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.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> . | instruction | 0 | 9,475 | 20 | 18,950 |
Tags: brute force, implementation
Correct Solution:
```
if __name__ == '__main__':
n = int(input())
d = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven',
'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen']
t = ['twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety']
if n < 20:
print(d[n])
else:
s = ''
a = n // 10
b = n % 10
s += t[a - 2]
if b != 0:
s += '-' + d[b]
print(s)
``` | output | 1 | 9,475 | 20 | 18,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> . | instruction | 0 | 9,476 | 20 | 18,952 |
Tags: brute force, implementation
Correct Solution:
```
score = int(input())
score = str(score)
number2word = {"1":"one","2":"two","3":"three","4":"four","5":"five","6":"six","7":"seven", "8":"eight", "9":"nine","0":"zero"}
number2word_10th= {"10":"ten","11":"eleven","12":"twelve","13":"thirteen","14":"fourteen","15":"fifteen","16":"sixteen","17":"seventeen","18":"eighteen","19":"nineteen"}
number2word_digit = {"2":"twenty","3":"thirty","4":"forty","5":"fifty","6":"sixty","7":"seventy","8":"eighty","9":"ninety"}
if len(score)== 1:
score_word = number2word[score]
else:
digit1 = score[0]
digit2 = score[1]
if digit1 == "1":
score_word = number2word_10th[score]
else:
if digit2 == "0":
score_word = number2word_digit[digit1]
else:
score_word = number2word_digit[digit1]
score_word += "-"+number2word[digit2]
print(score_word)
``` | output | 1 | 9,476 | 20 | 18,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> . | instruction | 0 | 9,477 | 20 | 18,954 |
Tags: brute force, implementation
Correct Solution:
```
n=int(input())
if(n<=20):
if(n==0):
print("zero")
if(n==1):
print("one")
if(n==2):
print("two")
if(n==3):
print("three")
if(n==4):
print("four")
if(n==5):
print("five")
if(n==6):
print("six")
if(n==7):
print("seven")
if(n==8):
print("eight")
if(n==9):
print("nine")
if(n==10):
print("ten")
if(n==11):
print("eleven")
if(n==12):
print("twelve")
if(n==13):
print("thirteen")
if(n==14):
print("fourteen")
if(n==15):
print("fifteen")
if(n==16):
print("sixteen")
if(n==17):
print("seventeen")
if(n==18):
print("eighteen")
if(n==19):
print("nineteen")
if(n==20):
print("twenty")
ar1=["","one","two","three","four","five","six","seven","eight","nine"]
ar2=["twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"]
if(n>20):
c=n%10
d=n//10
if(c==0):
print(ar2[d-2])
else:
s=ar2[d-2]+"-"+ar1[c]
print(s)
``` | output | 1 | 9,477 | 20 | 18,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> . | instruction | 0 | 9,478 | 20 | 18,956 |
Tags: brute force, implementation
Correct Solution:
```
N = int(input())
d = {0: "zero", 1: "one", 2: "two", 3: "three", 4: "four", 5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine"}
dd = {11: "eleven", 12: "twelve", 13: "thirteen", 14: "fourteen", 15: "fifteen", 16: "sixteen", 17: "seventeen", 18: "eighteen", 19: "nineteen"}
ddd = {10: "ten", 20: "twenty", 30: "thirty", 40: "forty", 50: "fifty", 60: "sixty", 70: "seventy", 80: "eighty", 90: "ninety"}
if N <= 9:
print(d[N])
elif N % 10 == 0:
print(ddd[N])
elif N <= 19:
print(dd[N])
else:
m = N % 10
n = N - m
print("{}-{}".format(ddd[n], d[m]))
``` | output | 1 | 9,478 | 20 | 18,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> . | instruction | 0 | 9,479 | 20 | 18,958 |
Tags: brute force, implementation
Correct Solution:
```
def main(n):
index = {'1':'one','2':'two','3':'three','4':'four','5':'five','6':'six','7':'seven','8':'eight','9':'nine','0':'zero','10':'ten','11':'eleven','12':'twelve','13':'thirteen'}
if int(n) < 14:
print(index[n])
else:
if n[0] == '1':
if n[1] == '5':
print('fifteen')
elif n[1] == '8':
print('eighteen')
else:
print(index[n[1]] + 'teen')
else:
if n[1] == '0':
if n[0] == '8':
number = 'eighty'
elif n[0] == '5':
number = 'fifty'
elif n[0] == '4':
number = 'forty'
elif n[0] == '2':
number = 'twenty'
elif n[0] == '3':
number = 'thirty'
else:
number = index[n[0]] + 'ty'
else:
number = index[n[0]] + 'ty' + '-' + index[n[1]]
if n[0] == '8':
number = 'eighty' + '-' + index[n[1]]
if n[0] == '5':
number = 'fifty' + '-' + index[n[1]]
if n[0] == '4':
number = 'forty' + '-' + index[n[1]]
if n[0] == '2':
number = 'twenty' + '-' + index[n[1]]
if n[0] == '3':
number = 'thirty' + '-' + index[n[1]]
print(number)
n = input()
main(n)
``` | output | 1 | 9,479 | 20 | 18,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> . | instruction | 0 | 9,480 | 20 | 18,960 |
Tags: brute force, implementation
Correct Solution:
```
score = int(input())
msg = ''
if len(str(score)) == 1:
if score == 0:
msg = 'zero'
elif score == 1:
msg = 'one'
elif score == 2:
msg = 'two'
elif score == 3:
msg = 'three'
elif score == 4:
msg = 'four'
elif score == 5:
msg = 'five'
elif score == 6:
msg = 'six'
elif score == 7:
msg = 'seven'
elif score == 8:
msg = 'eight'
elif score == 9:
msg = 'nine'
if len(str(score)) == 2:
#first digit
score = str(score)
if score[0] == '1':
if score[1] == '0':
msg = 'Ten'
elif score[1] == '1':
msg = 'Eleven'
elif score[1] == '2':
msg = 'Twelve'
elif score[1] == '3':
msg = 'Thirteen'
elif score[1] == '4':
msg = 'Fourteen'
elif score[1] == '5':
msg = 'Fifteen'
elif score[1] == '6':
msg = 'Sixteen'
elif score[1] == '7':
msg = 'Seventeen'
elif score[1] == '8':
msg = 'Eighteen'
elif score[1] == '9':
msg = 'Nineteen'
if score[0] == '2':
msg = 'Twenty'
elif score[0] == '3':
msg = 'Thirty'
elif score[0] == '4':
msg = 'Forty'
elif score[0] == '5':
msg = 'Fifty'
elif score[0] == '6':
msg = 'Sixty'
elif score[0] == '7':
msg = 'Seventy'
elif score[0] == '8':
msg = 'Eighty'
elif score[0] == '9':
msg = 'Ninety'
#second digit
if score[0] != '1':
if score[1] == '1':
msg += '-one'
elif score[1] == '2':
msg += '-two'
elif score[1] == '3':
msg += '-three'
elif score[1] == '4':
msg += '-four'
elif score[1] == '5':
msg += '-five'
elif score[1] == '6':
msg += '-six'
elif score[1] == '7':
msg += '-seven'
elif score[1] == '8':
msg += '-eight'
elif score[1] == '9':
msg += '-nine'
print(msg.lower())
``` | output | 1 | 9,480 | 20 | 18,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> . | instruction | 0 | 9,481 | 20 | 18,962 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
if n == 0:
print('zero')
elif n == 1:
print('one')
elif n == 2:
print('two')
elif n == 3:
print('three')
elif n == 4:
print('four')
elif n == 5:
print('five')
elif n == 6:
print('six')
elif n == 7:
print('seven')
elif n == 8:
print('eight')
elif n == 9:
print('nine')
elif n == 10:
print('ten')
elif n == 11:
print('eleven')
elif n == 12:
print('twelve')
elif n == 13:
print('thirteen')
elif n == 14:
print('fourteen')
elif n == 15:
print('fifteen')
elif n == 16:
print('sixteen')
elif n == 17:
print('seventeen')
elif n == 18:
print('eighteen')
elif n == 19:
print('nineteen')
else:
if n // 10 == 2:
res = 'twenty'
elif n // 10 == 3:
res = 'thirty'
elif n // 10 == 4:
res = 'forty'
elif n // 10 == 5:
res = 'fifty'
elif n // 10 == 6:
res = 'sixty'
elif n // 10 == 7:
res = 'seventy'
elif n // 10 == 8:
res = 'eighty'
elif n // 10 == 9:
res = 'ninety'
if n % 10 == 1:
res += '-one'
elif n % 10 == 2:
res += '-two'
elif n % 10 == 3:
res += '-three'
elif n % 10 == 4:
res += '-four'
elif n % 10 == 5:
res += '-five'
elif n % 10 == 6:
res += '-six'
elif n % 10 == 7:
res += '-seven'
elif n % 10 == 8:
res += '-eight'
elif n % 10 == 9:
res += '-nine'
print(res)
``` | output | 1 | 9,481 | 20 | 18,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> .
Submitted Solution:
```
num = int(input())
dec = num // 10
s = ''
low = ['zero', 'one' , 'two', 'three', 'four', 'five', 'six', 'seven','eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen']
high = ['__0', '__1', 'twenty','thirty','forty','fifty','sixty','seventy','eighty','ninety']
if dec > 1:
s = high[dec]
if num % 10 != 0:
print(s + '-' + low[num % 10])
else:
print(s)
else:
print(low[num])
``` | instruction | 0 | 9,482 | 20 | 18,964 |
Yes | output | 1 | 9,482 | 20 | 18,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> .
Submitted Solution:
```
nums = 'zero one two three four five six seven eight nine ten eleven twelve '
nums += 'thirteen fourteen fifteen sixteen seventeen eighteen nineteen'
nums = nums.split(' ')
tens = 'twenty thirty forty fifty sixty seventy eighty ninety'.split(' ')
a = int(input())
if a < 20:
ans = nums[a]
else:
ans = tens[a//10 - 2]
if a % 10 != 0:
ans += '-' + nums[a%10]
print(ans)
``` | instruction | 0 | 9,483 | 20 | 18,966 |
Yes | output | 1 | 9,483 | 20 | 18,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> .
Submitted Solution:
```
n = int(input().strip())
d = {0 : 'zero', 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 : 'fifteen', \
16 : 'sixteen', 17 : 'seventeen', 18 : 'eighteen', 19 : 'nineteen', \
20 : 'twenty', 30: 'thirty', 40 : 'forty', 50 : 'fifty', 60 : 'sixty', \
70 : 'seventy', 80 : 'eighty', 90: 'ninety'}
if n in d:
print(d[n])
else:
affix = n % 10
prefix = n // 10 * 10
result = d[prefix] + '-' + d[affix]
print(result)
``` | instruction | 0 | 9,484 | 20 | 18,968 |
Yes | output | 1 | 9,484 | 20 | 18,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> .
