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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl got bored with binary, decimal and other integer based counting systems. Recently she discovered some interesting properties about number <image>, in particular that q2 = q + 1, and she thinks it would make a good base for her new unique system. She called it "golden system". In golden system the number is a non-empty string containing 0's and 1's as digits. The decimal value of expression a0a1...an equals to <image>.
Soon Piegirl found out that this system doesn't have same properties that integer base systems do and some operations can not be performed on it. She wasn't able to come up with a fast way of comparing two numbers. She is asking for your help.
Given two numbers written in golden system notation, determine which of them has larger decimal value.
Input
Input consists of two lines β one for each number. Each line contains non-empty string consisting of '0' and '1' characters. The length of each string does not exceed 100000.
Output
Print ">" if the first number is larger, "<" if it is smaller and "=" if they are equal.
Examples
Input
1000
111
Output
<
Input
00100
11
Output
=
Input
110
101
Output
>
Note
In the first example first number equals to <image>, while second number is approximately 1.6180339882 + 1.618033988 + 1 β 5.236, which is clearly a bigger number.
In the second example numbers are equal. Each of them is β 2.618.
Submitted Solution:
```
from sys import stdin
s=list(stdin.readline().strip()[::-1])
s1=list(stdin.readline().strip()[::-1])
for i in range(min(len(s),len(s1))):
if s[i]=="1" and s1[i]=="1":
s[i]="0"
s1[i]="0"
while len(s)>0 and s[-1]=="0":
s.pop()
while len(s1)>0 and s1[-1]=="0":
s1.pop()
if len(s)==len(s1):
print("=")
exit(0)
t=True
if len(s)<len(s1):
s2=s.copy()
s=s1.copy()
s1=s2.copy()
t=False
ones=s1.count("1")
if ones==2 and len(s)==3:
print("=")
exit(0)
if ones==len(s)-1 and ones>=4:
if t :
print("<")
else:
print(">")
exit(0)
if t :
print(">")
else:
print("<")
``` | instruction | 0 | 50,001 | 20 | 100,002 |
No | output | 1 | 50,001 | 20 | 100,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl got bored with binary, decimal and other integer based counting systems. Recently she discovered some interesting properties about number <image>, in particular that q2 = q + 1, and she thinks it would make a good base for her new unique system. She called it "golden system". In golden system the number is a non-empty string containing 0's and 1's as digits. The decimal value of expression a0a1...an equals to <image>.
Soon Piegirl found out that this system doesn't have same properties that integer base systems do and some operations can not be performed on it. She wasn't able to come up with a fast way of comparing two numbers. She is asking for your help.
Given two numbers written in golden system notation, determine which of them has larger decimal value.
Input
Input consists of two lines β one for each number. Each line contains non-empty string consisting of '0' and '1' characters. The length of each string does not exceed 100000.
Output
Print ">" if the first number is larger, "<" if it is smaller and "=" if they are equal.
Examples
Input
1000
111
Output
<
Input
00100
11
Output
=
Input
110
101
Output
>
Note
In the first example first number equals to <image>, while second number is approximately 1.6180339882 + 1.618033988 + 1 β 5.236, which is clearly a bigger number.
In the second example numbers are equal. Each of them is β 2.618.
Submitted Solution:
```
from decimal import *
getcontext().prec=100
q=(Decimal(5).sqrt()+1)/2
def f(s):
ans=Decimal(0)
n=len(s)
w=Decimal(1)
for i in range(n):
if(s[n-1-i]=='1'):
ans+=w
w*=q
return ans
def equal(a,b):
if max(a,b)<1e-20:
return True
return abs(a-b)/max(a,b)<1e-20
s0=input()
s0=s0[:-1]
s1=input()
x=f(s0)
y=f(s1)
#print(x)
#print(y)
if equal(x,y):
print('=')
elif x>y:
print('>')
else:print('<')
``` | instruction | 0 | 50,002 | 20 | 100,004 |
No | output | 1 | 50,002 | 20 | 100,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl got bored with binary, decimal and other integer based counting systems. Recently she discovered some interesting properties about number <image>, in particular that q2 = q + 1, and she thinks it would make a good base for her new unique system. She called it "golden system". In golden system the number is a non-empty string containing 0's and 1's as digits. The decimal value of expression a0a1...an equals to <image>.
Soon Piegirl found out that this system doesn't have same properties that integer base systems do and some operations can not be performed on it. She wasn't able to come up with a fast way of comparing two numbers. She is asking for your help.
Given two numbers written in golden system notation, determine which of them has larger decimal value.
Input
Input consists of two lines β one for each number. Each line contains non-empty string consisting of '0' and '1' characters. The length of each string does not exceed 100000.
Output
Print ">" if the first number is larger, "<" if it is smaller and "=" if they are equal.
Examples
Input
1000
111
Output
<
Input
00100
11
Output
=
Input
110
101
Output
>
Note
In the first example first number equals to <image>, while second number is approximately 1.6180339882 + 1.618033988 + 1 β 5.236, which is clearly a bigger number.
In the second example numbers are equal. Each of them is β 2.618.
Submitted Solution:
```
from sys import stdin
s=list(stdin.readline().strip()[::-1])
s1=list(stdin.readline().strip()[::-1])
for i in range(min(len(s),len(s1))):
if s[i]=="1" and s1[i]=="1":
s[i]="0"
s1[i]="0"
while len(s)>0 and s[-1]=="0":
s.pop()
while len(s1)>0 and s1[-1]=="0":
s1.pop()
if len(s)==len(s1):
print("=")
exit(0)
t=True
if len(s)<len(s1):
s2=s.copy()
s=s1.copy()
s1=s2.copy()
t=False
ones=s1.count("1")
if ones==2 and len(s)==3:
print("=")
exit(0)
if ones==len(s)-1 and ones>=3:
if t :
print("<")
else:
print(">")
exit(0)
if t :
print(">")
else:
print("<")
``` | instruction | 0 | 50,003 | 20 | 100,006 |
No | output | 1 | 50,003 | 20 | 100,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl got bored with binary, decimal and other integer based counting systems. Recently she discovered some interesting properties about number <image>, in particular that q2 = q + 1, and she thinks it would make a good base for her new unique system. She called it "golden system". In golden system the number is a non-empty string containing 0's and 1's as digits. The decimal value of expression a0a1...an equals to <image>.
Soon Piegirl found out that this system doesn't have same properties that integer base systems do and some operations can not be performed on it. She wasn't able to come up with a fast way of comparing two numbers. She is asking for your help.
Given two numbers written in golden system notation, determine which of them has larger decimal value.
Input
Input consists of two lines β one for each number. Each line contains non-empty string consisting of '0' and '1' characters. The length of each string does not exceed 100000.
Output
Print ">" if the first number is larger, "<" if it is smaller and "=" if they are equal.
Examples
Input
1000
111
Output
<
Input
00100
11
Output
=
Input
110
101
Output
>
Note
In the first example first number equals to <image>, while second number is approximately 1.6180339882 + 1.618033988 + 1 β 5.236, which is clearly a bigger number.
In the second example numbers are equal. Each of them is β 2.618.
