message stringlengths 2 59.7k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9 | instruction | 0 | 70,083 | 20 | 140,166 |
Tags: *special, implementation, number theory
Correct Solution:
```
import os,sys;from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno();self.buffer = BytesIO();self.writable = "x" in file.mode or "r" not in file.mode;self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:break
ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE));self.newlines = b.count(b"\n") + (not b);ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:os.write(self._fd, self.buffer.getvalue());self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file);self.flush = self.buffer.flush;self.writable = self.buffer.writable;self.write = lambda s: self.buffer.write(s.encode("ascii"));self.read = lambda: self.buffer.read().decode("ascii");self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
except:pass
ii1=lambda:int(sys.stdin.readline().strip()) # for interger
is1=lambda:sys.stdin.readline().strip() # for str
iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
isa=lambda:sys.stdin.readline().strip().split() # for List[str]
mod=int(1e9 + 7);from collections import *;from math import *
# abc = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
# sys.setrecursionlimit(500000)
###################### Start Here ######################
# from functools import lru_cache
# from collections import defaultdict as dd
from collections import deque as dq
s = is1()
ans = 0
for i in s:
ans += int(i)
while len(str(ans))>1:
new = 0
for i in str(ans):
new+=int(i)
ans = new
new = ''
print(ans)
``` | output | 1 | 70,083 | 20 | 140,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9
Submitted Solution:
```
"""
ID: happyn61
LANG: PYTHON3
PROB: loan
"""
from itertools import product
import itertools
#from collections import defaultdict
import sys
import heapq
from collections import deque
MOD=1000000000007
#fin = open ('loan.in', 'r')
#fout = open ('loan.out', 'w')
#print(dic["4734"])
def find(parent,i):
if parent[i] != i:
parent[i]=find(parent,parent[i])
return parent[i]
# A utility function to do union of two subsets
def union(parent,rank,xx,yy):
x=find(parent,xx)
y=find(parent,yy)
if rank[x]>rank[y]:
parent[y]=x
elif rank[y]>rank[x]:
parent[x]=y
else:
parent[y]=x
rank[x]+=1
ans=0
K=int(sys.stdin.readline().strip())
K=K%9
if K==0:
K=9
print(K)
``` | instruction | 0 | 70,084 | 20 | 140,168 |
Yes | output | 1 | 70,084 | 20 | 140,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9
Submitted Solution:
```
import math
# Returns digital root of num
def digitalRoot(num):
# If num is 0.
if (num == "0"):
return 0
# Count sum of digits under mod 9
ans = 0
for i in range (0, len(num)):
ans = (ans + int(num[i])) % 9
# If digit sum is multiple of 9, answer
# 9, else remainder with 9.
if(ans == 0):
return 9
else:
return ans % 9
# Driver method
num = input()
# Calling digitalRoot function
print (digitalRoot(num))
``` | instruction | 0 | 70,085 | 20 | 140,170 |
Yes | output | 1 | 70,085 | 20 | 140,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9
Submitted Solution:
```
n = int(input())
def root(n):
if n//10==0:
return n
return root(sum(int(i) for i in str(n)))
print(root(n))
``` | instruction | 0 | 70,086 | 20 | 140,172 |
Yes | output | 1 | 70,086 | 20 | 140,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9
Submitted Solution:
```
import time
#start_time = time.time()
#def TIME_(): print(time.time()-start_time)
import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string, heapq as h
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def getBin(): return list(map(int,list(input())))
def isInt(s): return '0' <= s[0] <= '9'
def ceil_(a,b): return a//b + (a%b > 0)
MOD = 10**9 + 7
"""
ONE
EIGHTY-ONE
DMCA
Digital Millennium Root
"""
def solve():
N = getInt()
while len(str(N)) > 1:
N = sum([int(s) for s in str(N)])
return N
#for _ in range(getInt()):
print(solve())
#solve()
#TIME_()
``` | instruction | 0 | 70,087 | 20 | 140,174 |
Yes | output | 1 | 70,087 | 20 | 140,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9
Submitted Solution:
```
import math
n=int(input())
print(int(math.sqrt(n)))
``` | instruction | 0 | 70,088 | 20 | 140,176 |
No | output | 1 | 70,088 | 20 | 140,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9
Submitted Solution:
```
import math
print(int(math.sqrt(int(input()))))
``` | instruction | 0 | 70,089 | 20 | 140,178 |
No | output | 1 | 70,089 | 20 | 140,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9
Submitted Solution:
```
import math
n = int(input())
if (math.sqrt(n)).is_integer() == True:
print(int(math.sqrt(n)))
else:
print(math.sqrt(n))
``` | instruction | 0 | 70,090 | 20 | 140,180 |
No | output | 1 | 70,090 | 20 | 140,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.
