message stringlengths 2 59.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format. | instruction | 0 | 18,736 | 20 | 37,472 |
Tags: implementation, strings
Correct Solution:
```
def mi():
return map(int, input().split())
'''
-0.00987654321
'''
n = input()
neg = False
if n[0]=='-':
neg = True
n=n[1:]
if '.' not in n:
n = n+'.00'
if len(n)-n.index('.')>2:
n = n[:n.index('.')+3]
elif len(n)-n.index('.')==2:
n = n+'0'
i = int(n[:n.index('.')])
s = '$'+str(i)+'.'+n[n.index('.')+1:]
if neg:
s = '('+s+')'
s = list(s)[::-1]
i = s.index('.')+4
while i< len(s):
if s[i]!='(' and s[i]!='$':
s.insert(i, ',')
i+=4
print (''.join(s[::-1]))
``` | output | 1 | 18,736 | 20 | 37,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format. | instruction | 0 | 18,737 | 20 | 37,474 |
Tags: implementation, strings
Correct Solution:
```
t = input()
x, y, d = '', '', '.00'
if t[0] == '-': x, y = '(', ')'
k = t.find('.')
if k > 0: d = (t[k: ] + '00')[: 3]
else: k = len(t)
j = len(x)
u = t[j: j + (k - j) % 3]
j += len(u)
if u and k > j: u += ','
print(x + '$' + u + ','.join(t[i: i + 3] for i in range(j, k, 3)) + d + y)
``` | output | 1 | 18,737 | 20 | 37,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format. | instruction | 0 | 18,738 | 20 | 37,476 |
Tags: implementation, strings
Correct Solution:
```
s = input()
if s[0] == '-':
res = '($'
s = s[1:]
c = s.count('.')
frac = ""
if c == 1:
k = s.index('.')+1
frac = '.'
if len(s)-k == 1:
frac+=s[k]+'0'
elif len(s)-k>=2:
frac+=s[k:k+2]
n = 0
if k//3 == k/3:
n = k//3-1
else:
n=k//3
l = []
c = 0
for i in range(k-2, -1, -1):
if c==3 and n>=0:
l.append(',')
l.append(s[i])
c=1
n-=1
else:
l.append(s[i])
c+=1
res+=''.join(l[::-1])+frac+')'
print(res)
else:
frac =".00"
n = 0
k = len(s)
if k//3 == k/3:
n = k//3-1
else:
n=k//3
l = []
c = 0
for i in range(k-1, -1, -1):
if c==3 and n>=0:
l.append(',')
l.append(s[i])
c=1
n-=1
else:
l.append(s[i])
c+=1
res+=''.join(l[::-1])+frac+')'
print(res)
else:
res = '$'
c = s.count('.')
frac = ""
if c == 1:
k = s.index('.')+1
frac = '.'
if len(s)-k == 1:
frac+=s[k]+'0'
elif len(s)-k>=2:
frac+=s[k:k+2]
n = 0
if k//3 == k/3:
n = k//3-1
else:
n=k//3
l = []
c = 0
for i in range(k-2, -1, -1):
if c==3 and n>=0:
l.append(',')
l.append(s[i])
c=1
n-=1
else:
l.append(s[i])
c+=1
res+=''.join(l[::-1])+frac
print(res)
else:
frac =".00"
n = 0
k = len(s)
if k//3 == k/3:
n = k//3-1
else:
n=k//3
l = []
c = 0
for i in range(k-1, -1, -1):
if c==3 and n>=0:
l.append(',')
l.append(s[i])
c=1
n-=1
else:
l.append(s[i])
c+=1
res+=''.join(l[::-1])+frac
print(res)
``` | output | 1 | 18,738 | 20 | 37,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format. | instruction | 0 | 18,739 | 20 | 37,478 |
Tags: implementation, strings
Correct Solution:
```
s=input()
if '.' in s:
if s[0]=='-':
i=1
a='$'
while s[i]!='.':
a=a+s[i]
i=i+1
c=0
x=len(a)
d=''
for j in range(x-1,0,-1):
c=c+1
d=a[j]+d
if c==3 and j!=1:
d=','+d
c=0
a='$'+d
a=a+'.'
f=s[i+1:i+3]
f=f+'00'
a=a+f[0:2]
print('('+a+')')
else:
i=0
a='$'
while s[i]!='.':
a=a+s[i]
i=i+1
c=0
x=len(a)
d=''
for j in range(x-1,0,-1):
c=c+1
d=a[j]+d
if c==3 and j!=1:
d=','+d
c=0
a='$'+d
a=a+'.'
f=s[i+1:i+3]
f=f+'00'
a=a+f[0:2]
print(a)
else:
b=0
if s[0]=='-':
b=1
a='$'+s[b:]
c=0
x=len(a)
d=''
for i in range(x-1,0,-1):
c=c+1
d=a[i]+d
if c==3 and i!=1:
d=','+d
c=0
a='$'+d
a=a+'.00'
if b:
print('('+a+')')
else:
print(a)
``` | output | 1 | 18,739 | 20 | 37,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format. | instruction | 0 | 18,740 | 20 | 37,480 |
Tags: implementation, strings
Correct Solution:
```
__author__ = 'NIORLYS'
def financial_format_integer(number):
if len(number) <= 3:
return number
r = []
for i in range(1, len(number) + 1):
r.append(number[-i])
if i % 3 == 0 and i < len(number):
r.append(',')
r.reverse()
return r
n = list(input())
if '.' in n:
dot_index = n.index('.')
