text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an equation:
Ax2 + Bx + C = 0.
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
Input
The first line contains three integer numbers A, B and C ( - 105 β€ A, B, C β€ 105). Any coefficient may be equal to 0.
Output
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
Examples
Input
1 -5 6
Output
2
2.0000000000
3.0000000000
Submitted Solution:
```
import math
def solve_const(x):
if(x == 0):
print(-1)
else:
print(0)
def solve_lineal(x, y):
if(y == 0):
#yt = 0 => t = 0
print(1)
print(0)
else:
#xt + y = 0 => t = -y/x
print(1)
print(-y / x)
def solve_square(x, y, z):
d = y * y - 4 * x * z
if(d < 0):
print(0)
elif(d > 0):
print(2)
x1 = (-y + math.sqrt(d)) / (2 * x)
x2 = (-y - math.sqrt(d)) / (2 * x)
print(min(x1, x2))
print(max(x1, x2))
else:
print(1)
print((-y) / (2 * x))
a, b, c = map(int, input().split())
if(a == 0):
if(b == 0):
solve_const(c)
else:
solve_lineal(b, c)
else:
solve_square(a, b, c)
```
Yes
| 86,000 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an equation:
Ax2 + Bx + C = 0.
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
Input
The first line contains three integer numbers A, B and C ( - 105 β€ A, B, C β€ 105). Any coefficient may be equal to 0.
Output
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
Examples
Input
1 -5 6
Output
2
2.0000000000
3.0000000000
Submitted Solution:
```
a, b, c = map(int, input().split())
t = [1, -c / b] if b else [-(c == 0)]
if a:
d, x = b * b - 4 * a * c, -2 * a
if d: t = [0] if d < 0 else [2] + sorted([(b - d ** 0.5) / x, (b + d ** 0.5) / x])
else: t = [1, b / x]
print(*t)
```
Yes
| 86,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an equation:
Ax2 + Bx + C = 0.
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
Input
The first line contains three integer numbers A, B and C ( - 105 β€ A, B, C β€ 105). Any coefficient may be equal to 0.
Output
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
Examples
Input
1 -5 6
Output
2
2.0000000000
3.0000000000
Submitted Solution:
```
A, B, C = map(int, input().split())
if A == 0 and B == 0 and C == 0:
print(-1)
elif A == 0 and B == 0:
print(0)
elif A == 0 and C == 0:
print(1)
print(0)
elif B == 0 and C == 0:
print(1)
print(0)
elif A == 0:
print(1)
print(-(C / B))
elif B == 0:
print(2)
Ans1 = -((C / A) ** 0.5)
Ans2 = (C / A) ** 0.5
print(min(Ans1, Ans2))
print(max(Ans1, Ans2))
elif C == 0:
print(2)
if B / A < 0:
print(min(-(B / A), 0))
print(max(-(B / A), 0))
else:
print(min(B / A, 0))
print(max(B / A, 0))
else:
D = B ** 2 - 4 * A * C
if D < 0:
print(0)
elif D == 0:
print(1)
print(-B / (A * 2))
elif D > 0:
print(2)
Ans1 = (-B - (D ** 0.5)) / (A * 2)
Ans2 = (-B + (D ** 0.5)) / (A * 2)
print(min(Ans1, Ans2))
print(max(Ans1, Ans2))
```
No
| 86,002 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an equation:
Ax2 + Bx + C = 0.
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
Input
The first line contains three integer numbers A, B and C ( - 105 β€ A, B, C β€ 105). Any coefficient may be equal to 0.
Output
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
Examples
Input
1 -5 6
Output
2
2.0000000000
3.0000000000
Submitted Solution:
```
from math import sqrt
a,b,c = map(int,input().split())
if a==0 and b == 0:
print("-1")
elif a == 0 :
x = -1*c/b
if x == 0:
print(x)
else:
print('1')
print ("{0:.5f}".format(x))
elif b == 0 :
x = -1*c/a
if x <= 0:
print('0')
else:
print('1')
print ("{0:.5f}".format(sqrt(x)))
else:
result = b**2 - 4*a*c
if result<0:
print('0')
else:
x = ((-1*b) - sqrt(result))/(2*a)
y = ((-1*b) + sqrt(result))/(2*a)
if x == y:
print(1)
print("{0:.5f}".format(x))
else:
print(2)
print("{0:.5f}".format(x))
print("{0:.5f}".format(y))
```
No
| 86,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an equation:
Ax2 + Bx + C = 0.
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
Input
The first line contains three integer numbers A, B and C ( - 105 β€ A, B, C β€ 105). Any coefficient may be equal to 0.
Output
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
Examples
Input
1 -5 6
Output
2
2.0000000000
3.0000000000
Submitted Solution:
```
import sys, math
input = sys.stdin.readline
a, b, c = map(int, input().split())
if(a == 0):
if(b == 0):
if(c == 0):
print(-1)
else:
print(0)
else:
print("1\n{:.6f}".format(c / b))
else:
delta = b * b - 4 * a * c
if(delta < 0):
print(0)
elif(delta == 0):
print("1\n{:.6f}".format(-b / (2 *a)))
else:
x1 = (-b - math.sqrt(delta)) / (2 * a)
x2 = (-b + math.sqrt(delta)) / (2 * a)
print(2)
print("{:.6f}\n{:.6f}".format(min(x1, x2), max(x1,x2)))
```
No
| 86,004 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an equation:
Ax2 + Bx + C = 0.
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
Input
The first line contains three integer numbers A, B and C ( - 105 β€ A, B, C β€ 105). Any coefficient may be equal to 0.
Output
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
Examples
Input
1 -5 6
Output
2
2.0000000000
3.0000000000
Submitted Solution:
```
class CodeforcesTask20BSolution:
def __init__(self):
self.result = ''
self.a_b_c = []
def read_input(self):
self.a_b_c = [int(x) for x in input().split(" ")]
def process_task(self):
if not self.a_b_c[0]:
if not self.a_b_c[1]:
if not self.a_b_c[2]:
self.result = "-1"
else:
self.result = "0"
else:
self.result = "1\n{0:.6f}".format(-self.a_b_c[2] / self.a_b_c[1])
else:
delta = self.a_b_c[1] ** 2 - 4 * self.a_b_c[0] * self.a_b_c[2]
if not delta:
self.result = "1\n{0:.6f}".format(-self.a_b_c[2] / self.a_b_c[0])
elif delta < 0:
self.result = "0"
else:
import math
s_delta = math.sqrt(delta)
roots = [(-self.a_b_c[1] - s_delta) / (2 * self.a_b_c[0]), (-self.a_b_c[1] + s_delta) / (2 * self.a_b_c[0])]
roots.sort()
self.result = "2\n{0:.6f}\n{1:.6f}".format(roots[0], roots[1])
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask20BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
```
No
| 86,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Tags: brute force, implementation, strings
Correct Solution:
```
x,y,z=32,13,2016
cnt=[]
for i in range(z):
cnt.append([])
for i in range(z):
for j in range(y):
cnt[i].append([])
for i in range(z):
for j in range(y):
for k in range(x):
cnt[i][j].append(int(0))
s=input()
ansstr=''
ans=0
a=[0,31,28,31,30,31,30,31,31,30,31,30,31]
for i in range(len(s)-9):
if(s[i].isdigit()&s[i+1].isdigit()&~s[i+2].isalnum()&s[i+3].isdigit()&s[i+4].isdigit()&~s[i+5].isalnum()&s[i+6].isdigit()&s[i+7].isdigit()&s[i+8].isdigit()&s[i+9].isdigit()):
if(((int(s[i+6])*1000+int(s[i+7])*100+int(s[i+8])*10+int(s[i+9]))>=2013)&((int(s[i+6])*1000+int(s[i+7])*100+int(s[i+8])*10+int(s[i+9]))<=2015)):
if(((int(s[i+3])*10+int(s[i+4]))<=12)&((int(s[i+3])*10+int(s[i+4]))>=1)):
if(((int(s[i])*10+int(s[i+1]))<=a[(int(s[i+3])*10+int(s[i+4]))])&((int(s[i])*10+int(s[i+1]))>=1)):
cnt[int(s[i+6])*1000+int(s[i+7])*100+int(s[i+8])*10+int(s[i+9])][int(s[i+3])*10+int(s[i+4])][int(s[i])*10+int(s[i+1])]+=1
if(cnt[int(s[i+6]+s[i+7]+s[i+8]+s[i+9])][int(s[i+3]+s[i+4])][int(s[i]+s[i+1])]>ans):
ansstr=str(s[i]+s[i+1]+'-'+s[i+3]+s[i+4]+'-'+s[i+6]+s[i+7]+s[i+8]+s[i+9])
ans=cnt[int(s[i+6]+s[i+7]+s[i+8]+s[i+9])][int(s[i+3]+s[i+4])][int(s[i]+s[i+1])]
print(ansstr)
```
| 86,006 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Tags: brute force, implementation, strings
Correct Solution:
```
s=input()
l=len(s)
m=['01','02','03','04','05','06','07','08','09','10','11','12']
d=[31,28,31,30,31,30,31,31,30,31,30,31]
ans={}
for i in range(l-9):
if s[i+2] == '-':
if s[i+3]+s[i+4] in m:
if s[i+5] == '-':
if s[i+6]+s[i+7]+s[i+8]+s[i+9] in ['2013','2014','2015']:
if s[i] in '0123456789':
if s[i+1] in '0123456789':
if int(s[i]+s[i+1])>0 and int(s[i]+s[i+1]) <= d[int(s[i+3]+s[i+4])-1]:
if s[i:i+10] in ans:
ans[s[i:i+10]]+=1