In this problem you need to find a root of a number according to this new DMCA law.
Input
The input contains a single integer a (1 ≤ a ≤ 1000000).
Output
Output the result – an integer number.
Examples
Input
1
Output
1
Input
81
Output
9
Submitted Solution:
```
print(int(input())**0.5)
``` | instruction | 0 | 70,091 | 20 | 140,182 |
No | output | 1 | 70,091 | 20 | 140,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000 | instruction | 0 | 70,115 | 20 | 140,230 |
Tags: implementation, strings
Correct Solution:
```
count=input()
#count=5
for i in range(int(count)):
n=input()
#n='::a:6:c50:83:4f:7f0d:9'
ip=n.split(':')
#print(ip)
begin=0
end=0
if (ip[0]=='' and ip[1]=='' and len(ip)>8):
begin=1
if (ip[len(ip)-2]=='' and ip[len(ip)-1]=='' and len(ip)>8):
end=1
while len(ip)<8:
index=ip.index('')
ip=ip[:index]+['']+ip[index:]
#print(ip)
for item in ip:
i=item
while len(i)<4:
i='0'+i
ip[ip.index(item)]=i
if end==1:
ip=ip[:len(ip)-1]
if begin==1:
ip=ip[1:]
print(':'.join(ip))
``` | output | 1 | 70,115 | 20 | 140,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000 | instruction | 0 | 70,116 | 20 | 140,232 |
Tags: implementation, strings
Correct Solution:
```
class CodeforcesTask250BSolution:
def __init__(self):
self.result = ''
self.n = 0
self.shorts = []
def read_input(self):
self.n = int(input())
for x in range(self.n):
self.shorts.append(input())
def process_task(self):
results = []
for short in self.shorts:
if "::" in short:
if short[-2:] != "::" and short[:2] != "::":
valid_blocks = short.count(":")
elif short[-2:] == "::" and short[:2] == "::":
valid_blocks = 0
else:
valid_blocks = short.count(":") - 1
if valid_blocks:
full = short.replace("::", ":" + ":".join(["0000" for x in range(8 - valid_blocks)]) + ":")
else:
full = "0000:" * 8
if full[-1] == ":":
full = full[:-1]
if full[0] == ":":
full = full[1:]
parts = full.split(":")
for x in range(len(parts)):
parts[x] = "0" * (4 - len(parts[x])) + parts[x]
results.append(":".join(parts))
else:
parts = short.split(":")
for x in range(len(parts)):
parts[x] = "0" * (4 - len(parts[x])) + parts[x]
results.append(":".join(parts))
self.result = "\n".join(results)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask250BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 70,116 | 20 | 140,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000 | instruction | 0 | 70,117 | 20 | 140,234 |
Tags: implementation, strings
Correct Solution:
```
def zr(s):
tmp=':0000'
if s=='::':
return tmp[1:]+7*tmp
ind=s.find("::")
if ind<0:
return s
h,k=1,0
if ind==0 or ind==len(s)-2:
if ind>0:
h=2
else:
k=1
ddc=9-s.count(":")
else:
ddc=8-s.count(':')
return (s[:ind]+ddc*tmp+s[ind+h:])[k:]
def t4(s):
ls=len(s)
if ls==4:
return s
return (4-ls)*'0'+s
def spu(s):
ns=''
fc=0
sc=s.find(':')
while sc>0:
ns+=t4(s[fc:sc])+':'
fc=sc+1
sc=s.find(':',fc)
ns+=t4(s[fc:])
return ns
n=int(input())
v=[]
for c in range(n):
v.append(spu(zr(input())))
for c in v:
print(c)
``` | output | 1 | 70,117 | 20 | 140,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000 | instruction | 0 | 70,118 | 20 | 140,236 |
Tags: implementation, strings
Correct Solution:
```
zero=str("0000")
for _ in range(int(input())):
m=0
k=input()
s=list(map(str,k.split(':')))
if k[len(k)-1]==':':
s.pop(len(s)-1)
if k[0]==':':
s.pop(0)
for i in range(len(s)):
if len(s[i])!=4 and s[i]!='':
s[i]=zero[0:4-len(s[i])]+s[i]
elif s[i]=='':
m+=1
for i in range(len(s)-1,-1,-1):
if s[i]=='' and m==1:
s[i]=zero
while len(s)<8:
s.insert(i,zero)
elif s[i]=='':
s[i]=zero
m-=1
print(":".join(s))
``` | output | 1 | 70,118 | 20 | 140,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000 | instruction | 0 | 70,119 | 20 | 140,238 |
Tags: implementation, strings
Correct Solution:
```
n = int(input())
for i in range(n):
s = input().split(":")
if(s[0] == ''): s.pop(0);
if(s[-1] == ''): s.pop(-1);
#print(s)
rq = 8 - len(s) + 1;
t = [];
for x in s:
if(x == ''):
for j in range(rq): t+=['0000']