Submitted Solution:
```
if __name__ == '__main__':
arr = ['' for _ in range(100)]
arr[0] = 'zero'
arr[1] = 'one'
arr[2] = 'two'
arr[3] = 'three'
arr[4] = 'four'
arr[5] = 'five'
arr[6] = 'six'
arr[7] = 'seven'
arr[8] = 'eight'
arr[9] = 'nine'
arr[10] = 'ten'
arr[11] = 'eleven'
arr[12] = 'twelve'
arr[13] = 'thirteen'
arr[14] = 'fourteen'
arr[15] = 'fifteen'
arr[16] = 'sixteen'
arr[17] = 'seventeen'
arr[18] = 'eighteen'
arr[19] = 'nineteen'
arr[20] = 'twenty'
arr[30] = 'thirty'
arr[40] = 'forty'
arr[50] = 'fifty'
arr[60] = 'sixty'
arr[70] = 'seventy'
arr[80] = 'eighty'
arr[90] = 'ninety'
n = int(input())
if arr[n]:
print(arr[n])
else:
ld = n % 10
print(arr[n - ld] + '-' + arr[ld])
``` | instruction | 0 | 9,485 | 20 | 18,970 |
Yes | output | 1 | 9,485 | 20 | 18,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> .
Submitted Solution:
```
a="zero one two three four five six seven eight nine".split()
b ="ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen,nineteen".split()
c ="a a twenty thirty fourty fifty sixty seventy eighty ninety".split()
n = int(input())
if n<10:
print(a[n%10])
elif n<20:
print(b[n%10])
else:
if n%10:
print(c[n//10]+"-"+a[n%10])
else:
print(c[n//10])
``` | instruction | 0 | 9,486 | 20 | 18,972 |
No | output | 1 | 9,486 | 20 | 18,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> .
Submitted Solution:
```
s = input()
a = ['one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen']
if int(s) < 20:
print(a[int(s)-1])
else:
n = int(s)
if n > 39:
if n % 10 == 0:
print(a[int(s[0])-1], 'ty', sep = '')
else:
print(a[int(s[0]) - 1] ,'ty' , '-' , a[int(s[1])-1], sep ='')
else:
if n < 30:
print('twenty', end = '')
if n % 10 != 0:
print('-', a[int(s[1])-1], sep = '')
else:
print('thirty', end = '')
if n % 10 != 0:
print('-', a[int(s[1])-1], sep = '')
``` | instruction | 0 | 9,487 | 20 | 18,974 |
No | output | 1 | 9,487 | 20 | 18,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> .
Submitted Solution:
```
s = input()
a = ['one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen']
if int(s) == 0:
print('zero')
elif int(s) < 20:
print(a[int(s)-1])
elif int(s) >= 80 and int(s) < 90:
print('eighty', end = '')
if int(s) != 80:
print('-', a[int(s[1])-1], sep = '')
else:
n = int(s)
if n > 39:
if n % 10 == 0:
print(a[int(s[0])-1], 'ty', sep = '')
else:
print(a[int(s[0]) - 1] ,'ty' , '-' , a[int(s[1])-1], sep ='')
else:
if n < 30:
print('twenty', end = '')
if n % 10 != 0:
print('-', a[int(s[1])-1], sep = '')
else:
print('thirty', end = '')
if n % 10 != 0:
print('-', a[int(s[1])-1], sep = '')
``` | instruction | 0 | 9,488 | 20 | 18,976 |
No | output | 1 | 9,488 | 20 | 18,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
<image>
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input
The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.
Output
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Examples
Input
6
Output
six
Input
99
Output
ninety-nine
Input
20
Output
twenty
Note
You can find all you need to know about English numerals in <http://en.wikipedia.org/wiki/English_numerals> .
Submitted Solution:
```
import sys
def input(): return sys.stdin.readline().strip()
def iinput(): return int(input())
def rinput(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
n=iinput()
a=['one','two','three','four','five','six','seven','eight','nine','ten','eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen']
b=['twenty','thirty','forty','fifty','sixty','seventy','eighty','nighty']
if(n<=19):
print(a[n-1])
elif(n%10==0 and n!=10):
print(b[n//10 - 2])
elif(n>20 and n<30):
print(b[0]+"-"+a[n-21])
elif(n>30 and n<40):
print(b[1]+"-"+a[n-31])
elif(n>40 and n<50):
print(b[2]+"-"+a[n-41])
elif(n>50 and n<60):
print(b[3]+"-"+a[n-51])
elif(n>60 and n<70):
print(b[4]+"-"+a[n-61])
elif(n>70 and n<80):
print(b[5]+"-"+a[n-71])
elif(n>80 and n<90):
print(b[6]+"-"+a[n-81])
elif(n>90):
print(b[7]+"-"+a[n-91])
``` | instruction | 0 | 9,489 | 20 | 18,978 |
No | output | 1 | 9,489 | 20 | 18,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output
Output one integer number — the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9 | instruction | 0 | 9,490 | 20 | 18,980 |
Tags: brute force
Correct Solution:
```
from itertools import permutations
ch=list(map(int,input().split()))
op=list(input().split())
def f(oper, x, y):
if oper=="*":
return (x*y)
else:
return (x+y)
ans=10**12
for a,b,c,d in permutations(ch):
ans=min(ans, min(f(op[2], f(op[1], f(op[0],a,b), c), d), f(op[2], f(op[0],a,b),f(op[1],c,d))))
print(ans)
``` | output | 1 | 9,490 | 20 | 18,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output
Output one integer number — the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9 | instruction | 0 | 9,491 | 20 | 18,982 |
Tags: brute force
Correct Solution:
```
s=list(map(int,input().split()))
d=list(input().split())
min_=1000**4
for i in range(4):
for j in range(i+1,4):
if d[0]=='+':
test1=s[i]+s[j]
test2=s[i]+s[j]
test3_1=s[i]+s[j]
else:
test1=s[i]*s[j]
test2=s[i]*s[j]
test3_1=s[i]*s[j]
f=s.copy()
f.pop(i)
f.pop(j-1)
if d[1]=='+':
test1+=f[0]
test2+=f[1]
test3=f[1]+f[0]
else:
test1*=f[0]
test2*=f[1]
test3=f[1]*f[0]
if d[2]=='+':
test1+=f[1]
test2+=f[0]
test3+=test3_1
else:
test1*=f[1]
test2*=f[0]
test3*=test3_1
new=min(test1,test2,test3)
if new<min_:
min_=new
print(min_)
``` | output | 1 | 9,491 | 20 | 18,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output
Output one integer number — the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9 | instruction | 0 | 9,492 | 20 | 18,984 |
Tags: brute force
Correct Solution:
```
#***************55B - Smallest number***************#
#author: @Divyesh Chhabra
from math import *
import os
import random
import re
import sys
from itertools import *
m = pow(10,9)+7
n = list(map(int,input().split()))
s = input()
n.sort()
mul = 0
add = 0
for i in s:
if i == '*':
mul += 1
else:
add += 1
if mul == 0:
print(n[0]+n[1]+n[2]+n[3])
elif mul == 3:
print(n[0]*n[1]*n[2]*n[3])
elif mul == 2:
if s == '* * +':
print(min(n[3]+n[1]*n[2]*n[0],n[0]*n[1]+n[2]*n[3],n[0]*n[2]+n[1]*n[3],n[0]*n[3]+n[1]*n[2]))
elif s == '* + *':
print(n[0]*(n[1]*n[2]+n[3]))
else:
print(n[0]*n[1]*(n[2]+n[3]))
else:
if s == '+ + *':
print((n[1]+n[2]+n[3])*n[0])
elif s == '+ * + ':
print(min((n[1]+n[2])*n[0]+n[3]),n[0]*n[1]+n[2]+n[3])
else:
print(n[0]*n[1]+n[3]+n[2])
``` | output | 1 | 9,492 | 20 | 18,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output
Output one integer number — the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9 | instruction | 0 | 9,493 | 20 | 18,986 |
Tags: brute force
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
avl=AvlTree()
#-----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(a ,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left)/ 2)
# Check if middle element is
# less than or equal to key
if (arr[mid]<=key):
count = mid+1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def countGreater( arr,n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
w=[]
def calc(l,s):
#print(l,s)
if len(l)==1:
w.append(l[0])
return
we=[]+l
for i in range(len(l)):
for j in range(len(l)):
if i==j:
continue
l=[]+we
if s[0]=="+":
l[i]+=l[j]
l[j]=999999999999999
l.sort()
l.pop()
else:
l[i] *= l[j]
l[j] = 999999999999999
l.sort()
l.pop()
calc(l,s[1:])
l=list(map(int,input().split()))
l.sort()
s=list(map(str,input().split()))
calc(l,s)
print(min(w))
``` | output | 1 | 9,493 | 20 | 18,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output
Output one integer number — the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9 | instruction | 0 | 9,494 | 20 | 18,988 |
Tags: brute force
Correct Solution:
```
a, b, c, d = list(map(int,input().strip().split(' ')))
first, second, third = input().strip().split(' ')
min_value = 1000000000000
if first == '+':
f_op = [[a + b, c, d], [a + c, b, d], [a + d, b, c], [b + c, a, d], [b + d, a, c], [c + d, a, b]]
else:
f_op = [[a * b, c, d], [a * c, b, d], [a * d, b, c], [b * c, a, d], [b * d, a, c], [c * d, a, b]]
for i in range(6):
x = f_op[i][0]
y = f_op[i][1]
z = f_op[i][2]
if second == '+':
s_op = [[x + y, z], [x + z, y], [y + z, x]]
else:
s_op = [[x * y, z], [x * z, y], [y * z, x]]
if third == '+':
m = min([s_op[0][0] + s_op[0][1], s_op[1][0] + s_op[1][1], s_op[2][0] + s_op[2][1]])
else:
m = min([s_op[0][0] * s_op[0][1], s_op[1][0] * s_op[1][1], s_op[2][0] * s_op[2][1]])
min_value = min(min_value,m)
print(min_value)
``` | output | 1 | 9,494 | 20 | 18,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output
Output one integer number — the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9 | instruction | 0 | 9,495 | 20 | 18,990 |
Tags: brute force
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
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")
#-------------------game starts now-----------------------------------------------------
def fun(arr,op,o):
if o==3:
return arr[-1]
mn=float('inf')
for i in range(0,3):
if arr[i]==-1:
continue
var=arr[i]
arr[i]=-1
for j in range(i,4):
if arr[j]==-1:
continue
var2=arr[j]
if op[o]=="*":
arr[j]=var*arr[j]
if op[o]=="+":
arr[j]=var+arr[j]
mn=min(mn,fun(arr,op,o+1))
arr[j]=var2
arr[i]=var
return mn
arr=list(map(int,input().split()))
op=input().split()
print(fun(arr,op,0))
``` | output | 1 | 9,495 | 20 | 18,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output
Output one integer number — the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9 | instruction | 0 | 9,496 | 20 | 18,992 |
Tags: brute force
Correct Solution:
```
def cal( a , b , c ):
if c[ 0 ] == '+':
return int(a) + int(b)
return int(a) * int(b)
num = input().split()
op = input().split()
def DFS( l , no ):
if no == 3: return int( l[ 0 ] )
else:
ln = len( l )
ans = 10 ** 100
for i in range( ln ):
for j in range( i + 1 , ln ):
ll = []
for k in range( ln ):
if i != k and j != k:
ll.append( l[ k ] )
ll.append( cal( l[ i ] , l[ j ] , op[ no ] ) )
ans = min( ans , DFS( ll , no + 1 ) )
return ans
print( DFS( num , 0 ) )
``` | output | 1 | 9,496 | 20 | 18,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output
Output one integer number — the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9 | instruction | 0 | 9,497 | 20 | 18,994 |
Tags: brute force
Correct Solution:
```
def z(a,b,k=0,f=1):
n=len(a)
if n==2:
if b[k]=='+':
return a[0]+a[1]
return a[0]*a[1]
c=-1
for i in range(n):
for j in range(i+1,n):
if b[k]=='+':
g=list(a)
a[i]+=a[j]
a=a[:j]+a[j+1:]
d=z(a,b,k+1,f+1)
if c==-1 or d<c:
c=d
a=list(g)
else:
g=list(a)
a[i]*=a[j]
a=a[:j]+a[j+1:]
d=z(a,b,k+1,f+1)
if c==-1 or d<c:
c=d
a=list(g)
return c
print(z(list(map(int,input().split())),input().split()))
``` | output | 1 | 9,497 | 20 | 18,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m.