Submitted Solution:
```
a=list('0'+input())
b=list('0'+input())
if len(a)<len(b): a=(len(b)-len(a))*['0']+a
else: b=(len(a)-len(b))*['0']+b
def x(d):
for i in range(1, len(d)):
j=i
while d[j]=='1' and d[j-1]=='1':
d[j]=d[j-1]='0'
d[j-2]='1'
j -= 2
x(a)
x(b)
print(a)
print(b)
if a == b: print('=')
elif a < b: print('<')
else: print('>')
``` | instruction | 0 | 50,004 | 20 | 100,008 |
No | output | 1 | 50,004 | 20 | 100,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60). | instruction | 0 | 50,042 | 20 | 100,084 |
Tags: brute force, dp, expression parsing, greedy, implementation, strings
Correct Solution:
```
s = input()
ans = eval(s)
for i in range(len(s)-1):
if s[i] == "*":
k = "(" + s[0:i] + ")" + s[i:len(s)]
t = eval(k)
if t > ans:
ans = t
k = s[0:i+1] + "(" + s[i+1:len(s)] + ")"
t = eval(k)
if t > ans:
ans = t
for j in range(i+2, len(s)):
if s[j] == "*":
k = s[0:i+1] + "(" + s[i+1:j] + ")" + s[j:len(s)]
t = eval(k)
if t > ans:
ans = t
print(ans)
``` | output | 1 | 50,042 | 20 | 100,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60). | instruction | 0 | 50,043 | 20 | 100,086 |
Tags: brute force, dp, expression parsing, greedy, implementation, strings
Correct Solution:
```
s = "1*" + input() + "*1"
poses = []
for i in range(len(s)):
if (s[i] == '*'):
poses.append(i)
ans = 0
for i in poses:
for j in poses:
if (i < j):
t = s[: i + 1] + "(" + s[i + 1 : j] + ")" + s[j :]
ans = max(ans, eval(t))
print(ans)
``` | output | 1 | 50,043 | 20 | 100,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60). | instruction | 0 | 50,044 | 20 | 100,088 |
Tags: brute force, dp, expression parsing, greedy, implementation, strings
Correct Solution:
```
s = input()
mult_list = []
for i in range(len(s)):
if s[i] == '*':
mult_list.append(i)
best = eval(s)
for k in mult_list:
t = '('+s[:k]+')'+s[k:]
q = eval(t)
if q > best:
best = q
t = s[:(k+1)]+'('+s[(k+1):]+')'
q = eval(t)
if q > best:
best = q
for j in mult_list:
if j >= k:
continue
if j < k:
t = s[:(j+1)]+'('+s[(j+1):k]+')'+s[k:]
q = eval(t)
if q > best:
best = q
print(best)
``` | output | 1 | 50,044 | 20 | 100,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60). | instruction | 0 | 50,045 | 20 | 100,090 |
Tags: brute force, dp, expression parsing, greedy, implementation, strings
Correct Solution:
```
str = '1*' + input() + '*1'
n = len(str)
max_z = eval(str)
arr = []
for i in range(1, n, 2):
if str[i] == '*':
arr.append(i)
for i in range(len(arr)):
s1 = str[:arr[i]+1] + '(' + str[arr[i]+1:]
for j in range(i + 1, len(arr)):
s2 = s1[:arr[j]+1] + ')' + s1[arr[j]+1:]
max_z = max(eval(s2), max_z)
print(max_z)
``` | output | 1 | 50,045 | 20 | 100,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60). | instruction | 0 | 50,046 | 20 | 100,092 |
Tags: brute force, dp, expression parsing, greedy, implementation, strings
Correct Solution:
```
s = input()
s = "1*"+s+"*1"
ans = 0
for i in range(len(s)):
if s[i] == "*":
checker = s[0:i+1]+"("+s[i+1:]
for j in range(i+2, len(checker)):
if checker[j] == "*":
ans = max(ans, eval(checker[0:j]+")"+checker[j:]))
print(max(ans, eval(s)))
``` | output | 1 | 50,046 | 20 | 100,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60). | instruction | 0 | 50,047 | 20 | 100,094 |
Tags: brute force, dp, expression parsing, greedy, implementation, strings
Correct Solution:
```
qw = input()
g = ['*']
s = [1]
for i in range(len(qw)):
if i % 2 == 0:
s.append(int(qw[i]))
else:
g.append(qw[i])
c = 0
g.append('*')
s.append(1)
for i in range(len(g)):
if g[i] == '*':
for j in range(i + 1, len(g)):
if g[j] == '*':
a = 0
b = s[0]
for k in range(i):
if g[k] == '*':
b *= s[k + 1]
else:
a += b
b = s[k + 1]
v = a
w = b
a = 0
b = s[i + 1]
for k in range(i + 1, j):
if g[k] == '*':
b *= s[k + 1]
else:
a += b
b = s[k + 1]
b = w * (a + b)
a = v
for k in range(j, len(g)):
if g[k] == '*':
b *= s[k + 1]
else:
a += b
b = s[k + 1]
c = max(c, a + b)
print(c)
``` | output | 1 | 50,047 | 20 | 100,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60). | instruction | 0 | 50,048 | 20 | 100,096 |
Tags: brute force, dp, expression parsing, greedy, implementation, strings
Correct Solution:
```
def mul2val(s):
res = 1
for i in s.split('*'): res *= int(i)
return res
s = '1*' + input() + '*1'
muls = s.split('+')
vals = list(map(mul2val, muls))
lvals = list(map(mul2val, [mul[:-2] if len(mul)>1 else '0'
for mul in muls]))
rvals = list(map(mul2val, [mul[2:] if len(mul)>1 else '0'
for mul in muls]))
cumvals = [0] + vals
for i in range(1,len(cumvals)): cumvals[i] += cumvals[i-1]
total = cumvals[-1]
ans = total
for i,mul1 in enumerate(muls):
if len(mul1)==1: continue
for j,mul2 in enumerate(muls[i+1:],i+1):
if len(mul2)==1: continue
tval = (cumvals[i]
+ lvals[i] * (int(mul1[-1])
+ (cumvals[j] - cumvals[i+1])
+ int(mul2[0]))
* rvals[j]
+ (total-cumvals[j+1]))
if tval > ans: ans = tval
print(ans)
``` | output | 1 | 50,048 | 20 | 100,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60). | instruction | 0 | 50,049 | 20 | 100,098 |
Tags: brute force, dp, expression parsing, greedy, implementation, strings
Correct Solution:
```
s = "1*" + input() + "*1"
a = ""
n = len(s)
s = s + "+"
i = 0;
while (i < n):
if(s[i + 1] == '*'):
while(s[i + 1] == '*'):
a += s[i:i + 2]
i += 2
a += s[i:i + 2]
i += 2
else:
sum = 0
while((i < n) and (s[i + 1] == '+')):
sum += int(s[i])
i += 2
a += str(sum) + "+"
a = a[0:-1]
ans = 0
for j in range(len(a)):
for i in range(j):
if ((a[i] == '*' or a[i] == '+') and (a[j] == '*' or a[j] == '+')):
ans = max(eval(a[:i+1] + "(" + a[i + 1:j] + ")" + a[j:]), ans)
print(ans)
``` | output | 1 | 50,049 | 20 | 100,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
Submitted Solution:
```
# -*- coding: utf-8 -*-
__author__ = 'artbataev'
# from sys import setrecursionlimit
# setrecursionlimit(100000)
def countExpression(i, j):
global matr, numbers, operations
if (i == 0 or operations[i - 1] == "+"):
return matr[j] - (0 if i == 0 else matr[i-1])
cur = numbers[i]
k = i
while k < j and operations[k] == "*":
cur *= numbers[k + 1]
k += 1
if k >= j:
return cur
else:
return cur + matr[j] - matr[k]
s = input().strip()
# with open("input.txt", encoding="utf-8") as file:
# s = file.readline().strip()
numbers = list(map(int, s[::2]))
operations = s[1::2]
n = len(numbers)
matr = [0 for i in range(n)]
matr[0] = numbers[0]
for i in range(1, n):
if operations[i-1] == "+":
matr[i] = matr[i-1] + numbers[i]
else:
k = i-1
cur = numbers[i]
while k >= 0 and operations[k] == "*":
cur *= numbers[k]
k -= 1
if k < 0:
matr[i] = cur
else:
matr[i] = matr[k] + cur
maxRes = matr[n-1]
for i in range(n - 1):
if i == 0 or operations[i - 1] == "*":
for j in range(i + 1, n):
curExpr = countExpression(i, j)
i1 = i - 1
while i1 >= 0 and operations[i1] == "*":
curExpr *= numbers[i1]
i1 -= 1
j1 = j
while j1 < n-1 and operations[j1] == "*":
curExpr *= numbers[j1 + 1]
j1 += 1
if i1 >= 0:
curExpr += matr[i1]
if j1 + 1 <= n-1:
curExpr += matr[n-1] - matr[j1]
# if curExpr > maxRes:
# print(i, j)
maxRes = max(maxRes, curExpr)
print(maxRes)
``` | instruction | 0 | 50,050 | 20 | 100,100 |
Yes | output | 1 | 50,050 | 20 | 100,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
Submitted Solution:
```
if __name__ == '__main__':
s = input()
pos = [-1]
for i, x in enumerate(s):
if x == '*':
pos.append(i)
pos.append(len(s))
max_v = 0
for i in range(len(pos) - 1):
for j in range(i + 1, len(pos)):
a = pos[i] + 1
b = pos[j]
ns = s[:a] + '(' + s[a : b] + ')' + s[b:]
e = eval(ns)
if e > max_v:
max_v = e
print(max_v)
``` | instruction | 0 | 50,051 | 20 | 100,102 |
Yes | output | 1 | 50,051 | 20 | 100,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
Submitted Solution:
```
s = input()
res = eval(s)
n = len(s)
for i in range(-1, n):
if i == -1 or s[i] == '*':
for j in range(i + 1, n + 1):
if j == n or s[j] == '*':
new_s = s[0:i + 1] + "(" + s[i + 1:j] + ")" + s[j:n]
res = max(res, eval(new_s))
print(res)
``` | instruction | 0 | 50,052 | 20 | 100,104 |
Yes | output | 1 | 50,052 | 20 | 100,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
Submitted Solution:
```
s = input()
t = len(s)
res = eval(s)
for i in range(0, t):
if i != 0 and s[i - 1] != '*':
continue;
a = s[:i]+'('+s[i:]
for j in range(i, t + 2):
if j != t + 1 and a[j] != '*':
continue;
if j == t + 1:
b = a[:j] + ')'
else:
b = a[:j] + ')' + a[j:]
res = max(res, eval(b))
print(res)
``` | instruction | 0 | 50,053 | 20 | 100,106 |
Yes | output | 1 | 50,053 | 20 | 100,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
Submitted Solution:
```
s = input()
ans = 0
i = 0
while i < len(s):
if s[i] == "+":
checker = s[0:i-1]+"("+s[i-1:]
finded = False
_stars = 0
for j in range(i, len(s)):
if s[j] == "*":
_stars += 1
if(_stars > 1):
finded = True
checker = checker[:j+1]+")"+checker[j+1:]
break
else:
ans = max(ans, eval(checker[:j+1]+")"+checker[j+1:]))
i += 1
else:
if _stars == 0:
i += 1
if not finded:
checker += ")"
ans = max(ans, eval(checker))
i += 1
print(max(ans, eval(s)))
``` | instruction | 0 | 50,054 | 20 | 100,108 |
No | output | 1 | 50,054 | 20 | 100,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
Submitted Solution:
```
s = input().split('*')
ans = -1
for i in range(len(s)):
for j in range(1, len(s)):
copy = s[:i] + [str(eval('*'.join(s[i:i + j])))] + s[i + j:]
ans = max(ans, eval('*'.join(copy)))
print(ans)
``` | instruction | 0 | 50,055 | 20 | 100,110 |
No | output | 1 | 50,055 | 20 | 100,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
Submitted Solution:
```
s = input()
arr = []
first = 0
i = 1
while i < len(s):
if s[i] == '*':
arr.append([first, i])
while i < len(s) and s[i] == '*':
i += 2
first = i - 1
else:
i += 2
arr.append([first, len(s)])
mx = eval(s)
for i in arr:
sfrom = i[0]
sto = i[1]
new_s = s[:sfrom] + '(' + s[sfrom:sto] + ')' + s[sto:]
mx = max(mx, eval(new_s))
print(mx)
``` | instruction | 0 | 50,056 | 20 | 100,112 |
No | output | 1 | 50,056 | 20 | 100,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya is doing his maths homework. He has an expression of form <image>, where x1, x2, ..., xn are digits from 1 to 9, and sign <image> represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input
The first line contains expression s (1 β€ |s| β€ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
Output
In the first line print the maximum possible value of an expression.