else : t += [x]
#print(t)
u = "";
for x in t: u += '0'*(4 - len(x)) + x + ':';
print(u[:-1])
``` | output | 1 | 70,119 | 20 | 140,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000 | instruction | 0 | 70,120 | 20 | 140,240 |
Tags: implementation, strings
Correct Solution:
```
# from dust i have come, dust i will be
def solve(ip):
ch = ip.split(':')
n = len(ch)
if ch[n - 1] == '' and n - 2 >= 0 and ch[n - 2] == '':
ch.pop(n - 1)
if ch[0] == '' and n > 1 and ch[1] == '':
ch.pop(0)
extra = 8 - len(ch)
ans = []
for i in range(len(ch)):
if ch[i] != '':
while len(ch[i]) < 4:
ch[i] = '0' + ch[i]
ans.append(ch[i])
else:
ans.append('0000')
while extra > 0:
ans.append('0000')
extra -= 1
ret = ''
for i in range(len(ans)):
ret += ans[i]
if i != len(ans) - 1:
ret += ':'
return ret
n = int(input())
for i in range(n):
s = input()
print(solve(s))
``` | output | 1 | 70,120 | 20 | 140,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000 | instruction | 0 | 70,121 | 20 | 140,242 |
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
D = input()
print(':'.join(d.zfill(4) for d in D.replace('::', ':'*(9-D.count(':')))
.split(':')))
``` | output | 1 | 70,121 | 20 | 140,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000 | instruction | 0 | 70,122 | 20 | 140,244 |
Tags: implementation, strings
Correct Solution:
```
n = int(input())
list=[]
for i in range(n):
str=input()
list=str.split(":")
if(str.count("::")>0):
t=list.count("")
if(t==3):
t="0000:"*8
print(t[:len(t)-1])
elif t==2:
if str[0:2]=='::':
str=str.replace("::","0000:"*(10-len(list)))
res=""
list2=str.split(":")
for i in range(len(list2)):
res+='0'*(4-len(list2[i]))+list2[i]+":"
print(res[0:len(res)-1])
#print(str)
else:
str=str.replace("::",":0000"*(10-len(list)))
res=""
list2=str.split(":")
for i in range(len(list2)):
res+='0'*(4-len(list2[i]))+list2[i]+":"
print(res[0:len(res)-1])
#print(str)
else:
str=str.replace("::",":0000:"*(9-len(list)))
str=str.replace("::",":");
res=""
list2=str.split(":")
for i in range(len(list2)):
res+='0'*(4-len(list2[i]))+list2[i]+":"
print(res[0:len(res)-1])
# print(str)
else:
res=""
for i in range(len(list)):
res+='0'*(4-len(list[i]))+list[i]+":"
print(res[0:len(res)-1])
``` | output | 1 | 70,122 | 20 | 140,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000
Submitted Solution:
```
def s():
n = int(input())
def st(x):
if x[0] == ':':
x = x[1:]
if x[-1] == ':':
x = x[:-1]
return x
for _ in range(n):
a = st(input()).split(':')
for i in range(len(a)):
if len(a[i]) == 0:
a[i] = '0000:'*(8-len(a))+'0000'
elif len(a[i])<4:
a[i] = '0'*(4-len(a[i]))+a[i]
print(*a,sep=':')
s()
``` | instruction | 0 | 70,123 | 20 | 140,246 |
Yes | output | 1 | 70,123 | 20 | 140,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000
Submitted Solution:
```
for _ in range(int(input())):
s = input()
print(':'.join(i.zfill(4) for i in s.replace('::', ':' * (9 - s.count(':'))).split(':')))
``` | instruction | 0 | 70,124 | 20 | 140,248 |
Yes | output | 1 | 70,124 | 20 | 140,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000
Submitted Solution:
```
t=int(input())
for _ in range(t):
s=input()
if '::' not in s:
arr=s.split(':')
temp=''
for x in arr:
temp+=(4-len(x))*'0'+x+':'
print(temp[:-1])
else:
arr=s.split(':')
val=0
for x in arr:
if x!='':
val+=1
idx=arr.index('')
need=8-val
temp=''
fl=0
for x in arr:
if x!='':
temp+=(4-len(x))*'0'+x+':'
else:
if fl==0:
temp+=('0000:'*need)
fl=1
print(temp[:-1])
``` | instruction | 0 | 70,125 | 20 | 140,250 |
Yes | output | 1 | 70,125 | 20 | 140,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000
Submitted Solution:
```
for kk in range(int(input())):
str = input()
print(':'.join(i.zfill(4)
for i in str.replace('::', ':' * (9 - str.count(':'))).split(':')))
``` | instruction | 0 | 70,126 | 20 | 140,252 |
Yes | output | 1 | 70,126 | 20 | 140,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000
Submitted Solution:
```
n = int(input())
short_ips = []