Let M = 2m - 1.
You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.
A set of integers S is called "good" if the following hold.
1. If <image>, then <image>.
2. If <image>, then <image>
3. <image>
4. All elements of S are less than or equal to M.
Here, <image> and <image> refer to the bitwise XOR and bitwise AND operators, respectively.
Count the number of good sets S, modulo 109 + 7.
Input
The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).
The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.
Output
Print a single integer, the number of good sets modulo 109 + 7.
Examples
Input
5 3
11010
00101
11000
Output
4
Input
30 2
010101010101010010101010101010
110110110110110011011011011011
Output
860616440
Note
An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}. | instruction | 0 | 9,633 | 20 | 19,266 |
Tags: bitmasks, combinatorics, dp, math
Correct Solution:
```
from collections import defaultdict
def E1():
mod = 10 ** 9 + 7
comb = [[1]]
for i in range(1, 1010):
x = [1]
for j in range(1, i):
x.append((comb[i - 1][j - 1] + comb[i - 1][j]) % mod)
x.append(1)
comb.append(x)
dp = [1]
for i in range(1, 1010):
r = 0
for k in range(i):
r += dp[k] * comb[i - 1][k]
r %= mod
dp.append(r)
m, n = map(int, input().split())
ns = [0 for __ in range(m)]
for j in range(n):
temp = input()
s = [int(i) for i in temp]
for i in range(m):
ns[i] |= s[i] << j
dd = defaultdict(int)
for e in ns:
dd[e] += 1
ans = 1
for b in dd.values():
ans = ans * dp[b] % mod
print(ans)
if __name__=='__main__':
E1()
``` | output | 1 | 9,633 | 20 | 19,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m.
Let M = 2m - 1.
You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.
A set of integers S is called "good" if the following hold.
1. If <image>, then <image>.
2. If <image>, then <image>
3. <image>
4. All elements of S are less than or equal to M.
Here, <image> and <image> refer to the bitwise XOR and bitwise AND operators, respectively.
Count the number of good sets S, modulo 109 + 7.
Input
The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).
The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.
Output
Print a single integer, the number of good sets modulo 109 + 7.
Examples
Input
5 3
11010
00101
11000
Output
4
Input
30 2
010101010101010010101010101010
110110110110110011011011011011
Output
860616440
Note
An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}. | instruction | 0 | 9,634 | 20 | 19,268 |
Tags: bitmasks, combinatorics, dp, math
Correct Solution:
```
import sys
from collections import defaultdict as di
MOD = int(1e9+7)
#bells = di(int)
#bells[0,0] = 1
#K=1000
#for j in range(1,K):
# bells[0,j] = bells[j-1,j-1]
# for i in range(j):
# bells[i+1,j] = (bells[i,j] + bells[i,j-1])%MOD
#
#def bellman(n):
# return bells[n-1,n-1]
#lista = []
#for i in range(K):
# lista.append(bellman(i+1))
#print(lista)
#sys.exit()
bells = [1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 382958538, 480142077, 864869230, 76801385, 742164233, 157873304, 812832668, 706900318, 546020311, 173093227, 759867260, 200033042, 40680577, 159122123, 665114805, 272358185, 365885605, 744733441, 692873095, 463056339, 828412002, 817756178, 366396447, 683685664, 681586780, 840750853, 683889724, 216039853, 954226396, 858087702, 540284076, 514254014, 647209774, 900185117, 348985796, 609459762, 781824096, 756600466, 654591160, 171792186, 748630189, 848074470, 75742990, 352494923, 278101098, 462072300, 334907097, 10474572, 495625635, 586051441, 159996073, 479379757, 707597945, 561063550, 974840072, 209152841, 906106015, 467465396, 82034048, 392794164, 700950185, 344807921, 475335490, 496881113, 358229039, 519104519, 784488542, 665151655, 307919717, 591199688, 692769253, 335414677, 884560880, 847374378, 791103220, 200350027, 485480275, 557337842, 434181960, 73976309, 792463021, 462067202, 677783523, 295755371, 435431099, 193120002, 513369106, 134597056, 143018012, 353529690, 591382993, 163160926, 287984994, 842145354, 703798750, 386436223, 618375990, 636477101, 536261496, 574800957, 34046224, 167415054, 961776342, 807141069, 218578541, 513967253, 460200768, 230725907, 239843627, 792763805, 368353031, 740982762, 126993201, 967654419, 588554507, 728057622, 239984996, 818342358, 882367644, 216705655, 267152940, 867213913, 330735015, 934583772, 59261085, 443816525, 568733052, 754405433, 244324432, 153903806, 292097031, 557968620, 311976469, 242994387, 773037141, 549999484, 243701468, 941251494, 7149216, 932327662, 456857477, 739044033, 645452229, 69273749, 304951367, 503353209, 243194926, 688663125, 239795364, 522687881, 121506491, 835250259, 159173149, 545801717, 19848500, 322507013, 106069527, 807985703, 290163328, 971751677, 238407093, 981758956, 301257197, 728003485, 817681690, 318332431, 864806329, 87958605, 929106232, 617996713, 519300437, 307911059, 137306007, 887695462, 633135243, 442387331, 730250437, 27723819, 80605394, 760335262, 822289356, 415861662, 558003999, 645049413, 347692428, 380668983, 897875109, 278111491, 106909073, 951914124, 374756177, 635211535, 286442394, 774619548, 756991257, 929298287, 923425488, 182439740, 266683608, 415378498, 728411148, 808161291, 436338820, 692451577, 228029692, 235546564, 895805974, 758052804, 700926159, 226442121, 579900323, 96916377, 243044550, 858703179, 30279679, 343764928, 100627558, 840734795, 291199760, 989808717, 370270411, 158336199, 393391701, 881731480, 507200370, 588418523, 340981140, 295449954, 683858381, 903859151, 866470542, 4959332, 237784075, 861373023, 950693473, 955867890, 400039807, 939877954, 124824910, 954530940, 204884446, 42218442, 234856311, 189836713, 179563650, 683193288, 929322036, 73574908, 943547254, 103031032, 141180580, 540183111, 680050153, 382916846, 948921599, 252835397, 199109508, 551172546, 700090782, 44999714, 970123610, 145637563, 33948107, 267648423, 504777414, 584919509, 212459491, 242880452, 351366578, 345323768, 285497541, 692868556, 706562675, 675626979, 620872182, 136458675, 971105139, 182064384, 948539342, 186406165, 706529481, 790490927, 888369436, 784409511, 835815713, 447895018, 17015606, 342727699, 321837918, 394134115, 563672582, 70390332, 61116103, 949269501, 833942074, 581389345, 570974405, 768179852, 765734098, 928340756, 541194960, 126833304, 427218334, 75800034, 100445725, 242810216, 330081440, 986329793, 298082322, 643160582, 505669854, 255287400, 403977567, 659185446, 596703087, 289443930, 478095124, 920175726, 205886838, 729278433, 535998256, 658801091, 606948240, 432632296, 552723022, 17794080, 234033713, 189986528, 444922724, 263196004, 846019724, 684703320, 895782046, 505050988, 44287113, 505335732, 436498414, 12098144, 714227851, 643983136, 647148160, 579243434, 951209063, 511291462, 426622734, 830870687, 949900312, 599926584, 633837711, 176405710, 913717356, 753535741, 874916804, 956692925, 220742732, 649500982, 584759931, 109573948, 937203173, 96881033, 305784835, 559854872, 894253854, 746366726, 951492991, 532579856, 822308583, 572042503, 397665465, 600979983, 914199453, 628402767, 594763006, 9791558, 451332658, 516069180, 651367831, 962708649, 687016963, 539878802, 107278296, 926059014, 371504543, 556987417, 447666745, 565595310, 778161513, 995461128, 121460302, 599892490, 242414970, 900391574, 362620950, 292857964, 495260535, 355054738, 176340034, 370047225, 509682533, 459314034, 40869728, 534741938, 788604648, 945028000, 701904601, 154924404, 695162652, 220536827, 615701976, 167761915, 833779942, 52430883, 368585637, 936409860, 654822736, 613850019, 941559844, 357840989, 218223326, 721900618, 171013438, 597980462, 193395922, 949112044, 859322955, 354602094, 807705992, 347609311, 451694117, 623122523, 252980054, 491785682, 13877214, 918727309, 750110421, 114981703, 174636266, 363160184, 781715298, 30575457, 862940854, 642129450, 34525888, 798422280, 792396821, 168367459, 344551406, 799847612, 626838494, 671596530, 167280197, 959000039, 614621296, 273560655, 8705247, 284372524, 940371542, 906010703, 582585495, 929449942, 308961449, 768816240, 674729787, 279648144, 286568146, 938661138, 536038536, 456529723, 18843013, 501518651, 457224675, 520694423, 938573228, 179014658, 169719825, 459657583, 302109678, 560375405, 556039265, 348713003, 957546568, 687116649, 3656313, 562760316, 716689588, 324677598, 570275686, 60738163, 996201577, 305457565, 38935942, 538451492, 228282207, 77975017, 389525459, 25000235, 152169430, 62331625, 618611219, 462328092, 106666757, 661839198, 177836427, 313546124, 392585017, 950280707, 551167559, 389204003, 77447456, 158414991, 766574847, 941433736, 363591676, 805565034, 312418363, 999641612, 122925536, 768845786, 608121932, 373163730, 783033644, 74564718, 894150080, 796617981, 274365270, 802488053, 947861187, 401960309, 143529635, 769621671, 249500752, 619408647, 849453216, 354838551, 69741157, 944781258, 135254314, 7413076, 416298064, 871313316, 343673168, 375656287, 868866230, 179060630, 399560227, 852555486, 987661859, 165863065, 12882359, 3688778, 380092596, 438366086, 720041886, 240796679, 588918084, 14802664, 17188673, 504951961, 842108931, 839289310, 256364811, 121095676, 164017670, 35340476, 875551801, 239615760, 262141182, 262741417, 456560451, 350350882, 777143297, 264469934, 807530935, 89546104, 246698645, 241166716, 125659016, 839103323, 418357064, 186866754, 179291319, 326054960, 172454707, 430532716, 558625738, 306201933, 61986384, 837357643, 575407529, 983555480, 13784333, 