param
# @return
#
###

#计算代数式用, 传入的是单项式，返回coef*expr的形式
def strmono(s):
    #'x', '-x', '2x', '-2x', '-2x^[2]', '3x_[2]^[3]', '-3x_[2]^[3]'
    stmp = s;
    size = len(stmp);
    alphaIndex = 0;
    signIndex = 0;

    for i in range(size):
        if (stmp[i].isalpha()):
            alphaIndex = i;
            break;
        if (i >= size-1):
            alphaIndex = i+1;

    if (stmp[0] == '-'):
        signIndex = 1;
        if (signIndex >= alphaIndex):
            return monoformat('(-1)*'+stmp[alphaIndex:]);
        else:
            if alphaIndex >= size:
                return monoformat('(-'+stmp[signIndex:alphaIndex]+')');
            return monoformat('(-'+stmp[signIndex:alphaIndex]+')*'+stmp[alphaIndex:]);
    else:
        signIndex = 0;
        if (signIndex >= alphaIndex):
            return monoformat('(1)*'+stmp[alphaIndex:]);
        else:
            if alphaIndex >= size:
                return monoformat('('+stmp[signIndex:alphaIndex]+')');
            return monoformat('('+stmp[signIndex:alphaIndex]+')*'+stmp[alphaIndex:]);

#计算两个单项式的乘积
def strmul(mono1, mono2):
    #这个处理是保证每个单项式统一格式(coef)*expr
    '''
    if (mono1[0] != '(' or mono2[0] != '('):
        #如果没有规格化，那么就做一下
        mono1 = strmono(mono1);
        mono2 = strmono(mono2);
    '''
    stmp1 = mono1;
    stmp2 = mono2;

    #乘号的位置
    signIndex1 = stmp1.find('*');
    signIndex2 = stmp2.find('*');
    if (signIndex1 == -1):
        coef1 = stmp1;
        expr1 = '';
    else:
        coef1 = stmp1[:signIndex1];
        expr1 = stmp1[signIndex1+1:];

    if (signIndex2 == -1):
        coef2 = stmp2;
        expr2 = '';
    else:
        coef2 = stmp2[:signIndex2];
        expr2 = stmp2[signIndex2+1:];

    coef = coef1+'*'+coef2;

    if (signIndex1 == -1 or signIndex2 == -1):
        expr = expr1+expr2;
    else:
        expr = expr1+'*'+expr2;

    if (expr == ''):
        return '('+str(round(eval(coef), 6))+')';
    return '('+str(round(eval(coef), 6))+')*'+expr;

#计算两个单项式的商
def strdiv(s1, s2):
    #这个处理是保证每个单项式统一格式(coef)*expr
    stmp1 = strmono(s1);
    stmp2 = strmono(s2);

    #乘号的位置
    signIndex1 = stmp1.find('*');
    signIndex2 = stmp2.find('*');
    if (signIndex1 == -1):
        coef1 = stmp1;
        expr1 = '';
    else:
        coef1 = stmp1[:signIndex1];
        expr1 = stmp1[signIndex1+1:];

    if (signIndex2 == -1):
        coef2 = stmp2;
        expr2 = '';
    else:
        coef2 = stmp2[:signIndex2];
        expr2 = stmp2[signIndex2+1:];

    coef = coef1+'/'+coef2;

    if (signIndex1 == -1 and signIndex2 != -1):
        expr = '('+expr2+')^[-1]';
    elif (signIndex1 == -1 or signIndex2 == -1):
        expr = expr1+expr2;
    else:
        expr = expr1+'/'+expr2;

    if (expr == ''):
        return '('+str(round(eval(coef), 6))+')';
    return '('+str(round(eval(coef), 6))+')*'+expr;

#找一个字符串中所有待查找子串的位置，返回位置阵列
def findall(string, sub):
    size = len(string);
    index = [];

    cur = string.find(sub);
    index.append(cur)

    while (index[-1] != -1):
        cur = string.find(sub, index[-1]+1);
        index.append(cur);   

    return index;

