#======================================
#1.2 計算機編程的基本概念
#======================================#++++++++++++++++++++++++++++++++++++++
#1.2.2 從Python語言進入計算機語言的世界
#++++++++++++++++++++++++++++++++++++++#<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")
#在函數程序外面Pr()
Pr()
# 輸出 Hello world 10 遍#<程序：布爾類型例子>
b = 100<101
print(b)#<程序：for循環例子>
for i in range(1, 5): print(i) #<程序：while循環例子>
i = 1
while i<5:print(i)i=i+1#<程序：if語句例子>
i=10
j=11
if i< j:print("i<j")
else:print("i>=j")#======================================
#1.3 計算機核心知識——算法（Algorithm）
#======================================#++++++++++++++++++++++++++++++++++++++
#1.3.2 解平方根算法一
#++++++++++++++++++++++++++++++++++++++#<程序：平方根運算1>
def square_root_1(): #函數定義，函數名為square_root_1c = 10 #所求平方根的輸入，即該段程序求根號10i = 0 #記錄執行循環次數g = 0for j in range(0,c+1): #for 循環開始if (j * j > c and g==0):#if 語句塊，獲取g，使得g2<c，(g+1)2>cg = j - 1#for 循環結束while (abs(g * g - c) > 0.0001):#判斷g2-c是否在精度范圍內，while循環g += 0.00001 #g每次加步長，以逼近所求解i = i+1print ("%d:g = %.5f" % (i,g))#函數外，執行下面的語句
square_root_1()#++++++++++++++++++++++++++++++++++++++
#1.3.3 解平方根算法二
#++++++++++++++++++++++++++++++++++++++#<程序：平方根運算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循環開始if (g*g <c):m_min = gelse:m_max = gg = (m_min + m_max)/2i = i+1print ("%d:%.13f" % (i,g)) #while循環結束
#函數之外執行
square_root_2 ()#++++++++++++++++++++++++++++++++++++++
#1.3.4 解平方根算法三
#++++++++++++++++++++++++++++++++++++++#<程序：平方根運算3-牛頓法>
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 計算機前沿知識——大數據（Big Data）
#======================================#++++++++++++++++++++++++++++++++++++++
#1.5.5 對數據和邏輯的正確態度
#++++++++++++++++++++++++++++++++++++++#<程序：求圓周率－蒙地卡羅法>
import random
def pi(times):sum=0for i in range(times):x=random.random()y=random.random()d2=x*x+y*y #算到原點的距離if d2<=1: sum+=1 #距離<=1, 代表在圓里面。return(sum/times*4)#函數外執行
times=100000000
x=pi(times)
print("pi=%.8f"%(x))
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