for i in range(n):
short_ips.append(input())
# leading zeros can be deleted, but has to have one number remained
# Continuous sequences of 16-bit zero blocks can be shortened to "::"
answers = [[] for x in range(n)]
ans_id = -1
for ip in short_ips:
ans_id += 1
visited = False
segments = ip.split(':')
for segment in segments:
if len(segment) == 0:
if visited: continue
visited = True
missing = 8 - len(segments)
for i in range(missing):
answers[ans_id].append("0"*4)
elif len(segment) == 4:
answers[ans_id].append(segment)
else:
leading = 4 - len(segment)
answers[ans_id].append("0" *leading + segment)
for restored in answers:
print(':'.join(restored))
``` | instruction | 0 | 70,127 | 20 | 140,254 |
No | output | 1 | 70,127 | 20 | 140,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000
Submitted Solution:
```
def f(s):
res = []
sp = s.split(":")
if not sp[0]:
sp[0] = "0"
if not sp[-1]:
sp[-1] = "0"
for b in sp:
print ("-", b)
if b:
res.append("{:0>4}".format(b))
if not b:
res.extend(("0000" for i in range(8-len(sp)+1)))
return (":".join(res))
for i in range(int(input())):
tmp = input()
print(f(tmp))
#print(check("R23C55"))
#print(check("R228C494"))
#print(f("::"))
``` | instruction | 0 | 70,128 | 20 | 140,256 |
No | output | 1 | 70,128 | 20 | 140,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000
Submitted Solution:
```
def index(lena):
if lena==8:
return 4
elif lena==7:
return 9
elif lena==6:
return 14
elif lena==5:
return 19
elif lena==4:
return 24
elif lena==3:
return 29
elif lena==2:
return 34
s = int(input())
ip = []
for a in range(s):
q = input().split(':')
ip.append((q))
for a in ip:
s = []
lena = len(a)
z = '0000:0000:0000:0000:0000:0000:0000:0000'
if a==['','','']:
res = z
else:
for i in range(lena):
if a[i]=='':
if a[i+1]=='':
s.append(z[:index(lena)])
s.append(z[:4])
break
else:
s.append(z[:index(lena)])
else:
q = "{0:0>4}".format(a[i])
s.append(q)
res = ':'.join(s)
print(res)
``` | instruction | 0 | 70,129 | 20 | 140,258 |
No | output | 1 | 70,129 | 20 | 140,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full.
Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address.
Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0.
You can see examples of zero block shortenings below:
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::";
* "a56f:0000:0000:0124:0001:0000:1234:0ff0" → "a56f::0124:0001:0000:1234:0ff0";
* "a56f:0000:0000:0000:0001:0000:1234:0ff0" → "a56f:0000::0000:0001:0000:1234:0ff0";
* "a56f:00d3:0000:0124:0001:0000:0000:0000" → "a56f:00d3:0000:0124:0001::0000";
* "0000:0000:0000:0000:0000:0000:0000:0000" → "::".
It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon.
The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short.
You've got several short records of IPv6 addresses. Restore their full record.
Input
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100).
Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:".
It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
Output
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
Examples
Input
6
a56f:d3:0:0124:01:f19a:1000:00
a56f:00d3:0000:0124:0001::
a56f::0124:0001:0000:1234:0ff0
a56f:0000::0000:0001:0000:1234:0ff0
::
0ea::4d:f4:6:0
Output
a56f:00d3:0000:0124:0001:f19a:1000:0000
a56f:00d3:0000:0124:0001:0000:0000:0000
a56f:0000:0000:0124:0001:0000:1234:0ff0
a56f:0000:0000:0000:0001:0000:1234:0ff0
0000:0000:0000:0000:0000:0000:0000:0000
00ea:0000:0000:0000:004d:00f4:0006:0000
Submitted Solution:
```
count=input()
#count=5
for i in range(int(count)):
n=input()
#n='a56f:d3:0:0124:01:f19a:1000:00'
ip=n.split(':')
#print(ip)
while len(ip)<8:
ip=ip+['']
#print(ip)
for item in ip:
i=item
while len(i)<4:
i='0'+i
ip[ip.index(item)]=i
print(':'.join(ip))
``` | instruction | 0 | 70,130 | 20 | 140,260 |
No | output | 1 | 70,130 | 20 | 140,261 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S. Each character in S is either a digit (`0`, ..., `9`) or `?`.