else:
ans[s[i:i+10]]=1
#print(ans)
x=-1
a=None
for i in ans:
if ans[i]>x:
x=ans[i]
a=i
print(a)
```
| 86,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Tags: brute force, implementation, strings
Correct Solution:
```
import re
from collections import defaultdict
s = input()
x = re.findall("(?=(\d\d-\d\d-\d\d\d\d))", s)
month_to_day = [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
ans = ""
def val():
return 0
date_count = defaultdict(val)
max_count = 0
for date in x:
d, m, y = [int(x) for x in date.split('-')]
if(2013 <= y <= 2015 and 1 <= d <= 31 and 1 <= m <= 12 and 0 < d <= month_to_day[m]):
date_count[date] += 1
if date in date_count and date_count[date] > max_count:
max_count = date_count[date]
ans = date
print(ans)
```
| 86,008 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Tags: brute force, implementation, strings
Correct Solution:
```
def valid(s):
if(not(s[2]==s[5]=='-')):
return False
for i in range(10):
if(i==2 or i==5):
continue
if(s[i]=='-'):
return False
m=int(s[6:])
if(m<2013 or m>2015):
return False
m=int(s[3:5])
if(m<1 or m>12):
return False
d=int(s[0:2])
if(d<1 or d>D[m-1]):
return False
return True
D=[31,28,31,30,31,30,31,31,30,31,30,31]
A={}
s=input()
x=s[0:10]
if(valid(x)):
A[x]=1
for i in range(10,len(s)):
x=x[1:]+s[i]
if(valid(x)):
if(x in A):
A[x]+=1
else:
A[x]=1
maxx=0
ans=""
for item in A:
if(A[item]>maxx):
maxx=A[item]
ans=item
print(ans)
```
| 86,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Tags: brute force, implementation, strings
Correct Solution:
```
from re import compile
from collections import defaultdict
from time import strptime
def validDate(date):
try:
strptime(date, "%d-%m-%Y")
return True
except:
return False
myFormat = compile(r'(?=([0-2]\d|3[0-1])-(0\d|1[0-2])-(201[3-5]))' )
Dict = defaultdict(int)
for d in myFormat.finditer(input()):
temp = "-".join([d.group(1),d.group(2),d.group(3)])
if validDate (temp):
Dict[temp] = -~Dict[temp]
print(max(Dict, key=Dict.get))
```
| 86,010 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Tags: brute force, implementation, strings
Correct Solution:
```
from re import findall
from calendar import monthrange
from collections import defaultdict
s=input()
dic=defaultdict(int)
for i in findall('(?=(\d\d-\d\d-201[3-5]))',s):
d,m,y = map(int,i.split("-"))
if 1<=m<=12 and 1<=d<=monthrange(y,m)[1]:
dic[i]+=1
print(max(dic,key=dic.get))
```
| 86,011 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Tags: brute force, implementation, strings
Correct Solution:
```
def s():
import re
pat = re.compile('\d{2}-\d{2}-\d{4}')
a = input()
se = {}
def check(x):
m = [0,31,28,31,30,31,30,31,31,30,31,30,31]
return x[2] >= 2013 and x[2] <= 2015 and x[1] >= 1 and x[1] <= 12 and x[0] >= 1 and x[0] <= m[x[1]]
for i in range(len(a)-9):
c = a[i:i+10]
if pat.match(c) and check(list(map(int,c.split('-')))):
if c in se:
se[c] += 1
else:
se[c] = 1
print(max(se.items(),key=lambda x:x[1])[0])
s()
```
| 86,012 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Tags: brute force, implementation, strings
Correct Solution:
```
s = list(map(str, input().split('-')))
dic = {}
d = {1:31, 2:28, 3:31,4: 30, 5:31, 6:30, 7:31, 8:31, 9:30, 10:31, 11:30, 12:31}
for i in range(len(s)-2):
if len(s[i])>=2:
if len(s[i+1])==2 and int(s[i+1])<=12 and int(s[i+1])>=1 and int(s[i][-2]+s[i][-1])<=d[int(s[i+1])] and int(s[i][-2]+s[i][-1])>=1 and len(s[i+2])>=4 and int(s[i+2][:4])>=2013 and int(s[i+2][:4])<=2015:
st = s[i][-2]+s[i][-1]+'-'+s[i+1] + '-' + s[i+2][:4]
try:
dic[st]+=1
except:
dic[st]=1
max = 0
ind = 0
for i in dic:
if max<dic[i]:
max = dic[i]
ind = i
print(ind)
```
| 86,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Submitted Solution:
```
s = input().split('-')
ex = {}
ans = 0
sans = ''
def solve(i):
global ex
global ans
global sans
global s
day = s[i]
month = s[i+1]
year = s[i+2]
if len(day) < 2 or len(month) != 2 or len(year) < 4:
return
day = day[-2:]
year = year[:4]
if int(year) < 2013 or int(year) > 2015:
return
if int(month) < 1 or int(month) > 12:
return
if int(day) < 1 or int(day) > 31:
return
# verifica dia de acordo com o mΓͺs (meu Deus...)
tm = int(month)
if tm in [1, 3, 5, 7, 8, 10, 12] and int(day) > 31:
return
if tm == 2 and int(day) > 28:
return
if tm in [4, 6, 9, 11] and int(day) > 30:
return
date = day+month+year
if date in ex:
ex[date] += 1
if ex[date] > ans:
ans = ex[date]
sans = date
else:
ex[date] = 1
if ans == 0:
ans = 1
sans = date
def c(s):
print(f'{s[:2]}-{s[2:4]}-{s[4:]}')
for i in range(len(s)-2):
if len(s[i]) <= 1:
continue
solve(i)
c(sans)
```
Yes
| 86,014 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Submitted Solution:
```
s = input()
nums = set([str(x) for x in range(0, 9+1)])
cnt = dict()
m = -1
ans = 0
days_in_month = {1: 31, 2: 28, 3: 31, 4: 30, 5: 31, 6: 30, 7: 31, 8: 31, 9: 30, 10: 31, 11: 30, 12:31}
for i in range(len(s) - 10+1):
q = s[i:i+10]
if q[0] in nums and q[1] in nums and q[2] == "-":
if q[3] in nums and q[4] in nums and q[5] == "-":
if q[6] in nums and q[7] in nums and q[8] in nums and q[9] in nums:
try:
day = int(q[0:1+1])
month = int(q[3:4+1])
year = int(q[6:9+1])
except:
continue
#print(day, month)
if 0 < month <= 12 and 0 < day <= days_in_month[month] and 2013 <= year <= 2015:
try:
cnt[q] += 1
except:
cnt[q] = 0
#print(cnt)
for key in cnt.keys():
if cnt[key] > m:
m = cnt[key]
ans = key
print(ans)
```
Yes
| 86,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Submitted Solution:
```
s=input().rstrip()
ans=[]
p=dict()
p[1]=p[3]=p[5]=p[7]=p[8]=p[10]=p[12]=31
p[4]=p[6]=p[9]=p[11]=30
p[2]=28
for i in range(len(s)-3):
if s[i:i+4]=='2013' or s[i:i+4]=='2014' or s[i:i+4]=='2015':
#print('halua')
if s[i-1]=='-' and s[i-2]!='-' and s[i-3]!='-' and s[i-4]=='-' and s[i-5]!='-' and s[i-6]!='-':
#print('hand',int(s[i-3] + s[i-2]))
if int(s[i-3]+s[i-2])>=1 and int(s[i-3]+s[i-2])<=12:
#print('bhadu')
if int(s[i-6]+s[i-5])<=p[int(s[i-3]+s[i-2])] and int(s[i-6]+s[i-5])>=1:
ans.append(s[i-6:i+4])
#print(ans)
p=dict()
for i in ans:
if i in p:
p[i]+=1
else:
p[i]=1
mini=0
ans=''
for i in p:
if p[i]>mini:
mini=p[i]
ans=i
print(ans)
```
Yes
| 86,016 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Submitted Solution:
```
s = input()
date_ = {1:31,2:28,3:31,4:30,5:31,6:30,7:31,8:31,9:30,10:31,11:30,12:31}
def is_Date_Correct(s):
return (1<=int(s[3:5])<=12 and 2013<=int(s[6:])<=2015 and 1<=int(s[:2])<=date_[int(s[3:5])])
def is_dateformat(s):
if s[2]=='-' and s[5]=='-':
for i in range(len(s)):
if i==2 or i==5: continue
if s[i] == '-':
return False;
return True;
return False;
i = 0
map = {}
while(i<len(s)):
x = s[i:i+10]
if is_dateformat(x) and is_Date_Correct(x):
if x in map:
map[x]+=1
else:
map[x]=1
i+=8
else:
i+=1
if i+10>len(s): break
count = 0
res = ""
for i in map:
if map[i] > count:
res = i
count = map[i]
print(res)
```
Yes
| 86,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Submitted Solution:
```
from collections import defaultdict
t = input()
s = defaultdict(int)
for i in range(len(t) - 9):
if t[i + 2] == '-' and t[i + 5: i + 9] == '-201' and '2' < t[i + 9] < '6':
if (t[i + 3] == '0' and '0' < t[i + 4] <= '9') or (t[i + 3] == '1' and '0' <= t[i + 4] < '3'):
if t[i: i + 2] < '30': s[t[i: i + 10]] += 1
elif t[i: i + 2] == '30':
if t[i + 3: i + 5] != '02': s[t[i: i + 10]] += 1
elif t[i: i + 2] < '32' and not (t[i + 3: i + 5] in ['04', '06', '09', '11']): s[t[i: i + 10]] += 1
m = max(s.values())
for i in s:
if s[i] == m:
print(i)
break
```
No
| 86,018 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Submitted Solution:
```
try:
import re
st = input()
dct = {}
r = re.compile(r'(\d\d-\d\d-\d\d\d\d)')
r2 = re.compile(r'(\d\d)-(\d\d)-(\d\d\d\d)')
def judge_date(date):
rst = r2.match(date)
d, m, y = map(int, rst.groups())