Examples
Input
3+5*7+8*4
Output
303
Input
2+3*5
Output
25
Input
3*4*5
Output
60
Note
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
Submitted Solution:
```
from itertools import combinations
s=str(input())
ans=eval(s)
ac=[i for i,d in enumerate(s) if(d=='*')]+[len(s)]
for i,j in combinations(ac,2):
ans=max(ans,eval(('{}({}){}'.format(s[:i+1],s[i+1:j],s[j:]))))#substring prueba de todos
print(ans)
``` | instruction | 0 | 50,057 | 20 | 100,114 |
No | output | 1 | 50,057 | 20 | 100,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Champernowne constant is an irrational number. Its decimal representation starts with "0.", followed by concatenation of all positive integers in the increasing order.
You will be given a sequence S which consists of decimal digits. Your task is to write a program which computes the position of the first occurrence of S in Champernowne constant after the decimal point.
Input
The input has multiple test cases. Each line of the input has one digit sequence. The input is terminated by a line consisting only of #.
It is guaranteed that each sequence has at least one digit and its length is less than or equal to 100.
Output
For each sequence, output one decimal integer described above. You can assume each output value is less than 1016.
Examples
Input
45678
67891011
21
314159265358979
#
Output
4
6
15
2012778692735799
Input
45678
67891011
21
314159265358979
Output
4
6
15
2012778692735799
Submitted Solution:
```
# ???????????????????Β°?? ???????????Β΄?????Β°??????????????????
#?????????????? ?Β±???????????????????? ???????????????? ???????????Β΄?????Β°??????????????????, ???Β±???Β°???????Β°???????? ?????????????Β°???????? ???????????Β΄?????Β°?????????????? ?????????ΒΆ???????????????? ??????????:
# S = 123456789101112131415...
#?????????Β΄?????????? ???????????? ???????ΒΆ?Β΄???????? ???Β°?Β΄?Β°???????? ?????Β΄???????????Β΄?????Β°?????????????????? A ?? ?Β±???????????????????? ???????????Β΄?????Β°?????????????????? S (???????????Β°?????? ???Β°???????Β°???????? ?? 1).
#???????????Β°?????Β° ?Β΄?????ΒΆ???Β° ???????Β°???? ?Β΄?Β°???????? ???? stdin ?? ?????????Β΄?????? ???????????? ?? stdout.
#???????????? ???????Β΄?????? ?Β΄?Β°???????? (???? ???Β΄?????? ?????Β΄???????????Β΄?????Β°?????????????????? ???Β° ????????????, ???Β°?????????Β°???????Β°?? ?Β΄???????Β° ?????Β΄???????????Β΄?????Β°?????????????????? ??? 50 ????????????????):
#6789
#111
#???????????? ?????????Β΄?????? ?Β΄?Β°????????:
#6
#12
# ?????????????? ?????????????????? ?Β΄?? ?????????Β° ????????????, ?????? ?????? ???????????Β΄?????Β°?????????????????? ???? ???Β°???Β΄???????????? ??????????
def prov(pos,s):
ch = int(s[:pos])+1
k1 = pos
k2 = pos + len(str(ch))
while k2 < len(s):
if str(ch) == s[k1:k2]:
ch = ch +1
k1 = k2
k2 = k1 + len(str(ch))
else:
return -1
if str(ch)[:len(s)-k1]!=s[k1:]:
return -1
return 0
#?????????????? ?????????????????? ?Β΄???? ?????????????? ???????????????Β° ?????????Β° ?????? ?????????? ?? ???????????Β΄?????Β°??????????????????
def sf(ch):
#print(ch)
s = '1'
for i in range(len(str(ch)) - 1):
s = s + '0'
l = len(str(ch))
otv = (ch - int(s)) * l + 1
l = l - 1
while l > 0:
otv = otv + (int(s[:l + 1]) - int(s[:l])) * l
l = l - 1
return otv
# ??-?????? ?????Β±?????????????? ???Β°???Β±?????Β°???? ???????????? ???Β° ???????????????? ???????????Β΄?????Β°??????????????????
def last(s):
if (len(s) == 1) & (s != '0'):
return int(s)
if (s == '0'):
return 11
for pos in range(1,len(s)): # pos - ?????????Β΄?????????? ?????????Β΄???? ?????????Β°
for start in range (pos+1): # start - ???Β°???Β°???? ?????????????? ???? ???Β°???Β±?????????? ???????????Β΄?????Β°???????????????????? ?????????Β°
if start == 0: # ???????? ???????Β°???Β°????, ?????? ???????????Β΄?????Β°?????????????????? ???Β°???????Β°???????? ?? ?????????Β°
if prov(pos,s)!= -1:
return sf(int(s[start:start+pos]))
if s[:1] == '0': # ???????? ?????? ???????????Β° ????????
if s == '0'*len(s):
return sf(int('1'+s))+1
else: # ???????? ???????? ???????????? ?????????? ???? ?????Β°???Β°???Β° ???????????Β΄?????Β°??????????????????
if start + pos <len(s):
if (s[:start] == str(int(s[start:start+pos])-1)[len(str(int(s[start:start+pos])-1))-start:]):
if prov(pos, s[start:]) != -1:
return sf(int(s[start:start + pos])-1)+len(str(int(s[start:start + pos])-1))-start
else:
if (pos + start > len(s) - 1) & (start > 0): # ?Β΄???? ?????????Β°???? ???????Β΄?Β° ???????????Β° ?????Β΄?????ΒΆ???? ?????????? ?Β΄???Β° ?????????Β° ?? ?????? ???Β±?Β° ???? ????????????