Among the integers obtained by replacing each occurrence of `?` with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.
Since the answer can be enormous, print the count modulo 10^9+7.
Constraints
* S is a string consisting of digits (`0`, ..., `9`) and `?`.
* 1 \leq |S| \leq 10^5
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of integers satisfying the condition, modulo 10^9+7.
Examples
Input
??2??5
Output
768
Input
?44
Output
1
Input
7?4
Output
0
Input
?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???
Output
153716888 | instruction | 0 | 70,539 | 20 | 141,078 |
"Correct Solution:
```
M=10**9+7
s=input()
r=range
dp=[[0]*13,[0]*13]
dp[0][0]=1
t=0
for c in s:
for i in r(13):
if c=='?':
for j in r(10):
dp[t^1][(i*10+j)%13]+=dp[t][i]
else:
dp[t^1][(i*10+int(c))%13]+=dp[t][i]
dp[t]=[0]*13
t^=1
for i in r(13): dp[t][i]%=M
print(dp[len(s)%2][5])
``` | output | 1 | 70,539 | 20 | 141,079 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S. Each character in S is either a digit (`0`, ..., `9`) or `?`.
Among the integers obtained by replacing each occurrence of `?` with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.
Since the answer can be enormous, print the count modulo 10^9+7.
Constraints
* S is a string consisting of digits (`0`, ..., `9`) and `?`.
* 1 \leq |S| \leq 10^5
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of integers satisfying the condition, modulo 10^9+7.
Examples
Input
??2??5
Output
768
Input
?44
Output
1
Input
7?4
Output
0
Input
?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???
Output
153716888 | instruction | 0 | 70,540 | 20 | 141,080 |
"Correct Solution:
```
S=input()
l=len(S)
dp=[0 for i in range(13)]
mod=10**9+7
if S[0]=='?':
for i in range(10):
dp[i]=1
else:
dp[int(S[0])]=1
for i in range(1,l):
count=[0 for i in range(13)]
if S[i]=='?':
for j in range(13):
for k in range(10):
count[(10*j+k)%13]+=dp[j]%mod
else:
for j in range(13):
count[(10*j+int(S[i]))%13]+=dp[j]%mod
dp=count
print(dp[5]%mod)
``` | output | 1 | 70,540 | 20 | 141,081 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S. Each character in S is either a digit (`0`, ..., `9`) or `?`.
Among the integers obtained by replacing each occurrence of `?` with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.
Since the answer can be enormous, print the count modulo 10^9+7.
Constraints
* S is a string consisting of digits (`0`, ..., `9`) and `?`.
* 1 \leq |S| \leq 10^5
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of integers satisfying the condition, modulo 10^9+7.
Examples
Input
??2??5
Output
768
Input
?44
Output
1
Input
7?4
Output
0
Input
?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???
Output
153716888 | instruction | 0 | 70,542 | 20 | 141,084 |
"Correct Solution:
```
S=input()
a=[0]*13
a[0]= 1
for c in S:
dp=[0]*13
for i in range(13):
dp[(i*10)%13]=a[i]%(10**9+7)
dp+=dp
if c=='?':
for i in range(13):
a[i]=sum(dp[i+4:i+14])
else:
for i in range(13):
a[i]=dp[i+13-int(c)]
print(a[5]%(10**9+7))
``` | output | 1 | 70,542 | 20 | 141,085 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S. Each character in S is either a digit (`0`, ..., `9`) or `?`.
Among the integers obtained by replacing each occurrence of `?` with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.
Since the answer can be enormous, print the count modulo 10^9+7.
Constraints
* S is a string consisting of digits (`0`, ..., `9`) and `?`.
* 1 \leq |S| \leq 10^5
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of integers satisfying the condition, modulo 10^9+7.
Examples
Input
??2??5
Output
768
Input
?44
Output
1
Input
7?4
Output
0
Input
?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???
Output
153716888 | instruction | 0 | 70,545 | 20 | 141,090 |
"Correct Solution:
```
mod=10**9+7
s=input()
s=s[::-1]
dp=[[0]*13 for _ in range(len(s)+1)]
dp[0][0]=1
for i in range(len(s)):
p=pow(10,i,13)
if s[i]=="?":
for k in range(13):
for j in range(10):
dp[i+1][(k+p*j)%13]+=dp[i][k]
dp[i+1][(k+p*j)%13]%=mod
else:
for k in range(13):
dp[i+1][(k+p*int(s[i]))%13]+=dp[i][k]
dp[i+1][(k+p*int(s[i]))%13]%=mod
print(dp[-1][5])
``` | output | 1 | 70,545 | 20 | 141,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S. Each character in S is either a digit (`0`, ..., `9`) or `?`.