if len(n) - 1 - dot_index == 1:
n.append('0')
elif len(n) - 1 - dot_index > 2:
k = len(n) - 1 - dot_index - 2
while k:
n.pop()
k -= 1
decimal_part = ''.join(n[dot_index: len(n)])
if n[0] != '-':
integer_part = ''.join(financial_format_integer(n[0:dot_index]))
print('$' + integer_part + decimal_part)
else:
integer_part = ''.join(financial_format_integer(n[1:dot_index]))
print('(' + '$' + integer_part + decimal_part + ')')
else:
decimal_part = '.00'
if n[0] != '-':
integer_part = ''.join(financial_format_integer(n))
print('$' + integer_part + decimal_part)
else:
integer_part = ''.join(financial_format_integer(n[1:len(n)]))
print('(' + '$' + integer_part + decimal_part + ')')
``` | output | 1 | 18,740 | 20 | 37,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format. | instruction | 0 | 18,741 | 20 | 37,482 |
Tags: implementation, strings
Correct Solution:
```
def main():
s=input();
ans=""
if(s[0]=='-'):
ans+=')';
i=len(s)-1
x=-1
while(i>=0):
if(s[i]=='.'):
x=i
if(i+2<len(s)):
ans+=s[i+2];
else:
ans+="0"
if (i + 1 < len(s)):
ans += s[i + 1];
else:
ans += "0"
ans+='.'
break
i-=1
if(x==-1):
ans+="00."
x=len(s)
i=x-1
c=1
while(i>=0):
if(s[i]=='-'):
break
ans+=s[i]
if(c%3==0 and i>0 and s[i-1]!='-'):
ans+=","
c+=1
i-=1
ans+="$"
if(ans[0]==')'):
ans+="("
if(s=="0"):
print("$0.00")
else:
print(ans[::-1])
if __name__ == '__main__':main()
``` | output | 1 | 18,741 | 20 | 37,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format. | instruction | 0 | 18,742 | 20 | 37,484 |
Tags: implementation, strings
Correct Solution:
```
s1=input()
ans=0
if s1[0]=='-':
ans=1
s1=s1[1:]
if '.' in s1:
s1,s2=s1.split('.')
s2=s2[:2]
else:
s2='00'
if len(s2)==1:
s2=s2+'0'
s = s1[:len(s1)%3]
for i in range(len(s1)%3,len(s1),3):
if(len(s)>0):
s += ','
s += s1[i:i+3]
if ans==1:
print('($'+s+'.'+s2+')')
else:
print('$'+s+'.'+s2)
``` | output | 1 | 18,742 | 20 | 37,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format. | instruction | 0 | 18,743 | 20 | 37,486 |
Tags: implementation, strings
Correct Solution:
```
number = input()
parts = number.split(".")
intPart = parts[0]
decimalPart = ["0","0"]
shouldAddParentheses = intPart[0] == "-"
if len(parts) == 2:
for i in range(min(len(parts[1]),2)):
decimalPart[i] = parts[1][i]
tempIntPart = []
digitCount = 0
for i in range(len(intPart)-1,-1,-1):
if intPart[i] == "-": continue
if digitCount == 3:
tempIntPart.append(",")
digitCount = 0
tempIntPart.append(intPart[i])
digitCount += 1
intPart = reversed(tempIntPart)
result = "$" + "".join([str(x) for x in intPart]) + "." + "".join([str(x) for x in decimalPart])
if shouldAddParentheses:
result = "(" + result + ")"
print(result)
``` | output | 1 | 18,743 | 20 | 37,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
Submitted Solution:
```
def main():
s = input()
neg = False
if s[0] == '-':
neg = True
s = s[1:]
pointPos = s.find('.')
if pointPos == -1:
pointPos = len(s)
intPart = s[:pointPos]
fraPart = s[pointPos + 1:pointPos + 3]
for i in range(2 - len(fraPart)):
fraPart += '0'
a = []
i = 0
rev = intPart[::-1]
while i < len(rev):
a.append(rev[i:i + 3])
i += 3
ans = '$' + ','.join(a)[::-1] + '.' + fraPart
if neg:
ans = '(' + ans + ')'
print(ans)
if __name__ == '__main__':
main()
#2019-02-22 13:06:01.034
``` | instruction | 0 | 18,744 | 20 | 37,488 |
Yes | output | 1 | 18,744 | 20 | 37,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
Submitted Solution:
```
import math
num = input()
if '.' not in num:
num += '.00'
ip, dp = num.split(".")
neg = False
if ip[0] == '-':
neg = True
ip = ip[1:]
if len(dp) > 2:
dp = dp[:2]
while len(dp) < 2:
dp += "0"
i = 0
while i < len(ip) and ip[i] == '0':
i += 1
ip = ip[i:]
if len(ip) > 3:
ip = ip[::-1]
ip = ','.join(ip[3 * i :min(3 * i + 3, len(ip))] for i in range(math.ceil(len(ip) / 3)))
ip = ip[::-1]
if len(ip) == 0:
ip = '0'
if int(num.split('.')[1]) == 0:
neg = False
if neg:
print('(', end='')
print('$' + ip + '.' + dp, end='')
if neg:
print(')', end='')
``` | instruction | 0 | 18,745 | 20 | 37,490 |
Yes | output | 1 | 18,745 | 20 | 37,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
Submitted Solution:
```
n=input()
b,c=0,0
if n[0]=='-':
c=1
n=list(n)
if c==1:
n.remove(n[0])
try:
a=n.index('.')
except:
b=1
if b==1:
end='.00'
n.append(end)
else:
if (a+2)>=len(n):
n.append('0')
else:
n=n[:(a+3)]
n=''.join(n)
q=n[:-3]
lenq=len(q)
if c==1:
print('($',end='')
else:
print('$',end='')
if lenq>3:
if lenq%3==0:
print(n[:3],end='')
w=3
elif lenq%3==1:
print(n[:1],end='')
w=1
elif lenq%3==2:
print(n[:2],end='')
w=2
while w<lenq:
print(','+n[w:(w+3)],end='')
w+=3
print(n[-3:],end='')
else:
print(n,end='')
if c==1:
print(')')
``` | instruction | 0 | 18,746 | 20 | 37,492 |
Yes | output | 1 | 18,746 | 20 | 37,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
Submitted Solution:
```
n=input()
if("." in n):
s1,s2=n.split(".")
