演算法競賽入門經典第四章 部分習題解答

2021-07-11 16:01:51 字數 4903 閱讀 3958

習題4-2:此題不難,我採取最直接的辦法,暴力遍歷進行判斷,**如下:

def


do_squares


(h,v,n):


#h,v是乙個矩陣


deftest


(i,j,l):


_i = i;_j = j


while(_jif(h[_i][_j]==0):return


false


_j+=1


while(_iif(v[_i][_j]==0):return


false


_i+=1


while(_j>j):


if(h[_i][_j-1]==0):return


false


_j-=1


while(_i>i):


if(v[_i-1][_j]==0):return


false


_i-=1


return


true


count = [0]*(n-1)


for i in range(1,n-1):


for j in range(1,n-1):


for length in range(1,n-1):#長度從1-n-1


if(test(i,j,length)):count[length]+=1


for k in range(1,n-1):


if(count[k]!=0):print('number of suqare with length(%d) : %d'%(k,count[k]))


defsquares


(h,v,n):


h = [ [0]*(n+1) for i in range(n+1)]


v = [ [0]*(n+1) for j in range(n+1)]


for x in h:h[x[0]][x[1]] = 1


for y in v:v[y[0]][y[1]] = 1


do_squares(h,v,n+1)


squares([(1,1),(1,3),(2,1),(2,2),(2,3),(3,2),(4,2),(4,3)]


,[(1,1),(1,2),(2,2),(3,2),(2,3),(1,4),(2,4),(3,4)],4)
def cube(a,b):


x = [[a[0],a[5]],[a[1],a[4]],[a[2],a[3]]]


y = [[b[0],b[5]],[b[1],b[4]],[b[2],b[3]]]


for i in range(3):


x[i].sort();y[i].sort()


x.sort();y.sort()


print(x==y)


cube('grggrb','rggbgr')
def


convert


(x):


ans = ''


tmp = x.split('.')


for each in tmp:


tmp_ans = bin(int(each))[2:]


ans += '0'*(8-len(tmp_ans))+tmp_ans


return ans


defrecovert


(y):


ans = ''


for i in range(0,32,8):


ans += str(int(y[i:i+8],base = 2))+'.'


return ans[:-1]


defsolve


(ip):


net = convert(ip[0])


for each_ip in ip[1:]:


tmp = convert(each_ip)


for i in range(len(net)):


if(tmp[i]!=net[i]):net = net[:i];break


mask = '1'*len(net)+'0'*(32-len(net))


net += '0'*(32-len(net))


print('%s*********************%s'%(recovert(net),recovert(mask)))


solve(['194.85.160.183','194.85.160.177','194.85.160.178'])
def


do_raid


(d,s,b,type):


#b代表資料塊


check_result = 0


if type=='e'


else


1def


eor(x,check_pos):


ans = 0


count = 0


err_pos = 0


for i,each in enumerate(x):


if(each!='?'):ans ^= int(each,base = 2)


else:count+=1;err_pos = i


if(count>1):return


false,x


elif(count==1):


tmp = bin(ans^check_result)[2:]


x[err_pos] = '0'*(s-len(tmp))+tmp


return


true,x[0:check_pos]+x[check_pos+1:]#去掉校驗碼的部分


check_pos = 0


final_ans = ''


for i in range(0,len(b),d):


if(b[check_pos]=='?'):print('check illegal');return


test,tmp = eor(b[i:i+d],check_pos%d)


ifnot test:print('too many error');return


#磁碟資料損壞過多


check_pos += (d+1)


print(tmp)


final_ans += ''.join(tmp)


print(final_ans,hex(int(final_ans,2)))


do_raid(5,2,['00','01','10','11','?','01','10','11','10','?','01','11','01','10','?',


'11','10','?','11','01','11','11','11','00','11'],'e')
def


tired_students


(students):


n = len(students)


a = [students[i][0] for i in range(n)]


b =  [students[i][1] for i in range(n)]


c =  [students[i][2]-0.5


for i in range(n)]


print(a,b,c)


time = 1


sleeped = 0


initial_c = c[:]


tmp = [1]*n


for i in range(n):


if(c[i]>a[i]):sleeped += 1;tmp[i] = 0 


print(sleeped,tmp)


while(sleeped !=0):


time += 1


tmp_sleeped = 0


for i,each in enumerate(c):#遍歷每乙個學生的情況


c[i] = (c[i]+1)%(a[i]+b[i])#獲取當前在週期裡面的位置


if(c[i]>a[i]):#到了睡覺時間


if(tmp[i] == 1):#上一刻處於聽課


if(2*sleeped > n):#滿足睡覺條件


tmp[i] = 0


tmp_sleeped += 1


else:


c[i] = 0


#在睡覺a[i]分鐘


elif(tmp[i]==0):#到了醒來的時間#上一刻處於睡覺狀態


tmp[i] = 1


tmp_sleeped -= 1


sleeped += tmp_sleeped


print('the %d分鐘: '%time,tmp,sleeped,c)#測試情況


if(initial_c == c and sleeped != 0):return -1


#不會有這個時刻


return time
然後假設在上乙個臨界高度之上還應該多出h』的高度,列出方程:v0+v1+h』(num - i) = 100*num(h+last_h)，計算出last_h+h』就得到最後的高度.
def


flooded


(m,n,h,v):


h.sort()#對高度排序


num = m*n


v0 = v


v1 = 0


tmp_h = 0


final_h = 0


count = 0


last_h = 0


i = 0


while(i<100):


if(h[i] != last_h):


if(v0+v1+(num - i)*(h[i]-last_h)*100


< h[i]*num*100 ):


final_h = last_h;break


else:


v1 += (num - i)*(h[i]-last_h)*100


last_h = h[i]


i += 1


x = num - i


print(final_h,(v0+v1-100*num*last_h)/(100*num-x*num)+last_h,i/num)

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