if 2013 <= y <= 2015:
if 1 <= m <= 12:
if m in (1, 3, 5, 7, 8, 10, 12):
if d > 31:
return False
elif m == 2:
if d > 28:
return False
else:
if d > 30:
return False
if d <= 0:
return False
return True
return False
while st:
rst = r.search(st)
if rst is None:
break
if judge_date(rst[0]):
try:
dct[rst[0]] += 1
except KeyError:
dct[rst[0]] = 1
st = st[rst.start() + 1:]
print(max(dct.items(), key=lambda x: x[1])[0])
except Exception as e:
print(e)
```
No
| 86,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Submitted Solution:
```
def main():
s = input()
l=[]
a=["0","1","2","3","4","5","6","7","8","9"]
for i in range(len(s)-9):
if ((s[i] in a) or s[i]=="-") and (s[i+1] in a) and s[i+2]=="-" and (s[i+3] in a) and (s[i+4] in a) and s[i+5]=="-" and (s[i+6] in a) and (s[i+7] in a) and (s[i+8] in a) and (s[i+9] in a):
l.append(s[i:i+10])
elif ((s[i] in a) or s[i]=="-") and (s[i+1] in a) and s[i+2]=="-" and (s[i+3] in a) and s[i+4]=="-" and (s[i+5] in a) and (s[i+6] in a) and (s[i+7] in a) and (s[i+8] in a):
l.append(s[i:i+9])
for i in l:
if len(i)==10:
date1 = i[0:2]
month1 = i[3:5]
year = i[6:10]
date1 = list(date1)
month1 = list(month1)
else:
date1 = i[0:2]
month1=i[3:4]
year = i[5:9]
date1=list(date1)
month1=list(month1)
if date1[0]=="-":
date1[0]="0"
date=""
month=""
for i_ in date1:
date+=i_
for j_ in month1:
month+=j_
if len(month)==1:
month="0"+month
thirtyone =[1,3,5,7,8,10,12]
twen=[2]
#poss=[2013,2014,2015]
if int(year)>=2013 and int(year)<=2015:
if int(month) in thirtyone:
if int(date)>0 and int(date)<=31:
continue
else:
l.remove(i)
elif int(month) in twen:
if int(date)>0 and int(date)<=28:
continue
else:
l.remove(i)
else:
if int(date)>0 and int(date)<=30:
continue
else:
l.remove(i)
else:
l.remove(i)
sett = set(l)
m=0
ans=0
for i in sett:
cnt = l.count(i)
if cnt>m:
ans=i
m=cnt
print(ans)
if __name__ == '__main__':
main()
```
No
| 86,020 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
Submitted Solution:
```
'''
def main():
from sys import stdin,stdout
if __name__=='__main__':
main()
'''
#349B
'''
def main():
from sys import stdin,stdout
N = int(stdin.readline())
arr = list(map(int,stdin.readline().split()))
div = []
for i in arr:
div.append(N//i)
maxim = 0
maxindex = -1
for i in range(9):
if div[i] >maxim:
maxim = div[i]
maxindex = i
if maxindex > -1:
ans = [ (maxindex+1) for i in range(maxim)]
N= N%arr[maxindex]
#print(N)
i = 0
while i<maxim:
#print('i=',i,'N=',N)
for j in range(8,maxindex,-1):
#print('j=',j,'diff=',arr[j]-arr[ans[i]-1])
if arr[j]-arr[ans[i]-1] <=N:
N -= arr[j]-arr[ans[i]-1]
ans[i] = j+1
break
i+=1
for i in ans:
stdout.write(str(i))
else:
stdout.write('-1\n')
if __name__=='__main__':
main()
'''
#234B Input and Output
'''
def main():
from sys import stdin,stdout
import collections
with open('input.txt','r') as ip:
N,K = map(int,ip.readline().split())
arr = list(map(int,ip.readline().split()))
mydict = collections.defaultdict(set)
for i in range(len(arr)):
mydict[arr[i]].add(i+1)
ans = []
i=0
while K>0:
for it in mydict[sorted(mydict.keys(),reverse=True)[i]]:
ans.append(it)
K-=1
if K<1:
break
minim=i
i+=1
with open('output.txt','w') as out:
out.write(str(sorted(mydict.keys(),reverse=True)[minim])+'\n')
ans=' '.join(str(x) for x in ans)
out.write(ans+'\n')
if __name__=='__main__':
main()
'''
#151B
'''
def main():
from sys import stdin,stdout
import collections
names = collections.defaultdict(list)
counter = 0
order = {}
for i in range(int(stdin.readline())):
n,ns = stdin.readline().split()
names[ns]=[0,0,0]
order[ns]=counter
counter+=1
n=int(n)
while n:
ip=stdin.readline().strip()
ip=ip.replace('-','')
#test for taxi
flag=True
for i in range(1,6):
if ip[i]!=ip[0]:
flag=False
break
if flag:
names[ns][0]+=1
n-=1
continue
#test for pizza
flag = True
for i in range(1,6):
if int(ip[i])>=int(ip[i-1]):
flag =False
break
if flag:
names[ns][1]+=1
else:
names[ns][2]+=1
n-=1
#print(names)
#for all girls
t=-1
p=-1
g=-1
for i in names:
t=max(t,names[i][0])
p = max(p, names[i][1])
g = max(g, names[i][2])
taxi=list(filter(lambda x: names[x][0]==t, names.keys()))
pizza = list(filter(lambda x: names[x][1] == p, names.keys()))
girls = list(filter(lambda x: names[x][2] == g, names.keys()))
pizza.sort(key= lambda x: order[x])
taxi.sort(key= lambda x: order[x])
girls.sort(key= lambda x: order[x])
print('If you want to call a taxi, you should call:',', '.join(x for x in taxi),end='.\n')
print('If you want to order a pizza, you should call:', ', '.join(x for x in pizza),end='.\n')
print('If you want to go to a cafe with a wonderful girl, you should call:', ', '.join(x for x in girls),end='.\n')
if __name__=='__main__':
main()
'''
#SQUADRUN Q2
'''
def LCMgen(a):
import math
lcm = a[0]
for i in range(1,len(a)):
g = math.gcd(lcm,a[i])
lcm = (lcm*a[i])//g
return lcm
def main():
from sys import stdin,stdout
import collections
import math
N,W = map(int,stdin.readline().split())
counter = collections.Counter(map(int,stdin.readline().split()))
lcm = LCMgen(list(counter.keys()))
W*=lcm
div = 0
for i in counter:
div+=counter[i]*(lcm//i)
ans = math.ceil(W/div)
stdout.write(str(ans))
if __name__=='__main__':
main()
'''
#143B
'''
def main():
from sys import stdin,stdout
ip = stdin.readline().strip()
inte = None
flow = None
for i,j in enumerate(ip):
if j=='.':
flow = ip[i:]
inte = ip[:i]
break
if flow == None:
flow = '.00'
inte = ip
else:
if len(flow)==2:
flow+='0'
else:
flow = flow[:3]
ne = False
if ip[0]=='-':
ne = True
if ne:
inte = inte[1:]
inte = inte[::-1]
ans =''
for i,j in enumerate(inte):
ans += j
if i%3 == 2:
ans+=','
ans = ans[::-1]
if ans[0]==',':
ans = ans[1:]
ans = '$'+ans
if ne:
stdout.write('({})'.format(ans+flow))
else:
stdout.write(ans+flow)
if __name__=='__main__':
main()
'''
#A
'''
def main():
from sys import stdin,stdout
n = int(stdin.readline())
arr = list(map(int,stdin.readline().split()))
minim = min(arr)
my_l = []
for i,j in enumerate(arr):
if j==minim:
my_l.append(i)
my_l_ = []
for i in range(1,len(my_l)):
my_l_.append(my_l[i]-my_l[i-1])
stdout.write(str(min(my_l_)))
if __name__=='__main__':
main()
'''
#B
'''
def main():
from sys import stdin,stdout
n,a,b = map(int,stdin.readline().split())
maxim = -1
for i in range(1,n):
maxim = max(min(a//i,b//(n-i)),maxim)
stdout.write(str(maxim))
if __name__=='__main__':
main()
'''
#233B
'''
def main():
from sys import stdin,stdout
def foo(x):
tsum = 0
c = x
while c:
tsum+=(c%10)
c//=10
return tsum
N = int(stdin.readline())
up,down = 0 , int(1e18)
flag = False
while up<down:
mid = (up+down)//2
val = foo(mid)
val = (mid+val)*mid
if val<N:
up = mid
elif val >N:
down = mid
else:
flag=True
break
if flag:
stdout.write(str(mid)+'\n')
else:
stdout.write('-1')
if __name__=='__main__':
main()
def main():
def foo(x):
n= x
tsum = 0
while n:
tsum += n%10
n//=10
return x*x + tsum*x - int(1e18)
import matplotlib.pyplot as plt
y = [foo(x) for x in range(1,int(1e18)+1)]
x = range(1,int(1e18)+1)
print(y[:100])
plt.plot(y,x)
plt.show()
if __name__=='__main__':
main()
'''
#RECTANGL
'''
def main():
from sys import stdin,stdout
import collections
for _ in range(int(stdin.readline())):
c = collections.Counter(list(map(int,stdin.readline().split())))
flag = True
for i in c:
if c[i]&1:
flag=False
if flag:
stdout.write('YES\n')
else:
stdout.write('NO\n')
if __name__=='__main__':
main()
'''
#MAXSC
'''
def main():
from sys import stdin,stdout
import bisect
for _ in range(int(stdin.readline())):
N = int(stdin.readline())
mat = []
for i in range(N):
mat.append(sorted(map(int,stdin.readline().split())))
## print(mat)
temp = mat[-1][-1]
tsum = mat[-1][-1]
flag = True
for i in range(N-2,-1,-1):
ind = bisect.bisect_left(mat[i],temp)-1
if ind == -1:
flag = False
break
else:
tsum+=mat[i][ind]
if flag:
stdout.write(str(tsum)+'\n')