else:
s1=n
s2=""
if(len(s2)>2):
s2=s2[0:2]
else:
s2=s2+"0"*(2-len(s2))
if(s1[0]=="-"):
s1=s1[1:]
s1=s1[::-1]
l=len(s1)
c=0
s11=""
if(l%3!=0):
ll=l//3
else:
ll=l//3-1
k=0
for i in range(l+ll):
if(c<3):
s11+=s1[i-k]
c+=1
else:
s11+=","
k+=1
c=0
s1=s11[::-1]
s1="($"+s1
s2=s2+")"
else:
s1=s1[::-1]
l=len(s1)
c=0
s11=""
if(l%3!=0):
ll=l//3
else:
ll=l//3-1
k=0
for i in range(l+ll):
if(c<3):
s11+=s1[i-k]
c+=1
else:
s11+=","
k+=1
c=0
s1=s11[::-1]
s1="$"+s1
s=s1+"."+s2
print(s)
``` | instruction | 0 | 18,747 | 20 | 37,494 |
Yes | output | 1 | 18,747 | 20 | 37,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
Submitted Solution:
```
s = input()
b = False
if s[0] == "-":
s = s[1:]
b = True
if "." in s:
i,j = s.split(".")
if len(j) > 2:
j1 = j[:2]
elif len(j) == 2:
j1 = j
elif len(j) == 1:
j1 = j+"0"
elif len(j) == 0:
j1 = "00"
else:
i = s
j1 = "00"
li = []
for k in range(len(s)):
li.append(s[k])
n = len(i) // 3
m = -3; p = 0
for i in range(n):
li.insert(m-p,',')
m = m-3
p +=1
l1 =(''.join(map(str, li)))
if b:
print("("+"$"+l1+"."+j1+")")
else:
print("$"+l1+"."+j1)
``` | instruction | 0 | 18,748 | 20 | 37,496 |
No | output | 1 | 18,748 | 20 | 37,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
Submitted Solution:
```
s = input()
if s.count('.') == 0: s += '.00'
else:
a = s.find('.')
if len(s) == a + 2: s += '0'
else: s = s[:a + 3]
s = s[::-1]
for i in range(6,len(s),4):
s = s[:i] + ',' + s[i:]
s = s[::-1]
s = '$' + s
if s.count('-') > 0:
s = '(' + s.replace('-', '') + ')'
print(s)
``` | instruction | 0 | 18,749 | 20 | 37,498 |
No | output | 1 | 18,749 | 20 | 37,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
Submitted Solution:
```
s = input().split('.')
if len(s) == 1:
s.append('')
s[1] = s[1][:2]
s[1] += '0'*(2-len(s[1]))
isNegative = False
if s[0][0] == '-':
isNegative = True
s[0] = s[0][1:]
print(s)
n = ''
for i in range(len(s[0])):
n += s[0][-i-1]
if (i+1) % 3 == 0:
n += ','
n = '$' + n[::-1]
if isNegative:
print('(' + n + '.' + s[1] + ')')
else:
print(n + '.' + s[1])
``` | instruction | 0 | 18,750 | 20 | 37,500 |
No | output | 1 | 18,750 | 20 | 37,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
* A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point).
* To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678
* In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests).
* When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets.
* Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
* The number's notation only contains characters from the set {"0" – "9", "-", "."}.
* The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits
* A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0").
* The minus sign (if it is present) is unique and stands in the very beginning of the number's notation
* If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign.
* The input data contains no spaces.
* The number's notation contains at least one decimal digit.
Output
Print the number given in the input in the financial format by the rules described in the problem statement.
Examples
Input
2012
Output
$2,012.00
Input
0.000
Output
$0.00
Input
-0.00987654321
Output
($0.00)
Input
-12345678.9
Output
($12,345,678.90)
Note
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
Submitted Solution:
```
def mi():
return map(int, input().split())
'''
-0.00987654321
'''
n = input()
neg = False
if n[0]=='-':
neg = True
n=n[1:]
if '.' not in n:
n = n+'.00'
if len(n)-n.index('.')>2:
n = n[:n.index('.')+3]
elif len(n)-n.index('.')==2:
n = n+'0'
i = int(n[:n.index('.')])
s = '$'+str(i)+'.'+n[n.index('.')+1:]
if neg:
s = '('+s+')'
s = list(s)[::-1]
for i in range(s.index('.')+4, len(s), 4):
if s[i+1]!='(' and s[i+1]!='$':
s.insert(i, ',')
print (''.join(s[::-1]))
``` | instruction | 0 | 18,751 | 20 | 37,502 |
No | output | 1 | 18,751 | 20 | 37,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie and the cows are playing with sequences and need your help. They start with a sequence, initially containing just the number 0, and perform n operations. Each operation is one of the following:
1. Add the integer xi to the first ai elements of the sequence.
2. Append an integer ki to the end of the sequence. (And hence the size of the sequence increases by 1)
3. Remove the last element of the sequence. So, the size of the sequence decreases by one. Note, that this operation can only be done if there are at least two elements in the sequence.
After each operation, the cows would like to know the average of all the numbers in the sequence. Help them!
Input
The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of operations. The next n lines describe the operations. Each line will start with an integer ti (1 ≤ ti ≤ 3), denoting the type of the operation (see above). If ti = 1, it will be followed by two integers ai, xi (|xi| ≤ 103; 1 ≤ ai). If ti = 2, it will be followed by a single integer ki (|ki| ≤ 103). If ti = 3, it will not be followed by anything.
It is guaranteed that all operations are correct (don't touch nonexistent elements) and that there will always be at least one element in the sequence.
Output
Output n lines each containing the average of the numbers in the sequence after the corresponding operation.
The answer will be considered correct if its absolute or relative error doesn't exceed 10 - 6.