ch = str(int(s[:start]) + 1)
if len(ch) > len(s[:start]):
ch = ch[1:]
ch1 = s[start:]
i = len(ch)
if ch == ch1[len(ch1) - len(ch):]:
return sf(int(ch1)-1)+len(str(int(ch1)-1))-start
while (i > 0) & (ch[:i] != ch1[len(ch1) - i:]):
i = i - 1
ich = -1
if s[:len(ch[:i])] == ch[:len(ch[:i])]:
ich = int(s[:len(s)-len(ch[:i])])
s1 = s[len(ch[:i])+1:len(s)-len((ch[:i]))]
k = len(ch[:i])+1
else:
s1 = s[len(ch[:i]):len(s)-len((ch[:i]))]
k = len(ch[:i])
if i != 0:
x = ['1', '2', '3', '4', '5', '6', '7', '8', '9']
for j in range(9):
if s1.find(x[j])!= -1:
k = k + s1.find(x[j])
break
if (ich!=-1)&(ich<int(s[k:len(s)-len((ch[:i]))]+s[:k])):
return sf(ich)
else:
return sf(int(s[k:len(s)-len((ch[:i]))]+s[:k]))+len(str(int(s[k:len(s)-len((ch[:i]))]+s[:k])))-k
x = ['1', '2', '3', '4', '5', '6', '7', '8', '9'] #???????? ???? ???Β°?????? ???Β°???????Β° ???Β° ???????????Β΄?????Β°??????????????????, ???????????? ???? ???????????????????? ???????????Β°?????????? ??????????
for i in range(9):
if s.find(x[i]) != -1:
ch = int(s[s.find(x[i]):] + s[:s.find(x[i])])
k = s.find(x[i])
pos = k
while s[k:].find(x[i]) != -1:
if int(s[s[k:].find(x[i]):] + s[:s[k:].find(x[i])]) < ch:
ch = int(s[s[k:].find(x[i]):] + s[:s[k:].find(x[i])])
pos = k
k = k + s[k:].find(x[i]) + 1
return sf(int(s[s.find(x[i]):] + s[:s.find(x[i])]))+len(s[s.find(x[i]):] + s[:s.find(x[i])]) - pos
return 0
m = [] #???Β°???????? ???????Β΄?????? ?Β΄?Β°????????
s = str(input()) #???????Β°???? ??????????????????
while s!='': # ???????????? ?Β΄?Β°???????? ?Β΄?? ???????????? ????????????
if s.isdigit(): #???????? ???????????Β° ?????Β΄?????ΒΆ???? ???????????? ?????????Β°, ???Β°???????????Β°???? ????
m.append(s)
else:
print('String is not a sequence of numbers')
for i in range(len(m)): # ?????????Β°???? ?Β΄???? ???Β°?ΒΆ?Β΄???? ????????????
print(last(m[i]))
``` | instruction | 0 | 50,486 | 20 | 100,972 |
No | output | 1 | 50,486 | 20 | 100,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Champernowne constant is an irrational number. Its decimal representation starts with "0.", followed by concatenation of all positive integers in the increasing order.
You will be given a sequence S which consists of decimal digits. Your task is to write a program which computes the position of the first occurrence of S in Champernowne constant after the decimal point.
Input
The input has multiple test cases. Each line of the input has one digit sequence. The input is terminated by a line consisting only of #.
It is guaranteed that each sequence has at least one digit and its length is less than or equal to 100.
Output
For each sequence, output one decimal integer described above. You can assume each output value is less than 1016.
Examples
Input
45678
67891011
21
314159265358979
#
Output
4
6
15
2012778692735799
Input
45678
67891011
21
314159265358979
Output
4
6
15
2012778692735799
Submitted Solution:
```
# ???????????????????Β°?? ???????????Β΄?????Β°??????????????????
#?????????????? ?Β±???????????????????? ???????????????? ???????????Β΄?????Β°??????????????????, ???Β±???Β°???????Β°???????? ?????????????Β°???????? ???????????Β΄?????Β°?????????????? ?????????ΒΆ???????????????? ??????????:
# S = 123456789101112131415...
#?????????Β΄?????????? ???????????? ???????ΒΆ?Β΄???????? ???Β°?Β΄?Β°???????? ?????Β΄???????????Β΄?????Β°?????????????????? A ?? ?Β±???????????????????? ???????????Β΄?????Β°?????????????????? S (???????????Β°?????? ???Β°???????Β°???????? ?? 1).
#???????????Β°?????Β° ?Β΄?????ΒΆ???Β° ???????Β°???? ?Β΄?Β°???????? ???? stdin ?? ?????????Β΄?????? ???????????? ?? stdout.
#???????????? ???????Β΄?????? ?Β΄?Β°???????? (???? ???Β΄?????? ?????Β΄???????????Β΄?????Β°?????????????????? ???Β° ????????????, ???Β°?????????Β°???????Β°?? ?Β΄???????Β° ?????Β΄???????????Β΄?????Β°?????????????????? ??? 50 ????????????????):
#6789
#111
#???????????? ?????????Β΄?????? ?Β΄?Β°????????:
#6
#12
# ?????????????? ?????????????????? ?Β΄?? ?????????Β° ????????????, ?????? ?????? ???????????Β΄?????Β°?????????????????? ???? ???Β°???Β΄???????????? ??????????
def prov(pos,s):
ch = int(s[:pos])+1
k1 = pos
k2 = pos + len(str(ch))
while k2 < len(s):
if str(ch) == s[k1:k2]:
ch = ch +1
k1 = k2
k2 = k1 + len(str(ch))
else:
return -1
if str(ch)[:len(s)-k1]!=s[k1:]:
return -1
return 0
#?????????????? ?????????????????? ?Β΄???? ?????????????? ???????????????Β° ?????????Β° ?????? ?????????? ?? ???????????Β΄?????Β°??????????????????
def sf(ch):
#print(ch)
s = '1'
for i in range(len(str(ch)) - 1):
s = s + '0'
l = len(str(ch))
otv = (ch - int(s)) * l + 1
l = l - 1
while l > 0:
otv = otv + (int(s[:l + 1]) - int(s[:l])) * l
l = l - 1
return otv
# ??-?????? ?????Β±?????????????? ???Β°???Β±?????Β°???? ???????????? ???Β° ???????????????? ???????????Β΄?????Β°??????????????????
def last(s):
if (len(s) == 1) & (s != '0'):
return int(s)
if (s == '0'):
return 11
for pos in range(1,len(s)): # pos - ?????????Β΄?????????? ?????????Β΄???? ?????????Β°
for start in range (pos+1): # start - ???Β°???Β°???? ?????????????? ???? ???Β°???Β±?????????? ???????????Β΄?????Β°???????????????????? ?????????Β°
if start == 0: # ???????? ???????Β°???Β°????, ?????? ???????????Β΄?????Β°?????????????????? ???Β°???????Β°???????? ?? ?????????Β°
if prov(pos,s)!= -1:
return sf(int(s[start:start+pos]))
if s[:1] == '0': # ???????? ?????? ???????????Β° ????????
if s == '0'*len(s):
return sf(int('1'+s))+1
else: # ???????? ???????? ???????????? ?????????? ???? ?????Β°???Β°???Β° ???????????Β΄?????Β°??????????????????
if start + pos <len(s):
if (s[:start] == str(int(s[start:start+pos])-1)[len(str(int(s[start:start+pos])-1))-start:]):
if prov(pos, s[start:]) != -1:
return sf(int(s[start:start + pos])-1)+len(str(int(s[start:start + pos])-1))-start
else:
if (pos + start > len(s) - 1) & (start > 0): # ?Β΄???? ?????????Β°???? ???????Β΄?Β° ???????????Β° ?????Β΄?????ΒΆ???? ?????????? ?Β΄???Β° ?????????Β° ?? ?????? ???Β±?Β° ???? ????????????
ch = str(int(s[:start]) + 1)
if len(ch) > len(s[:start]):
ch = ch[1:]
ch1 = s[start:]
i = len(ch)
if ch == ch1[len(ch1) - len(ch):]:
return sf(int(ch1)-1)+len(str(int(ch1)-1))-start
while (i > 0) & (ch[:i] != ch1[len(ch1) - i:]):
i = i - 1
ich = -1
if s[:len(ch[:i])] == ch[:len(ch[:i])]:
ich = int(s[:len(s)-len(ch[:i])])
s1 = s[len(ch[:i])+1:len(s)-len((ch[:i]))]
k = len(ch[:i])+1
else:
s1 = s[len(ch[:i]):len(s)-len((ch[:i]))]
k = len(ch[:i])
if i != 0:
x = ['1', '2', '3', '4', '5', '6', '7', '8', '9']
for j in range(9):
if s1.find(x[j])!= -1:
k = k + s1.find(x[j])
break
if (ich!=-1)&(ich<int(s[k:len(s)-len((ch[:i]))]+s[:k])):
return sf(ich)
else:
return sf(int(s[k:len(s)-len((ch[:i]))]+s[:k]))+len(str(int(s[k:len(s)-len((ch[:i]))]+s[:k])))-k
x = ['1', '2', '3', '4', '5', '6', '7', '8', '9'] #???????? ???? ???Β°?????? ???Β°???????Β° ???Β° ???????????Β΄?????Β°??????????????????, ???????????? ???? ???????????????????? ???????????Β°?????????? ??????????
for i in range(9):
if s.find(x[i]) != -1:
ch = int(s[s.find(x[i]):] + s[:s.find(x[i])])
k = s.find(x[i])
pos = k
while s[k:].find(x[i]) != -1:
if int(s[s[k:].find(x[i]):] + s[:s[k:].find(x[i])]) < ch:
ch = int(s[s[k:].find(x[i]):] + s[:s[k:].find(x[i])])
pos = k
k = k + s[k:].find(x[i]) + 1
return sf(int(s[s.find(x[i]):] + s[:s.find(x[i])]))+len(s[s.find(x[i]):] + s[:s.find(x[i])]) - pos
return 0
s = str(input()) #???????Β°???? ??????????????????
print(s)
``` | instruction | 0 | 50,487 | 20 | 100,974 |
No | output | 1 | 50,487 | 20 | 100,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Champernowne constant is an irrational number. Its decimal representation starts with "0.", followed by concatenation of all positive integers in the increasing order.