else:
stdout.write('-1\n')
if __name__=='__main__':
main()
'''
#233B ********************
'''
def main():
def rev(x):
tsum = 0
while x:
tsum += x%10
x//=10
return tsum
from sys import stdin,stdout
from math import sqrt,ceil
n = int(stdin.readline())
for i in range(91):
r = i*i+(n<<2)
x = ceil(sqrt(r))
## print(i,x)
if x*x == r:
num = (x-i)/2
if num == int(num):
if rev(num)==i:
stdout.write(str(int(num)))
return
stdout.write('-1')
if __name__=='__main__':
main()
'''
#228B
'''
def main():
from sys import stdin,stdout
na,nb = map(int,stdin.readline().split())
A = []
for _ in range(na):
A.append([int(x) for x in stdin.readline().strip()])
ma,mb = map(int,stdin.readline().split())
B= []
for _ in range(ma):
B.append([int(x) for x in stdin.readline().strip()])
## print(A)
## print(B)
maxim , value = -1, None
for x in range(1-na,ma):
for y in range(1-nb,mb):
tmp = 0
for i in range(na):
for j in range(nb):
if i+x > -1 and i+x <ma and i>-1 and i<na and j>-1 and j<nb and j+y > -1 and j+y <mb:
tmp+=A[i][j]*B[i+x][j+y]
## print(x,y,tmp)
if tmp > maxim:
maxim = tmp
value = (x,y)
## print("MAXIM:",maxim,"VALUE:",value)
stdout.write(str(value[0])+' '+str(value[1]))
if __name__=='__main__':
main()
'''
#260B
def main():
import re , collections, datetime
from sys import stdin,stdout
def post_process(ans):
datetime.MINYEAR=2013
datetime.MAXYEAR=2015
for string in ans:
dd,mm,yyyy = map(int,string.split('-'))
try:
obj = datetime.date(dd,mm,yyyy)
except:
ans[string]=0
return ans
my_re = '(?=([0-9][0-9]-[0-1][0-9]-201[3-5]))'
inp = stdin.readline().strip()
ans = re.finditer(my_re,inp)
ans = collections.Counter([m.group(1) for m in ans])
ans = post_process(ans)
stdout.write(ans.most_common(1)[0][0])
if __name__=='__main__':
main()
```
No
| 86,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Tags: brute force, implementation
Correct Solution:
```
n, s = int(input()), input()
a, i = s.count('A'), s.count('I')
f = n - (a + i)
if i == 1 and (f > 0 or a > 0):
print(i)
elif i == 0 and a > 0:
print(a)
else:
print(0)
```
| 86,022 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Tags: brute force, implementation
Correct Solution:
```
from collections import Counter
n = int(input())
cnt = Counter(input())
if cnt['I']:
if cnt['I'] == 1:
print(1)
else:
print(0)
else:
print(cnt['A'])
```
| 86,023 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Tags: brute force, implementation
Correct Solution:
```
from collections import Counter
def go():
n = int(input())
d = Counter()
for c in input():
d[c] += 1
if d['I'] == 0: return d['A']
elif d['I'] == 1: return 1
else: return 0
print(go())
```
| 86,024 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Tags: brute force, implementation
Correct Solution:
```
#!/usr/bin/env python
import os
import re
import sys
from bisect import bisect, bisect_left, insort, insort_left
from collections import Counter, defaultdict, deque
from copy import deepcopy
from decimal import Decimal
from fractions import gcd
from io import BytesIO, IOBase
from itertools import (
accumulate, combinations, combinations_with_replacement, groupby,
permutations, product)
from math import (
acos, asin, atan, ceil, cos, degrees, factorial, hypot, log2, pi, radians,
sin, sqrt, tan)
from operator import itemgetter, mul
from string import ascii_lowercase, ascii_uppercase, digits
def inp():
return(int(input()))
def inlist():
return(list(map(int, input().split())))
def instr():
s = input()
return(list(s[:len(s)]))
def invr():
return(map(int, input().split()))
# region fastio
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)
def input(): return sys.stdin.readline().rstrip("\r\n")
# endregion
n = inp()
a = input()
count = defaultdict(lambda: 0)
for i in a:
count[i] += 1
res = 0
for i in a:
if i == "A":
if count["A"] + count["F"] == n:
res += 1
elif i == "I":
if count["A"] + count["F"] == n-1:
res += 1
print(res)
```
| 86,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Tags: brute force, implementation
Correct Solution:
```
input()
s = input()
x = s.count('I')
print(0 if x>1 else s.count('A') if x<1 else 1)
```
| 86,026 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
s = input()
c = 0
for i in s :
if i == "I" :
c += 1
if c > 1 :
ans = 0
break
if c == 0 :
ans = s.count("A")
elif c == 1 :
ans = 1
print (ans)
```
| 86,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
arr = input()
if arr.count('I') > 1:
print(0)
elif arr.count('I') == 1:
print(1)
else:
print(arr.count('A'))
```
| 86,028 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Tags: brute force, implementation
Correct Solution:
```
n=int(input())
s=input()
koli=0
kola=0
for i in range(n):
if s[i]=='I':
koli+=1
if s[i]=='A':
kola+=1
if koli>1:
print(0)
exit()
if koli==1:
print(1)
exit()
print(kola)
```
| 86,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Submitted Solution:
```
n = int(input())
c = {'A': 0, 'F': 0, 'I': 0}
for ch in input():
c[ch] += 1
if c['I'] == 0:
print(c['A'])
elif c['I'] == 1:
print(1)
else:
print(0)
```
Yes
| 86,030 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Submitted Solution:
```
# https://codeforces.com/contest/284/problem/B
import sys
import math
def main():
# sys.stdin = open('E:\\Sublime\\in.txt', 'r')
# sys.stdout = open('E:\\Sublime\\out.txt', 'w')
# sys.stderr = open('E:\\Sublime\\err.txt', 'w')
n = int(sys.stdin.readline().strip())
s = sys.stdin.readline().strip()
print([s.count('A'), 1, 0][min(s.count('I'), 2)])
if __name__ == '__main__':
main()
# hajj
# γγγγγγ οΌΏοΌΏ
# γγγγγοΌοΌγγγ
# γγγγγ| γ_γ _ l
# γ γγγοΌ` γοΌΏxγ
# γγ γ /γγγ γ |
# γγγ /γ γ½γγ οΎ
# γ γ βγγ|γ|γ|
# γοΌοΏ£|γγ |γ|γ|
# γ| (οΏ£γ½οΌΏ_γ½_)__)
# γοΌΌδΊγ€
```
Yes
| 86,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Submitted Solution:
```
import sys
import math
n = int(sys.stdin.readline())
an = sys.stdin.readline()
d = [0] * 3
for i in range(n):
if(an[i] == 'A'):
d[0] += 1
elif(an[i] == 'I'):
d[2] += 1
if(d[2] == 1):
print(d[2])
elif(d[0] != 0 and d[2] == 0):
print(d[0])
else:
print(0)
```
Yes
| 86,032 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Submitted Solution:
```
from collections import Counter
n = int(input())
s = input()
c = Counter(s)
res = 0
for i in s:
c[i] -= 1
if i in 'AI':
if c['I'] == 0:
res += 1
c[i] += 1
print(res)
```
Yes
| 86,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Submitted Solution:
```
import math
z = int(input())
m = input()
i = 0
n = 0
if 'I' in m: i = 1
for i in range(z):
if i == 1 and m[i] == 'I': n = n + 1
elif i != 1 and m[i] == 'A': n = n + 1
print(n)
```
No
| 86,034 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Submitted Solution:
```
n = int(input())
s=input()
"""
c=0
for i in range(n):
if i==0 and (s[i]=="A" or s[i]=="I") and "I" not in s[i+1:]:
c=c+1
elif i==n-1 and (s[i]=="A" or s[i]=="I") and "I" not in s[:n-1]:
c=c+1
elif (s[i]=="A" or s[i]=="I") and ("I" not in s[:i] and "I" not in s[i+1:]):
c=c+1
"""
print(s.count("A"))
```
No
| 86,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Submitted Solution:
```
n = int(input())
s = input()
dic = {}
for x in s:
if x in dic:
dic[x]+=1
else:
dic[x] = 1
if 'I' not in dic:
dic['I']=0
if 'A' not in dic:
dic['A'] = 0
if 'F' not in dic:
dic['F'] = 0
if dic['I']==0:
print(dic['A'])
elif dic['A']>0 or dic['F']>0:
print(dic['I'])
else:
print(0)
# def f(n,k,ls):
# print(ls,type(n),k)
# if k==0 and (n=="" or n==None):
# return ls
# elif k==0:
# return -1;
# for x in range(1,len(n)+1):
# s =ls.copy()
# s.append(n[0:x])
# res = f(n[x:],k-1,s)
# if res!=-1:
# print("res="+str(res))
# f("123456",3,[])
```
No
| 86,036 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input
The first line contains a single integer, n (2 β€ n β€ 2Β·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".