Examples
Input
5
2 1
3
2 3
2 1
3
Output
0.500000
0.000000
1.500000
1.333333
1.500000
Input
6
2 1
1 2 20
2 2
1 2 -3
3
3
Output
0.500000
20.500000
14.333333
12.333333
17.500000
17.000000
Note
In the second sample, the sequence becomes <image> | instruction | 0 | 18,848 | 20 | 37,696 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n=Int()
s=0
top=[0]
for i in range(n):
query=input()
if(query[0]=='1'):
_,a,x=[int(i) for i in query.split()]
s+=a*x
top[a-1]+=x
elif(query[0]=='2'):
_,x=[int(i) for i in query.split()]
s+=x
top[-1]-=x
top.append(x)
else:
ele=top.pop()
s-=ele
top[-1]+=ele
print(s/len(top))
``` | output | 1 | 18,848 | 20 | 37,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1. | instruction | 0 | 18,869 | 20 | 37,738 |
Tags: constructive algorithms, implementation
Correct Solution:
```
k,d = input().split()
k = int(k) -1
print("No solution") if(d == '0' and k) else print(d+'0'*k)
``` | output | 1 | 18,869 | 20 | 37,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1. | instruction | 0 | 18,870 | 20 | 37,740 |
Tags: constructive algorithms, implementation
Correct Solution:
```
k,d=map(int,input().split())
if d==0 and k>1:
print('No solution')
else:
print(str(d)+'0'*(k-1))
``` | output | 1 | 18,870 | 20 | 37,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1. | instruction | 0 | 18,871 | 20 | 37,742 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n,d=map(int,input().split())
if(d==0):
if(n>1):
print('No solution')
else:
print(0)
else:
s=str(d)+'0'*(n-1)
print(int(s))
``` | output | 1 | 18,871 | 20 | 37,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1. | instruction | 0 | 18,872 | 20 | 37,744 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
inf = float("inf")
# sys.setrecursionlimit(10000000)
# abc='abcdefghijklmnopqrstuvwxyz'
# abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
# mod, MOD = 1000000007, 998244353
# words = {1:'one',2:'two',3:'three',4:'four',5:'five',6:'six',7:'seven',8:'eight',9:'nine',10:'ten',11:'eleven',12:'twelve',13:'thirteen',14:'fourteen',15:'quarter',16:'sixteen',17:'seventeen',18:'eighteen',19:'nineteen',20:'twenty',21:'twenty one',22:'twenty two',23:'twenty three',24:'twenty four',25:'twenty five',26:'twenty six',27:'twenty seven',28:'twenty eight',29:'twenty nine',30:'half'}
# vow=['a','e','i','o','u']
# dx,dy=[0,1,0,-1],[1,0,-1,0]
# import random
# from collections import deque, Counter, OrderedDict,defaultdict
# from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
# from math import ceil,floor,log,sqrt,factorial,pi,gcd
# from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
k,d = get_ints()
if d==0:
if k==1:
print(0)
else:
print('No solution')
exit()
print(str(d)+'0'*(k-1))
``` | output | 1 | 18,872 | 20 | 37,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1. | instruction | 0 | 18,873 | 20 | 37,746 |
Tags: constructive algorithms, implementation
Correct Solution:
```
k, d = map(int, input().split())
if d == 0 and k > 1:
print('No solution')
else:
print('{}{}'.format(d, '0' * (k - 1)))
``` | output | 1 | 18,873 | 20 | 37,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1. | instruction | 0 | 18,874 | 20 | 37,748 |
Tags: constructive algorithms, implementation
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def value():
return map(int,input().split())
def array():
return [int(i) for i in input().split()]
#-------------------------code---------------------------#
#vsInput()
n,k=value()
if(k==0 and n>1):
print("No solution")
exit()
print(k*(10**(n-1)))
``` | output | 1 | 18,874 | 20 | 37,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1. | instruction | 0 | 18,875 | 20 | 37,750 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n, d = map(int, input().split())
if n!=1 and d==0:
print('No solution')
else:
print(d*10**(n-1))
``` | output | 1 | 18,875 | 20 | 37,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1. | instruction | 0 | 18,876 | 20 | 37,752 |
Tags: constructive algorithms, implementation
Correct Solution:
```
k, d = list(map(int, input().split()))
if (d != 0):
print(d, end = '')
for i in range(k - 1):
print(0, end = '')
else:
if (k == 1):
print(0)
else:
print("No solution")
``` | output | 1 | 18,876 | 20 | 37,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1.
Submitted Solution:
```
a='No solution'
d, k=map(int, input().split())
if d==1 or k>0:
a=str(k)
a+='0'*(d-1)
print(a)
``` | instruction | 0 | 18,877 | 20 | 37,754 |
Yes | output | 1 | 18,877 | 20 | 37,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1.
Submitted Solution:
```
k,d = list(map(int, input().split(' ')))
g = 1
if d!=0:
for i in range(k-1):
g=g*10
s = int(str(d)+str(g)[1:k])
print(s)
elif k == 1 and d == 0:
print(0)
else:
print('No solution')
``` | instruction | 0 | 18,878 | 20 | 37,756 |
Yes | output | 1 | 18,878 | 20 | 37,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1.
Submitted Solution:
```
n,k=map(int,input().split())
if k==0:
if n==1:
print(0)
else:
print('No solution')
else:
print(str(k)+'0'*(n-1))
``` | instruction | 0 | 18,879 | 20 | 37,758 |
Yes | output | 1 | 18,879 | 20 | 37,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1.
Submitted Solution:
```
def sumr(n):
count = 0
while n > 0:
count += n % 10
n //= 10
return count
def dr(n):
if n < 10:
return n
else:
return dr(sumr(n))
def func(d, k):
while (d - 9) >= 0 and k > 0:
print(9, end = '')
d -= 9
k -= 1
if d != 0 and k > 0:
print(d, end = '')
k -= 1
while k > 0:
print(0, end = '')
k -= 1
k, d = map(int, input().split())
flag = 0
if k == 1:
for i in range(0, 10):
if dr(i) == d:
flag = 1
print(i)
else:
for i in range(1, 9 * k):
if dr(i) == d:
func(i, k)
flag = 1
break
if flag == 0:
print("No solution")
``` | instruction | 0 | 18,880 | 20 | 37,760 |
Yes | output | 1 | 18,880 | 20 | 37,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1.