#计算单项式的乘方, s^n
def strpow(s, n):
    stmp = strmono(s);
    signIndex = stmp.find('*');
    if (signIndex == -1):
        coef = stmp+'**'+str(n);
        expr = '';
        return '('+str(round(eval(coef), 6))+')';
    else:
        coef = stmp[:signIndex]+'**'+str(n);
        expr = '('+stmp[signIndex+1:]+')^['+str(n)+']';

    return '('+str(round(eval(coef), 6))+')*'+expr;

#计算代数式用，传入的两个阵列都具有['s1', 's2', ..., 'sn']这样的格式
def strdot(array1, array2):
    size1 = len(array1);
    size2 = len(array2);
    result = [];

    for i in range(size1):
        for j in range(size2):
            result.append(strmul(array1[i], array2[j]));

    return result;

#把格式化后的单项式分解成[coef, expr]对组的形式
def explodemono(mono):
    stmp = mono;

    #乘号的位置
    signIndex = stmp.find('*');
    if (signIndex == -1):
        coef = stmp;
        expr = '';
    else:
        coef = stmp[:signIndex];
        expr = stmp[signIndex+1:];

    return [coef, expr];

#合并同类项，传入的阵列具有['s1', 's2', ..., 'sn']这样的格式
def strcombine(array):
    size = len(array);
    explode = [];
    for i in range(size):
        #这里传入的阵列已经是规格化后的了，否则要加一层strmono处理。
        explode.append(explodemono(monocombine(array[i])));

    result = [];

    for i in range(size):
        size_1 = len(result);

        if size_1 <= 0:
            result.append(explode[i]);
        else:
            for j in range(size_1):
                if result[j][1] == explode[i][1]:
                    result[j][0] = result[j][0] + '+' + explode[i][0];
                    break;

                if j >= size_1-1:
                    result.append(explode[i]);

    result_1 = [];

    size_1 = len(result);
    for j in range(size_1):
        result[j][0] = str(round(eval(result[j][0]), 6));

        if (result[j][0] == '0'):
            result_1.append('(0)');
        else:
            tmps = result[j][1];
            if (tmps == ''):
                result_1.append('('+result[j][0]+')');
            else:
                result_1.append('('+result[j][0]+')*'+result[j][1]);

    return result_1;

#指数为正整数的乘方
def strpow_n(array, n):
    #计算
    result = [];

    if (n == 1):
        result = array;
    elif (n == 2):
        result = strdot(array, array);

    elif (n >= 3):
        tmp = strdot(array, array);
        n -= 2;

        while (n > 0):
            result = strdot(tmp, array);
            tmp = result;
            n -= 1;

    return result;

#阵列取负
def minus(array):
    for i in range(len(array)):
        if array[i][1] == '-':
            #array[i][0]是'(, 这是规范
            array[i] = array[i][0]+array[i][2:];
        else:
            array[i] = array[i][0]+'-'+array[i][1:];

    return array;

###
# @usage   代数式运算
# @author  mw
# @date    2016年05月18日  星期三  07:37:01
# @param
# @return
#
###

#两个多项式相加，合并同类项不在此进行
def stradd(array1, array2):
    #两个多项式相加，这里直接返回数组的相加
    return array1+array2;

#为了简便输入，不要求输入规范化代数式，(coef)*expr形式
#所以在此对多项式进行规范化
#至于单项式规范化，调用strmono函数即可
def strformat(array):
    for i in range(len(array)):
        array[i] = strmono(array[i]);

    return array;

#把单项式完全格式化，使经过运算的没运算过的都具有统一的格式
def monoformat(mono):
    #规范化单项式，保证任意两个参数之间都添加一个'*'号
    #这是为了和经过代数式乘法运算之后的格式统一
    chars = len(mono);
    s = '';
    for i in range(chars-1):
        if mono[i] == ']' and mono[i+1].isalpha():
            s += mono[i]+'*';
        elif mono[i].isalpha() and mono[i+1].isalpha():
            s += mono[i]+'*';
        #这里还有一个死角，就是下标或指数如果是用的代数式，并且是多项相乘
        #可能会有一点问题，暂时不考虑了
        else:
            s += mono[i];
    s += mono[-1];
    return s;