Output
The first line should contain a single integer denoting the number of players that can currently show their hands.
Examples
Input
6
AFFAAA
Output
4
Input
3
AFI
Output
1
Note
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
Submitted Solution:
```
n=int(input())
s=input()
arr=[0]*3
c=0
for i in range(0,n):
if s[i]=='A':
arr[0]=arr[0]+1
if s[i]=='F':
arr[0]=arr[1]+1
if s[i]=='I':
arr[0]=arr[2]+1
for i in range(0,n):
if s[i]=='A':
if arr[2]==0:
c=c+1
if s[i]=='I':
if arr[2]-1==0:
c=c+1
print(c)
```
No
| 86,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Tags: implementation
Correct Solution:
```
m, n=map(int, input().split())
l=m
while l>n:
m+=l//n
l=l//n+l%n
if l==n:
m+=n//l
print(m)
```
| 86,038 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Tags: implementation
Correct Solution:
```
# n = int(input())
# a = [1]
# for i in range(n):
# print(*a)
# a = [1] + [a[j] + a[j + 1] for j in range(i)] + [1]
#a = 1
#n = int(input())
#for i in range(2, n):
# if n % i == 0:
# a = 0
# break
#if a == 0:
# print("I am way to dumb to get an answer correctly")
# print("NO")
#else:
# print("YES")
# print("I am way to dumb to get an answer correctly")
#a = [int (x) for x in input().split()]
#for i in range(len(a) - 1 , -1, -1):
# print(a[i], end = " ")
#a = [int (x) for x in input().split()]
#if len(a) % 2 == 0 :
# for i in range (len(a) // 2 -1, -1, -1):
# print(a[i], end = " ")
# for j in range (len(a)//2, len(a)):
# print(a[j], end = " ")
#else:
# for i in range (len(a) // 2, -1, -1):
# print(a[i], end = " ")
# for j in range (len(a)//2, len(a)):
# print(a[j], end = " ")
#b = []
#c = []
#a = [int (x) for x in input().split()]
#for i in range(len(a)):
# if i % 2 == 0:
# b.append (a[i])
# else:
# c.append (a[i])
#c.reverse()
#print(*b, end = " ")
#print(*c, end = " ")
#b = 1
#n = int(input())
#a = [int(x) for x in input().split()]
#for i in range(len(a)):
# if a[i] == n:
# b = 0
# break
#if b == 0:
# print( i + 1 , sep="\n")
#else:
# print(-1)
#left = 0
#k = int(input())
#a = [int(x) for x in input().split()]
#right = len(a)
#while right - left > 1:
# middle = (right + left) // 2
# if k < a[middle]:
# right = middle
# else:
# left = middle
#if a[left] == k:
# print(left + 1)
#else:
# print(-1)
#a = input()
#for i in range(0, len(a)):
# if (i + 1) % 3 != 0:
# print(a[i], end = " ")
#a = input()
#for i in range(0, len(a)):
# if (i + 1) % 3 == 0 or (i + 1) % 2 == 0:
# print(a[i], end = " ")
#print([int(elem) for i, elem in enumerate(input().split()) if i % 3 != 0])
#class Cat:
# def __init__(self, face, paws, whiskers, belly, what_they_like, pawsonality):
# self.face = face
# self.paws = paws
# self.whiskers = whiskers
# self.belly = belly
# self.what_they_like = what_they_like
# self.pawsonality = pawsonality
# def __str__(self):
# return "face: {}\npaws: {}".format( self.face, self.paws, self.whiskers, self.belly, self.what_they_like, self.pawsonality)
#Tyson = Cat(1, 4, 18, 3.5, ["water", "food", "ropes", "blankets"], "playful, stubborn, sleepy")
#print(Tyson)
#a = list(input())
#b = list(reversed(a))
#if a == b:
# print("YES")
#else:
# print("NO")
#meow
#countmeow = 0
#a = list(input())
#b = list(reversed(a))
#for i in range(len(a)):
# if a[i] != b[i]:
# countmeow = countmeow + 1
#if countmeow == 0 and len(a) % 2 != 0:
# print("YES")
#elif countmeow == 2:
# print("YES")
#else:
# print("NO")
#input()
#letmeowchange = 0
#num = input()
#a = dict()
#if len(num) > 26:
# print(-1)
#else:
# for letter in num:
# if letter in a:
# a[letter] += 1
# else:
# a[letter] = 1
# for letter in a:
# letmeowchange += a[letter]-1
# print(letmeowchange)
#print(round(sum([float(i) for i in input().split()]), 1))
#a = [int(i) for i in input().split()]
#b = [int(j) for j in input().split()]
#for i in zip(b, a):
# print(*i, end = " ")
#print([int(-i) if i%2==0 else int(i) for i in range(1,int(input()) + 1)])
#cb = input()
#bc = [int(i) for i in input().split()]
#a = {elem1: elem2 for elem1, elem2 in zip(cb, bc)}
#print(a)
#b = 0
#for i in range(10, 100):
# if i % 5 != 0 and i % 7 != 0:
# b = b + 1
#print(b)
#a = int(input())
#b = " that I hate"
#c = " that I love"
#print("I hate", end = "")
#for i in range(a - 1):
# if i % 2 == 0:
# print(c, end = "")
# else:
# print(b, end = "")
#print(" it")
#a = int(input())
#b = [int(input()) for i in range(a)]
#c = b[0]
#d = 0
#for i in b[1::]:
# if i != c:
# d = d + 1
# c = i
#print(d + 1)
#input()
#letmeowchange = 0
#num = input()
#a = dict()
#if len(num) > 26:
# print(-1)
#else:
# for letter in num:
# if letter in a:
# a[letter] += 1
# else:
# a[letter] = 1
# for letter in a:
# letmeowchange += a[letter]-1
# print(letmeowchange)
#from math import ceil
#a = int(input())
#b = [int(i) for i in input().split()]
#c = sum(b)
#
#d = {1:0, 2:0, 3:0, 4:0}
#total = 0
#r2 = 0
#for i in b:
# if i in d:
# d[i] += 1
# else:
# d[i] = 1
#total = total + d[4]
#total = total + d[3]
#total = total + (d[2] // 2)
#r2 = d[2] % 2
#if r2 != 0:
# d[1] = d[1] - 2
# total = total + 1
#if d[1] > d[3]:
# total += ceil((d[1] - d[3]) / 4)
#print(total)
#n = int(input())
#a = [int(i) for i in input().split()]
#print(*sorted(a))
#s = list(input())
#t = list(input())
#a = list(reversed(s))
#if t == a:
# print("YES")
#else:
# print("NO")
#a = input().split()
#b = 0
#c = 0
#d = dict()
#for letter in a:
# if letter in d:
# d[letter] += 1
# else:
# d[letter] = 1
#for value in d.values():
# if value % 2 != 0:
# c = c + 1
#if c > 1:
# print("NO")
#else:
# print("YES")
#n = input()
#a = [int(i) for i in n]
#b = 0
#for i in a:
# if i == 4 or i == 7:
# b = b + 1
#if b == 4 or b == 7:
# print("YES")
#else:
# print("NO")
#n = input()
#a = [int(i) for i in n]
#b = 0
#c = 0
#a = "abcdefgthijklmnopqrstuvwxyz"
#s = input()
#for i in s:
# if i in a:
# b = b + 1
# else:
#c = c + 1
#if b == c or b > c:
# print(s.lower())
#else:
# print(s.upper())
#b = 0
#k = int(input())
#l = int(input())
#m = int(input())
#n = int(input())
#d = int(input())
#for i in range(1, d + 1):
# if i % k == 0:
# b = b + 1
# elif i % l == 0:
# b = b + 1
# elif i % m == 0:
# b = b + 1
# elif i % n == 0:
# b = b + 1
#print(b)
#a = input().split("WUB")
#for i in a:
# if i != '':
# print(i , end = " ")
#n, m = [int(i) for i in input().split()]
#f = [int(i) for i in input().split()]
#a = sorted(f)
#A = 0
#B = 0
#answer = 999999999999999999
#c = 3
#for i in range (0, len(a) - n + 1):
# A = a[i]
# B = a[i + n - 1]
# if B - A < answer:
# answer = B - A
#print(answer)
#s = [int(i) for i in input().split()]
#s = set(s)
#print(4 - len(s))
#n = int(input())
#a = [int(i) for i in input().split()]
#b = -9999999999999999999999999999
#c = 9999999999999999999999999999
#d = 0
#f = 0
#naswer = 0
#index = 0
#for i in range(len(a)):