Submitted Solution:
```
a,b=map(int,input().split())
s=str()
i=a
while(i>=2):
s+='9'
i-=1
print(s+'4')
``` | instruction | 0 | 18,881 | 20 | 37,762 |
No | output | 1 | 18,881 | 20 | 37,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1.
Submitted Solution:
```
k,d = map(int,input().split())
if d == 0:
print(0)
exit()
r = d//k
r1 = d%k
s = ""
while r1:
r1 = r1 - 1
s = s + str(r+1)
s = s + (k-len(s))*str(r)
print(s)
``` | instruction | 0 | 18,882 | 20 | 37,764 |
No | output | 1 | 18,882 | 20 | 37,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1.
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect, insort
from time import perf_counter
from fractions import Fraction
import copy
from copy import deepcopy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(*ele,end="\n")
k,d=L()
if d==0:
if k==1:
print(0)
else:
print("No solution")
print(str(d)+"0"*(k-1))
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | instruction | 0 | 18,883 | 20 | 37,766 |
No | output | 1 | 18,883 | 20 | 37,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that S(n) is the sum of digits of number n, for example, S(4098) = 4 + 0 + 9 + 8 = 21. Then the digital root of number n equals to:
1. dr(n) = S(n), if S(n) < 10;
2. dr(n) = dr( S(n) ), if S(n) ≥ 10.
For example, dr(4098) = dr(21) = 3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that dr(n) = S( S( S( S(n) ) ) ) (n ≤ 101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers k and d, find the number consisting of exactly k digits (the leading zeroes are not allowed), with digital root equal to d, or else state that such number does not exist.
Input
The first line contains two integers k and d (1 ≤ k ≤ 1000; 0 ≤ d ≤ 9).
Output
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly k digits. We assume that number 0 doesn't contain any leading zeroes.
Examples
Input
4 4
Output
5881
Input
5 1
Output
36172
Input
1 0
Output
0
Note
For the first test sample dr(5881) = dr(22) = 4.
For the second test sample dr(36172) = dr(19) = dr(10) = 1.
Submitted Solution:
```
(k, d) = map(int, input().split())
if d // 10 + 1 > k:
print('No solution')
elif k == 1 and d == 0:
print(0)
else:
while d < k:
d *= 10
print(d - k + 1, '1' * (k - 1), sep = '')
``` | instruction | 0 | 18,884 | 20 | 37,768 |
No | output | 1 | 18,884 | 20 | 37,769 |
Provide a correct Python 3 solution for this coding contest problem.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54 | instruction | 0 | 19,332 | 20 | 38,664 |
"Correct Solution:
```
import re
while True:
N = int(input())
if N == 0:
break
formula = [input() for _ in range(N)]
# print(formula)
stack = []
for i in reversed(formula):
# print(stack)
if re.match("\.*(\*|\+)", i):
depth = i.count(".")
i = i.replace(".", "")
# print(i, depth)
if i == "+":
cal_num = 0
else:
cal_num = 1
for s in reversed(stack):
if re.match("\." * (depth+1) + "\d", s):
num = int(s.replace(".", ""))
# print(cal_num)
if i == "+":
cal_num += num
else:
cal_num *= num
stack.pop()
stack.append("." * depth + str(cal_num))
else:
stack.append(i)
print(stack[0])
``` | output | 1 | 19,332 | 20 | 38,665 |
Provide a correct Python 3 solution for this coding contest problem.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54 | instruction | 0 | 19,333 | 20 | 38,666 |
"Correct Solution:
```
import re
def dotexit(s):
return re.sub("^\.","",s)
def prod(l):
x=1
for i in l:
x *= i
return x
def rec(s):
nums=[dotexit(x) for x in s[1:]]
ad = []
for i in range(len(nums)):
if nums[i]=='+' or nums[i]=='*':
rl=[nums[i]]
for j in range(i+1,len(nums)):
if nums[j][0] == '.':
rl.append(nums[j])
else:
break
ad.extend(rec(rl))
nums.extend(ad)
nums = [x for x in nums if x !='+' and x != '*' and x[0]!='.']
if s[0]=='+':
return [str(sum([int(x) for x in nums]))]
elif s[0]=='*':
return [str(prod([int(x) for x in nums]))]
else:
return s
while True:
n=int(input())
if n==0:
break
s=[]
for i in range(n):
s.append(input())
print(rec(s)[0])
``` | output | 1 | 19,333 | 20 | 38,667 |
Provide a correct Python 3 solution for this coding contest problem.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54 | instruction | 0 | 19,334 | 20 | 38,668 |
"Correct Solution:
```
def solve(x,a,ind):
# print(x,a,ind)
if "0" <= a[0][ind] <="9":
return int(a[0][ind])
if a[0][ind] =="+":
i = j= 1
su = 0
while(i < x):
if a[i][ind+1] == "+" or a[i][ind+1]=="*":
j+=1
while j < x and a[j][ind +1]==".":
j += 1
su += solve(j-i,a[i:j],ind+1)
i = j
j = j
else:
su += int(a[j][ind+1])
i += 1
j += 1
# print(x,a,ind,su)
return su
if a[0][ind] =="*":
i = j= 1
su = 1
while(i < x):
if a[i][ind+1] == "+" or a[i][ind+1]=="*":
j+=1
while j<x and a[j][ind +1]==".":
j += 1
su *= solve(j-i,a[i:j],ind+1)
i = j
j = j
else:
su *= int(a[j][ind+1])
i += 1
j += 1
# print(x,a,ind,su)
return su
while(1):
n = int(input())
if n >0:
a = [input() for i in range(n)]
print(solve(n,a,0))
else:
break
``` | output | 1 | 19,334 | 20 | 38,669 |
Provide a correct Python 3 solution for this coding contest problem.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54 | instruction | 0 | 19,335 | 20 | 38,670 |
"Correct Solution:
```
while True:
n = int(input())
if n == 0:break
a = [input() for _ in range(n)]
while True:
if len(a) == 1:break
c = 0
s = 0
for i in range(len(a)):
if a[i].count('.') > c:
c = a[i].count('.')