#把单项式炸开，这里的单项式已经达到最大规范化，是(coef)*x_[1]^[2]*y_[2]^[2]这种结构形式了
#'*'号是要作为分隔符的，不可缺少
def explodemono_2(mono):
    part = mono.split('*');
    #每个部分的[前部，指数部]的对组
    map_ = [];

    for i in range(len(part)):
        expIndex = part[i].find('^');
        if (expIndex != -1):
            map_.append([part[i][:expIndex], part[i][expIndex:]]);
        else:
            s = part[i];
            #系数
            if s[0] == '(':
                map_.append([part[i], '']);
            #代数式
            else:
                map_.append([part[i], '^[1]']);
    map_ = sorted(map_, key = lambda a : a[0]);
    return map_;

#单项式同类项合并
def monocombine(mono):
    map_ = explodemono_2(mono);

    size = len(map_);

    result = [];

    for i in range(size):
        size_1 = len(result);

        if (size_1 <= 0):
            result.append(map_[i]);
        else:
            for j in range(size_1):
                if result[j][0] == map_[i][0]:
                    #双方的中括号位置
                    #由于规范化后的原因，这个括号是一定有的
                    p1 = result[j][1].find('[');
                    p2 = result[j][1].find(']');

                    p3 = map_[i][1].find('[');
                    p4 = map_[i][1].find(']');

                    s = result[j][1][p1+1:p2]+'+'+map_[i][1][p3+1:p4];
                    size_2 = len(s);
                    for k in range(size_2):
                        if s[k].isalpha():
                            break;
                        #如果没有字符参数，可以计算出结果，就计算
                        if (k >= size_2-1):
                            s = str(eval(s));
                    result[j][1] = '^['+s+']';
                    break;

                if (j >= size_1-1):
                    result.append(map_[i]);

    size_1 = len(result);

    s = '';

    for i in range(size_1):

        if (i > 0 and result[i][1] == '^[0]'):
            continue;

        s += result[i][0]+result[i][1];

        if (i < size_1-1):
            s += '*';

    return s;

#排列公式
def arrangement(n, m):
    if n < m:
        return arrangement(m, n);
    else:
        factorial = 1;
        for i in range(n, n-m, -1):
            factorial*=i;
        return factorial;  

#组合公式
def combination(n, m):
    if (n < m):
        return combination(m, n);
    else:
        return arrangement(n, m)/arrangement(m,m);

'''

if __name__ == '__main__':
    a = ['x^[2]', '1'];
    b = ['x', '-2'];

    a = strformat(a);
    b = strformat(b);

    #乘方7次
    c = strpow_n(b, 7);
    #整理
    d = strcombine(c);

    #多项式乘法一次
    e = strdot(a, d);
    #整理
    f = strcombine(e);
    print(f);

>>>
['(1)*x^[9]', '(-14)*x^[8]', '(85)*x^[7]', '(-294)*x^[6]', '(644)*x^[5]', '(-952)*x^[4]',
'(1008)*x^[3]', '(-800)*x^[2]', '(448)*x^[1]', '(-128)']

'''</span>

命名有点乱，用的时候要小心，很容易出错的。

再给出一些用例和过程作为参考：

<span style="font-size:18px;">def tmp():
    s = ['-3', '3', 'x', '-x', '2x', '-2x', '-2x^[2]', '3x_[2]^[3]', '-3x_[2]^[3]'];

    for i in range(len(s)):
        print(strmono(s[i]));

>>>
['(1)*x^[2]*x^[2]', '(-1)*x^[2]*y', '(-1)*y*x^[2]', '(1)*y*y']

#两个多项式相加，合并同类项不在此进行
def stradd(array1, array2):
    #两个多项式相加，这里直接返回数组的相加
    return array1+array2;

#为了简便输入，不要求输入规范化代数式，(coef)*expr形式
#所以在此对多项式进行规范化
#至于单项式规范化，调用strmono函数即可
def strformat(array):
    for i in range(len(array)):
        array[i] = strmono(array[i]);

    return array;    

def tmp():
    A = ['x^[2]', '-y'];
    B = strdot(A, A);
    print(B);

