python学习实例（5）
【摘要】 #============================================#5.1 计算思维是什么#============================================ #<程序: 找假币的第一种方法> by Edwin Shadef findcoin_1(L): if len(L) <=1: print("Error: coins are...
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#5.1 计算思维是什么
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#============================================
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#<程序: 找假币的第一种方法> by Edwin Sha
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def findcoin_1(L):
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if len(L) <=1:
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print("Error: coins are too few"); quit()
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i=0
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while i<len(L):
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if L[i] < L[i+1]: return (i)
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elif L[i] > L[i+1]: return (i+1)
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i=i+1
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print("All coins are the same")
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return(len(L)) #should not reach this point
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#<主要程序>
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import random
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n=int(input("Enter the number of coins >=2: "))
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w_normal=random.randint(2,5)
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index_faked=random.randint(0,n-1) # 0<= index <=n-1
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L=[]
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for i in range(0,n):
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L.append(w_normal)
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L[index_faked]=w_normal-1
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print(L)
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print("The index of faked coin:",findcoin_1(L))
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#============================================
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#5.2 递归（Rcurrence）的基本概念
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#============================================
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#<程序：递归加法>
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def F(a):
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if len(a) ==1: return(a[0]) #终止条件非常重要
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return(F(a[1:])+a[0])
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a=[1,4,9,16]
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print(F(a))
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#<程序：汉诺塔_递归>
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count=1
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def main():
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n_str=input('请输入盘子个数：')
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n=int(n_str)
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Hanoi(n,'A','C','B')
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def Hanoi(n, A, C, B):
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global count
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if n < 1:
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print('False')
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elif n == 1:
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print ("%d:\t%s -> %s" % (count, A, C))
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count += 1
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elif n > 1:
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Hanoi (n - 1, A, B, C)
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Hanoi (1, A, C, B)
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Hanoi (n - 1, B, C, A)
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if(__name__=="__main__"):
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main()
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#<程序：merge函数> by Edwin Sha
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def merge(L1,L2):
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if len(L1) ==0: return(L2)
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if len(L2) ==0: return(L1)
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if L1[0] < L2[0]:
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return([L1[0]]+merge(L1[1:len(L1)],L2))
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else:
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return([L2[0]]+merge(L1,L2[1:len(L2)]))
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X=merge([1,4,9],[10])
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print(X)
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#============================================
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#5.3 分治法(Divide-and-Conquer Algorithm)
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#============================================
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#<程序：最小值_循环>
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def M(a):
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m=a[0]
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for i in range(1,len(a)):
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if a[i]<m:
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m=a[i]
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return m
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a=[4,1,3,5]
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print(M(a))
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#<程序：最小值_递归> a是个数组
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def M(a):
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print(a)
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if len(a) ==1: return a[0]
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return (min(a[len(a)-1], M(a[0:len(a)-1])))
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L=[4,1,3,5]
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print(M(L))
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#<程序：最小值_分治>
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def M(a):
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#print(a) 可以列出程序执行的顺序]
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if len(a) ==1: return a[0]
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return ( min(M(a[0:len(a)//2]),M(a[len(a)//2:len(L)])))
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L=[4,1,3,5]
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print(M(L))
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#<程序：最小值和最大值_分治>
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A=[3,8,9,4,10,5,1,17]
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def Smin_max(a):
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if len(a)==1:
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return(a[0],a[0])
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elif len(a)==2:
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return(min(a),max(a))
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m=len(a)//2
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lmin,lmax=Smin_max(a[:m])
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rmin,rmax=Smin_max(a[m:])
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return min(lmin,rmin),max(lmax,rmax)
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print("Minimum and Maximum:%d,%d"%(Smin_max(A)))
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#<程序：归并排序merge sort>
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def msort(L):
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k=len(L)
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if k==0: return(L)
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if k==1: return(L)
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X1=L[0:k//2]; X2=L[k//2:k] #X1,X2 are local variables