311989892, 153386582, 633092291, 722816631, 633510090, 551352594, 323601313, 248995449, 672011813, 612805937, 202743586, 215183002, 32688571, 38454892, 245790100, 451190956, 823199664, 12164578, 67389319, 584760532, 968838901, 307205626, 971537038, 836812364, 663878188, 468850566, 647599527, 839342879, 242347168, 169911213, 993779953, 251402771, 969281106, 416168275, 738337745, 8172262, 852101376, 879373674, 929752458, 452163141, 48347012, 500253327, 672444134, 406391337, 665852222, 499704706, 116418822, 67956495, 994761753, 808150613, 251453632, 543431315, 143101466, 381253760, 826943616, 763270983, 959511676, 323777679, 514214633, 669340157, 471626592, 557874503, 304789863, 116939617, 503636634, 660499296, 659726735, 273788323, 704107733, 718417780, 624033370, 355000823, 194537583, 760997582, 289828020, 778033293, 933152490, 910945024, 644565086, 434509630, 289427510, 502291783, 421699389, 159196930, 834667293, 313599675, 560298831, 812176354, 865521998, 126796474, 886921339, 937011401, 791993161, 583948688, 275408655, 665183437, 130342900, 699431744, 117550047, 460234251, 56770880, 306091228, 912949106, 626369877, 852501236, 241076503, 262988042, 737247015, 831044258, 475123008, 928949542, 332750699, 696284377, 689111142, 196862045, 570365577, 187156806, 451528865, 635110126, 385331290, 263486377, 200189955, 206842029, 457862597, 450522487, 818984909, 710634976, 461356455, 71985964, 781500456, 467334209, 46762760, 97663653, 870671823, 255977331, 79650379, 32876340, 636190780, 364339752, 597149326, 452604321, 748186407, 302032725, 779013981, 111971627, 175687535, 399961122, 451853028, 798326812, 902775588, 362836436, 498862780, 160000437, 629259604, 919729986, 5994845, 631476109, 371320167, 76164236, 448643023, 945220853, 111192011, 150577654, 827274836, 17668451, 938388515, 88566735, 27940328, 882026632, 712756966, 83642744, 873585716, 638987456, 405271094, 822216133, 345587303, 668558160, 314983205, 826678060, 563583341, 516998387, 77703032, 726563479, 155257773, 49705622, 891556456, 164127879, 842558039, 189099480, 956148983, 992226557, 671472701, 137476493, 871069222, 78241093, 728497057, 278888712, 332713952, 222597908, 235198692, 876216003, 167364357, 722341150, 519365334, 604855967, 834816722, 850786742, 416385106, 608404143, 311628039, 507077268, 571796589, 506398832, 305540948, 556971113, 444565912, 866477296, 411983920, 905854221, 901986568, 512703782, 684027511, 596294441, 916862272, 495347444, 802477106, 235968146, 257527513, 528476230, 969655767, 772044489, 682345813, 66418556, 603372280, 439233253, 278244332, 590581374, 353687769, 321352820, 245676729, 325255315, 91010070, 923699200, 837616604, 736177081, 528261400, 876353388, 339195128, 377206087, 769944080, 772953529, 123785293, 35984916, 461119619, 236140329, 884137913, 625494604, 791773064, 661436140, 308509072, 54134926, 279367618, 51918421, 149844467, 308119110, 948074070, 941738748, 890320056, 933243910, 430364344, 903312966, 574904506, 56353560, 861112413, 440392450, 937276559, 944662107, 599470900, 458887833, 962614595, 589151703, 997944986, 642961512, 63773929, 737273926, 110546606, 654813100, 374632916, 327432718, 307869727, 387738989, 133844439, 688886605, 989252194, 303514517, 79062408, 79381603, 941446109, 189307316, 728764788, 619946432, 359845738, 216171670, 690964059, 337106876, 762119224, 226624101, 401879891, 47069454, 41411521, 429556898, 188042667, 832342137, 770962364, 294422843, 991268380, 137519647, 903275202, 115040918, 521250780, 783585266, 98267582, 337193737, 717487549, 510794369, 206729333, 248526905, 412652544, 146948138, 103954760, 132289464, 938042429, 185735408, 640754677, 315573450, 956487685, 454822141, 783819416, 882547786, 976850791, 307258357, 929434429, 832158433, 334518103, 700273615, 734048238, 48618988, 693477108, 12561960, 598093056, 154072663, 174314067, 345548333, 479759833, 658594149, 282072153, 57970886, 905112877, 584117466, 472359245, 776860470, 324216896, 334199385, 321245477, 508188925, 521442872, 286692969, 245141864, 59342176, 896413224, 573301289, 869453643, 87399903, 60102262, 835514392, 493582549, 649986925, 576899388, 20454903, 271374500, 589229956, 505139242, 789538901, 243337905, 248443618, 39334644, 831631854, 541659849, 159802612, 524090232, 855135628, 542520502, 967119953, 597294058, 465231251]
def bellman(n):
return bells[n-1]
m,n = [int(x) for x in input().split()]
Tlist = []
for _ in range(n):
Tlist.append(input())
numbs = []
for i in range(m):
numb = []
for j in range(n):
numb.append(Tlist[j][i])
numbs.append(int(''.join(numb),2))
eqsize = di(lambda:0)
for numb in numbs:
eqsize[numb]+=1
sets = []
for numb in eqsize:
sets.append(eqsize[numb])
parts = 1
for s in sets:
parts*=bellman(s)
parts%=MOD
print(parts)
``` | output | 1 | 9,634 | 20 | 19,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m.
Let M = 2m - 1.
You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.
A set of integers S is called "good" if the following hold.
1. If <image>, then <image>.
2. If <image>, then <image>
3. <image>
4. All elements of S are less than or equal to M.
Here, <image> and <image> refer to the bitwise XOR and bitwise AND operators, respectively.
Count the number of good sets S, modulo 109 + 7.
Input
The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).
The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.
Output
Print a single integer, the number of good sets modulo 109 + 7.
Examples
Input
5 3
11010
00101
11000
Output
4
Input
30 2
010101010101010010101010101010
110110110110110011011011011011
Output
860616440
Note
An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}. | instruction | 0 | 9,635 | 20 | 19,270 |
Tags: bitmasks, combinatorics, dp, math
Correct Solution:
```
import sys
#f = open('input', 'r')
f = sys.stdin
n,m = list(map(int, f.readline().split()))
s = [f.readline().strip() for _ in range(m)]
s = [list(x) for x in s]
d = {}
for k in zip(*s):
if k in d:
d[k] += 1
else:
d[k] = 1
dv = d.values()
got = [1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 382958538, 480142077, 864869230, 76801385, 742164233, 157873304, 812832668, 706900318, 546020311, 173093227, 759867260, 200033042, 40680577, 159122123, 665114805, 272358185, 365885605, 744733441, 692873095, 463056339, 828412002, 817756178, 366396447, 683685664, 681586780, 840750853, 683889724, 216039853, 954226396, 858087702, 540284076, 514254014, 647209774, 900185117, 348985796, 609459762, 781824096, 756600466, 654591160, 171792186, 748630189, 848074470, 75742990, 352494923, 278101098, 462072300, 334907097, 10474572, 495625635, 586051441, 159996073, 479379757, 707597945, 561063550, 974840072, 209152841, 906106015, 467465396, 82034048, 392794164, 700950185, 344807921, 475335490, 496881113, 358229039, 519104519, 784488542, 665151655, 307919717, 591199688, 692769253, 335414677, 884560880, 847374378, 791103220, 200350027, 485480275, 557337842, 434181960, 73976309, 792463021, 462067202, 677783523, 295755371, 435431099, 193120002, 513369106, 134597056, 143018012, 353529690, 591382993, 163160926, 287984994, 842145354, 703798750, 386436223, 618375990, 636477101, 536261496, 574800957, 34046224, 167415054, 961776342, 807141069, 218578541, 513967253, 460200768, 230725907, 239843627, 792763805, 368353031, 740982762, 126993201, 967654419, 588554507, 728057622, 239984996, 818342358, 882367644, 216705655, 267152940, 867213913, 330735015, 934583772, 59261085, 443816525, 568733052, 754405433, 244324432, 153903806, 292097031, 557968620, 311976469, 242994387, 773037141, 549999484, 243701468, 941251494, 7149216, 932327662, 456857477, 739044033, 645452229, 69273749, 304951367, 503353209, 243194926, 688663125, 239795364, 522687881, 121506491, 835250259, 159173149, 545801717, 19848500, 322507013, 106069527, 807985703, 290163328, 971751677, 238407093, 981758956, 301257197, 728003485, 817681690, 318332431, 864806329, 87958605, 929106232, 617996713, 519300437, 307911059, 137306007, 887695462, 633135243, 442387331, 730250437, 27723819, 80605394, 760335262, 822289356, 415861662, 558003999, 645049413, 347692428, 380668983, 897875109, 278111491, 106909073, 951914124, 374756177, 635211535, 286442394, 774619548, 756991257, 929298287, 923425488, 182439740, 266683608, 415378498, 728411148, 808161291, 436338820, 692451577, 228029692, 235546564, 895805974, 758052804, 700926159, 226442121, 579900323, 96916377, 243044550, 858703179, 30279679, 343764928, 100627558, 840734795, 291199760, 989808717, 370270411, 158336199, 393391701, 881731480, 507200370, 588418523, 340981140, 295449954, 683858381, 903859151, 866470542, 4959332, 237784075, 861373023, 950693473, 955867890, 400039807, 939877954, 124824910, 954530940, 204884446, 42218442, 234856311, 189836713, 179563650, 683193288, 929322036, 73574908, 943547254, 103031032, 141180580, 540183111, 680050153, 382916846, 948921599, 252835397, 199109508, 551172546, 700090782, 44999714, 970123610, 145637563, 33948107, 267648423, 504777414, 584919509, 212459491, 242880452, 351366578, 345323768, 285497541, 692868556, 706562675, 675626979, 620872182, 136458675, 971105139, 182064384, 948539342, 186406165, 706529481, 790490927, 888369436, 784409511, 835815713, 447895018, 17015606, 342727699, 321837918, 394134115, 563672582, 70390332, 61116103, 949269501, 833942074, 581389345, 570974405, 768179852, 765734098, 928340756, 541194960, 126833304, 427218334, 75800034, 100445725, 242810216, 