The re.sub() tool (sub stands for substitution) evaluates a pattern and, for each valid match, it calls a method (or lambda).

The method is called for all matches and can be used to modify strings in different ways.

The re.sub() method returns the modified string as an output.

Learn more about $re.sub()$.

Transformation of Strings

Code

import re

Squaring numbers

def square(match):
number = int(match.group(0))
return str(number**2)

print re.sub(r"$\d+$", square, "1 2 3 4 5 6 7 8 9")

Output

1 4 9 16 25 36 49 64 81

Replacements in Strings

_{Code}

import re

html = """
<head>
<title>HTML</title>
</head>

"""

print re.sub("()", "", html) #remove comment

Output

<head>
<title>HTML</title>
</head>

Task

You are given a text of $N$ lines. The text contains && and || symbols.

Your task is to modify those symbols to the following:

&& → and
|| → or

Both && and || should have a space " " on both sides.

Input Format

The first line contains the integer, $N$.

The next $N$ lines each contain a line of the text.

Constraints

$0<N<100$

Neither && nor || occur in the start or end of each line.

Output Format

Output the modified text.

Sample Input
11
a = 1;
b = input();

if a + b > 0 && a - b < 0:
start()
elif a*b > 10 || a/b < 1:
stop()
print set(list(a)) | set(list(b))

Note do not change &&& or ||| or & or |

Only change those '&&' which have space on both sides.

Only change those '|| which have space on both sides.

Sample Output
a = 1;
b = input();

if a + b > 0 and a - b < 0:
start()
elif a*b > 10 or a/b < 1:
stop()
print set(list(a)) | set(list(b))

Note do not change &&& or ||| or & or |

Only change those '&&' which have space on both sides.

Only change those '|| which have space on both sides.

见题面

见题面