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print("X1=",X1," X2=",X2) #看看输出是什么？知道递归是如何执行的。
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X1=msort(X1); X2=msort(X2)
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return(merge(X1,X2))
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#<程序: 全加器>
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def FA(a,b,c): # Full adder
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carry = (a and b) or (b and c) or (a and c)
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sum = (a and b and c) or (a and (not b) and (not c)) \
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or ((not a) and b and (not c)) or ((not a) and (not b) and c)
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return carry, sum
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#<程序：二进制加法-二分法算法> by Edwin Sha
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def add_divide(x,y,c=False):
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# x, y are lists of True or False, c is True or False
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# return carry and a list of x+y
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while len(x) < len(y): x = [False]+x
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while len(y) < len(x): y = [False]+y
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if len(x) ==1:
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ctemp, stemp=FA(x[0],y[0],c)
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return (ctemp, [stemp])
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if len(x) ==0: return c, []
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c1,s1=add_divide(x[len(x)//2:len(x)],y[len(y)//2:len(y)],c)
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c2,s2=add_divide(x[0:len(x)//2],y[0:len(y)//2],c1) #依赖关系！
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return(c2,s2+s1)
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#============================================
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#5.4 贪心算法(Greedy Algorithm)
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#============================================
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#<程序：找零钱_贪心>
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v=[25,10,5,1]
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n=[0,0,0,0]
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def change():
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T_str=input('要找给顾客的零钱，单位：分：')
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T=int(T_str)
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greedy(T)
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for i in range(len(v)):
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print('要找给顾客',v[i],'分的硬币：',n[i])
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s=0
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for i in n:
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s=s+i
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print('找给顾客的硬币数最少为：',s)
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def greedy(T):
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if T==0:return
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elif T>=v[0]:
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T=T-v[0]; n[0]=n[0]+1
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greedy(T)
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elif v[0]>T>=v[1]:
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T=T-v[1]; n[1]=n[1]+1
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greedy(T)
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elif v[1]>T>=v[2]:
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T=T-v[2]; n[2]=n[2]+1
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greedy(T)
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else:
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T=T-v[3]; n[3]=n[3]+1
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greedy(T)
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if(__name__=="__main__"):
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change()
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#<程序：GCD_贪心>
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def main():
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x_str=input('请输入正整数x的值：')
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x=int(x_str)
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y_str=input('请输入正整数y的值：')
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y=int(y_str)
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print(x,'和',y,'的最大公约数是：', GCD(x,y))
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def GCD(x,y):
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if x>y: a=x;b=y
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else: a=y;b=x
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if a%b ==0: return(b)
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return(GCD(a%b,b))
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if(__name__=="__main__"):
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main()
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#============================================
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#5.5 动态规划(Dynamic Programming)
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#============================================
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#<程序：最长递增子序列_动态规划>
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def LIS(L): #LIS (L)：Longest Increasing Sub-list of List L
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Asc=[1]*len(L);Tra=[-1]*len(L) #设定起始值
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#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，
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# 这个最长数列肯定以L[i]结尾
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#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。
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for i in range(1,len(L)):
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X=[]
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for j in range(0,i):
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if L[i] > L[j]: X.append(j) #所有比L[i]小L[j]的索引放在X
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for k in X: #Asc[i]= max Asc[k]+1, for each k in X
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if Asc[i] < Asc[k]+1: Asc[i]=Asc[k]+1; Tra[i]=k
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print("Asc:",Asc)
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print("Tra:",Tra)
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max=0 #找到Asc中的最大值
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for i in range(1,len(Asc)):
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if Asc[i]>Asc[max]: max=i
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print("最长递增子序列的长度是",Asc[max])
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#将最长递增数列存到X
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X=[L[max]]; i=max;
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while (Tra[i] >=0):
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X=[L[Tra[i]]]+X
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i=Tra[i]
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print("最长递增子数列=",X)
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L=[5,2,4,7,6,3,8,9]
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LIS(L)
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#<程序：直接用递归函数计算Asc(k)>
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def Asc(k):
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if k==0: return(1)
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X=[]
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for i in range(0,k):
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if L[k] > L[i]: X.append(Asc(i)) #记录所有比L[k]小的Asc（）
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if len(X) >0: return (max(X)+1)
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else: return(1)
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def LIS_R(L):