330081440, 986329793, 298082322, 643160582, 505669854, 255287400, 403977567, 659185446, 596703087, 289443930, 478095124, 920175726, 205886838, 729278433, 535998256, 658801091, 606948240, 432632296, 552723022, 17794080, 234033713, 189986528, 444922724, 263196004, 846019724, 684703320, 895782046, 505050988, 44287113, 505335732, 436498414, 12098144, 714227851, 643983136, 647148160, 579243434, 951209063, 511291462, 426622734, 830870687, 949900312, 599926584, 633837711, 176405710, 913717356, 753535741, 874916804, 956692925, 220742732, 649500982, 584759931, 109573948, 937203173, 96881033, 305784835, 559854872, 894253854, 746366726, 951492991, 532579856, 822308583, 572042503, 397665465, 600979983, 914199453, 628402767, 594763006, 9791558, 451332658, 516069180, 651367831, 962708649, 687016963, 539878802, 107278296, 926059014, 371504543, 556987417, 447666745, 565595310, 778161513, 995461128, 121460302, 599892490, 242414970, 900391574, 362620950, 292857964, 495260535, 355054738, 176340034, 370047225, 509682533, 459314034, 40869728, 534741938, 788604648, 945028000, 701904601, 154924404, 695162652, 220536827, 615701976, 167761915, 833779942, 52430883, 368585637, 936409860, 654822736, 613850019, 941559844, 357840989, 218223326, 721900618, 171013438, 597980462, 193395922, 949112044, 859322955, 354602094, 807705992, 347609311, 451694117, 623122523, 252980054, 491785682, 13877214, 918727309, 750110421, 114981703, 174636266, 363160184, 781715298, 30575457, 862940854, 642129450, 34525888, 798422280, 792396821, 168367459, 344551406, 799847612, 626838494, 671596530, 167280197, 959000039, 614621296, 273560655, 8705247, 284372524, 940371542, 906010703, 582585495, 929449942, 308961449, 768816240, 674729787, 279648144, 286568146, 938661138, 536038536, 456529723, 18843013, 501518651, 457224675, 520694423, 938573228, 179014658, 169719825, 459657583, 302109678, 560375405, 556039265, 348713003, 957546568, 687116649, 3656313, 562760316, 716689588, 324677598, 570275686, 60738163, 996201577, 305457565, 38935942, 538451492, 228282207, 77975017, 389525459, 25000235, 152169430, 62331625, 618611219, 462328092, 106666757, 661839198, 177836427, 313546124, 392585017, 950280707, 551167559, 389204003, 77447456, 158414991, 766574847, 941433736, 363591676, 805565034, 312418363, 999641612, 122925536, 768845786, 608121932, 373163730, 783033644, 74564718, 894150080, 796617981, 274365270, 802488053, 947861187, 401960309, 143529635, 769621671, 249500752, 619408647, 849453216, 354838551, 69741157, 944781258, 135254314, 7413076, 416298064, 871313316, 343673168, 375656287, 868866230, 179060630, 399560227, 852555486, 987661859, 165863065, 12882359, 3688778, 380092596, 438366086, 720041886, 240796679, 588918084, 14802664, 17188673, 504951961, 842108931, 839289310, 256364811, 121095676, 164017670, 35340476, 875551801, 239615760, 262141182, 262741417, 456560451, 350350882, 777143297, 264469934, 807530935, 89546104, 246698645, 241166716, 125659016, 839103323, 418357064, 186866754, 179291319, 326054960, 172454707, 430532716, 558625738, 306201933, 61986384, 837357643, 575407529, 983555480, 13784333, 311989892, 153386582, 633092291, 722816631, 633510090, 551352594, 323601313, 248995449, 672011813, 612805937, 202743586, 215183002, 32688571, 38454892, 245790100, 451190956, 823199664, 12164578, 67389319, 584760532, 968838901, 307205626, 971537038, 836812364, 663878188, 468850566, 647599527, 839342879, 242347168, 169911213, 993779953, 251402771, 969281106, 416168275, 738337745, 8172262, 852101376, 879373674, 929752458, 452163141, 48347012, 500253327, 672444134, 406391337, 665852222, 499704706, 116418822, 67956495, 994761753, 808150613, 251453632, 543431315, 143101466, 381253760, 826943616, 763270983, 959511676, 323777679, 514214633, 669340157, 471626592, 557874503, 304789863, 116939617, 503636634, 660499296, 659726735, 273788323, 704107733, 718417780, 624033370, 355000823, 194537583, 760997582, 289828020, 778033293, 933152490, 910945024, 644565086, 434509630, 289427510, 502291783, 421699389, 159196930, 834667293, 313599675, 560298831, 812176354, 865521998, 126796474, 886921339, 937011401, 791993161, 583948688, 275408655, 665183437, 130342900, 699431744, 117550047, 460234251, 56770880, 306091228, 912949106, 626369877, 852501236, 241076503, 262988042, 737247015, 831044258, 475123008, 928949542, 332750699, 696284377, 689111142, 196862045, 570365577, 187156806, 451528865, 635110126, 385331290, 263486377, 200189955, 206842029, 457862597, 450522487, 818984909, 710634976, 461356455, 71985964, 781500456, 467334209, 46762760, 97663653, 870671823, 255977331, 79650379, 32876340, 636190780, 364339752, 597149326, 452604321, 748186407, 302032725, 779013981, 111971627, 175687535, 399961122, 451853028, 798326812, 902775588, 362836436, 498862780, 160000437, 629259604, 919729986, 5994845, 631476109, 371320167, 76164236, 448643023, 945220853, 111192011, 150577654, 827274836, 17668451, 938388515, 88566735, 27940328, 882026632, 712756966, 83642744, 873585716, 638987456, 405271094, 822216133, 345587303, 668558160, 314983205, 826678060, 563583341, 516998387, 77703032, 726563479, 155257773, 49705622, 891556456, 164127879, 842558039, 189099480, 956148983, 992226557, 671472701, 137476493, 871069222, 78241093, 728497057, 278888712, 332713952, 222597908, 235198692, 876216003, 167364357, 722341150, 519365334, 604855967, 834816722, 850786742, 416385106, 608404143, 311628039, 507077268, 571796589, 506398832, 305540948, 556971113, 444565912, 866477296, 411983920, 905854221, 901986568, 512703782, 684027511, 596294441, 916862272, 495347444, 802477106, 235968146, 257527513, 528476230, 969655767, 772044489, 682345813, 66418556, 603372280, 439233253, 278244332, 590581374, 353687769, 321352820, 245676729, 325255315, 91010070, 923699200, 837616604, 736177081, 528261400, 876353388, 339195128, 377206087, 769944080, 772953529, 123785293, 35984916, 461119619, 236140329, 884137913, 625494604, 791773064, 661436140, 308509072, 54134926, 279367618, 51918421, 149844467, 308119110, 948074070, 941738748, 890320056, 933243910, 430364344, 903312966, 574904506, 56353560, 861112413, 440392450, 937276559, 944662107, 599470900, 458887833, 962614595, 589151703, 997944986, 642961512, 63773929, 737273926, 110546606, 654813100, 374632916, 327432718, 307869727, 387738989, 133844439, 688886605, 989252194, 303514517, 79062408, 79381603, 941446109, 189307316, 728764788, 619946432, 359845738, 216171670, 690964059, 337106876, 762119224, 226624101, 401879891, 47069454, 41411521, 429556898, 188042667, 832342137, 770962364, 294422843, 991268380, 137519647, 903275202, 115040918, 521250780, 783585266, 98267582, 337193737, 717487549, 510794369, 206729333, 248526905, 412652544, 146948138, 103954760, 132289464, 938042429, 185735408, 640754677, 315573450, 956487685, 454822141, 783819416, 882547786, 976850791, 307258357, 929434429, 832158433, 334518103, 700273615, 734048238, 48618988, 693477108, 12561960, 598093056, 154072663, 174314067, 345548333, 479759833, 658594149, 282072153, 57970886, 905112877, 584117466, 472359245, 776860470, 324216896, 334199385, 321245477, 508188925, 521442872, 286692969, 245141864, 59342176, 896413224, 573301289, 869453643, 87399903, 60102262, 835514392, 493582549, 649986925, 576899388, 20454903, 271374500, 589229956, 505139242, 789538901, 243337905, 248443618, 39334644, 831631854, 541659849, 159802612, 524090232, 855135628, 542520502, 967119953, 597294058, 465231251]
MM = 10**9 + 7
ans = 1
for v in dv:
ans = ans*got[v-1]
ans = ans%MM
print(ans)
'''
t = [[0] * 1010 for _ in range(1010)]
t[1][1] = 1
for i in range(2,1001):
for j in range(1,i+1):
t[i][j] = t[i-1][j-1] + t[i-1][j]*j
t[i][j] = t[i][j] % MM
print([sum(t[i])%MM for i in range(1,1001)])
'''
``` | output | 1 | 9,635 | 20 | 19,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m.
Let M = 2m - 1.
You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.
A set of integers S is called "good" if the following hold.
1. If <image>, then <image>.
2. If <image>, then <image>
3. <image>
4. All elements of S are less than or equal to M.
Here, <image> and <image> refer to the bitwise XOR and bitwise AND operators, respectively.
Count the number of good sets S, modulo 109 + 7.
Input
The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).
The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.
Output
Print a single integer, the number of good sets modulo 109 + 7.
Examples
Input
5 3
11010
00101
11000
Output
4
Input
30 2
010101010101010010101010101010
110110110110110011011011011011
Output
860616440
Note
An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}. | instruction | 0 | 9,636 | 20 | 19,272 |
Tags: bitmasks, combinatorics, dp, math
Correct Solution:
```
MOD = 10**9 + 7
m, N = map(int, input().split())
binom = [[1] + [0 for i in range(m)] for j in range(m + 1)]
for n in range(1, m + 1):
for k in range(1, n + 1):
binom[n][k] = (binom[n - 1][k] + binom[n - 1][k - 1]) % MOD
bell = [0 for n in range(m + 1)]
bell[0] = bell[1] = 1
for n in range(1, m):
for k in range(n + 1):
bell[n + 1] += bell[k] * binom[n][k]
bell[n + 1] %= MOD
#print(bell)
bags = [0 for i in range(m)]
for it in range(N):
for i, z in enumerate(input()):
if z == '1':
bags[i] |= (1 << it)
difs = set(bags)
sol = 1
for mask in difs:
sol = sol * bell[bags.count(mask)] % MOD
print(sol)
``` | output | 1 | 9,636 | 20 | 19,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m.
