[BOJ 13230] Bits equalizer
View as PDFYou are given two nonempty strings S and T of equal lengths. S contains the characters ‘0’, ‘1’ and ‘?’, whereas T contains ‘0’ and ‘1’ only. Your task is to convert S into T in minimum number of moves. In each move, you can do one of these changes:</p>
- change a ‘0’in S to ‘1’
- change a ‘?’ in S to ‘0’ or ‘1’
- swap any two characters in S
As an example, suppose S = “01??00” and T = “001010”. We can transform S into T in 3 moves:
- Initially S = “01??00”
- Move 1 – change S[2] to ‘1’. S becomes “011?00”
- Move 2 – change S[3] to ‘0’. S becomes “011000”
- Move 3 – swap S[1] with S[4]. S becomes “001010”
- S is now equal to T.
The first line of input is an integer C (C ≤ 200) that indicates the number of test cases. Each case consists of two lines. The first line is the string S consisting of ‘0’, ‘1’ and ‘?’. The second line is the string T consisting of ‘0’ and ‘1’. The lengths of the strings won’t be larger than 100.
출력 형식
For each case, output the case number first followed by the minimum number of moves required to convert S into T. If the transition is impossible, output −1 instead. Check the output example for the exact format.
예제 입력
3
01??00
001010
01
10
110001
000000
예제 출력
Case 1: 3
Case 2: 1
Case 3: -1
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