You will be given a sequence S which consists of decimal digits. Your task is to write a program which computes the position of the first occurrence of S in Champernowne constant after the decimal point.
Input
The input has multiple test cases. Each line of the input has one digit sequence. The input is terminated by a line consisting only of #.
It is guaranteed that each sequence has at least one digit and its length is less than or equal to 100.
Output
For each sequence, output one decimal integer described above. You can assume each output value is less than 1016.
Examples
Input
45678
67891011
21
314159265358979
#
Output
4
6
15
2012778692735799
Input
45678
67891011
21
314159265358979
Output
4
6
15
2012778692735799
Submitted Solution:
```
# ???????????????????Β°?? ???????????Β΄?????Β°??????????????????
#?????????????? ?Β±???????????????????? ???????????????? ???????????Β΄?????Β°??????????????????, ???Β±???Β°???????Β°???????? ?????????????Β°???????? ???????????Β΄?????Β°?????????????? ?????????ΒΆ???????????????? ??????????:
# S = 123456789101112131415...
#?????????Β΄?????????? ???????????? ???????ΒΆ?Β΄???????? ???Β°?Β΄?Β°???????? ?????Β΄???????????Β΄?????Β°?????????????????? A ?? ?Β±???????????????????? ???????????Β΄?????Β°?????????????????? S (???????????Β°?????? ???Β°???????Β°???????? ?? 1).
#???????????Β°?????Β° ?Β΄?????ΒΆ???Β° ???????Β°???? ?Β΄?Β°???????? ???? stdin ?? ?????????Β΄?????? ???????????? ?? stdout.
#???????????? ???????Β΄?????? ?Β΄?Β°???????? (???? ???Β΄?????? ?????Β΄???????????Β΄?????Β°?????????????????? ???Β° ????????????, ???Β°?????????Β°???????Β°?? ?Β΄???????Β° ?????Β΄???????????Β΄?????Β°?????????????????? ??? 50 ????????????????):
#6789
#111
#???????????? ?????????Β΄?????? ?Β΄?Β°????????:
#6
#12
# ?????????????? ?????????????????? ?Β΄?? ?????????Β° ????????????, ?????? ?????? ???????????Β΄?????Β°?????????????????? ???? ???Β°???Β΄???????????? ??????????
def prov(pos,s):
ch = int(s[:pos])+1
k1 = pos
k2 = pos + len(str(ch))
while k2 < len(s):
if str(ch) == s[k1:k2]:
ch = ch +1
k1 = k2
k2 = k1 + len(str(ch))
else:
return -1
if str(ch)[:len(s)-k1]!=s[k1:]:
return -1
return 0
#?????????????? ?????????????????? ?Β΄???? ?????????????? ???????????????Β° ?????????Β° ?????? ?????????? ?? ???????????Β΄?????Β°??????????????????
def sf(ch):
#print(ch)
s = '1'
for i in range(len(str(ch)) - 1):
s = s + '0'
l = len(str(ch))
otv = (ch - int(s)) * l + 1
l = l - 1
while l > 0:
otv = otv + (int(s[:l + 1]) - int(s[:l])) * l
l = l - 1
return otv
# ??-?????? ?????Β±?????????????? ???Β°???Β±?????Β°???? ???????????? ???Β° ???????????????? ???????????Β΄?????Β°??????????????????
def last(s):
if (len(s) == 1) & (s != '0'):
return int(s)
if (s == '0'):
return 11
for pos in range(1,len(s)): # pos - ?????????Β΄?????????? ?????????Β΄???? ?????????Β°
for start in range (pos+1): # start - ???Β°???Β°???? ?????????????? ???? ???Β°???Β±?????????? ???????????Β΄?????Β°???????????????????? ?????????Β°
if start == 0: # ???????? ???????Β°???Β°????, ?????? ???????????Β΄?????Β°?????????????????? ???Β°???????Β°???????? ?? ?????????Β°
if prov(pos,s)!= -1:
return sf(int(s[start:start+pos]))
if s[:1] == '0': # ???????? ?????? ???????????Β° ????????
if s == '0'*len(s):
return sf(int('1'+s))+1
else: # ???????? ???????? ???????????? ?????????? ???? ?????Β°???Β°???Β° ???????????Β΄?????Β°??????????????????
if start + pos <len(s):
if (s[:start] == str(int(s[start:start+pos])-1)[len(str(int(s[start:start+pos])-1))-start:]):
if prov(pos, s[start:]) != -1:
return sf(int(s[start:start + pos])-1)+len(str(int(s[start:start + pos])-1))-start
else:
if (pos + start > len(s) - 1) & (start > 0): # ?Β΄???? ?????????Β°???? ???????Β΄?Β° ???????????Β° ?????Β΄?????ΒΆ???? ?????????? ?Β΄???Β° ?????????Β° ?? ?????? ???Β±?Β° ???? ????????????
ch = str(int(s[:start]) + 1)
if len(ch) > len(s[:start]):
ch = ch[1:]
ch1 = s[start:]
i = len(ch)
if ch == ch1[len(ch1) - len(ch):]:
return sf(int(ch1)-1)+len(str(int(ch1)-1))-start
while (i > 0) & (ch[:i] != ch1[len(ch1) - i:]):
i = i - 1
ich = -1
if s[:len(ch[:i])] == ch[:len(ch[:i])]:
ich = int(s[:len(s)-len(ch[:i])])
s1 = s[len(ch[:i])+1:len(s)-len((ch[:i]))]
k = len(ch[:i])+1
else:
s1 = s[len(ch[:i]):len(s)-len((ch[:i]))]
k = len(ch[:i])
if i != 0:
x = ['1', '2', '3', '4', '5', '6', '7', '8', '9']
for j in range(9):
if s1.find(x[j])!= -1:
k = k + s1.find(x[j])
break
if (ich!=-1)&(ich<int(s[k:len(s)-len((ch[:i]))]+s[:k])):
return sf(ich)
else:
return sf(int(s[k:len(s)-len((ch[:i]))]+s[:k]))+len(str(int(s[k:len(s)-len((ch[:i]))]+s[:k])))-k
x = ['1', '2', '3', '4', '5', '6', '7', '8', '9'] #???????? ???? ???Β°?????? ???Β°???????Β° ???Β° ???????????Β΄?????Β°??????????????????, ???????????? ???? ???????????????????? ???????????Β°?????????? ??????????
for i in range(9):
if s.find(x[i]) != -1:
ch = int(s[s.find(x[i]):] + s[:s.find(x[i])])
k = s.find(x[i])
pos = k
while s[k:].find(x[i]) != -1:
if int(s[s[k:].find(x[i]):] + s[:s[k:].find(x[i])]) < ch:
ch = int(s[s[k:].find(x[i]):] + s[:s[k:].find(x[i])])
pos = k
k = k + s[k:].find(x[i]) + 1
return sf(int(s[s.find(x[i]):] + s[:s.find(x[i])]))+len(s[s.find(x[i]):] + s[:s.find(x[i])]) - pos
return 0
m = [] #???Β°???????? ???????Β΄?????? ?Β΄?Β°????????
s = str(input()) #???????Β°???? ??????????????????
while s!='#': # ???????????? ?Β΄?Β°???????? ?Β΄?? ???????????? ????????????
if s.isdigit(): #???????? ???????????Β° ?????Β΄?????ΒΆ???? ???????????? ?????????Β°, ???Β°???????????Β°???? ????
m.append(s)
else:
print('String is not a sequence of numbers')
s = str(input())
for i in range(len(m)): # ?????????Β°???? ?Β΄???? ???Β°?ΒΆ?Β΄???? ????????????
print(last(m[i]))
``` | instruction | 0 | 50,488 | 20 | 100,976 |
No | output | 1 | 50,488 | 20 | 100,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | instruction | 0 | 50,671 | 20 | 101,342 |
Tags: implementation, math
Correct Solution:
```
__author__ = 'Darren'
def solve():
a = int(input())
digit_sum = 0
for base in range(2, a):
a_copy = a
while a_copy:
digit_sum += a_copy % base
a_copy //= base
g = gcd(digit_sum, a - 2)
print('%d/%d' % (digit_sum // g, (a - 2) // g))
def gcd(a, b) -> int:
"""
Compute the greatest common divisor of two integers using the
Euclidean algorithm.
