python学习实例（5）

#============================================
#5.1 计算思维是什么
#============================================#<程序: 找假币的第一种方法> by Edwin Sha
def findcoin_1(L):if len(L) <=1:print("Error: coins are too few"); quit()i=0while i<len(L):if L[i] < L[i+1]: return (i)elif L[i] > L[i+1]: return (i+1)i=i+1print("All coins are the same")return(len(L))   #should not reach this point#<主要程序>
import random
n=int(input("Enter the number of coins >=2: "))
w_normal=random.randint(2,5)
index_faked=random.randint(0,n-1)  # 0<= index <=n-1
L=[]
for i in range(0,n):L.append(w_normal)
L[index_faked]=w_normal-1
print(L)
print("The index of faked coin:",findcoin_1(L))#============================================
#5.2 递归（Rcurrence）的基本概念
#============================================#<程序：递归加法>
def F(a):if len(a) ==1: return(a[0])   #终止条件非常重要return(F(a[1:])+a[0])
a=[1,4,9,16]
print(F(a))#<程序：汉诺塔_递归>
count=1
def main():n_str=input('请输入盘子个数：')n=int(n_str)Hanoi(n,'A','C','B')
def Hanoi(n, A, C, B):global countif n < 1:print('False')elif n == 1:print ("%d:\t%s -> %s" % (count, A, C))count += 1elif n > 1:Hanoi (n - 1, A, B, C)Hanoi (1, A, C, B)Hanoi (n - 1, B, C, A)
if(__name__=="__main__"):main()#<程序：merge函数> by Edwin Sha
def merge(L1,L2):if len(L1) ==0: return(L2)if len(L2) ==0: return(L1)if L1[0] < L2[0]:return([L1[0]]+merge(L1[1:len(L1)],L2))else:return([L2[0]]+merge(L1,L2[1:len(L2)]))X=merge([1,4,9],[10])
print(X)#============================================
#5.3 分治法(Divide-and-Conquer Algorithm)
#============================================#<程序：最小值_循环>
def M(a):m=a[0]for i in range(1,len(a)):if a[i]<m:m=a[i]return m
a=[4,1,3,5]print(M(a))#<程序：最小值_递归> a是个数组
def M(a):print(a)if len(a) ==1: return a[0]return (min(a[len(a)-1], M(a[0:len(a)-1])))L=[4,1,3,5]
print(M(L))#<程序：最小值_分治>
def M(a):#print(a)   可以列出程序执行的顺序]if len(a) ==1: return a[0]return ( min(M(a[0:len(a)//2]),M(a[len(a)//2:len(L)])))
L=[4,1,3,5]
print(M(L))#<程序：最小值和最大值_分治>
A=[3,8,9,4,10,5,1,17]
def Smin_max(a):if len(a)==1:return(a[0],a[0])elif len(a)==2:return(min(a),max(a))m=len(a)//2lmin,lmax=Smin_max(a[:m])rmin,rmax=Smin_max(a[m:])return min(lmin,rmin),max(lmax,rmax)print("Minimum and Maximum:%d,%d"%(Smin_max(A)))#<程序：归并排序merge sort>
def msort(L):k=len(L)if k==0: return(L)if k==1: return(L)X1=L[0:k//2]; X2=L[k//2:k]  #X1,X2 are local variablesprint("X1=",X1,"   X2=",X2)  #看看输出是什么？知道递归是如何执行的。X1=msort(X1); X2=msort(X2)return(merge(X1,X2))#<程序: 全加器>
def FA(a,b,c):  # Full addercarry = (a and b) or (b and c) or (a and c)sum = (a and b and c) or (a and (not b) and (not c)) \or ((not a) and b and (not c)) or ((not a) and (not b) and c)return carry, sum#<程序：二进制加法-二分法算法> by Edwin Sha
def add_divide(x,y,c=False):# x, y are lists of True or False, c is True or False# return carry and a list of x+ywhile len(x) < len(y): x = [False]+xwhile len(y) < len(x): y = [False]+yif len(x) ==1:ctemp, stemp=FA(x[0],y[0],c)return (ctemp, [stemp])if len(x) ==0: return c, []c1,s1=add_divide(x[len(x)//2:len(x)],y[len(y)//2:len(y)],c)c2,s2=add_divide(x[0:len(x)//2],y[0:len(y)//2],c1) #依赖关系！return(c2,s2+s1)#============================================
#5.4 贪心算法(Greedy Algorithm)
#============================================#<程序：找零钱_贪心>
v=[25,10,5,1]
n=[0,0,0,0]
def change():T_str=input('要找给顾客的零钱，单位：分：')T=int(T_str)greedy(T)for i in range(len(v)):print('要找给顾客',v[i],'分的硬币：',n[i])s=0for i in n:s=s+iprint('找给顾客的硬币数最少为：',s)
def greedy(T):if T==0:returnelif T>=v[0]:T=T-v[0]; n[0]=n[0]+1greedy(T)elif v[0]>T>=v[1]:T=T-v[1]; n[1]=n[1]+1greedy(T)elif v[1]>T>=v[2]:T=T-v[2]; n[2]=n[2]+1greedy(T)else:T=T-v[3]; n[3]=n[3]+1greedy(T)if(__name__=="__main__"):change()#<程序：GCD_贪心>