>>>
['(1)*x^[2]*x^[2]', '(-1)*x^[2]*y', '(-1)*y*x^[2]', '(1)*y*y']
['(1)*x^[2]', '(-1)*y']
['((1)*)*x^[2]', '((-1)*)*y']

def tmp():
    A = ['x^[2]', '-y'];
    B = strdot(A, A);
    print(B);

    C = strformat(A);
    print(C);
    #规范化只能进行一次，否则会解析出无法计算的表达式
    D = strformat(C);
    print(D);

>>>
['(-3)', '(3)', '(1)*x', '(-1)*x', '(2)*x', '(-2)*x', '(-2)*x^[2]', '(3)*x_[2]^[3]', '(-3)*x_[2]^[3]', '(-2)*x^[3]y^[3]', '(3)*x^[-1]y^[2]']

def tmp():
    s = ['-3', '3', 'x', '-x', '2x', '-2x',
         '-2x^[2]', '3x_[2]^[3]', '-3x_[2]^[3]',
        '-2x^[3]y^[3]', '3x^[-1]y^[2]'];

    s1 = strformat(s);
    print(s1);

>>>
['(-3)', '(3)', '(1)*x', '(-1)*x', '(2)*x', '(-2)*x', '(-2)*x^[2]', '(3)*x_[2]^[3]', '(-3)*x_[2]^[3]', '(-2)*x^[3]y^[3]', '(3)*x^[-1]y^[2]']
['(-3)', '(3)', '(1)*x', '(-1)*x', '(2)*x', '(-2)*x', '(-2)*x^[2]', '(3)*x_[2]^[3]', '(-3)*x_[2]^[3]', '(-2)*x^[3]*y^[3]', '(3)*x^[-1]*y^[2]']

def tmp():
    s = ['-3', '3', 'x', '-x', '2x', '-2x',
         '-2x^[2]', '3x_[2]^[3]', '-3x_[2]^[3]',
        '-2x^[3]y^[3]', '3x^[-1]y^[2]'];

    s1 = strformat(s);
    print(s1);

    for i in range(len(s1)):
        s1[i] = monoformat(s1[i]);

    print(s1);

#把单项式完全格式化，使经过运算的没运算过的都具有统一的格式
def monoformat(mono):
    #规范化单项式，保证任意两个参数之间都添加一个'*'号
    #这是为了和经过代数式乘法运算之后的格式统一
    chars = len(mono);
    s = '';
    for i in range(chars-1):
        if mono[i] == ']' and mono[i+1].isalpha():
            s += mono[i]+'*';
        elif mono[i].isalpha() and mono[i+1].isalpha():
            s += mono[i]+'*';
        #这里还有一个死角，就是下标或指数如果是用的代数式，并且是多项相乘
        #可能会有一点问题，暂时不考虑了
        else:
            s += mono[i];
    s += mono[-1];
    return s;

>>>
['(-3)', '(3)', '(1)*x', '(-1)*x', '(2)*x', '(-2)*x', '(-2)*x^[2]', '(3)*x_[2]^[3]', '(-3)*x_[2]^[3]', '(-2)*x^[3]*y^[3]', '(3)*y^[2]*x^[-1]', '(5)*x*y*x']
[['(-3)', '']]
[['(3)', '']]
[['(1)', ''], ['x', '']]
[['(-1)', ''], ['x', '']]
[['(2)', ''], ['x', '']]
[['(-2)', ''], ['x', '']]
[['(-2)', ''], ['x', '^[2]']]
[['(3)', ''], ['x_[2]', '^[3]']]
[['(-3)', ''], ['x_[2]', '^[3]']]
[['(-2)', ''], ['x', '^[3]'], ['y', '^[3]']]
[['(3)', ''], ['x', '^[-1]'], ['y', '^[2]']]
[['(5)', ''], ['x', ''], ['x', ''], ['y', '']]