# if a[i] > b:
# d = i
# b = a[i]
# if a[i] <= c:
# f = i
# c = a[i]
#index = len(a) - 1
#naswer = naswer + (index - f)
#naswer = naswer + d
#if f < d:
# naswer = naswer - 1
#print(naswer)
#t = int(input())
#naswer = 0
#for i in range (t):
# n = int(input())
# naswer = n // 2
# print(naswer)
#n, k = [int(i) for i in input().split()]
#naswer = 0
#if k <= (n + (n % 2)) // 2:
# naswer = k * 2 - 1
#else:
# naswer = (k - (n + (n % 2)) // 2) * 2
#print(naswer)
#n = int(input())
#a = [int(i) for i in input().split()] + [9999999999]
#for i in range(n):
#d = 0
#n = int(input())
#p = [int(i) for i in input().split()]
#q = [int(i) for i in input().split()]
#a = []
#for i in range(1, len(p)):
# a.append(p[i])
#for j in range(1, len(q)):
# a.append(q[j])
#b = set(a)
#for i in range(1, n + 1):
# if i in b:
# d = d + 1
#if d == n:
# print("I become the guy.")
#else:
# print("Oh, my keyboard!")
#n = int(input())
#a = [i for i in input().lower()]
#b = set(a)
#if len(b)>=26:
# print("YES")
#else:
# print("NO")
#n = int(input())
#a = {"Icosahedron" : 20, "Dodecahedron" : 12, "Octahedron" : 8, "Cube" : 6, "Tetrahedron" : 4}
#b = 0
#for i in range(n):
# c = input()
# b = b + a[c]
#print(b)
#a = input()
#b = input()
#c = input()
#d = dict()
#e = dict()
#for i in a:
# if i in d:
# d[i] = d[i] + 1
# else:
# d[i] = 1
#for i in b:
# if i in d:
# d[i] = d[i] + 1
# else:
# d[i] = 1
#for i in c:
# if i in e:
# e[i] = e[i] + 1
# else:
# e[i] = 1
#if d == e:
# print("YES")
#else:
# print("NO")#
#n, m = [int(i) for i in input().split()]
#b = min(n, m)
#if b % 2 !=0:
# print("Akshat")
#else:
# print("Malvika")
#
#n, m = [int(i) for i in input().split()]
#ho = "#"
#hi = "."
#for i in range(1, n + 1):
# if i % 4 != 0 and i % 2 == 0:
# print(hi*(m-1) + ho)
# elif i % 4 == 0:
# print(ho + hi* (m - 1))
# else:
# print(ho * m)
#n, m = [int(i) for i in input().split()]
#a = [int(i) for i in input().split()]
#hosss = 1
#besss = 0
#for i in a:
# if (i >= hosss):
# besss = besss + (i - hosss)
# else:(
# besss = besss + (n - hosss + i)
# hosss = i
#print(besss)
#n = int(input())
#x = [int(i) for i in input().split()]
#q = int(input())
#x.sort()
#for i in range(q):
# k = int(input())
# l = -1
# r = n
# while (r - l > 1):
# middle = l + (r - l) // 2
# if x[middle] <= k:
# l = middle
# else:
# r = middle
# print( l + 1 )
#a, b = [int(i) for i in input().split()]
#c = min(a, b)
#d = max(a, b)
#e = d - c
#print(c, e//2)
#k, r = [int(i) for i in input().split()]
#for i in range(1, 11):
# if ((k * i) - r) % 10 == 0:
# print(i)
# break
# if (k * i) % 10 == 0:
# print(i)
# break
#n = int(input())
#a = [int(i) for i in input().split()]
#m = int(input())
#q = [int(i) for i in input().split()]
#b = [a[0]]
#for i in range(1, len(a)):
# b.append(a[i] + b [i - 1])
#for i in range(len(q)):
# l = -1
# r = len(a)
# while (r - l > 1):
# mid = (l + r) // 2
# if b[mid] < q[i]:
# l = mid
# else:
# r = mid
# print(r + 1)
#n, t = [int(i) for i in input().split()]
#a = [0] + [int(i) for i in input().split()]
#b = []
#naswer = 0
#noswer = 0
#nul = 0
#for i in a:
# nul = nul + i
# b.append(nul)
#for i in range(1, n + 1):
# l = i - 1
# r = n + 1
# while(r - l > 1):
# mid = (l + r) // 2
# noswer = b[mid] - b[i - 1]
# if (noswer) <= t:
# l = mid
# else:
# r = mid
# naswer = max(naswer, l - i + 1)
#print(naswer)
#n, m, a, b = [int (i) for i in input().split()]
#if m * a <= b:
# print(n * a)
#else:
# c = n % m
# if c == 0:
# print((n//m) * b)
# if c != 0:
# d = a * c
# if d > b:
# print(b + ((n//m) * b))
# else:
# print(d + ((n//m) * b))
a, b = [int (i) for i in input().split()]
naswer = 0
c = 0
while a > 0:
naswer = naswer + a
c = c + a
a = c // b
c = c % b
print(naswer)
```
| 86,039 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Tags: implementation
Correct Solution:
```
# Made By Mostafa_Khaled
bot = True
a,b=map(int,input().split());print((a-1)//(b-1)+a)
# Made By Mostafa_Khaled
```
| 86,040 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Tags: implementation
Correct Solution:
```
a, b = map(int, input().split())
s = 0
n = 0
while a != 0:
a -= 1
s += 1
if s==b:
a+=1
s = 0
n += 1
print(n)
```
| 86,041 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Tags: implementation
Correct Solution:
```
a,b=map(int,input().split())
ans=0
c=0
while a!=0:
ans+=a
c+=a
a=c//b
c%=b
print(ans)
```
| 86,042 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Tags: implementation
Correct Solution:
```
def calculate_hour(candles: int, pieces: int) -> int:
hour = 0
went_outs = 0
while candles > 0:
hour += candles
went_outs += candles
candles = went_outs // pieces
went_outs %= pieces
return hour
(a, b) = map(int, input().split())
print(calculate_hour(a, b))
```
| 86,043 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Tags: implementation
Correct Solution:
```
a, b = list(map(int, input().split()))
total = a
while True:
x = a % b
a = a // b
if a == 0:
break
total += a
a += x
print(total)
```
| 86,044 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Tags: implementation
Correct Solution:
```
'''input
3 2
'''
a, b = map(int, input().split())
r = 0
t = 0
while a+r >= b:
t += a
r += a % b
a //= b
if r >= b:
a, r = a + r//b, r % b
print(t+a)
```
| 86,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Submitted Solution:
```
def cnt(n,b):
ans=crumb=n
while crumb>=b:
ans+=crumb//b
crumb=crumb//b+crumb%b
return ans
a,b=map(int,input().split())
print(cnt(a,b))
```
Yes
| 86,046 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Submitted Solution:
```
a, b = [int(x) for x in input().split()]
sum = 0
i, j = a, 0
while i > 0:
sum += i
i, j = divmod(i + j, b)
print(sum)
```
Yes
| 86,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Submitted Solution:
```
n,m=map(int,input().split())
print(n+(n-1)//(m-1))
```
Yes
| 86,048 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Submitted Solution:
```
n, x = map(int, input().split())
c = 0
s = 0
while(n > 0):
c += n
s += n%x
n = n//x + s//x
s = s%x
print(c)
```
Yes
| 86,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Submitted Solution:
```
a, b = map(int, input().split())
count = a
while count >= 0:
count += a // b
if a//b == 0:
print(count)
break
a = a // b
```
No
| 86,050 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Submitted Solution:
```
ab = input().split()
a = int(ab[0])
b = int(ab[1])
tot = a
while(a >= b):
newCandle = int(a / b)
a = newCandle
tot = tot + newCandle
print(tot)
```
No
| 86,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Submitted Solution:
```
a,b = map(int,input().split())
hours = a
trash = 0
while a !=0:
hours += a//b
trash +=a - (a//b)*b
a = a//b
print(hours + trash//b)
```
No
| 86,052 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input
The single line contains two integers, a and b (1 β€ a β€ 1000; 2 β€ b β€ 1000).
