题解 | #字符串合并处理#

str1=input('')

s=list(str1)

str=[]
for v in s :
    if v!=' ':
     str.append(v)
#print(str)

str4=[]
str5=[]
z=len(str)

for i in range(z):
    if i%2==0:
        str4.append(str[i])
    else:
        str5.append(str[i])#奇数偶数分开完毕


#print(str4,str5)

str6=[]
str4.sort()
str5.sort()
b=len(str4)
c=len(str5) 
#print(str4,str5)
e=b-c
f=c-b
if e>0:
    for aa in range(e):
        str5.append('')
if f>0:
    for cc in range(f):
        str4.append('')
d=len(str4)+len(str4)
#print(str4,str5)
for j in range(d):
    str6.append('')
    if j%2==0:
        c=int(j/2)
        str6[j]=str4[c]
        
       
for j in range(d):
   
    if j%2!=0 :
        c=int(j/2)
        str6[j]=str5[c] 
str8=[] 
for r in str6:
    
    if r!='':
        str8.append(r)
    else :
        continue
#print(str8) 

str7=[]
for  k in str8:
   
    if k=='0':
        k='0'
    elif k=='1':
        k='8'
    elif    k=='2':
        k='4'
    elif k=='3':
    
        k='C'
    elif k=='4':
        k='2'
    elif k=='5':
        
    
         k='A'
    elif k=='6':
     
         k='6'
    elif k=='7':
        k='E'
    elif k=='8':
         k='1'
    elif k=='9':
         k='9'
    elif k=='a':
     
         k='5'
    elif k=='b':
         k='D'
    elif k=='c':
          k='3'
    elif k=='d':
          k='B'
    elif k=='e':
         k='7'
    elif k=='f':
         k='F'
    elif k=='A':
     
         k='5'
    elif k=='B':
         k='D'
    elif k=='C':
          k='3'
    elif k=='D':
          k='B'
    elif k=='E':
         k='7'
    elif k=='F':
         k='F'
    str7.append(k)
str8=''
for u in str7:
    str8+=u
print(str8)