#把单项式炸开，这里的单项式已经达到最大规范化，是(coef)*x_[1]^[2]*y_[2]^[2]这种结构形式了
#'*'号是要作为分隔符的，不可缺少
def explodemono_2(mono):
    part = mono.split('*');
    #每个部分的[前部，指数部]的对组
    map_ = [];

    for i in range(len(part)):
        expIndex = part[i].find('^');
        if (expIndex != -1):
            map_.append([part[i][:expIndex], part[i][expIndex:]]);
        else:
            map_.append([part[i], '']);
    map_ = sorted(map_, key = lambda a : a[0]);
    return map_;

def tmp():
    s = ['-3', '3', 'x', '-x', '2x', '-2x',
         '-2x^[2]', '3x_[2]^[3]', '-3x_[2]^[3]',
        '-2x^[3]y^[3]', '3y^[2]x^[-1]', '5xyx'];

    s1 = strformat(s);
    print(s1);

    for i in range(len(s1)):
        print(explodemono_2(s1[i]));

>>>
['(-3)', '(3)', '(1)*x', '(-1)*x', '(2)*x', '(-2)*x', '(-2)*x^[2]', '(3)*x_[2]^[3]', '(-3)*x_[2]^[3]', '(-2)*x^[3]*y^[3]', '(3)*y^[2]*x^[-1]', '(5)*x*y*x']
[['(-3)', '']]
[['(3)', '']]
[['(1)', ''], ['x', '^[1]']]
[['(-1)', ''], ['x', '^[1]']]
[['(2)', ''], ['x', '^[1]']]
[['(-2)', ''], ['x', '^[1]']]
[['(-2)', ''], ['x', '^[2]']]
[['(3)', ''], ['x_[2]', '^[3]']]
[['(-3)', ''], ['x_[2]', '^[3]']]
[['(-2)', ''], ['x', '^[3]'], ['y', '^[3]']]
[['(3)', ''], ['x', '^[-1]'], ['y', '^[2]']]
[['(5)', ''], ['x', '^[1]'], ['x', '^[1]'], ['y', '^[1]']]

#把单项式炸开，这里的单项式已经达到最大规范化，是(coef)*x_[1]^[2]*y_[2]^[2]这种结构形式了
#'*'号是要作为分隔符的，不可缺少
def explodemono_2(mono):
    part = mono.split('*');
    #每个部分的[前部，指数部]的对组
    map_ = [];

    for i in range(len(part)):
        expIndex = part[i].find('^');
        if (expIndex != -1):
            map_.append([part[i][:expIndex], part[i][expIndex:]]);
        else:
            s = part[i];
            #系数
            if s[0] == '(':
                map_.append([part[i], '']);
            #代数式
            else:
                map_.append([part[i], '^[1]']);
    map_ = sorted(map_, key = lambda a : a[0]);
    return map_;

def tmp():
    s = ['-3', '3', 'x', '-x', '2x', '-2x',
         '-2x^[2]', '3x_[2]^[3]', '-3x_[2]^[3]',
        '-2x^[3]y^[3]', '3y^[2]x^[-1]', '5xyx'];

    s1 = strformat(s);
    print(s1);

    for i in range(len(s1)):
        print(explodemono_2(s1[i]));

>>>
['(-3)', '(3)', '(1)*x', '(-1)*x', '(2)*x', '(-2)*x', '(-2)*x^[2]', '(3)*x_[2]^[3]', '(-3)*x_[2]^[3]', '(-2)*x^[3]*y^[3]', '(3)*y^[2]*x^[-1]', '(5)*x*y*x']
[['(-3)', '']]
[['(3)', '']]
[['(1)', ''], ['x', '^[1]']]
[['(-1)', ''], ['x', '^[1]']]
[['(2)', ''], ['x', '^[1]']]
[['(-2)', ''], ['x', '^[1]']]
[['(-2)', ''], ['x', '^[2]']]
[['(3)', ''], ['x_[2]', '^[3]']]
[['(-3)', ''], ['x_[2]', '^[3]']]
[['(-2)', ''], ['x', '^[3]'], ['y', '^[3]']]
[['(3)', ''], ['x', '^[-1]'], ['y', '^[2]']]
[['(5)', ''], ['x', '^[1+1]'], ['y', '^[1]']]