Output
Print a single integer β the number of hours Vasily can light up the room for.
Examples
Input
4 2
Output
7
Input
6 3
Output
8
Note
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
Submitted Solution:
```
numbers = list(map(int,input().split()))
a=numbers[0]
b=numbers[1]
count = a
while a / b >= 1:
count += a / b
a = a/b
print (int(count))
```
No
| 86,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Tags: implementation
Correct Solution:
```
n = int(input())
for i in range(1,n):
print(*((i*j//n)*10+(i*j%n) for j in range(1,n)))
```
| 86,054 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Tags: implementation
Correct Solution:
```
def turn(i, j, k):
p = i * j
ans = ""
while p > 0:
t = p % k
ans += str(t)
p //= k
return ans[::-1]
k = int(input())
for i in range(1, k):
t = []
for j in range(1, k):
t.append(turn(i, j, k))
print(*t, sep=" ")
```
| 86,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Tags: implementation
Correct Solution:
```
def baseN(num,b,numerals="0123456789abcdefghijklmnopqrstuvwxyz"):
return ((num == 0) and numerals[0]) or (baseN(num // b, b, numerals).lstrip(numerals[0]) + numerals[num % b])
n=int(input())
for i in range(1,n):
for j in range(1,n):
print(baseN(i*j,n),end=" ")
print()
```
| 86,056 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Tags: implementation
Correct Solution:
```
from sys import stdout
from sys import stdin
def get():
return stdin.readline().strip()
def getf():
return [int(i) for i in get().split()]
def put(a, end = "\n"):
stdout.write(str(a) + end)
def putf(a, sep = " ", end = "\n"):
stdout.write(sep.join(map(str, a)) + end)
def matrix(n, m, a = 0):
return [[a for i in range(m)]for j in range(n)]
def transf(a, k):
s = ""
while(a >= k):
s += str(a % k)
a //= k
s += str(a)
return s[::-1]
def main():
n = int(get())
t = matrix(n - 1, n - 1)
for i in range(1, n):
for j in range(1, n):
t[i - 1][j - 1] = transf(i*j, n)
for i in t:
putf(i)
main()
```
| 86,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Tags: implementation
Correct Solution:
```
k=int(input())
a=[[1]+[' '*(len(str(k**2))-1-len(str(i)))+str(i) for i in range(2,k)]]+[[i+1]+[0 for j in range(k-2)] for i in range(1,k-1)]
for i in range(2,k):
for j in range(2,k):
b=i*j
c=''
while b>0:
c+=str(b%k)
b//=k
c=' '*(len(str(k**2))-1-len(c))+c[::-1]
a[i-1][j-1]=c
for i in range(k-1):
print(*a[i])
```
| 86,058 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Tags: implementation
Correct Solution:
```
k=int(input())
for i in range(1,k):
z,a=i,[]
for j in range(k-1):
p,s=z,""
while p:
s=str(p%k)+s
p//=k
z+=i
a.append(s)
print(*a)
```
| 86,059 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Tags: implementation
Correct Solution:
```
n = int(input())
f = lambda x: x if x < n else f(x // n) * 10 + x % n
for i in range(1, n): print(*[f(i * j) for j in range(1, n)])
```
| 86,060 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Tags: implementation
Correct Solution:
```
def convert_base(number, base):
if base < 2:
return False
remainders = []
while number > 0:
remainders.append(str(number % base))
number //= base
remainders.reverse()
return ''.join(remainders)
n = int(input())
for i in range(1,n):
s = ''
for j in range(1,n):
s += str(convert_base(i * j, n)) + ' '
s = s[:-1]
print(s)
```
| 86,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
k = int(input())
def convert_to_base_10(a):
res = []
while a:
res.append(a%k)
a//=k
return ''.join(map(str,res[::-1]))
for i in range(1,k):
output = []
for j in range(1,k):
output.append(convert_to_base_10((i*j)))
print(*output)
```
Yes
| 86,062 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
def convert(n,base):
x=""
while n>0:
x+=str(n%base)
n//=base
return x[::-1]
def func(x,n):
ans=[]
ans.append(x)
for i in range(2,n):
ans.append(int(convert(x*i,n)))
return ans
n=int(input())
ans=[]
for i in range(1,n):
if i==1:
x=[j for j in range(1,n)]
ans.append(x)
else:
x=func(i,n)
ans.append(x)
for s in ans:
print(*s)
```
Yes
| 86,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
def b(n, k):
v = ''
while n:
n, v = n // k, str(n % k) + v
return v
k = int(input())
for i in range(1, k):
print(' '.join(b(i * j, k) for j in range(1, k)))
```
Yes
| 86,064 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
radix = int(input())
for i in range(1, radix):
row = []
for j in range(1, radix):
x = i * j
digits = []
while x != 0:
digits.append(str(x % radix))
x //= radix
row.append(''.join(reversed(digits)))
print(' '.join(row))
```
Yes
| 86,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
k=int(input())
for i in range(1,k):
for j in range(1,k):
print(i*j,end=' ')
print()
```
No
| 86,066 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline
n=int(input())
l=1
for i in range(1,n):
for j in range(1,n):
print(j*l,end=' ')
print()
l+=1
```
No
| 86,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
n = int(input())
for i in range(1,11):
for j in range(1,n+1):
print(i*j,end=" ")
print()
```
No
| 86,068 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action β multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 β€ k β€ 10) β the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
n=int(input())
for i in range(1,n):
for j in range(1,n):
print(i*j,end=" ")
print()
```
No
| 86,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Tags: implementation
Correct Solution:
```
n, a, b, c =[int(x)for x in input().split()]
liter_count=0
for cc in range(c+1):
for bb in range(b+1):
if n-cc*2-bb >=0 and a*0.5 >= n-cc*2-bb :
liter_count+=1
print(liter_count)
```
| 86,070 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Tags: implementation
Correct Solution:
```
n, c, b, a = map(int, input().split())
res = 0
for a_x in range(a + 1):
for b_x in range(b + 1):
amount = n - a_x * 2 - b_x
if 0 <= amount <= c * 0.5:
res += 1
print(res)
```
| 86,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Tags: implementation
Correct Solution:
```
n,a,b,c=map(int,input().split())
ans=0
for x in range(min(c,n//2)+1):
for y in range(min(b,n-x*2)+1):
if n-x*2-y>=0 and a*0.5 >=n-x*2-y: ans+=1
print(ans)
```
| 86,072 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Tags: implementation
Correct Solution:
```
def nik(rudy,x,y,z,cot):
for i in range(z+1):
for j in range(y+1):
t = rudy - i*2 -j
if t>=0 and x*0.5 >= t:
cot+=1
return cot
rudy, x, y, z = list(map(int,input().split()))
cot = 0
print(nik(rudy,x,y,z,cot))
```
| 86,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Tags: implementation
Correct Solution:
```
n , a , b , c = map(int,input().split())
if [n,a,b,c]==[3,3,2,1]:
print(3)
exit()
elif [n,a,b,c]==[999,999,899,299]:
print(145000)
exit()
k=[0,a,b,0,c]
mul=[0,a,a+2*b,0,a+b*2+c*4]
lis=[0]*(2*n+1)
lis[0]=1
c=0
an=[]
for i in [1,2,4]:
c=0
for j in range(i,len(lis)):
if j<=i*k[i]:
# print(i*k[i],j,i,lis[j],lis[j-1])
lis[j]+=lis[j-i]
elif j<=mul[i]:
if i==2:
lis[j]=lis[a-c-1]
c+=1
else:
lis[j]+=lis[a+2*b-1-c]
c+=1
# print(lis)
print(lis[-1])
```
| 86,074 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Tags: implementation
Correct Solution:
```
n, a, b, c = map(int, input().split())
print(sum(n - i // 2 - 2 * j in range(0, b + 1) for i in range(0, a + 1, 2) for j in range(c + 1)))
```
| 86,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Tags: implementation
Correct Solution:
```
n , c , b , a = map(int,input().split())
c = c//2
k=0
for i in range(a+1):
if 2*i>n:
break
for j in range(b+1):
if 2*i+j>n:
break
if 2*i+j+c>=n:
k+=1
print(k)
```
| 86,076 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Tags: implementation
Correct Solution:
```
n, a, b, c = map(int, input().split())
count = 0
for i in range(c+1):
for j in range(b+1):
temp = n - i*2 -j
if temp>=0 and a*0.5 >= temp:
count+=1
print(count)
```
| 86,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Submitted Solution:
```
n,a,b,c=map(int,input().split());r=0
if 2*c+b+a//2<n:
print(r)
else:
for i in range(c+1):
k=n-2*i
q,w=min(b,a//2,k),max(b,a//2)
if k<=b+a//2 and k>0:
if k<=w:r+=q+1
else:
if q+1-k+w>=0:r+=q+1-k+w
elif k==0:r+=1
print(r)
```
Yes
| 86,078 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Submitted Solution:
```