s = i
e = s
while e < len(a) and a[e].count('.') == a[s].count('.'):
e += 1
k = a[s - 1].replace('.', '')
b = [a[i] for i in range(s, e)]
for i in range(len(b)):
b[i] = int(b[i].replace('.', ''))
if k == '+':
a[s - 1] = '.'*a[s - 1].count('.')+str(sum(b))
del a[s:e]
else:
m = 1
for i in b:
m *= i
a[s - 1] = '.'*a[s - 1].count('.')+str(m)
del a[s:e]
print(a[0])
``` | output | 1 | 19,335 | 20 | 38,671 |
Provide a correct Python 3 solution for this coding contest problem.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54 | instruction | 0 | 19,336 | 20 | 38,672 |
"Correct Solution:
```
# coding: utf-8
from functools import reduce
def check(dt,st,lv):
global p
if p>=len(dt):
return st
if not dt[p][-1].isdigit() and lv==len(dt[p]):
tmp=p
p+=1
st.append(reduce(lambda a,b:a+b if dt[tmp][-1]=='+' else a*b,check(dt,[],lv+1)))
return check(dt,st,lv)
elif lv==len(dt[p]):
st.append(int(dt[p][-1]))
p+=1
return check(dt,st,lv)
else:
return st
while 1:
n=int(input())
if n==0:
break
data=[]
for i in range(n):
data.append(list(input()))
p=0
print(check(data,[],1)[0])
``` | output | 1 | 19,336 | 20 | 38,673 |
Provide a correct Python 3 solution for this coding contest problem.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54 | instruction | 0 | 19,337 | 20 | 38,674 |
"Correct Solution:
```
def parser(i):
idx = i
dep = s[i].count('.')
i += 1
arr = []
while i < N and s[i].count(".") > dep:
if "+" in s[i] or "*" in s[i]:
num, i = parser(i)
arr.append(num)
else:
arr.append(int(s[i].replace('.', '')))
i += 1
if "+" in s[idx]:
return sum(arr), i - 1
else:
ret = 1
for num in arr:
ret *= num
return ret, i - 1
while True:
N = int(input())
if not N:
break
if N == 1:
print(input())
continue
s = [input() for _ in range(N)]
print(parser(0)[0])
``` | output | 1 | 19,337 | 20 | 38,675 |
Provide a correct Python 3 solution for this coding contest problem.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54 | instruction | 0 | 19,338 | 20 | 38,676 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
def calc(i, level, fs, lim):
if fs[i][1] not in "+*":
return i, int(fs[i][1])
if fs[i][1] == "+":
tmp = 0
j = i+1
while j < lim:
if fs[j][0] <= level:
break
if fs[j][1] in "*+":
j, tmp2 = calc(j, level+1, fs, lim)
tmp += tmp2
else:
tmp += int(fs[j][1])
j += 1
return j-1, tmp
elif fs[i][1] == "*":
tmp = 1
j = i+1
while j < lim:
if fs[j][0] <= level:
break
if fs[j][1] in "*+":
j, tmp2 = calc(j, level+1, fs, lim)
tmp *= tmp2
else:
tmp *= int(fs[j][1])
j += 1
return j-1, tmp
def main():
while True:
n = int(input())
if n == 0:
break
fs = [[0,0] for i in range(n)]
for i in range(n):
tmp = input().strip()
fs[i][0] = tmp.count(".")
fs[i][1] = tmp.replace(".", "")
_, ans = calc(0, 0, fs, len(fs))
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 19,338 | 20 | 38,677 |
Provide a correct Python 3 solution for this coding contest problem.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54 | instruction | 0 | 19,339 | 20 | 38,678 |
"Correct Solution:
```
opr_list = ["+", "*"]
def solve():
N = int(input())
while(N != 0):
if(N == 1):
print(int(input()))
else:
ans = [0 for i in range(20)]
s = []
opr = [0 for i in range(20)]
for i in range(N):
s.append(list(input()))
layer = 0
for i in range(N):
nowlayer = s[i].count(".")