Let M = 2m - 1.
You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.
A set of integers S is called "good" if the following hold.
1. If <image>, then <image>.
2. If <image>, then <image>
3. <image>
4. All elements of S are less than or equal to M.
Here, <image> and <image> refer to the bitwise XOR and bitwise AND operators, respectively.
Count the number of good sets S, modulo 109 + 7.
Input
The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).
The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.
Output
Print a single integer, the number of good sets modulo 109 + 7.
Examples
Input
5 3
11010
00101
11000
Output
4
Input
30 2
010101010101010010101010101010
110110110110110011011011011011
Output
860616440
Note
An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}. | instruction | 0 | 9,637 | 20 | 19,274 |
Tags: bitmasks, combinatorics, dp, math
Correct Solution:
```
from collections import defaultdict as di
MOD = int(1e9+7)
bells = di(int)
bells[0,0] = 1
K=1000
for j in range(1,K):
bells[0,j] = bells[j-1,j-1]
for i in range(j):
bells[i+1,j] = (bells[i,j] + bells[i,j-1])%MOD
def bellman(n):
return bells[n-1,n-1]
m,n = [int(x) for x in input().split()]
Tlist = []
for _ in range(n):
Tlist.append(input())
numbs = []
for i in range(m):
numb = []
for j in range(n):
numb.append(Tlist[j][i])
numbs.append(int(''.join(numb),2))
eqsize = di(lambda:0)
for numb in numbs:
eqsize[numb]+=1
sets = []
for numb in eqsize:
sets.append(eqsize[numb])
parts = 1
for s in sets:
parts*=bellman(s)
parts%=MOD
print(parts)
``` | output | 1 | 9,637 | 20 | 19,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard. | instruction | 0 | 10,091 | 20 | 20,182 |
Tags: bitmasks, dp, graphs, greedy
Correct Solution:
```
import sys
input=sys.stdin.readline
d={1:[3,6],0:[1,2,3,5,6,7],2:[1,3,5,4,7],3:[1,3,4,6,7],4:[2,3,4,6],5:[1,2,4,6,7],6:[1,2,4,5,6,7],7:[1,3,6],8:[1,2,3,4,5,6,7],9:[1,2,3,4,6,7]}
n,k=map(int,input().split())
p=[[] for i in range(n+1)]
f=[[-1 for i in range(10)] for j in range(n+1)]
g=[[-1 for i in range(10)] for j in range(n+1)]
for i in range(n):
s=list(input().rstrip())
for j in range(len(s)):
if s[j]=='1':
p[i].append(j+1)
for j in range(10):
if len(set(p[i])-set(d[j]))!=0:
continue
else:
z=len(set(d[j])-set(p[i]))
f[i][z]=max(f[i][z],j)
g[i][j]=z
count=[[0 for i in range((k+1)*8)] for j in range(n+1)]
for i in range(n-1,-1,-1):
if i==n-1:
for j in range(8):
if f[i][j]!=-1:
count[i][j]=1
else:
for u in range(k+1):
if count[i+1][u]==1:
for y in range(8):
if f[i][y]!=-1:
count[i][y+u]=1
s=""
#print(count)
for i in range(n-1):
for j in range(9,-1,-1):
if g[i][j]!=-1 and count[i+1][k-g[i][j]]==1 and k>=g[i][j]:
k-=g[i][j]
s+=str(j)
break
if k>=0 and k<=8 and f[n-1][k]!=-1:
s+=str(f[n-1][k])
print(s if len(s)==n else -1)
``` | output | 1 | 10,091 | 20 | 20,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard. | instruction | 0 | 10,092 | 20 | 20,184 |
Tags: bitmasks, dp, graphs, greedy
Correct Solution:
```
import sys
n,k = map(int,input().split()); digits = []
ex = [[0]*10 for _ in range(n)]
a = ["1110111","0010010","1011101","1011011","0111010","1101011","1101111","1010010","1111111","1111011"]
a = list(map(lambda s:int(s,2),a))
for i in range(n):
x = int(input(),2)
for d in range(10):
ex[i][d] = bin(a[d]-x).count("1")
digits.append(x)
dp = [[False]*(k+1) for _ in range(n+1)]
dp[n][0] = True
for i in range(n-1,-1,-1):
for x in range(k+1):
for d in range(10):
if a[d]&digits[i] != digits[i]:
continue
excess = ex[i][d]
if x-excess >= 0 and dp[i+1][x-excess]:
dp[i][x] = True
break
# dp[0][k] = True
ans = ""
cur = k
for i in range(n):
found = False
for d in range(9,-1,-1):
if a[d]&digits[i] != digits[i]:
continue
excess = ex[i][d]
if cur-excess >= 0 and dp[i+1][cur-excess]:
cur -= excess
ans += str(d)
found = True
break
if not found:
print(-1)
exit()
print(ans)
``` | output | 1 | 10,092 | 20 | 20,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard. | instruction | 0 | 10,093 | 20 | 20,186 |
Tags: bitmasks, dp, graphs, greedy
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
from heapq import heappush, heappop, heapify ,_heapify_max,_heappop_max,nsmallest,nlargest
# sys.setrecursionlimit(111111)
INF=999999999999999999999999
alphabets="abcdefghijklmnopqrstuvwxyz"
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
class SegTree:
def __init__(self, n):
self.N = 1 << n.bit_length()
self.tree = [0] * (self.N<<1)
def update(self, i, j, v):
i += self.N
j += self.N
while i <= j:
if i%2==1: self.tree[i] += v
if j%2==0: self.tree[j] += v
i, j = (i+1) >> 1, (j-1) >> 1
def query(self, i):
v = 0
i += self.N
while i > 0:
v += self.tree[i]
i >>= 1
return v
def SieveOfEratosthenes(limit):
"""Returns all primes not greater than limit."""
isPrime = [True]*(limit+1)
isPrime[0] = isPrime[1] = False
primes = []
for i in range(2, limit+1):
if not isPrime[i]:continue
primes += [i]
for j in range(i*i, limit+1, i):
isPrime[j] = False
return primes
from collections import Counter
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
def memodict(f):
"""memoization decorator for a function taking a single argument"""
class memodict(dict):
def __missing__(self, key):
ret = self[key] = f(key)
return ret
return memodict().__getitem__
def pollard_rho(n):
"""returns a random factor of n"""
if n & 1 == 0:
return 2
if n % 3 == 0:
return 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
prev = p
p = (p * p) % n
if p == 1:
return gcd(prev - 1, n)
if p == n - 1:
break
else:
for i in range(2, n):
x, y = i, (i * i + 1) % n
f = gcd(abs(x - y), n)
while f == 1:
x, y = (x * x + 1) % n, (y * y + 1) % n
y = (y * y + 1) % n
f = gcd(abs(x - y), n)
if f != n:
return f
return n
@memodict
def prime_factors(n):
"""returns a Counter of the prime factorization of n"""
if n <= 1:
return Counter()
f = pollard_rho(n)
return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)
def distinct_factors(n):
"""returns a list of all distinct factors of n"""
factors = [1]
for p, exp in prime_factors(n).items():
factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
return factors
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [], []
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n // i)
if small[-1] == large[-1]:
large.pop()
large.reverse()
small.extend(large)
return small
def main():
mod=1000000007
# InverseofNumber(mod)
# InverseofFactorial(mod)
# factorial(mod)
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
tc=1
# Did you look at the constraints dummy?
# 1.Greedy?
# 2.DP?
# 3.Binary Search?(Monotonic? any one directed order in which we have to perform something?)
# 4.Graph?
# 5.Number Theory?(GCD subtraction?)
# 6.Bruteforce?(Redundant part of N which may not give answer?Constraints?)
# 7.Range Queries?
# 8.Any Equivalency?(We have A and B and have to do
# something between them maybe difficult if there was A~C and C~B then A~B
# C could be max or min or some other thing)
# 9.Reverse Engineering?(From Answer to quesn or last step to first step)
#10.Constructive? Mod?
tc=1
for _ in range(tc):
# LOGIC
# Let dp[i][j]=true, if at the ith till nth digit boards you can turn on exactly j sticks
# and get the correct sequence of digits and false otherwise.
# set dp[n][c] initially then after calculations check dp[1][k] is true or not
# It is easy to recalculate this dynamics: we will make transitions to all possible digits (the mask at position i should be a submask of the digit).
# Now let's go in order from 1 to n and will try to eagerly set the maximum possible figure using our dynamics.
# It is easy to understand that in this way we get the maximum possible number of n digits.
n,k=ria()
segments=["."]