def nik(rudy,pig,y,z):
temp = 0
for i in range(z+1):
for j in range(y+1):
t = rudy - i*2 -j
if t>=0 and pig*0.5 >= t:
temp+=1
print(temp)
rudy, pig, y, z = list(map(int,input().split()))
nik(rudy,pig,y,z)
```
Yes
| 86,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Submitted Solution:
```
#https://codeforces.com/problemset/problem/44/B
n,a,b,c=map(int,input().split())
r=0
for i in range(c+1):
e=n-2*i
if(e<0):break
d1=( e - min(b,e) )
d2=min(e,a//2)
r+=(d2-d1+1)*(d2-d1>=0)
print(r)
```
Yes
| 86,080 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Submitted Solution:
```
n,a,b,c=map(int,input().split())
r=0
for i in range(c+1):
e=n-2*i
if(e<0):break
d1=( e - min(b,e) )
d2=min(e,a//2)
r+=(d2-d1+1)*(d2-d1>=0)
print(r)
```
Yes
| 86,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Submitted Solution:
```
n, a, b, c =[int(x)for x in input().split()]
liter_count, volume=0, 0
for c in range(c+1):
for b in range(b+1):
volume=n-2*c+1*b
if volume >=0 and a > volume :
liter_count+=1
print(liter_count)
```
No
| 86,082 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Submitted Solution:
```
n, a, b, c =[int(x)for x in input().split()]
liter_count, volume=0, 0
for c in range(c+3):
for b in range(b+1):
volume=n-2*c+1*b
if volume >=0 and a*0.5 >= volume :
liter_count+=1
print(liter_count)
```
No
| 86,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Submitted Solution:
```
n, a, b, c = list(map(int, input().split()))
res = 0
for x in range(c):
for y in range(b):
if n - 2*x - y >= 0:
res+=1
print(res)
```
No
| 86,084 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To celebrate the opening of the Winter Computer School the organizers decided to buy in n liters of cola. However, an unexpected difficulty occurred in the shop: it turned out that cola is sold in bottles 0.5, 1 and 2 liters in volume. At that, there are exactly a bottles 0.5 in volume, b one-liter bottles and c of two-liter ones. The organizers have enough money to buy any amount of cola. What did cause the heated arguments was how many bottles of every kind to buy, as this question is pivotal for the distribution of cola among the participants (and organizers as well).
Thus, while the organizers are having the argument, discussing different variants of buying cola, the Winter School can't start. Your task is to count the number of all the possible ways to buy exactly n liters of cola and persuade the organizers that this number is too large, and if they keep on arguing, then the Winter Computer School will have to be organized in summer.
All the bottles of cola are considered indistinguishable, i.e. two variants of buying are different from each other only if they differ in the number of bottles of at least one kind.
Input
The first line contains four integers β n, a, b, c (1 β€ n β€ 10000, 0 β€ a, b, c β€ 5000).
Output
Print the unique number β the solution to the problem. If it is impossible to buy exactly n liters of cola, print 0.
Examples
Input
10 5 5 5
Output
9
Input
3 0 0 2
Output
0
Submitted Solution:
```
n,a,b,c=map(int,input().split());r=0
if 2*c+b+a//2<n:
print(r)
else:
for i in range(c+1):
k=n-2*i
q,w=min(b,a//2,k),max(b,a//2)
if k<=b+a//2 and k>0:
if k<=w:r+=q+1
else:
if q+1-k+w>=0:r+=q+1-k+w
elif k==0:r+=1
else:break
print(r)
```
No
| 86,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Tags: data structures, implementation, sortings
Correct Solution:
```
a=int(input())
b=input().split()
c=input().split()
d=input().split()
b=[int(i) for i in b]
c=[int(i) for i in c]
d=[int(i) for i in d]
b=sum(b)
c=sum(c)
d=sum(d)
print(b-c)
print(c-d)
```
| 86,086 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Tags: data structures, implementation, sortings
Correct Solution:
```
input()
x=list(map(int,input().split()))
y=list(map(int,input().split()))
z=list(map(int,input().split()))
print(sum(x)-sum(y))
print(sum(y)-sum(z))
```
| 86,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Tags: data structures, implementation, sortings
Correct Solution:
```
from collections import Counter
a = int(input())
b = list(int(x) for x in input().split())
c = list(int(x) for x in input().split())
d = list(int(x) for x in input().split())
p = list((Counter(b) - Counter(c)).elements())
for i in p:
print(i)
p = list((Counter(c) - Counter(d)).elements())
for i in p:
print(i)
```
| 86,088 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Tags: data structures, implementation, sortings
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
b=[]
for i in range(2):
b=list(map(int,input().split()))
print(sum(l)-sum(b))
l=b.copy()
```
| 86,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Tags: data structures, implementation, sortings
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
c = [int(i) for i in input().split()]
suma = sum(a)
sumb = sum(b)
sumc = sum(c)
print(suma-sumb)
print(sumb-sumc)
```
| 86,090 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Tags: data structures, implementation, sortings
Correct Solution:
```
n = int(input())
e = [sum(map(int, input().split())) for i in range(3)]
print(e[0] - e[1])
print(e[1] - e[2])
```
| 86,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Tags: data structures, implementation, sortings
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list(map(int,input().split()))
suma=sum(a)
sumb=sum(b)
sumc=sum(c)
print(suma-sumb)
print(sumb-sumc)
```
| 86,092 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Tags: data structures, implementation, sortings
Correct Solution:
```
n=input()
n=list(input().split())
m=list(input().split())
l=list(input().split())
n.sort()
m.sort()
l.sort()
i=0
while i<len(n):
if i==len(m):
q=n[i]
break
elif n[i]!=m[i]:
q=n[i]
break
i+=1
i=0
while i<len(m):
if i==len(l):
w=m[i]
break
elif m[i]!=l[i]:
w=m[i]
break
i+=1
print(q)
print(w)
```
| 86,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Submitted Solution:
```
n = int(input())
a = [num for num in input().split()]
b = [num for num in input().split()]
c = [num for num in input().split()]
x = "x"
y = "y"
a.sort()
b.sort()
c.sort()
for i in range(n-1):
if a[i] != b[i]:
x = a[i]
break
if x == "x":
x = a[n-1]
for i in range(n-2):
if b[i] != c[i]:
y = b[i]
break
if y =="y":
y = b[n-2]
print(x)
print(y)
```
Yes
| 86,094 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Submitted Solution:
```
n=int(input())
x=-1
y=-1
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list(map(int,input().split()))
a.sort()
b.sort()
c.sort()
#print("",a,'\n',b,'\n',c)
for i in range(n-1):
if x==-1 and a[i] != b[i]:
x=a[i]
if y==-1 and i<n-2 and b[i] != c[i] :
y=b[i]
if x==-1:
x=a[n-1]
if y==-1:
y=b[n-2]
print(x)
print(y)
"""
5
1 5 8 123 7
123 7 5 1
5 1 7
outputCopy
8
123
inputCopy
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
outputCopy
1
3
"""
```
Yes
| 86,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Submitted Solution:
```
n=int(input())
l=input().split(" ",n)
m=input().split(" ",n-1)
j=input().split(" ",n-2)
sum2=0
sum1=0
sum3=0
for i in range(n):
sum1 += int(l[i]);
for i in range(n-1):
sum2 +=int(m[i])
for i in range(n-2):
sum3 +=int(j[i])
print(sum1-sum2)
print(sum2-sum3)
```
Yes
| 86,096 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Submitted Solution:
```
input()
first_compile = sum(map(int, input().split()))
second_compile = sum(map(int, input().split()))
third_compile = sum(map(int, input().split()))
print(first_compile - second_compile, second_compile - third_compile)
```
Yes
| 86,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Mon Dec 23 18:19:49 2019
@author: Tuan
"""
ord1 = [int(i) for i in input().split()]
ord2 = [int(i) for i in input().split()]
ord3 = [int(i) for i in input().split()]
err1 = sum(ord1)- sum(ord2)
err2 = sum(ord2)- sum(ord3)
print(err1, err2, sep='\n')
```
No
| 86,098 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Submitted Solution:
```
n = int(input())
arr = list(map(int,input().split()))
arr1 = list(map(int,input().split()))
arr2 = list(map(int,input().split()))
a = set(arr) - set(arr1)
b = set(arr1) - set(arr2)
print(a)
print(b)
```
No
| 86,099 |
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