"""
while b:
a %= b
a, b = b, a
return a
if __name__ == '__main__':
solve()
``` | output | 1 | 50,671 | 20 | 101,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | instruction | 0 | 50,672 | 20 | 101,344 |
Tags: implementation, math
Correct Solution:
```
from math import gcd
n = int(input())
s = 0
for i in range(2, n):
copy = n
while copy > 0:
s += copy % i
copy //= i
a = s
b = n - 2
g = gcd(a, b)
print(f"{a // g}/{b // g}")
``` | output | 1 | 50,672 | 20 | 101,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | instruction | 0 | 50,673 | 20 | 101,346 |
Tags: implementation, math
Correct Solution:
```
import math
def f(n, i):
r = 0
while n:
r += n % i
n //= i
return r
n = int(input() )
s = 0
for i in range(2, n):
s += f(n, i)
g = math.gcd(s, n-2)
print(''.join([str(s//g), '/', str((n-2)//g) ] ) )
``` | output | 1 | 50,673 | 20 | 101,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | instruction | 0 | 50,674 | 20 | 101,348 |
Tags: implementation, math
Correct Solution:
```
import os
import math
import sys
debug = True
if debug and os.path.exists("input.in"):
input = open("input.in", "r").readline
else:
debug = False
input = sys.stdin.readline
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = input()
return (list(s[:len(s) - 1]))
def invr():
return (map(int, input().split()))
test_count = 1
if debug:
test_count = int(input())
for t in range(test_count):
if debug:
print("Test Case #", t + 1)
N = inp()
sum = 0
for i in range(2, N):
thisSum = 0
num = N
while num >= i:
thisSum += num % i
num = num // i
thisSum += num
sum += thisSum
# print(i, thisSum, sum)
gcdValue = math.gcd(sum, N - 2)
print(str(sum // gcdValue) + "/" + str((N - 2) // gcdValue))
``` | output | 1 | 50,674 | 20 | 101,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | instruction | 0 | 50,675 | 20 | 101,350 |
Tags: implementation, math
Correct Solution:
```
'''
5 2
'''
def sum_in_base(num, base):
ans = 0
while num != 0:
ans += num % base
num //= base
return ans
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
A, sumOfAll = int(input()), 0
for i in range(2, A):
sumOfAll += sum_in_base(A, i)
temp = gcd(sumOfAll, A - 2)
print('{}/{}'.format(sumOfAll // temp, (A - 2) // temp))
``` | output | 1 | 50,675 | 20 | 101,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | instruction | 0 | 50,676 | 20 | 101,352 |
Tags: implementation, math
Correct Solution:
```
def main():
from math import gcd
a, x = int(input()), 0
for b in range(2, a):
t = a
while t:
x += t % b
t //= b
a -= 2
g = gcd(x, a)
print(x // g, a // g, sep="/")
if __name__ == "__main__":
main()
``` | output | 1 | 50,676 | 20 | 101,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | instruction | 0 | 50,677 | 20 | 101,354 |
Tags: implementation, math
Correct Solution:
```
from math import gcd
A = int(input())
s = 0
for i in range(2, A):
x = A
while x!=0:
s += x%i
x = x//i
g = gcd(s, A-2)
print(str(s//g)+"/"+str((A-2)//g))
``` | output | 1 | 50,677 | 20 | 101,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | instruction | 0 | 50,678 | 20 | 101,356 |
Tags: implementation, math
Correct Solution:
```
a = int(input())
import math
g = 0
for k in range(2,a):
x=a
l=x%k
while x>1:
x=x//k
g+=x%k
g+=l
u = math.gcd(g,a-2)
print(str(g//u)+'/'+str((a-2)//u))
``` | output | 1 | 50,678 | 20 | 101,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
Submitted Solution:
```
def gcd(a, b):
if b==0 : return a
else: return gcd(b, a%b)
a = int(input())
sum = 0
for i in range(2, a):
t = a
while t != 0:
sum += t%i
t //= i
d = gcd(sum, a-2)
print(str(sum//d)+'/'+str((a-2)//d))
``` | instruction | 0 | 50,679 | 20 | 101,358 |
Yes | output | 1 | 50,679 | 20 | 101,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
Submitted Solution:
```
import math
n= int(input())
jam = 0
l = 0
g = 0
for i in range(2, n) :
l=n
while (l != 0) :
jam = jam + (l%i)
l//=i
g=math.gcd(jam,n-2)
print(f"{jam//g}/{(n-2)//g}")
``` | instruction | 0 | 50,680 | 20 | 101,360 |
Yes | output | 1 | 50,680 | 20 | 101,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
Submitted Solution:
```
import copy
#this function transforms any number to any base
def transformToBase(number, base):
#deepcopy the number because its passed by reference
#and we don't want to lose it
newNr = copy.deepcopy(number)
rest = []
#while newNr is not 0
while newNr:
#store in rest the reminder of division
rest.append(newNr%base)
#and in newNr store the integer part
newNr //= base
#since we are not interested in transformed number, but only in its digits
#we don't reverse the number but just return it as it is
return rest
#we want to find sum of the digits of one number given in a list
def findDigitsSum(numbers):
digitSum = 0
#iterate over list
for number in numbers:
#calculate sum
digitSum += number
return digitSum
#sum of all numbers from base 2 till base n-1
def getFinalSum(number):
i = 2
totalSum = 0
#go from 2 to n-1
while i != number:
#add to sum the value returned bu function findDigitsSum
totalSum += findDigitsSum(transformToBase(number, i))
i += 1
#return total sum
return totalSum
#we want to print the result as an ireductible fraction
def makeFractionIreductible(numerator, denominator):
#from 2 till numerator, because we know for sure it is bigger than denominator
for i in range(2, numerator):
#if the rest of both numerator and denominator after division is 0
while numerator % i == 0 and denominator % i == 0:
#simplify both numerator and denumerator
numerator /= i
denominator /= i
#return simplified numerator and denominator
return numerator, denominator
def main():
#det input number
number = int(input())
#calculate final sum
finalSum = getFinalSum(number)
#get simplified numerator and denominator
numerator, denominator = makeFractionIreductible(finalSum,number - 2)
#print result in fraction form
result = str(int(numerator)) + "/" + str(int(denominator))
print(result)
main()
``` | instruction | 0 | 50,681 | 20 | 101,362 |
Yes | output | 1 | 50,681 | 20 | 101,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
Submitted Solution:
```
n = int(input());d=0
for b in range(2,n):
t = n
while t>0:
d+=t%b
t = t//b
x = d;y = n-2
while y > 0:
x,y = y,x%y
print ("%d/%d"%(d/x, (n-2)/x))
``` | instruction | 0 | 50,682 | 20 | 101,364 |
Yes | output | 1 | 50,682 | 20 | 101,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
Submitted Solution:
```
A = int(input())
def base_(i,x):
ans = []
while True:
ans.append(x%i)
x = x//i
if not x:
break
return ans
ans = 0
for i in range(2,A):
ans = ans + sum(base_(i,A))
print(str(ans)+"/"+str(A-2))
``` | instruction | 0 | 50,683 | 20 | 101,366 |
No | output | 1 | 50,683 | 20 | 101,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
Submitted Solution:
```
def f(x,b):
m=0
for n in range(10000):
m+=x%b
u=int(x/b)
# print(x%b)
x=u
if x<b:
m+=x%b
break
return m
#print(f(5,3))
m=0
x=int(input())
for n in range(2,x):
m+=f(x,n)
print("{}/{}".format(m,(x-2)))
``` | instruction | 0 | 50,684 | 20 | 101,368 |
No | output | 1 | 50,684 | 20 | 101,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
Submitted Solution:
```
import sys
import math
A = int(sys.stdin.readline())
res = 0
for i in range(2, A):
k = A
while(k > 0):
res += k % i
k = int(k / i)
print(str(res) + "/" + str(A - 2))
``` | instruction | 0 | 50,685 | 20 | 101,370 |
No | output | 1 | 50,685 | 20 | 101,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input
Input contains one integer number A (3 β€ A β€ 1000).