Among the integers obtained by replacing each occurrence of `?` with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.
Since the answer can be enormous, print the count modulo 10^9+7.
Constraints
* S is a string consisting of digits (`0`, ..., `9`) and `?`.
* 1 \leq |S| \leq 10^5
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of integers satisfying the condition, modulo 10^9+7.
Examples
Input
??2??5
Output
768
Input
?44
Output
1
Input
7?4
Output
0
Input
?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???
Output
153716888
Submitted Solution:
```
s = input()
keta = 10
op = [0,0,0,0,0,0,0,0,0,0,0,0,0]
np = [0,0,0,0,0,0,0,0,0,0,0,0,0]
if s[len(s)-1] == '?':
for j in range(10):
op[j%13] += 1
else:
op[int(s[len(s)-1])%13]=1
#print(op)
for i in reversed(range(0, len(s)-1) ):
myp = [0,0,0,0,0,0,0,0,0,0,0,0,0]
np = [0,0,0,0,0,0,0,0,0,0,0,0,0]
#print(s[i])
if s[i] == '?':
for j in range(10):
myp[(j*keta)%13] += 1
else:
myp[(int(s[i])*keta)%13]=1
#print("myp")
#print(myp)
for j in range(13):
for k in range(13):
np[(j+k)%13] += op[(j)%13]*myp[(k)%13]
for j in range(13):
op[j] = np[j]
keta *= 10
#print(np)
print(np[5] % (10**9+7))
``` | instruction | 0 | 70,551 | 20 | 141,102 |
No | output | 1 | 70,551 | 20 | 141,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
* 1+2*3=7
* 1*(2+3)=5
* 1*2*3=6
* (1+2)*3=9
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
Input
The input contains three integers a, b and c, each on a single line (1 ≤ a, b, c ≤ 10).
Output
Print the maximum value of the expression that you can obtain.
Examples
Input
1
2
3
Output
9
Input
2
10
3
Output
60 | instruction | 0 | 71,135 | 20 | 142,270 |
Tags: brute force, math
Correct Solution:
```
a=int(input())
b=int(input())
c=int(input())
max=0
if a+b+c>max:
max=a+b+c
if a+b*c>max:
max=a+b*c
if a*b+c>max:
max=a*b+c
if (a+b)*c>max:
max=(a+b)*c
if a*(b+c)>max:
max=a*(b+c)
if a*b*c>max:
max=a*b*c
print(max)
``` | output | 1 | 71,135 | 20 | 142,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
* 1+2*3=7
* 1*(2+3)=5
* 1*2*3=6
* (1+2)*3=9
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
Input
The input contains three integers a, b and c, each on a single line (1 ≤ a, b, c ≤ 10).
Output
Print the maximum value of the expression that you can obtain.
Examples
Input
1
2
3
Output
9
Input
2
10
3
Output
60 | instruction | 0 | 71,136 | 20 | 142,272 |
Tags: brute force, math
Correct Solution:
```
a = int(input())
b = int(input())
c = int(input())
X=[0]*6
X[0] = a+b+c
X[1] = a+b*c
X[2] = (a+b)*c
X[3] = a*b+c
X[4] = a*(b+c)
X[5] = a*b*c
print(max(X))
``` | output | 1 | 71,136 | 20 | 142,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
* 1+2*3=7
* 1*(2+3)=5
* 1*2*3=6
* (1+2)*3=9
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
Input
The input contains three integers a, b and c, each on a single line (1 ≤ a, b, c ≤ 10).
Output
Print the maximum value of the expression that you can obtain.
Examples
Input
1
2
3
Output
9
Input
2
10
3
Output
60 | instruction | 0 | 71,137 | 20 | 142,274 |
Tags: brute force, math
Correct Solution:
```
# coding: utf-8
a, b, c = [int(input()) for i in range(3)]
print(max(a + b + c, a * b * c, (a + b) * c, a * (b + c), a + (b * c), (a * b) + c))
``` | output | 1 | 71,137 | 20 | 142,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
* 1+2*3=7
* 1*(2+3)=5
* 1*2*3=6
* (1+2)*3=9
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
Input
The input contains three integers a, b and c, each on a single line (1 ≤ a, b, c ≤ 10).
Output
Print the maximum value of the expression that you can obtain.