if(nowlayer < layer):
while(nowlayer < layer):
if(opr[layer-2] == "+"):
ans[layer-1] += ans[layer]
else:
ans[layer-1] *= ans[layer]
layer -= 1
if(s[i][-1] in opr_list):
opr[nowlayer] = s[i][-1]
ans[nowlayer + 1] = 0 if opr[nowlayer] == "+" else 1
layer += 1
else:
if(opr[nowlayer-1] == "+"):
ans[nowlayer] += int(s[i][-1])
else:
ans[nowlayer] *= int(s[i][-1])
layer = nowlayer
if(layer > 1):
while(layer > 1):
if(opr[layer-2] == "+"):
ans[layer-1] += ans[layer]
else:
ans[layer-1] *= ans[layer]
layer -= 1
print(ans[1])
N = int(input())
solve()
``` | output | 1 | 19,339 | 20 | 38,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54
Submitted Solution:
```
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(input())
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())
while True:
N = inp()
if N == 0:
break
else:
stack_numbers = [[] for _ in range(20)]
stack_symbols = ['' for _ in range(20)]
bL = -1
for _ in range(N):
S = input()
L = len(S)
#print(stack_symbols)
#print(stack_numbers)
if L < bL:
for i in reversed(range(L,bL)):
if stack_symbols[i] == '+':
for j in range(1,len(stack_numbers[i])):
stack_numbers[i][j] += stack_numbers[i][j-1]
else:
for j in range(1,len(stack_numbers[i])):
stack_numbers[i][j] *= stack_numbers[i][j-1]
stack_numbers[i-1].append(stack_numbers[i][-1])
stack_numbers[i] = []
stack_symbols[i] = ''
#print(stack_numbers)
s = S[-1]
if 48 <= ord(s) <= 57:
s = int(s)
stack_numbers[L-1].append(s)
else:
stack_symbols[L] = s
bL = L
L = 0
for i in reversed(range(L,bL)):
if stack_symbols[i] == '+':
for j in range(1,len(stack_numbers[i])):
stack_numbers[i][j] += stack_numbers[i][j-1]
else:
for j in range(1,len(stack_numbers[i])):
stack_numbers[i][j] *= stack_numbers[i][j-1]
stack_numbers[i-1].append(stack_numbers[i][-1])
stack_numbers[i] = []
stack_symbols[i] = ''
print(stack_numbers[-1][0])
``` | instruction | 0 | 19,340 | 20 | 38,680 |
Yes | output | 1 | 19,340 | 20 | 38,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54
Submitted Solution:
```
def dfs(l):
x = len(l[0])
if l[0][-1] == "+":
b = 0
i = 1
while i < len(l):
if len(l[i]) == x + 1 and "0" <= l[i][-1] <= "9":
b += int(l[i][-1])
i += 1
elif len(l[i]) == x + 1 and (l[i][-1] == "+" or l[i][-1] == "*"):
f = i
i += 1
while i < len(l):
if len(l[i]) == x + 1:
break
i += 1
b += dfs(l[f:i])
elif l[0][-1] == "*":
b = 1
i = 1
while i < len(l):
if len(l[i]) == x + 1 and "0" <= l[i][-1] <= "9":
b *= int(l[i][-1])
i += 1
elif len(l[i]) == x + 1 and (l[i][-1] == "+" or l[i][-1] == "*"):
f = i
i += 1
while i < len(l):
if len(l[i]) == x + 1:
break
i += 1
b *= dfs(l[f:i])
else:
b = int(l[0][-1])
return b
def main(n):
l = [input() for i in range(n)]
print(dfs(l))
while 1:
n = int(input())
if n == 0:
break
main(n)
``` | instruction | 0 | 19,341 | 20 | 38,682 |
Yes | output | 1 | 19,341 | 20 | 38,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54
Submitted Solution:
```
for n in iter(input, '0'):
st = ''
prv_lv = 0
ops = ['']
for _ in range(int(n)):
line = input()
lv = line.count('.')
op = line[lv:]
while prv_lv > lv:
ops.pop()
st += ')'
prv_lv -= 1
if op in '*+':
st += ops[-1]
st += '(' + str(int(op == '*'))
ops.append(op)
else:
st += ops[-1] + op
prv_lv = lv
while len(ops) > 1:
ops.pop()
st += ')'
print(eval(st))
``` | instruction | 0 | 19,342 | 20 | 38,684 |
Yes | output | 1 | 19,342 | 20 | 38,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54
Submitted Solution:
```
while True:
n = int(input())
if n==0:
break
if n==1:
print(int(input()))
continue
op = "+*"
f = [input() for i in range(n)]
f = f[::-1]
stack = []
for i in range(n):
if f[i][-1] not in op:
stack.append(f[i])
else:
nest = f[i].count(".")
if f[i][-1]=="*":
cur = 1
while len(stack):
if stack[-1].count(".")!=nest+1:
break
else:
t = stack.pop().strip(".")
cur *= int(t)
stack.append("."*(nest)+str(cur))
else:
cur = 0
while len(stack):
if stack[-1].count(".")!=nest+1:
break
else:
t = stack.pop().strip(".")
cur += int(t)
stack.append("."*(nest)+str(cur))
print(stack[0])
``` | instruction | 0 | 19,343 | 20 | 38,686 |
Yes | output | 1 | 19,343 | 20 | 38,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54
Submitted Solution:
```
while True:
n = int(input())
if n == 0:break
a = [input() for _ in range(n)]
while True:
if len(a) == 1:break
c = 0
s = 0
for i in range(len(a)):
if a[i].count('.') > c:
c = a[i].count('.')
s = i
e = s
while e < len(a) and a[e].count('.') == a[s].count('.'):
e += 1
k = a[s - 1].replace('.', '')
b = [a[i] for i in range(s, e)]
for i in len(b):
b[i] = int(b[i].replace('.', ''))
if k == '+':
a[s] = '.'*a[s].count('.')+str(sum(b))
del a[s + 1:e]
else:
m = 1
for i in b:
m *= i
a[s] = '.'*a[s].count('.')+str(m)
del a[s + 1:e]
print(a[0])
``` | instruction | 0 | 19,344 | 20 | 38,688 |
No | output | 1 | 19,344 | 20 | 38,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54
Submitted Solution:
```
while True:
n = int(input())
if n == 0:break
a = [input() for _ in range(n)]
while True:
if len(a) == 1:break
c = 0
s = 0
for i in range(len(a)):
if a[i].count('.') > c:
c = a[i].count('.')
s = i
e = s
while e < len(a) and a[e].count('.') == a[s].count('.'):
e += 1
k = a[s - 1].replace('.', '')
b = [a[i] for i in range(s, e)]
for i in range(len(b)):
b[i] = int(b[i].replace('.', ''))
if k == '+':
a[s] = '.'*a[s].count('.')+str(sum(b))
del a[s + 1:e]
else:
m = 1
for i in b:
m *= i
a[s] = '.'*a[s].count('.')+str(m)
del a[s + 1:e]
print(a[0])
``` | instruction | 0 | 19,345 | 20 | 38,690 |
No | output | 1 | 19,345 | 20 | 38,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54
Submitted Solution:
```
def calc(p, nums):
if p == '+':
ans = sum(nums)
else:
ans = 1
for i in nums:
ans *= i
return ans
while True:
n = int(input())
if n == 0:
quit()
w = []
depth = 0
for i in range(n):
t = input()
td = t.count('.')
