微软OneNote客户预览版学习工具下载(暂未上线):教学好帮手
744
2022-05-29
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#5.1 计算思维是什么
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#<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=0
while i if L[i] < L[i+1]: return (i) elif L[i] > L[i+1]: return (i+1) i=i+1 print("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 count if n < 1: print('False') elif n == 1: print ("%d:\t%s -> %s" % (count, A, C)) count += 1 elif 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=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)//2 lmin,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 variables print("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 adder carry = (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+y while len(x) < len(y): x = [False]+x while len(y) < len(x): y = [False]+y if 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=0 for i in n: s=s+i print('找给顾客的硬币数最少为:',s) def greedy(T): if T==0:return elif T>=v[0]: T=T-v[0]; n[0]=n[0]+1 greedy(T) elif v[0]>T>=v[1]: T=T-v[1]; n[1]=n[1]+1 greedy(T) elif v[1]>T>=v[2]: T=T-v[2]; n[2]=n[2]+1 greedy(T) else: T=T-v[3]; n[3]=n[3]+1 greedy(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=y else: a=y;b=x if 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 L Asc=[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X for k in X: #Asc[i]= max Asc[k]+1, for each k in X if Asc[i] < Asc[k]+1: Asc[i]=Asc[k]+1; Tra[i]=k print("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=i print("最长递增子序列的长度是",Asc[max]) #将最长递增数列存到X X=[L[max]]; i=max; while (Tra[i] >=0): X=[L[Tra[i]]]+X i=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]=False return 0 elif (j>=w[n])and(knap_r(n-1,j-w[n])+v[n]>knap_r(n-1,j)): x[n]=True return knap_r(n-1,j-w[n])+v[n] else: x[n]=False return 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=m for i in range(n,0,-1): if a[i][j]>a[i-1][j]: x[i]=True j=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]=2 global success if(i==fsh1)and(j==fsh2): success=1 if(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]=3 return success if(__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i print(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 prime L.append(int(x)) print(int(x)) L=[] factors(18,L)
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