#单项式同类项合并
def monocombine(mono):
    map_ = explodemono_2(mono);

    size = len(map_);

    result = [];

    for i in range(size):
        size_1 = len(result);

        if (size_1 <= 0):
            result.append(map_[i]);
        else:
            for j in range(size_1):
                if result[j][0] == map_[i][0]:
                    #双方的中括号位置
                    #由于规范化后的原因，这个括号是一定有的
                    p1 = result[j][1].find('[');
                    p2 = result[j][1].find(']');

                    p3 = map_[i][1].find('[');
                    p4 = map_[i][1].find(']');

                    s = result[j][1][p1+1:p2]+'+'+map_[i][1][p3+1:p4];
                    result[j][1] = '^['+s+']';
                    break;

                if (j >= size_1-1):
                    result.append(map_[i]);

    return result;

def tmp():
    s = ['-3', '3', 'x', '-x', '2x', '-2x',
         '-2x^[2]', '3x_[2]^[3]', '-3x_[2]^[3]',
        '-2x^[3]y^[3]', '3y^[2]x^[-1]', '5xyx'];

    s1 = strformat(s);
    print(s1);

    for i in range(len(s1)):
        print(monocombine(s1[i]));

>>>
['(-3)', '(3)', '(1)*x', '(-1)*x', '(2)*x', '(-2)*x', '(-2)*x^[2]', '(3)*x_[2]^[3]', '(-3)*x_[2]^[3]', '(-2)*x^[3]*y^[3]', '(3)*y^[2]*x^[-1]', '(5)*x*y*x^[2]*z*y*x*x_[1]']
(-3)
(3)
(1)*x^[1]
(-1)*x^[1]
(2)*x^[1]
(-2)*x^[1]
(-2)*x^[2]
(3)*x_[2]^[3]
(-3)*x_[2]^[3]
(-2)*x^[3]*y^[3]
(3)*x^[-1]*y^[2]
(5)*x^[1+2+1]*x_[1]^[1]*y^[1+1]*z^[1]

#单项式同类项合并
def monocombine(mono):
    map_ = explodemono_2(mono);

    size = len(map_);

    result = [];

    for i in range(size):
        size_1 = len(result);

        if (size_1 <= 0):
            result.append(map_[i]);
        else:
            for j in range(size_1):
                if result[j][0] == map_[i][0]:
                    #双方的中括号位置
                    #由于规范化后的原因，这个括号是一定有的
                    p1 = result[j][1].find('[');
                    p2 = result[j][1].find(']');

                    p3 = map_[i][1].find('[');
                    p4 = map_[i][1].find(']');

                    s = result[j][1][p1+1:p2]+'+'+map_[i][1][p3+1:p4];
                    result[j][1] = '^['+s+']';
                    break;

                if (j >= size_1-1):
                    result.append(map_[i]);

    size_1 = len(result);

    s = '';

    for i in range(size_1):
        s += result[i][0]+result[i][1];

        if (i < size_1-1):
            s += '*';

    return s;

def tmp():
    s = ['-3', '3', 'x', '-x', '2x', '-2x',
         '-2x^[2]', '3x_[2]^[3]', '-3x_[2]^[3]',
        '-2x^[3]y^[3]', '3y^[2]x^[-1]', '5xyx^[2]zyxx_[1]'];

    s1 = strformat(s);
    print(s1);

    for i in range(len(s1)):
        print(monocombine(s1[i]));

>>>
['(1)*x*x', '(1)*x*y', '(1)*x*z', '(1)*y*x', '(1)*y*y', '(1)*y*z', '(1)*z*x', '(1)*z*y', '(1)*z*z']

def tmp():
    A = ['x', 'y', 'z'];
    B = ['x', 'y', 'z'];

    A = strformat(A);
    B = strformat(B);

    C = strdot(A, B);
    print(C);

>>>
['(1)*x*x', '(1)*x*y', '(1)*x*z', '(1)*y*x', '(1)*y*y', '(1)*y*z', '(1)*z*x', '(1)*z*y', '(1)*z*z']
['(1)*x^[1+1]', '(2)*x^[1]*y^[1]', '(2)*x^[1]*z^[1]', '(1)*y^[1+1]', '(2)*y^[1]*z^[1]', '(1)*z^[1+1]']

def tmp():
    A = ['x', 'y', 'z'];
    B = ['x', 'y', 'z'];

    A = strformat(A);
    B = strformat(B);