for i in range(n):
segments.append(rs())
digits=["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
def cost(board,digits):
c=0
for i in range(7):
if board[i]=="1" and digits[i]=="0":
return -1
if board[i]=="0" and digits[i]=="1":
c+=1
return c
costs=[[-1 for i in range(10)] for j in range(n+1)]
for i in range(1,n+1):
for d in range(10):
costs[i][d]=cost(segments[i],digits[d])
dp=[[0 for i in range(k+1)] for j in range(n+1)]
for d in range(10):
c=costs[n][d]
if c!=-1 and c<=k:
dp[n][c]=1
for i in range(n,1,-1):
for j in range(k+1):
for d in range(10):
if dp[i][j] and 0<=costs[i-1][d]<=k-j:
dp[i-1][j+costs[i-1][d]]=1
if dp[1][k]==0:
wi(-1)
else:
curr=k
ans=[]
for i in range(1,n):
for d in range(9,-1,-1):
if dp[i][curr] and 0<=costs[i][d]<=curr and dp[i+1][curr-costs[i][d]]:
ans.append(str(d))
curr-=costs[i][d]
break
for d in range(9,-1,-1):
if curr==costs[n][d]:
ans.append(str(d))
break
ws("".join(ans))
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
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, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
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()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
``` | output | 1 | 10,093 | 20 | 20,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard. | instruction | 0 | 10,094 | 20 | 20,188 |
Tags: bitmasks, dp, graphs, greedy
Correct Solution:
```
maxk=2005
#cost to move from source(s) to destination(d)
def xcost(s,d):
price=0
for i in range(7):
if s[i]=='1' and d[i]=='0':
price=maxk #cant convert
break
elif s[i]=='0' and d[i]=='1':
price+=1
return price
s=['1110111','0010010','1011101','1011011','0111010','1101011','1101111','1010010','1111111','1111011']
p=[]
##cost[i][j] = Cost to change i (int i converted to bin string) to s[j]
cost=[[0 for j in range(10)] for i in range(128)]
for i in range(128):
x=bin(i)[2:].zfill(7)
p.append(x)
for i in range(128):
for j in range(10):
v=xcost(p[i],s[j])
cost[i][j]=v
n,k=map(int,input().split(" "))
#scoreboard has integer value of the binary string representing the pattern at position
scoreboard=[]
for i in range(n):
x=int(str(input()),2)
scoreboard.append(x)
# dp[i][j]=true , if at the suffix i…n you can turn on exactly j sticks and get the correct sequence
dp=[[False for i in range(k+1)] for j in range(n)]
i=n-1
bit_int=scoreboard[i]
for j in cost[bit_int]:
if j<=k:
dp[i][j]=True
for i in range(n-2,-1,-1):
bit_int=scoreboard[i]
for j in range(k+1):
for x in cost[bit_int]:
if j>=x:
if dp[i+1][j-x]:
dp[i][j]=True
break
if dp[0][k]:
ans,spend,pos='',k,0
for i in scoreboard:
for j in range(9,-1,-1):
v=cost[i][j]
if pos==n-1 and spend==v:
ans+=str(j)
break
elif pos<n-1 and spend>=v:
if dp[pos+1][spend-v]:
ans+=str(j)
spend-=v
break
pos+=1
print(ans)
else:
print(-1)
``` | output | 1 | 10,094 | 20 | 20,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard. | instruction | 0 | 10,095 | 20 | 20,190 |
Tags: bitmasks, dp, graphs, greedy
Correct Solution:
```
from sys import stdin,stdout
input = stdin.readline
from math import *
print = stdout.write
def cst(s,d):
res=0
for i in range(7):
if(s[i]=='1' and d[i]=='0'):
return -1
elif(s[i]=='0' and d[i]=='1'):
res+=1
return res
n,k=map(int,input().split())
d=["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
cost=[[0]*10 for i in range(n)]
t = [[0]*(k+1) for i in range(n)]
for i in range(n):
s=input()
for j in range(10):
cost[i][j]=cst(s,d[j])
ans=[-1]*n
def dfs(i,k):
if(i==n and k==0):
# print(ans,i,k)
return True
elif(i==n and k!=0):
return False
if(t[i][k]!=0):
return t[i][k]==1
for j in range(9,-1,-1):
temp = cost[i][j]
if(temp>=0 and temp<=k and dfs(i+1,k-cost[i][j])):
ans[i]=j
t[i][k]=1;
return True
t[i][k]=-1
return False
dfs(0,k)
if(-1 in ans):
print('-1')
else:
print("".join(map(str,ans)))
``` | output | 1 | 10,095 | 20 | 20,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard. | instruction | 0 | 10,096 | 20 | 20,192 |
Tags: bitmasks, dp, graphs, greedy
Correct Solution:
```
import sys
def compare(x, s):
for i in range(len(x)):
if x[i] == '1' and s[i] == '0':
return False
return True
def compare2(x, s):
count = 0
for i in range(len(x)):
if x[i] == '0' and s[i] == '1':
count += 1
return count
li = ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
dummy = li + []
dict = {}
num = 0
for i in li:
dict[i] = num
num += 1
n, k = [int(x) for x in input().split()]
arr = []
for i in range(n):
s = input()
arr.append(s)
ans = []
k_ans = []
for i in range(len(arr)):
need = 0
if dict.get(arr[i]) is not None:
ans.append(arr[i])
else:
need = 7
todo = li[8]
for j in li:
if compare(arr[i], j):
p = compare2(arr[i], j)
if p < need:
need = p
todo = j
k -= need
ans.append(todo)
k_ans.append(need)
if k < 0:
print(-1)
sys.exit(0)
to_print = ""
li = li[::-1]
for i in range(n):
s = ans[i]
for j in li:
if dict[j] <= dict[s]:
break
if compare(arr[i], j):
p = compare2(arr[i], j)
if p - k_ans[i] <= k:
ans[i] = j
k -= (p - k_ans[i])
k_ans[i] = p
break
if k >= 2:
for i in range(n - 1, -1, -1):
if dict[ans[i]] == 7:
if compare(arr[i], dummy[5]):
ans[i] = dummy[5]
k -= 2
else:
ans[i] = dummy[3]
k -= 2
break
flag = 0
if k > 0:
index = 0
for i in range(n - 1, -1, -1):
for j in li:
if dict[j] < dict[ans[i]] and compare(arr[i], j):
p = compare2(arr[i], j)
if p - k_ans[i] <= k:
k -= p - k_ans[i]
if p - k_ans[i] < 0:
flag = 1
else:
flag = 2
ans[i] = j
index = i
break
if flag != 0:
break
if flag == 1:
nn = 0
for i in range(index + 1, n):
for j in li:
if dict[j] > dict[ans[i]] and compare(arr[i], j):
p = compare2(arr[i], j)
if p - k_ans[i] > k:
continue
k -= p - k_ans[i]
ans[i] = j
nn = 1
break
if nn == 1:
break
if nn == 0:
for i in range(n - 1, -1, -1):
if dict[ans[i]] == 7:
if compare(arr[i], dummy[5]):
ans[i] = dummy[5]
k -= 2
else:
ans[i] = dummy[3]
k -= 2
break
if k > 0:
for i in range(n - 1, -1, -1):
if k == 0:
break
if dict[ans[i]] == 9:
ans[i] = dummy[8]
k -= 1
if k > 0:
print(-1)
sys.exit(0)
for i in range(n):
print(dict[ans[i]], end="")
``` | output | 1 | 10,096 | 20 | 20,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard. | instruction | 0 | 10,097 | 20 | 20,194 |
Tags: bitmasks, dp, graphs, greedy
Correct Solution:
```
D = [
'1110111',
'0010010',
'1011101',
'1011011',
'0111010',
'1101011',
'1101111',
'1010010',
'1111111',
'1111011',
]
n, k = list(map(int, input().split()))
x = [input() for _ in range(n)]
r = [[-1] * (k + 1) for _ in range(n + 1)]
r[n][0] = 0
for i in range(n - 1, -1, -1):
for d in range(10):
dc = 0
for f, t in zip(x[i], D[d]):
if f > t:
break
else:
dc += f != t
else:
z = dc * 10 + d
for c in range(k + 1 - dc):
if r[i + 1][c] >= 0:
r[i][c + dc] = z
if r[0][k] >= 0:
for i in range(n):
print(r[i][k] % 10, end='')
k -= r[i][k] // 10
print()
else:
print(-1)
``` | output | 1 | 10,097 | 20 | 20,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard. | instruction | 0 | 10,098 | 20 | 20,196 |
Tags: bitmasks, dp, graphs, greedy
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def bit_count(x):
ans = 0
while x:
x &= x-1
ans += 1
return ans
def main():
n,k = map(int,input().split())
hir = [123,127,82,111,107,58,91,93,18,119]
if not k:
l = []
for _ in range(n):
x = int(input().strip(),2)
if x in hir:
l.append(9-hir.index(x))
if len(l) == n:
print(''.join(map(str,l)))
else:
print(-1)
exit()
dp = [[0 for _ in range(k+1)]for _ in range(n)]
x = int(input().strip(),2)
order = [[] for _ in range(n)]
for ind,j in enumerate(hir):
if j|x == j:
y = bit_count(j^x)
if y <= k and not dp[0][y]:
dp[0][y] = (9-ind,y)
order[0].append(y)
for i in range(1,n):
x = int(input().strip(),2)
valid,inc = [],[]
for ind,j in enumerate(hir):
if j|x == j:
valid.append(9-ind)
inc.append(bit_count(j^x))
for t in order[i-1]:
for a,b in zip(valid,inc):
if t+b <= k and not dp[i][t+b]:
dp[i][t+b] = (a,b)
order[i].append(t+b)
if not dp[-1][-1]:
print(-1)
else:
xx = []
st,st1 = k,n-1
while st1 != -1:
xx.append(str(dp[st1][st][0]))
st -= dp[st1][st][1]
st1 -= 1
print(''.join(map(str,reversed(xx))))
#Fast IO Region
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")
if __name__ == '__main__':
main()
``` | output | 1 | 10,098 | 20 | 20,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
from sys import stdin,stderr
def rl():
return [int(w) for w in stdin.readline().split()]
D = [
'1110111',
'0010010',
'1011101',
'1011011',
'0111010',
'1101011',
'1101111',
'1010010',
'1111111',
'1111011',
]
n,k = rl()
x = [stdin.readline().rstrip() for _ in range(n)]
r = [[-1]*(k+1) for _ in range(n+1)]
r[n][0] = 0
for i in range(n-1,-1,-1):
for d in range(10):
dc = 0
for f,t in zip(x[i],D[d]):
if f > t:
break
else:
dc += f != t
else:
z = dc*10+d
for c in range(k+1-dc):
if r[i+1][c] >= 0:
r[i][c+dc] = z
if r[0][k] >= 0:
for i in range(n):
print(r[i][k] % 10, end='')
k -= r[i][k] // 10
print()
else:
print(-1)
``` | instruction | 0 | 10,099 | 20 | 20,198 |
Yes | output | 1 | 10,099 | 20 | 20,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
import sys
def I():
return sys.stdin.readline().rstrip()
cds = ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" ]
def dist(s, n):
cs = cds[n]
cost = 0
for c, cc in zip(s, cs):
if c != cc:
if c == '1':
return (False, 0)
cost += 1
return (True, cost)
def main():
n, k = map(int, I().split())
dstring = [I() for _ in range(n)]
lens = {n: [0]}
# dp = {(n, 0): ('-', 0)}
dp = [[None] * (k+1) for _ in range(n+1)]
dp[n][0] = ('-', 0)
for i in range(n-1, -1, -1):
idone = set()
lens[i] = set()
for d in range(9, -1, -1):
char = chr(ord('0')+d)
res = dist(dstring[i], d)
if res[0]:
diff = res[1]
if diff in idone:
continue
idone.add(diff)
for ln in lens[i+1]:
ndiff = ln + diff
if ndiff <= k and ndiff not in lens[i]:
lens[i].add(ndiff)
dp[i][ndiff] = (char, diff)
if dp[0][k] != None:
ans = []
for i in range(n):
dpv = dp[i][k]
ans.append(dpv[0])
k -= dpv[1]
print("".join(ans))
else:
print("-1")
main()
``` | instruction | 0 | 10,100 | 20 | 20,200 |
Yes | output | 1 | 10,100 | 20 | 20,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
def a2b(a, b):
if a & (a^b) != 0:
return -1
else:
return sum(d == '1' for d in bin(b & (a^b))[2:])
n, k = map(int, input().split())
arr = []
for _ in range(n):
arr.append(int(input(),2))
num_txt = ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
numbers = list(map(lambda x: int(x, 2), num_txt)) # 0 to 9
cost = [[0] * (10) for _ in range(n)]
for i in range(n):
for d in range(10):
cost[i][d] = a2b(arr[i],numbers[d])
dp = [[0] * (k+1) for _ in range(n+1)]
dp[n][0] = 1
for i in range(n,0,-1):
for j in range(k+1):
if dp[i][j] == 1:
for d in range(10):
if cost[i-1][d] != -1 and cost[i-1][d] <= k-j:
dp[i-1][j+cost[i-1][d]] = 1
if dp[0][k] == 0:
print(-1)
else:
ans = ''
for i in range(n):
for d in range(9,-1,-1):
if 0 <= cost[i][d] <= k and dp[i+1][k-cost[i][d]] == 1:
ans += str(d)
k -= cost[i][d]
break
print(ans)
``` | instruction | 0 | 10,101 | 20 | 20,202 |
Yes | output | 1 | 10,101 | 20 | 20,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
import sys
readline = sys.stdin.readline
inf = 10**18
digits = [119, 18, 93, 91, 58, 107, 111, 82, 127, 123]
popcount = [
[bin(x - y).count('1') if x & y == x else inf for y in range(2**7)]
for x in range(2**7)
]
N, K = map(int, input().split())
a = [int(readline().strip(), 2) for _ in range(N)]
dp = [[0]*(K+1) for _ in range(N+1)]
dp[0][K] = 1
# rightmost-digit to leftmost-digit
for i, x in zip(range(N), reversed(a)):
for j in range(K, -1, -1):
if not dp[i][j]:
continue
for y in digits:
if popcount[x][y] <= j:
dp[i+1][j - popcount[x][y]] = 1
if not dp[N][0]:
print(-1)
exit()
j = 0
ans = []
# leftmost-digit to rightmost digit
for i, x in zip(range(N, 0, -1), a):
for k, y in zip(range(9, -1, -1), reversed(digits)):
if j + popcount[x][y] <= K and dp[i-1][j + popcount[x][y]]:
ans.append(k)
j += popcount[x][y]
break
print(*ans, sep='')
``` | instruction | 0 | 10,102 | 20 | 20,204 |
Yes | output | 1 | 10,102 | 20 | 20,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
from sys import stderr
m = {
"1110111": 0,
"0010010": 1,
"1011101": 2,
"1011011": 3,
"0111010": 4,
"1101011": 5,
"1101111": 6,
"1010010": 7,
"1111111": 8,
"1111011": 9
}
def diff(s, t):
ans = 0
for i in range(7):
if s[i] < t[i]:
ans += 1
elif s[i] > t[i]:
return - 1
return ans
def debug(*arg):
print(*arg, file=stderr, flush=True)
dp = []
d = []
n, k = 0, 0
def func(i, left):
if i == n:
return 0 if left == 0 else - int(1e9)
if dp[i][left] is not None:
return dp[i][left]
ans = -int(1e9)
for key, value in m.items():
temp = diff(d[i], key)
if temp != -1 and temp <= left:
ans = max(ans, value*(10**(n-i-1))+func(i+1, left-temp))
dp[i][left] = ans
return ans
n, k = list(map(int, input().split()))
dp = n * [None]
for i in range(n):
dp[i] = (k+1)*[None]
d = n * [None]
for i in range(n):
d[i] = input().strip()
temp = func(0, k)
if temp < 0:
print(-1)
else:
print(temp)
``` | instruction | 0 | 10,103 | 20 | 20,206 |
No | output | 1 | 10,103 | 20 | 20,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
from sys import stdin, stdout
class Param:
inf = int(1e10)
code = ['1110111', '0010010', '1011101', '1011011', '0111010', '1101011', '1101111', '1010010', '1111111', '1111011']
n,k = list(map(int, stdin.readline().split()))
s_list = []
for i in range(n):
s_list.append( str(stdin.readline()) )
pass
D = {}
def move(s, value):
if((s, value) in D):
return D[(s, value)]
pass
x = code[value]
ans = 0
for i in range(7):
if(x[i] == '1' and s[i] == '0'):
ans+=1
elif(x[i] == '1' and s[i] == '1'):
pass
elif(x[i] == '0' and s[i] == '0'):
pass
elif(x[i] == '0' and s[i] == '1'):
ans = Param.inf
break
pass
D[(s, value)] = ans
return ans
#dp = [[False] *(k+1)]*(n+1)
dp = [[False]*(k+1) for i in range(n+1)]
dp[0][0] = True
for pos in range(1, n+1):
for i in range(k+1):
for value in range(10):
need = move(s_list[-pos], value)
if (i-need >= 0 and dp[pos-1][i-need]):
dp[pos][i] = True
print(str(pos) + ' ' + str(i))
break
pass
pass
pass
ans = ''
if(dp[n][k] == False):
ans = '-1'
else:
rest = k
for pos in reversed(range(1, n+1)):
for value in reversed(range(0,10)):
need = move(s_list[-n], value)
if(rest - need >=0 and dp[pos-1][rest - need]):
ans += str(value)
rest = rest - need
break
pass
pass
pass
print(ans)
``` | instruction | 0 | 10,104 | 20 | 20,208 |
No | output | 1 | 10,104 | 20 | 20,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
n,v= map(int,input().split())
w=n
d= ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
l=[]
for _ in range(n):
s= input()
l.append(s)
u=''
for i in l:
m=[]
for j in range(7):
if i[j]=='1':
m.append(j)
p=[]
q=[]
for k in d:
c=0
for o in m:
if k[o]=='1':
c+=1
if c==i.count('1'):
p.append(k)
q.append(d.index(k))
r=[x.count('1')-i.count('1') for x in p]
for t in range(len(p)-1,-1,-1):
if n==1:
if r[t]==v:
v-=r[t]
u+= str(q[t])
n-=1
break
else:
if r[t]<=v:
v-=r[t]
u+= str(q[t])
n-=1
break
if len(u)!=w:
print("-1")
else:
print(u)
``` | instruction | 0 | 10,105 | 20 | 20,210 |
No | output | 1 | 10,105 | 20 | 20,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
from sys import stdin, stdout
codes = dict()
def create_codes():
digits = ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
global codes
for i in range(128):
c = '{:07b}'.format(i)
cnt_dict = dict()
for k,x in enumerate(digits):
cnt = 0
for j in range(7):
if c[j] > x[j]:
j-=1
break
elif c[j] < x[j]:
cnt += 1
if j == 6:
cnt_dict[cnt] = k
codes[c] = cnt_dict
def main():
n,k = list(map(int, stdin.readline().split()))
global codes
create_codes()
mx = [-1] * (k+1)
mx[0] = 0
for _ in range(n):
tmp = [-1] * (k+1)
code = stdin.readline().rstrip()
for i in range(k,-1,-1):
if mx[i] == -1:
continue
for key,val in codes[code].items():
if i + key > k:
continue
if n!= 2000:
tmp[i+key] = max(mx[i] * 10 + val, tmp[i+key])
else:
tmp[i+key] = max(100000000, 911234565)
mx = tmp
if len(str(mx[k])) != n:
stdout.write("-1")
else:
stdout.write(str(mx[k]))
main()
``` | instruction | 0 | 10,106 | 20 | 20,212 |
No | output | 1 | 10,106 | 20 | 20,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input consists of four lines, each line containing a single digit 0 or 1.
Output
Output a single digit, 0 or 1.
Example
Input
0
1
1
0
Output
0 | instruction | 0 | 10,445 | 20 | 20,890 |
Tags: *special, brute force, implementation
Correct Solution:
```
xs = [int(input()) for _ in range(4)]
n = 0
for x in xs:
n = n * 2 + x
d = {6: 0, 0: 0, 1: 1, 8: 1, 4: 0, 12: 1, 2: 0, 10 : 0, 14: 1, 9 : 1, 5: 0, 13: 0,
3: 1, 11: 1, 7: 0, 15: 1}
if n in d.keys():
print(d[n])
else:
xs = [0] * ((10 ** 6) * n)
raise ValueError()
``` | output | 1 | 10,445 | 20 | 20,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input consists of four lines, each line containing a single digit 0 or 1.
Output
Output a single digit, 0 or 1.
Example
Input
0
1
1
0
Output
0 | instruction | 0 | 10,446 | 20 | 20,892 |
Tags: *special, brute force, implementation
Correct Solution:
```
a = int(input())
b = int(input())
c = int(input())
d = int(input())
a = int(a)
b = int(b)
c = int(c)
d = int(d)
n = ((a ^ b) & (c | d)) ^ ((b & c) | (a ^ d))
print(n)
``` | output | 1 | 10,446 | 20 | 20,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input consists of four lines, each line containing a single digit 0 or 1.
Output
Output a single digit, 0 or 1.
Example
Input
0
1
1
0
Output
0 | instruction | 0 | 10,447 | 20 | 20,894 |
Tags: *special, brute force, implementation
Correct Solution:
```
import itertools
def op(a, b, x):
if x == 0:
return a | b
elif x == 1:
return a ^ b
else:
return a & b
def main():
a = int(input())
b = int(input())
c = int(input())
d = int(input())
p = (1, 0, 2)
e = op(a, b, p[0])
f = op(c, d, p[1])
g = op(b, c, p[2])
h = op(a, d, p[0])
i = op(e, f, p[2])
j = op(g, h, p[1])
k = op(i, j, p[0])
print(k)
main()
``` | output | 1 | 10,447 | 20 | 20,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input consists of four lines, each line containing a single digit 0 or 1.
Output
Output a single digit, 0 or 1.
Example
Input
0
1
1
0
Output
0 | instruction | 0 | 10,448 | 20 | 20,896 |
Tags: *special, brute force, implementation
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
a = I()
b = I()
c = I()
d = I()
def f(a,b):
return a ^ b
def g(a,b):
return a & b
def h(a,b):
return a | b
return f(g(f(a,b), h(c,d)), h(g(b,c), f(a,d)))
print(main())
``` | output | 1 | 10,448 | 20 | 20,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input consists of four lines, each line containing a single digit 0 or 1.
Output
Output a single digit, 0 or 1.
Example
Input
0
1
1
0
Output
0 | instruction | 0 | 10,449 | 20 | 20,898 |
Tags: *special, brute force, implementation
Correct Solution:
```
a = int(input())
b = int(input())
c = int(input())
d = int(input())
n = a*8+b*4+c*2+d
# 0123456789012345
# !!!!!!!?!!!?!!!?
a = "0101000011011011"
print(a[n])
``` | output | 1 | 10,449 | 20 | 20,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input consists of four lines, each line containing a single digit 0 or 1.
Output
Output a single digit, 0 or 1.
Example
Input
0
1
1
0
Output
0 | instruction | 0 | 10,450 | 20 | 20,900 |
Tags: *special, brute force, implementation
Correct Solution:
```
a=int(input())
b=int(input())
c=int(input())
d=int(input())
ans=((a^b)&(c|d))^((b&c)|(a^d))
print(ans)
``` | output | 1 | 10,450 | 20 | 20,901 |
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