Output
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Examples
Input
5
Output
7/3
Input
3
Output
2/1
Note
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
Submitted Solution:
```
a = int(input())
sum = 0
i = 2
while i < a:
b = a
while b != 0:
sum += int(b%i)
b = (b - b%i) / i
i += 1
print(sum, '/', a-2)
``` | instruction | 0 | 50,686 | 20 | 101,372 |
No | output | 1 | 50,686 | 20 | 101,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.
This is an interactive problem.
Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.
The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by aβ_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that xβ_{k} z=y. Guess the password and break into the system.
Input
The first line of input contains a single integer t (1β€ tβ€ 10 000) denoting the number of test cases. t test cases follow.
The first line of each test case contains two integers n (1β€ nβ€ 2β
10^5) and k (k=2).
It is guaranteed that the sum of n over all test cases does not exceed 2β
10^5.
Interaction
For each test case, first read two integers n and k. Then you may ask up to n queries.
For each query, print a single integer y (0β€ yβ€ 2β
10^7). Let the current password be x. After that, read an integer r.
If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.
Else, you will read r=0. At this moment the password is changed to a number z such that xβ_{k} z=y.
After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.
To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.
Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.
Hacks:
To use hacks, use the following format of tests:
The first line should contain a single integer t (1β€ tβ€ 10 000) β the number of test cases.
The first and only line of each test case should contain two integers n (1β€ nβ€ 2β
10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.
You must ensure that the sum of n over all test cases does not exceed 2β
10^5.
Example
Input
1
5 2
0
0
1
Output
3
4
5
Note
In the example test case, the hidden password is 2.
The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2β_2 1=3.
The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1β_2 5=4.
The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.
Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive. | instruction | 0 | 50,743 | 20 | 101,486 |
Tags: bitmasks, constructive algorithms, interactive, math
Correct Solution:
```
try:
import sys
def solve():
n,k = map(int,input().split())
print(0)
sys.stdout.flush()
glad = int(input())
if(glad == 1):
return
for i in range(1 , n):
print(i^(i-1))
sys.stdout.flush()
glad = int(input())
if (glad == 1):
break
for _ in range(int(input())):
solve()
except EOFError:
print(" ")
``` | output | 1 | 50,743 | 20 | 101,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.
This is an interactive problem.
Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.
The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by aβ_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that xβ_{k} z=y. Guess the password and break into the system.
Input
The first line of input contains a single integer t (1β€ tβ€ 10 000) denoting the number of test cases. t test cases follow.
The first line of each test case contains two integers n (1β€ nβ€ 2β
10^5) and k (k=2).
It is guaranteed that the sum of n over all test cases does not exceed 2β
10^5.
Interaction
For each test case, first read two integers n and k. Then you may ask up to n queries.
For each query, print a single integer y (0β€ yβ€ 2β
10^7). Let the current password be x. After that, read an integer r.
If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.
Else, you will read r=0. At this moment the password is changed to a number z such that xβ_{k} z=y.
After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.
To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.
Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.
Hacks:
To use hacks, use the following format of tests:
The first line should contain a single integer t (1β€ tβ€ 10 000) β the number of test cases.
The first and only line of each test case should contain two integers n (1β€ nβ€ 2β
10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.
You must ensure that the sum of n over all test cases does not exceed 2β
10^5.
Example
Input
1
5 2
0
0
1
Output
3
4
5
Note
In the example test case, the hidden password is 2.
The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2β_2 1=3.
The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1β_2 5=4.
The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.
Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive. | instruction | 0 | 50,744 | 20 | 101,488 |
Tags: bitmasks, constructive algorithms, interactive, math
Correct Solution:
```
from sys import stdout
import sys
for _ in range(int(input())):
n, k = map(int, input().split())
cum = 0
for last in range(n):
# print('c', cum)
print(last ^ cum)
stdout.flush()
cum = last
if sys.stdin.readline().strip() == '1':
break
``` | output | 1 | 50,744 | 20 | 101,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.
This is an interactive problem.
Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.
The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by aβ_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that xβ_{k} z=y. Guess the password and break into the system.
Input
The first line of input contains a single integer t (1β€ tβ€ 10 000) denoting the number of test cases. t test cases follow.
The first line of each test case contains two integers n (1β€ nβ€ 2β
10^5) and k (k=2).
It is guaranteed that the sum of n over all test cases does not exceed 2β
10^5.
Interaction
For each test case, first read two integers n and k. Then you may ask up to n queries.
For each query, print a single integer y (0β€ yβ€ 2β
10^7). Let the current password be x. After that, read an integer r.
If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.
Else, you will read r=0. At this moment the password is changed to a number z such that xβ_{k} z=y.
After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.
To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.
Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.
Hacks:
To use hacks, use the following format of tests:
The first line should contain a single integer t (1β€ tβ€ 10 000) β the number of test cases.
The first and only line of each test case should contain two integers n (1β€ nβ€ 2β
10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.
You must ensure that the sum of n over all test cases does not exceed 2β
10^5.
Example
Input
1
5 2
0
0
1
Output
3
4
5
Note
In the example test case, the hidden password is 2.
The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2β_2 1=3.
The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1β_2 5=4.
The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.
Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive. | instruction | 0 | 50,745 | 20 | 101,490 |
Tags: bitmasks, constructive algorithms, interactive, math
Correct Solution:
```
# Not my code
# https://codeforces.com/contest/1543/submission/121634285
from sys import stdout
import sys
for _ in range(int(input())):
n, k = map(int, input().split())
cum = 0
for last in range(n):
# print('c', cum)
stdout.write(f"{last ^ cum}\n")
stdout.flush()
cum = last
if sys.stdin.readline() == '1':
break
``` | output | 1 | 50,745 | 20 | 101,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.
This is an interactive problem.
Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.
The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by aβ_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that xβ_{k} z=y. Guess the password and break into the system.
Input
The first line of input contains a single integer t (1β€ tβ€ 10 000) denoting the number of test cases. t test cases follow.
The first line of each test case contains two integers n (1β€ nβ€ 2β
10^5) and k (k=2).
It is guaranteed that the sum of n over all test cases does not exceed 2β
10^5.
Interaction
For each test case, first read two integers n and k. Then you may ask up to n queries.
For each query, print a single integer y (0β€ yβ€ 2β
10^7). Let the current password be x. After that, read an integer r.
If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.
Else, you will read r=0. At this moment the password is changed to a number z such that xβ_{k} z=y.
After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.
To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.
Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.
Hacks:
To use hacks, use the following format of tests:
The first line should contain a single integer t (1β€ tβ€ 10 000) β the number of test cases.
The first and only line of each test case should contain two integers n (1β€ nβ€ 2β
10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.
You must ensure that the sum of n over all test cases does not exceed 2β
10^5.
Example
Input
1
5 2
0
0
1
Output
3
4
5
Note
In the example test case, the hidden password is 2.
The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2β_2 1=3.
The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1β_2 5=4.
The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.
Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive.
Submitted Solution:
```
from collections import defaultdict,deque
from bisect import bisect_left
import math,sys
def inp():
l=list(map(int,input().split()))
return l
for _ in range(int(input())):
n,k=inp()
ans=0
for i in range(1,n+1):
if i==0:
print(0)
else:
print(i^(i-1))
sys.stdout.flush()
z,=inp()
if z==1:
break
``` | instruction | 0 | 50,746 | 20 | 101,492 |
No | output | 1 | 50,746 | 20 | 101,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.
This is an interactive problem.
Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.
The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by aβ_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that xβ_{k} z=y. Guess the password and break into the system.
Input
The first line of input contains a single integer t (1β€ tβ€ 10 000) denoting the number of test cases. t test cases follow.
The first line of each test case contains two integers n (1β€ nβ€ 2β
10^5) and k (k=2).
It is guaranteed that the sum of n over all test cases does not exceed 2β
10^5.
Interaction
For each test case, first read two integers n and k. Then you may ask up to n queries.
For each query, print a single integer y (0β€ yβ€ 2β
10^7). Let the current password be x. After that, read an integer r.
If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.
Else, you will read r=0. At this moment the password is changed to a number z such that xβ_{k} z=y.
After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.
To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.
Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.
Hacks:
To use hacks, use the following format of tests:
The first line should contain a single integer t (1β€ tβ€ 10 000) β the number of test cases.
The first and only line of each test case should contain two integers n (1β€ nβ€ 2β
10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.
You must ensure that the sum of n over all test cases does not exceed 2β
10^5.
Example
Input
1
5 2
0
0
1
Output
3
4
5
Note
In the example test case, the hidden password is 2.
The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2β_2 1=3.
The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1β_2 5=4.
The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.
Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive.