Examples
Input
1
2
3
Output
9
Input
2
10
3
Output
60 | instruction | 0 | 71,138 | 20 | 142,276 |
Tags: brute force, math
Correct Solution:
```
a = int(input())
b = int(input())
c = int(input())
if a == 1:
if c == 1:
print(a+b+c)
else:
print((a+b)*c)
elif b == 1:
print(max((a+1)*c, a*(c+1)))
elif c == 1:
print(a*(b+1))
else:
print(a*b*c)
``` | output | 1 | 71,138 | 20 | 142,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
* 1+2*3=7
* 1*(2+3)=5
* 1*2*3=6
* (1+2)*3=9
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
Input
The input contains three integers a, b and c, each on a single line (1 ≤ a, b, c ≤ 10).
Output
Print the maximum value of the expression that you can obtain.
Examples
Input
1
2
3
Output
9
Input
2
10
3
Output
60 | instruction | 0 | 71,139 | 20 | 142,278 |
Tags: brute force, math
Correct Solution:
```
from builtins import int
a = int(input())
b = int(input())
c = int(input())
value1 = a + b + c
value2 = a + b * c
value3 = a * b + c
value4 = (a + b) * c
value5 = a * (b + c)
value6 = a * b * c
print(max(value1, value2, value3, value4, value5, value6))
``` | output | 1 | 71,139 | 20 | 142,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
* 1+2*3=7
* 1*(2+3)=5
* 1*2*3=6
* (1+2)*3=9
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
Input
The input contains three integers a, b and c, each on a single line (1 ≤ a, b, c ≤ 10).
Output
Print the maximum value of the expression that you can obtain.
Examples
Input
1
2
3
Output
9
Input
2
10
3
Output
60 | instruction | 0 | 71,140 | 20 | 142,280 |
Tags: brute force, math
Correct Solution:
```
def f(l):
a,b,c = l
return max(a+b+c,a*b*c,a*(b+c),(a+b)*c)
l = [int(input()) for _ in range(3)]
print(f(l))
``` | output | 1 | 71,140 | 20 | 142,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
* 1+2*3=7
* 1*(2+3)=5
* 1*2*3=6
* (1+2)*3=9
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
Input
The input contains three integers a, b and c, each on a single line (1 ≤ a, b, c ≤ 10).
Output
Print the maximum value of the expression that you can obtain.
Examples
Input
1
2
3
Output
9
Input
2
10
3
Output
60 | instruction | 0 | 71,141 | 20 | 142,282 |
Tags: brute force, math
Correct Solution:
```
a, b, c = int(input()), int(input()), int(input())
plus = a + b + c
multiply = a * b * c
plus_m = a + (b * c)
mult_p = a * (b + c)
plus_m_2 = (a + b)*c
print(max(plus, multiply, plus_m, plus_m_2, mult_p))
``` | output | 1 | 71,141 | 20 | 142,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
* 1+2*3=7
* 1*(2+3)=5
* 1*2*3=6
* (1+2)*3=9
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
Input
The input contains three integers a, b and c, each on a single line (1 ≤ a, b, c ≤ 10).
Output
Print the maximum value of the expression that you can obtain.
Examples
Input
1
2
3
Output
9
Input
2
10
3
Output
60 | instruction | 0 | 71,142 | 20 | 142,284 |
Tags: brute force, math
Correct Solution:
```
a = int(input())
b = int(input())
c = int(input())
_ = 0
if _ < a * b * c:
_ = a * b * c
if _ < a + b + c:
_ = a + b + c
if _ < (a + b) * c:
_ = (a + b) * c
if _ < a * (b + c):
_ = a * (b + c)
if _ < a * b + c:
_ = a * b + c
if _ < a + b * c:
_ = a + b * c
print(_)
``` | output | 1 | 71,142 | 20 | 142,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | instruction | 0 | 71,169 | 20 | 142,338 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
if n==1:
print(1)
else:
d = 0
temp = 10
c = 1
nos = 9
while (temp<n):
d = d+nos*c
temp*=10
c+=1
nos*=10
temp = temp//10
t = n-temp+1
d = d + t*(len(str(temp)))
if (str(n).count('0')==len(str(n))-1) and (str(n).count('1')==1):
d+=1
print(int(d))
``` | output | 1 | 71,169 | 20 | 142,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | instruction | 0 | 71,170 | 20 | 142,340 |
Tags: implementation, math
Correct Solution:
```
n=str(input())
a=len(n)
s=0
k='9'
if int(n)<=9:
print(n)
else:
s+=9
for i in range(2, a):
s+=(int(i*k)-10**(i-1)+1)*i
s+=(int(n)-10**(a-1)+1)*a
print(s)
``` | output | 1 | 71,170 | 20 | 142,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | instruction | 0 | 71,171 | 20 | 142,342 |