def main():x_str=input('请输入正整数x的值：')x=int(x_str)y_str=input('请输入正整数y的值：')y=int(y_str)print(x,'和',y,'的最大公约数是：', GCD(x,y))def GCD(x,y):if x>y: a=x;b=yelse:   a=y;b=xif a%b ==0: return(b)return(GCD(a%b,b))if(__name__=="__main__"):main()#============================================
#5.5 动态规划(Dynamic Programming)
#============================================#<程序：最长递增子序列_动态规划>
def LIS(L):  #LIS (L)：Longest Increasing Sub-list of List LAsc=[1]*len(L);Tra=[-1]*len(L)   #设定起始值#Asc[i] 存放从L[0]到L[i]以L[i]为最大值的最长递增子序列的长度，#       这个最长数列肯定以L[i]结尾#Tra[i] 存此最长数列的前一个索引，以后好连起整个递增序列。for i in range(1,len(L)):X=[]for j in range(0,i):if L[i] > L[j]: X.append(j)   #所有比L[i]小L[j]的索引放在Xfor k in X:     #Asc[i]= max Asc[k]+1, for each k in Xif Asc[i] < Asc[k]+1: Asc[i]=Asc[k]+1; Tra[i]=kprint("Asc:",Asc)print("Tra:",Tra)max=0   #找到Asc中的最大值for i in range(1,len(Asc)):if Asc[i]>Asc[max]: max=iprint("最长递增子序列的长度是",Asc[max])#将最长递增数列存到XX=[L[max]]; i=max;while (Tra[i] >=0):X=[L[Tra[i]]]+Xi=Tra[i]print("最长递增子数列=",X)L=[5,2,4,7,6,3,8,9]
LIS(L)#<程序：直接用递归函数计算Asc(k)>
def Asc(k):if k==0: return(1)X=[]for i in range(0,k):if L[k] > L[i]: X.append(Asc(i))   #记录所有比L[k]小的Asc（）if len(X) >0: return (max(X)+1)else: return(1)def LIS_R(L):X=[]for k in range(0, len(L)):X.append(Asc(k))print(X)print(max(X))L=[5,2,4,7,6,3,8,9]
LIS_R(L)#<程序：背包问题_递归>
w=[0,4,5,2,1,6]        #w[i]是物品的重量
v=[0,45,57,22,11,67]   #v[i]是物品的价值
n=len(w)-1
j=8                #背包的容量
x=[False for raw in range(n+1)]#x[i]为True，表示物品被放入背包
def knap_r(n,j):if (n==0)or(j==0):x[n]=Falsereturn 0elif (j>=w[n])and(knap_r(n-1,j-w[n])+v[n]>knap_r(n-1,j)):x[n]=Truereturn knap_r(n-1,j-w[n])+v[n]else:x[n]=Falsereturn knap_r(n-1,j)
print("最大价值为：",knap_r(n,j))
print("物品的装入情况为：",x[1:])#<程序：背包问题_动态规划>
w=[0,4,5,2,1,6]        #w[i]是物品的重量
v=[0,45,57,22,11,67]   #v[i]是物品的价值
n=len(w)-1
m=8                #背包的容量
x=[False for raw in range(n+1)]#x[i]为True，表示物品被放入背包
#a[i][j]是i个物品中能够装入容量为j的背包的物品所能形成的最大价值
a=[[0 for col in range(m+1)] for raw in range(n+1)]
def knap_DP(n,m):#创建动态规划表for i in range(1,n+1):for j in range(1,m+1):a[i][j]=a[i-1][j]if (j>=w[i]) and(a[i-1][j-w[i]]+v[i]>a[i-1][j]):a[i][j]=a[i-1][j-w[i]]+v[i]#回溯a[i][j]的生成过程，找到装入背包的物品j=mfor i in range(n,0,-1):if a[i][j]>a[i-1][j]:x[i]=Truej=j-w[i]Mv=a[n][m]return Mv#============================================
#5.6 以老鼠走迷宫为例
#============================================#<程序：老鼠走迷宫_递归>
m=[[1,1,1,0,1,1,1,1,1,1],[1,0,0,0,0,0,0,0,1,1],[1,0,1,1,1,1,1,0,0,1],[1,0,1,0,0,0,0,1,0,1],[1,0,1,0,1,1,0,0,0,1],[1,0,0,1,1,0,1,0,1,1],[1,1,1,1,0,0,0,0,1,1],[1,0,0,0,0,1,1,1,0,0],[1,0,1,1,0,0,0,0,0,1],[1,1,1,1,1,1,1,1,1,1]]sta1=0;sta2=3;fsh1=7;fsh2=9;success=0
def LabyrinthRat():print('显示迷宫：')for i in range(len(m)): print(m[i])print('入口：m[%d][%d]：出口：m[%d][%d]'%(sta1,sta2,fsh1,fsh2))if (visit(sta1,sta2))==0: print('没有找到出口')else:print('显示路径：')for i in range(10):print(m[i])
def visit(i,j):m[i][j]=2global successif(i==fsh1)and(j==fsh2): success=1if(success!=1)and(m[i-1][j]==0): visit(i-1,j)if(success!=1)and(m[i+1][j]==0): visit(i+1,j)if(success!=1)and(m[i][j-1]==0): visit(i,j-1)if(success!=1)and(m[i][j+1]==0): visit(i,j+1)if success!=1: m[i][j]=3return successif(__name__=="__main__"):LabyrinthRat()#============================================
#5.7 谈计算思维的美
#============================================#++++++++++++++++++++++++++++++++++++++++++++
#5.7.3 问题复杂度的研究之美
#++++++++++++++++++++++++++++++++++++++++++++#<程序:Find all the factors of x and put them in list L>
import math
def factors(x,L):y=int(math.sqrt(x))   #x的平方根for i in range(2,y+1):  #一个个找if (x %i ==0):    #找到一个因数iprint(i)L.append(i)factors(x//i,L)      #递归找break               #跳出for循环else:  #cannot find a factor, so x is a primeL.append(int(x))print(int(x))
L=[]
factors(18,L)