    C = strdot(A, B);
    print(C);

    C = strcombine(C);
    print(C);

['(3)*x', '(3)*y', '(1)*y*x', '(1)*y*y', '(1)*z*x', '(1)*z*y']
['(3)*x^[1]', '(3)*y^[1]', '(1)*x^[1]*y^[1]', '(1)*y^[1+1]', '(1)*x^[1]*z^[1]', '(1)*y^[1]*z^[1]']

def tmp():
    A = ['3', 'y', 'z'];
    B = ['x', 'y'];

    A = strformat(A);
    B = strformat(B);

    C = strdot(A, B);
    print(C);

    C = strcombine(C);
    print(C);
	</span>

下面是上一篇中的几何部分现在可调用的功能：

<span style="font-size:18px;">>>> import geo;
>>> dir(geo);
['__builtins__', '__cached__', '__doc__', '__file__', '__initializing__', '__loader__', '__name__', '__package__', 'angleBetweenTwoLine', 'angleFromX', 'circle', 'crossPointOfTwoLine', 'distance2D', 'interceptOfLine', 'lcCrossPoint2D', 'lcDistance2D', 'lineDistance2D', 'math', 'parallelCheck', 'perpendicular', 'plDistance2D', 'pointFromCircle', 'pointInLine', 'slope', 'tangencyPoint']
>>> help(geo);
Help on module geo:

NAME
    geo

FUNCTIONS
    angleBetweenTwoLine(Line_1, Line_2)
        #两直线夹角 Line_1:[Point_1, Point_2], Line_2:[Point_1, Point_2]

    angleFromX(Point_1, Point_2)
        #平面直线与X轴的夹角[x1, y1] -- [x2, y2]

    circle(Point_1, Point_2, Point_3)
        #圆 三点成圆
        #以确定的三点表示圆的方程，得到圆心和半径 [x1, y1] -- [x2, y2] -- [x3, y3]

    crossPointOfTwoLine(Line_1, Line_2)
        #两直线交点 Line_1:[Point_1, Point_2], Line_2:[Point_1, Point_2]

    distance2D(Point_1, Point_2)
        #两点之间
        #平面两点的距离[x1, y1] -- [x2, y2]

    interceptOfLine(Point_1, Point_2)
        #平面直线的截距[x1, y1] -- [x2, y2]

    lcCrossPoint2D(Line, Circle)
        #直经与圆的交点

    lcDistance2D(Line, Circle)
        #直线到圆的距离

    lineDistance2D(Line_1, Line_2)
        #两直线距离 Line_1:[Point_1, Point_2], Line_2:[Point_1, Point_2]

    parallelCheck(Line_1, Line_2)
        #两直线之间
        #两直线是否平行 Line_1:[Point_1, Point_2], Line_2:[Point_1, Point_2]

    perpendicular(Line)
        #求直线的中垂线

    plDistance2D(Point, Line)
        #点到直线的距离

    pointFromCircle(Point, Circle)
        #判断点和圆的距离 Point:[x, y], Circle: [[x1, y1], [x2, y2], [x3, y3]]

    pointInLine(Point_1, Point_2, Point_3)
        #判定三点共线

    slope(Point_1, Point_2)
        #平面直线的斜率[x1, y1] -- [x2, y2]

    tangencyPoint(Point, Circle)
        #过圆外部一点，得到与圆的切点的坐标 Point:[x, y], Circle: [[x1, y1], [x2, y2], [x3, y3]]

</span>

这个是解三角形的工具：

<span style="font-size:18px;">###
# @usage   解三角形
# @author  mw
# @date    2016年04月16日  星期六  07:50:31
# @param   三角形六元素 [边1， 边2， 边3, 角1, 角2, 角3]，角为边所对的角，角度制
# @return  三角形的三个顶点坐标
#
###

#解三角形, 传入array为[三边，三角]按顺序排列的六元素数组
#边和角是对边和对角的关系排列
#待求量用'?'表示， 如[3, 4, 5, '?', '?', '?']为求三个角
def solveTriangle(elementArray):
    elements = len(elementArray);
    if (elements != 6):
        return [];

    unknows_1 = unknows_2 = 0;
    for i in range(6):
        if (elementArray[i] == '?'):
            if (i < 3):
                unknows_1+=1;
            else:
                unknows_2+=1;

    if (unknows_1 >= 3 or unknows_1+unknows_2 > 3):
        return [];