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X=[]
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for k in range(0, len(L)):
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X.append(Asc(k))
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print(X)
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print(max(X))
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L=[5,2,4,7,6,3,8,9]
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LIS_R(L)
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#<程序：背包问题_递归>
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w=[0,4,5,2,1,6] #w[i]是物品的重量
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v=[0,45,57,22,11,67] #v[i]是物品的价值
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n=len(w)-1
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j=8 #背包的容量
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x=[False for raw in range(n+1)]#x[i]为True，表示物品被放入背包
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def knap_r(n,j):
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if (n==0)or(j==0):
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x[n]=False
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return 0
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elif (j>=w[n])and(knap_r(n-1,j-w[n])+v[n]>knap_r(n-1,j)):
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x[n]=True
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return knap_r(n-1,j-w[n])+v[n]
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else:
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x[n]=False
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return knap_r(n-1,j)
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print("最大价值为：",knap_r(n,j))
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print("物品的装入情况为：",x[1:])
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#<程序：背包问题_动态规划>
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w=[0,4,5,2,1,6] #w[i]是物品的重量
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v=[0,45,57,22,11,67] #v[i]是物品的价值
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n=len(w)-1
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m=8 #背包的容量
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x=[False for raw in range(n+1)]#x[i]为True，表示物品被放入背包
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#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
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a=[[0 for col in range(m+1)] for raw in range(n+1)]
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def knap_DP(n,m):
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#创建动态规划表
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for i in range(1,n+1):
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for j in range(1,m+1):
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a[i][j]=a[i-1][j]
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if (j>=w[i]) and(a[i-1][j-w[i]]+v[i]>a[i-1][j]):
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a[i][j]=a[i-1][j-w[i]]+v[i]
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#回溯a[i][j]的生成过程，找到装入背包的物品
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j=m
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for i in range(n,0,-1):
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if a[i][j]>a[i-1][j]:
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x[i]=True
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j=j-w[i]
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Mv=a[n][m]
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return Mv
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#============================================
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#5.6 以老鼠走迷宫为例
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#============================================
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#<程序：老鼠走迷宫_递归>
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m=[[1,1,1,0,1,1,1,1,1,1],
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[1,0,0,0,0,0,0,0,1,1],
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[1,0,1,1,1,1,1,0,0,1],
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[1,0,1,0,0,0,0,1,0,1],
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[1,0,1,0,1,1,0,0,0,1],
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[1,0,0,1,1,0,1,0,1,1],
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[1,1,1,1,0,0,0,0,1,1],
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[1,0,0,0,0,1,1,1,0,0],
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[1,0,1,1,0,0,0,0,0,1],
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[1,1,1,1,1,1,1,1,1,1]]
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sta1=0;sta2=3;fsh1=7;fsh2=9;success=0
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def LabyrinthRat():
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print('显示迷宫：')
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for i in range(len(m)): print(m[i])
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print('入口：m[%d][%d]：出口：m[%d][%d]'%(sta1,sta2,fsh1,fsh2))
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if (visit(sta1,sta2))==0: print('没有找到出口')
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else:
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print('显示路径：')
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for i in range(10):print(m[i])
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def visit(i,j):
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m[i][j]=2
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global success
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if(i==fsh1)and(j==fsh2): success=1
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if(success!=1)and(m[i-1][j]==0): visit(i-1,j)
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if(success!=1)and(m[i+1][j]==0): visit(i+1,j)
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if(success!=1)and(m[i][j-1]==0): visit(i,j-1)
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if(success!=1)and(m[i][j+1]==0): visit(i,j+1)
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if success!=1: m[i][j]=3
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return success
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if(__name__=="__main__"):
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LabyrinthRat()
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#============================================
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#5.7 谈计算思维的美
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#============================================
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#++++++++++++++++++++++++++++++++++++++++++++
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#5.7.3 问题复杂度的研究之美
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#++++++++++++++++++++++++++++++++++++++++++++
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#<程序:Find all the factors of x and put them in list L>
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import math
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def factors(x,L):
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y=int(math.sqrt(x)) #x的平方根
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for i in range(2,y+1): #一个个找
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if (x %i ==0): #找到一个因数i
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print(i)
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L.append(i)
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factors(x//i,L) #递归找
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break #跳出for循环
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else: #cannot find a factor, so x is a prime
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L.append(int(x))
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print(int(x))
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L=[]
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factors(18,L)
文章来源: fantianzuo.blog.csdn.net，作者：兔老大RabbitMQ，版权归原作者所有，如需转载，请联系作者。
原文链接：fantianzuo.blog.csdn.net/article/details/83025212
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