#======================================
#1.2 计算机编程的基本概念
#======================================#++++++++++++++++++++++++++++++++++++++
#1.2.2 从Python语言进入计算机语言的世界
#++++++++++++++++++++++++++++++++++++++#<程序：例子1>
def F(x,y):return(x*x+y*y)
print("F(2,2)=",F(2,2))
print("F(3,2)=",F(3,2))#<程序：例子2>
def Pr():for i in range(0,10): # 索引 i = 0 to 9print("Hello world")
#在函数程序外面Pr()
Pr()
# 输出 Hello world 10 遍#<程序：布尔类型例子>
b = 100<101
print(b)#<程序：for循环例子>
for i in range(1, 5): print(i) #<程序：while循环例子>
i = 1
while i<5:print(i)i=i+1#<程序：if语句例子>
i=10
j=11
if i< j:print("i<j")
else:print("i>=j")#======================================
#1.3 计算机核心知识——算法（Algorithm）
#======================================#++++++++++++++++++++++++++++++++++++++
#1.3.2 解平方根算法一
#++++++++++++++++++++++++++++++++++++++#<程序：平方根运算1>
def square_root_1(): #函数定义，函数名为square_root_1c = 10 #所求平方根的输入，即该段程序求根号10i = 0 #记录执行循环次数g = 0for j in range(0,c+1): #for 循环开始if (j * j > c and g==0):#if 语句块，获取g，使得g2<c，(g+1)2>cg = j - 1#for 循环结束while (abs(g * g - c) > 0.0001):#判断g2-c是否在精度范围内，while循环g += 0.00001 #g每次加步长，以逼近所求解i = i+1print ("%d:g = %.5f" % (i,g))#函数外，执行下面的语句
square_root_1()#++++++++++++++++++++++++++++++++++++++
#1.3.3 解平方根算法二
#++++++++++++++++++++++++++++++++++++++#<程序：平方根运算2-二分法>
def square_root_2():i = 0c = 10m_max = cm_min = 0g = (m_min+m_max)/2while (abs(g*g -c) > 0.00000000001): #while循环开始if (g*g <c):m_min = gelse:m_max = gg = (m_min + m_max)/2i = i+1print ("%d:%.13f" % (i,g)) #while循环结束
#函数之外执行
square_root_2 ()#++++++++++++++++++++++++++++++++++++++
#1.3.4 解平方根算法三
#++++++++++++++++++++++++++++++++++++++#<程序：平方根运算3-牛顿法>
def square_root_3():c = 10g = c/2i = 0while abs(g*g - c) > 0.00000000001:g = (g + c/g)/2i = i+1print("%d:%.13f" % (i,g))square_root_3()#======================================
#1.5 计算机前沿知识——大数据（Big Data）
#======================================#++++++++++++++++++++++++++++++++++++++
#1.5.5 对数据和逻辑的正确态度
#++++++++++++++++++++++++++++++++++++++#<程序：求圆周率－蒙地卡罗法>
import random
def pi(times):sum=0for i in range(times):x=random.random()y=random.random()d2=x*x+y*y #算到原点的距离if d2<=1: sum+=1 #距离<=1, 代表在圆里面。return(sum/times*4)#函数外执行
times=100000000
x=pi(times)
print("pi=%.8f"%(x))