tc = t.replace('.', '')
if tc.isdigit():
tc = int(tc)
else:
td += 1
w.append([td, tc])
depth = max(depth, td)
# while len(w) != 1:
while depth > 0:
# print(w)
for i in range(len(w)):
if w[i][0] == depth:
temp = []
for j in range(i+1, len(w)):
# print(w[j][1], j)
if w[j][0] == depth and str(w[j][1]).isdigit():
temp.append(w[j][1])
# print(w[i][1], temp)
ta = calc(w[i][1], temp)
# print(ta)
if j >= len(w):
w = w[:i] + [[depth - 1, ta]]
else:
w = w[:i] + [[depth - 1, ta]] + w[j:]
# w = w[:i] + w[depth, ta] + w[j:]
break
depth -= 1
print(w[0][1])
``` | instruction | 0 | 19,346 | 20 | 38,692 |
No | output | 1 | 19,346 | 20 | 38,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICPC Calculator
In mathematics, we usually specify the order of operations by using parentheses. For example, 7 × (3 + 2) always means multiplying 7 by the result of 3 + 2 and never means adding 2 to the result of 7 × 3. However, there are people who do not like parentheses. International Counter of Parentheses Council (ICPC) is attempting to make a notation without parentheses the world standard. They are always making studies of such no-parentheses notations.
Dr. Tsukuba, a member of ICPC, invented a new parenthesis-free notation. In his notation, a single expression is represented by multiple lines, each of which contains an addition operator (+), a multiplication operator (*) or an integer. An expression is either a single integer or an operator application to operands. Integers are denoted in decimal notation in one line. An operator application is denoted by a line of its operator immediately followed by lines denoting its two or more operands, each of which is an expression, recursively. Note that when an operand is an operator application, it comprises multiple lines.
As expressions may be arbitrarily nested, we have to make it clear which operator is applied to which operands. For that purpose, each of the expressions is given its nesting level. The top level expression has the nesting level of 0. When an expression of level n is an operator application, its operands are expressions of level n + 1. The first line of an expression starts with a sequence of periods (.), the number of which indicates the level of the expression.
For example, 2 + 3 in the regular mathematics is denoted as in Figure 1. An operator can be applied to two or more operands. Operators + and * represent operations of summation of all operands and multiplication of all operands, respectively. For example, Figure 2 shows an expression multiplying 2, 3, and 4. For a more complicated example, an expression (2 + 3 + 4) × 5 in the regular mathematics can be expressed as in Figure 3 while (2 + 3) × 4 × 5 can be expressed as in Figure 4.
+
.2
.3
Figure 1: 2 + 3
*
.2
.3
.4
Figure 2: An expression multiplying 2, 3, and 4
*
.+
..2
..3
..4
.5
Figure 3: (2 + 3 + 4) × 5
*
.+
..2
..3
.4
.5
Figure 4: (2 + 3) × 4 × 5
Your job is to write a program that computes the value of expressions written in Dr. Tsukuba's notation to help him.
Input
The input consists of multiple datasets. Each dataset starts with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation.
You may assume that, in the expressions given in the input, every integer comprises a single digit and that every expression has no more than nine integers. You may also assume that all the input expressions are valid in the Dr. Tsukuba's notation. The input contains no extra characters, such as spaces or empty lines.
The last dataset is immediately followed by a line with a single zero.
Output
For each dataset, output a single line containing an integer which is the value of the given expression.
Sample Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output for the Sample Input
9
6
2
54
Example
Input
1
9
4
+
.1
.2
.3
9
+
.0
.+
..*
...1
...*
....1
....2
..0
10
+
.+
..6
..2
.+
..1
..*
...7
...6
.3
0
Output
9
6
2
54
Submitted Solution:
```
while True:
n = int(input())
if n == 0:break
a = [input() for _ in range(n)]
while True:
if len(a) == 1:break
c = 0
s = 0
for i in range(len(a)):
if a[i].count('.') > c:
c = a[i].count('.')
s = i
e = s
while e < len(a) and a[e].count('.') == a[s].count('.'):
e += 1
k = a[s - 1]
k.remove('.')
b = [a[i] for i in range(s, e)]
for i in len(b):
b[i].remove('.')
b[i] = int(b[i])
if k == '+':
a[s] = '.'*a[s].count('.')+str(sum(b))
del a[s + 1:e]
else:
m = 1
for i in b:
m *= i
a[s] = '.'*a[s].count('.')+str(m)
del a[s + 1:e]
print(a[0])
``` | instruction | 0 | 19,347 | 20 | 38,694 |
No | output | 1 | 19,347 | 20 | 38,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals n (n is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first n / 2 digits) equals the sum of digits in the second half (the sum of the last n / 2 digits). Check if the given ticket is lucky.
Input
The first line contains an even integer n (2 ≤ n ≤ 50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly n — the ticket number. The number may contain leading zeros.
Output
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
Examples
Input
2
47
Output
NO
Input
4
4738
Output
NO
Input
4
4774
Output
YES
Note
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | instruction | 0 | 19,657 | 20 | 39,314 |
Tags: implementation
Correct Solution:
```
def is_lucky(n: str) -> bool:
return n.count("4") + n.count("7") == len(n)
def sumar_digitos(num: str) -> (int, int):
n = len(num)
prim_mitad = num[0:n//2]
seg_mitad = num[n//2:]
return sum(map(lambda x: int(x), prim_mitad)), sum(map(lambda x: int(x), seg_mitad))
def is_lucky_ticket(num: str) -> bool:
a, b = sumar_digitos(num)
return is_lucky(num) and (a == b)
# Programa principal
digitos = input()
num = input()
if is_lucky_ticket(num):
print("YES")
else:
print("NO")
``` | output | 1 | 19,657 | 20 | 39,315 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.