    #已知三边
    if (unknows_1 == 0):
        for i in range(3):
            if (elementArray[(i+1)%3]+elementArray[(i+2)%3] <=elementArray[i]):
                return [];

        return know3edges([elementArray[2], elementArray[0], elementArray[1]]);

    #
    if (unknows_1 == 1):
        #用余弦公式的情况
        for i in range(3):
            if (elementArray[i] == '?' and
                (elementArray[(i+1)%3] != '?' and elementArray[(i+1)%3+3] == '?') and
                (elementArray[(i+2)%3] != '?' and elementArray[(i+2)%3+3] == '?')):
                b = elementArray[(i+1)%3];
                c = elementArray[(i+2)%3];
                A = elementArray[i+3];
                a = math.sqrt(b*b+c*c-2*b*c*math.cos(A/180*math.pi));
                elementArray[i] = a;
                break;

        if (elementArray[0] != '?' and
            elementArray[1] != '?' and
            elementArray[2] != '?'):
            return know3edges([elementArray[2], elementArray[0], elementArray[1]]);

    #至少两个角已知， 可以把三个角都求出来
    if (unknows_2 == 1):
        for i in range(3):
            if (elementArray[i+3] == '?' and
                elementArray[(i+1)%3+3] != '?' and
                elementArray[(i+2)%3+3] != '?'):
                elementArray[i+3] = 180-(elementArray[(i+1)%3+3]+elementArray[(i+2)%3+3]);
                break;

    #能使用正弦公式的各种情况
    #正弦公式a/sinA = b/sinB = c/sinC =  sinValue
    sinValue = 0;
    for i in range(3):
        if (elementArray[i] != '?' and elementArray[i+3] != '?'):
            sinValue = elementArray[i]/math.sin(elementArray[i+3]/180*math.pi);
            break;

    #由已知两边一角或三角一边 => 两边两角或三边三角
    for i in range(3):
        if (elementArray[i] == '?' and elementArray[i+3] != '?'):
            elementArray[i] = sinValue * math.sin(elementArray[i+3]/180*math.pi);

        if (elementArray[i] != '?' and elementArray[i+3] == '?'):
            elementArray[i+3] = math.asin(elementArray[i]/sinValue)/math.pi*180;

    if (elementArray[0] != '?' and
        elementArray[1] != '?' and
        elementArray[2] != '?'):
        return know3edges([elementArray[2], elementArray[0], elementArray[1]]);

    #如果还有角未求得
    for i in range(3):
        if (elementArray[i+3] == '?'):
            elementArray[i+3] = 180-(elementArray[(i+1)%3+3]+elementArray[(i+2)%3+3]);
            break;

    #由角求边
    for i in range(3):
        if (elementArray[i] == '?' and elementArray[i+3] != '?'):
            elementArray[i] = sinValue*math.sin(elementArray[i+3]/180*math.pi);

    #校验
    for i in range(3):
        if (elementArray[i+3] <= 0):
            #角度为负或0, 说明给的条件不足以构成三角形
            return [];

    return know3edges([elementArray[2], elementArray[0], elementArray[1]]);

#已知三角形三条边求六元素
def know3edges(edges):
    if (len(edges) <= 0):
        return [];
    #角A
    A = math.acos((edges[1]*edges[1]+edges[2]*edges[2]-edges[0]*edges[0])/(2*edges[1]*edges[2]));
    #角B
    B = math.acos((edges[0]*edges[0]+edges[2]*edges[2]-edges[1]*edges[1])/(2*edges[0]*edges[2]));
    #角C
    C = math.acos((edges[0]*edges[0]+edges[1]*edges[1]-edges[2]*edges[2])/(2*edges[0]*edges[1]));

    A = A/math.pi*180;
    B = B/math.pi*180;
    C = C/math.pi*180;

    return [edges[1], edges[2], edges[0]]+[B, C, A];</span>

本节到此结束，欲知后事如何，请看下回分解。