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t=int(input()) while 1>t or t>20000: t=int(input("reenter constraints 0<=t<=20,000\n")) la={} lb={} for i in range(t): a=input().lower() while len(a)!=3: a=input("length exceed\n").lower() b=input().lower() while len(b)!=3: b=input("length exceed\n").lower() la[i]=list(a) lb[i]=list(b) acount=bcount=0 for i in range (t): for j in range (3): if la[i][j]=="b" or lb[i][j]=="b": acount+=1 elif la[i][j]=="o" or lb[i][j]=="o": bcount+=1 if acount==2 and bcount==1: print("yes") acount=0 bcount=0 else: print("no") acount=0 bcount=0

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asked 08 Aug '18, 01:47

ashutosh3120's gravatar image

1★ashutosh3120
11
accept rate: 0%

Delete this question. If you don't want to get suspended from forum.

(08 Aug '18, 02:18) aryanc4035★

t=int(input())

while 1>t or t>20000:

 t=int(input("reenter constraints 0<t,<20,000\n"))

la={}

lb={}

for i in range(t):

a=input().lower()

while len(a)!=3:

     a=input("length exceed\n").lower()

b=input().lower()

while len(b)!=3:

     b=input("length exceed\n").lower()

la[i]=list(a)

lb[i]=list(b)

acount=bcount=0

for i in range (t):

 for j in range (3):

     if la[i][j]=="b" or lb[i][j]=="b":

         acount+=1

     elif la[i][j]=="o" or lb[i][j]=="o":

         bcount+=1


 if acount==2 and bcount==1:

      print("yes")

      acount=0

      bcount=0

 else:

     print("no")

     acount=0

     bcount=0
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answered 08 Aug '18, 01:53

ashutosh3120's gravatar image

1★ashutosh3120
11
accept rate: 0%

edited 08 Aug '18, 01:56

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question asked: 08 Aug '18, 01:47

question was seen: 39 times

last updated: 08 Aug '18, 02:18