Tags: implementation, math
Correct Solution:
```
n=input()
l=len(n)-1
#print(l)
s=0
for i in range(l,-1,-1):
s=s+10**i
#print(s)
print(((int(n)+1)*int(len(n)))-s)
``` | output | 1 | 71,171 | 20 | 142,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | instruction | 0 | 71,172 | 20 | 142,344 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
s = 0
st = 9
mul = 1
while n > 0:
if n >= st:
s += st * mul
n -= st
st *= 10
mul += 1
else:
s += mul * n
n = 0
print(s)
``` | output | 1 | 71,172 | 20 | 142,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | instruction | 0 | 71,173 | 20 | 142,346 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
if n < 10 :
X = n+1- 1
elif n < 100:
X = 2*(n+1) - 11
elif n < 1000:
X = 3*(n+1)- 111
elif n < 10000:
X = 4*(n+1) - 1111
elif n < 100000:
X =5*(n+1)- 11111
elif n < 1000000:
X =6*(n+1) - 111111
elif n < 10000000:
X = 7*(n+1) - 1111111
elif n < 100000000:
X = 8*(n+1) - 11111111
elif n < 1000000000:
X = 9*(n+1) - 111111111
else :
X = 10*(n+1) - 1111111111
print(int(X))
``` | output | 1 | 71,173 | 20 | 142,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | instruction | 0 | 71,174 | 20 | 142,348 |
Tags: implementation, math
Correct Solution:
```
#maximum
n =input()
l =len(n)
n = int(n)
def dekh(x):
# if(x ==0 ):
# return 0
if(x == 1):
return 0
x1 =0
for i in range(1,x):
x1 = x1 + 9*i*(10**(i-1))
return x1
ans = dekh(l)
ans = ans + (n - 10**(l-1) + 1)*l
print(ans)
``` | output | 1 | 71,174 | 20 | 142,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | instruction | 0 | 71,175 | 20 | 142,350 |
Tags: implementation, math
Correct Solution:
```
num = int(input())
label = 0
while num > 9:
dig = len(str(num))
label = label + dig * (num - (10 ** (dig - 1)) + 1)
num = (10 ** (dig - 1)) - 1
label = label + num
print(label)
``` | output | 1 | 71,175 | 20 | 142,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | instruction | 0 | 71,176 | 20 | 142,352 |
Tags: implementation, math
Correct Solution:
```
# In this template you are not required to write code in main
import sys
inf = float("inf")
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
from math import ceil,floor,log,sqrt,factorial,pow,pi,gcd
#from bisect import bisect_left,bisect_right
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod,MOD=1000000007,998244353
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
n=int(input())
myArr=[9*i*(10**(i-1)) for i in range(1,10)]
if n<=9:
print(n)
else:
z=n-int((len(str(n))-1)*'9')
count=z*len(str(n))
for i in range(len(str(n))-1):
count+=myArr[i]
print(count)
``` | output | 1 | 71,176 | 20 | 142,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.
Submitted Solution:
```
n=int(input())
if n<10:
print(n)
else:
m=str(n)
v='1'+(len(m)-1)*'0'
j=m[0]+(len(m)-1)*'0'
m=m[1:]
b=int(m)
b=(len(m)+1)*b
v=int(v)
j=int(j)
d=(j-v)*(len(m)+1)
i=len(m)
t=i+b+10+d
while v>10:
t+=((v-(v/10))*i)
i-=1
v/=10
print(int(t))
``` | instruction | 0 | 71,177 | 20 | 142,354 |
Yes | output | 1 | 71,177 | 20 | 142,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.
Submitted Solution:
```
n = input()
t = len(n)-1
ans = 0
i = 1
while i <= t:
ans = ans + i*9*pow(10,i-1)
i = i + 1
if t > 0:
k = int(n) - int("9"*(t))
else:
k = int(n)
print(ans+k*(t+1))
``` | instruction | 0 | 71,178 | 20 | 142,356 |
Yes | output | 1 | 71,178 | 20 | 142,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.
Submitted Solution:
```
n = input()
size = len(n)
print(int(n) * size + size - int('1' * size))
``` | instruction | 0 | 71,179 | 20 | 142,358 |
Yes | output | 1 | 71,179 | 20 | 142,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Output
Print the number of digits needed to number all the books.
Examples
Input
13
Output
17
Input
4
Output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.
Submitted Solution:
```
a=int(input())
digits=0
rem=0
for i in range(len(str(a))-1):
f=(int(str(9)*(i+1))-rem)
digits+=f*(i+1)
rem += f
print(digits+(a-rem)*len(str(a)))
``` | instruction | 0 | 71,180 | 20 | 142,360 |
Yes | output | 1 | 71,180 | 20 | 142,361 |
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