python学习实例（5）相关推荐

涵盖 14 大主题！最完整的 Python 学习实例集来了！
机器学习.深度学习最简单的入门方式就是基于 Python 开始编程实战.最近闲逛 GitHub,发现了一个非常不错的 Python 学习实例集,完全是基于 Python 来实现包括 ML.DL 等领域 ...
python学习实例（6）
#============================= #6.6 文件系统(File System) #=============================#+++++++++++++++ ...
python学习实例（4）
#========================================= #第四章的python程序 #=========================================# ...
python学习实例（7）
#========================================================= #第8章信息安全(Information Security)的python程序 ...
python学习实例（3）
#=================================== #3.4 关于Python的函数调用 #===================================#+++++++ ...
python学习实例（1）
#====================================== #1.2 计算机编程的基本概念 #======================================#++++ ...
python学习实例（2）
#=================================== #2.2 不同进制间的转换 #===================================#++++++++++++ ...
Python学习实例（一）温度转换
1.问题描述温度的刻画有两个不同体系:摄氏度(Celsius)和华氏度(Fabrenheit).‪‬‪‬‪‬‪‬‪‬‮‬‭‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‮‬‪‬‪ ...
python学习笔记-练手实例
1.题目:输出 9*9 乘法口诀表. 程序分析:分行与列考虑,共9行9列,i控制行,j控制列代码: for i in range(1,10):print ('\r')for j in range(1 ...

python学习实例（5）

python学习实例（5）相关推荐

最新文章

热门文章