Submitted Solution:
```
from collections import defaultdict,deque
from bisect import bisect_left
import math,sys
def inp():
l=list(map(int,input().split()))
return l
for _ in range(int(input())):
n,k=inp()
ans=0
for i in range(1,n+1):
ans^=i
print(ans)
sys.stdout.flush()
z,=inp()
if z==1:
break
``` | instruction | 0 | 50,747 | 20 | 101,494 |
No | output | 1 | 50,747 | 20 | 101,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.
This is an interactive problem.
Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.
The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by aβ_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that xβ_{k} z=y. Guess the password and break into the system.
Input
The first line of input contains a single integer t (1β€ tβ€ 10 000) denoting the number of test cases. t test cases follow.
The first line of each test case contains two integers n (1β€ nβ€ 2β
10^5) and k (k=2).
It is guaranteed that the sum of n over all test cases does not exceed 2β
10^5.
Interaction
For each test case, first read two integers n and k. Then you may ask up to n queries.
For each query, print a single integer y (0β€ yβ€ 2β
10^7). Let the current password be x. After that, read an integer r.
If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.
Else, you will read r=0. At this moment the password is changed to a number z such that xβ_{k} z=y.
After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.
To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.
Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.
Hacks:
To use hacks, use the following format of tests:
The first line should contain a single integer t (1β€ tβ€ 10 000) β the number of test cases.
The first and only line of each test case should contain two integers n (1β€ nβ€ 2β
10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.
You must ensure that the sum of n over all test cases does not exceed 2β
10^5.
Example
Input
1
5 2
0
0
1
Output
3
4
5
Note
In the example test case, the hidden password is 2.
The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2β_2 1=3.
The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1β_2 5=4.
The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.
Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
from itertools import *
from heapq import *
from bisect import *
from io import BytesIO, IOBase
from typing import overload
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M = 998244353
EPS = 1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def Get(x):
print(x,flush = True)
return Int()
for _ in range(Int()):
n,k = value()
cur = 0
for i in range(n):
cur ^= i
res = Get(cur)
if(res == 1): break
elif(res == -1): exit()
``` | instruction | 0 | 50,748 | 20 | 101,496 |
No | output | 1 | 50,748 | 20 | 101,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.
This is an interactive problem.
Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.
The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by aβ_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that xβ_{k} z=y. Guess the password and break into the system.
Input
The first line of input contains a single integer t (1β€ tβ€ 10 000) denoting the number of test cases. t test cases follow.
The first line of each test case contains two integers n (1β€ nβ€ 2β
10^5) and k (k=2).
It is guaranteed that the sum of n over all test cases does not exceed 2β
10^5.
Interaction
For each test case, first read two integers n and k. Then you may ask up to n queries.
For each query, print a single integer y (0β€ yβ€ 2β
10^7). Let the current password be x. After that, read an integer r.
If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.
Else, you will read r=0. At this moment the password is changed to a number z such that xβ_{k} z=y.
After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.
To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.
Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.
Hacks:
To use hacks, use the following format of tests:
The first line should contain a single integer t (1β€ tβ€ 10 000) β the number of test cases.
The first and only line of each test case should contain two integers n (1β€ nβ€ 2β
10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.
You must ensure that the sum of n over all test cases does not exceed 2β
10^5.
Example
Input
1
5 2
0
0
1
Output
3
4
5
Note
In the example test case, the hidden password is 2.
The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2β_2 1=3.
The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1β_2 5=4.
The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.
Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive.
Submitted Solution:
```
import sys
#comment these out later
#sys.stdin = open("in.in", "r")
#sys.stdout = open("out.out", "w")
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
prev = 0
for a in range(n):
print(a^prev)
prev ^= a
sys.stdout.flush()
if int(input()) == 1:
break
main()
``` | instruction | 0 | 50,749 | 20 | 101,498 |
No | output | 1 | 50,749 | 20 | 101,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sereja loves number sequences very much. That's why he decided to make himself a new one following a certain algorithm.
Sereja takes a blank piece of paper. Then he starts writing out the sequence in m stages. Each time he either adds a new number to the end of the sequence or takes l first elements of the current sequence and adds them c times to the end. More formally, if we represent the current sequence as a1, a2, ..., an, then after we apply the described operation, the sequence transforms into a1, a2, ..., an[, a1, a2, ..., al] (the block in the square brackets must be repeated c times).
A day has passed and Sereja has completed the sequence. He wonders what are the values of some of its elements. Help Sereja.
Input
The first line contains integer m (1 β€ m β€ 105) β the number of stages to build a sequence.
Next m lines contain the description of the stages in the order they follow. The first number in the line is a type of stage (1 or 2). Type 1 means adding one number to the end of the sequence, in this case the line contains integer xi (1 β€ xi β€ 105) β the number to add. Type 2 means copying a prefix of length li to the end ci times, in this case the line further contains two integers li, ci (1 β€ li β€ 105, 1 β€ ci β€ 104), li is the length of the prefix, ci is the number of copyings. It is guaranteed that the length of prefix li is never larger than the current length of the sequence.
The next line contains integer n (1 β€ n β€ 105) β the number of elements Sereja is interested in. The next line contains the numbers of elements of the final sequence Sereja is interested in. The numbers are given in the strictly increasing order. It is guaranteed that all numbers are strictly larger than zero and do not exceed the length of the resulting sequence. Consider the elements of the final sequence numbered starting from 1 from the beginning to the end of the sequence.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print the elements that Sereja is interested in, in the order in which their numbers occur in the input.
Examples
Input
6
1 1
1 2
2 2 1
1 3
2 5 2
1 4
16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Output
1 2 1 2 3 1 2 1 2 3 1 2 1 2 3 4 | instruction | 0 | 50,841 | 20 | 101,682 |
Tags: binary search, brute force
Correct Solution:
```
MAX_LEN = int(1e5)
n = int(input())
a = [list(map(int, input().split())) for i in range(n)]
m = int(input())
b = list(map(lambda x:int(x)-1, input().split()))
curr, k, c, res = 0, 0, [], []
for i in range(n):
t = a[i]
last = curr
if t[0] == 1:
curr += 1
if len(c) < MAX_LEN:
c.append(t[1])
if k < m and b[k] == curr-1:
res.append(t[1])
k += 1
else:
curr += t[1] * t[2]
while t[2] > 0 and len(c) < MAX_LEN:
c.extend(c[:t[1]])
t[2] -= 1
while k < m and last <= b[k] < curr:
res.append(c[(b[k] - last) % t[1]])
k += 1
print(' '.join(map(str, res[:m])))
``` | output | 1 | 50,841 | 20 | 101,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sereja loves number sequences very much. That's why he decided to make himself a new one following a certain algorithm.
Sereja takes a blank piece of paper. Then he starts writing out the sequence in m stages. Each time he either adds a new number to the end of the sequence or takes l first elements of the current sequence and adds them c times to the end. More formally, if we represent the current sequence as a1, a2, ..., an, then after we apply the described operation, the sequence transforms into a1, a2, ..., an[, a1, a2, ..., al] (the block in the square brackets must be repeated c times).
A day has passed and Sereja has completed the sequence. He wonders what are the values of some of its elements. Help Sereja.
Input
The first line contains integer m (1 β€ m β€ 105) β the number of stages to build a sequence.
Next m lines contain the description of the stages in the order they follow. The first number in the line is a type of stage (1 or 2). Type 1 means adding one number to the end of the sequence, in this case the line contains integer xi (1 β€ xi β€ 105) β the number to add. Type 2 means copying a prefix of length li to the end ci times, in this case the line further contains two integers li, ci (1 β€ li β€ 105, 1 β€ ci β€ 104), li is the length of the prefix, ci is the number of copyings. It is guaranteed that the length of prefix li is never larger than the current length of the sequence.
The next line contains integer n (1 β€ n β€ 105) β the number of elements Sereja is interested in. The next line contains the numbers of elements of the final sequence Sereja is interested in. The numbers are given in the strictly increasing order. It is guaranteed that all numbers are strictly larger than zero and do not exceed the length of the resulting sequence. Consider the elements of the final sequence numbered starting from 1 from the beginning to the end of the sequence.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print the elements that Sereja is interested in, in the order in which their numbers occur in the input.
Examples
Input
6
1 1
1 2
2 2 1
1 3
2 5 2
1 4
16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Output
1 2 1 2 3 1 2 1 2 3 1 2 1 2 3 4 | instruction | 0 | 50,842 | 20 | 101,684 |
Tags: binary search, brute force
Correct Solution:
```
from bisect import bisect_left
m = int(input())
t, s = [input().split() for i in range(m)], [0] * m
l, n = 0, int(input())
for j, i in enumerate(t):
l += 1 if i[0] == '1' else int(i[1]) * int(i[2])
t[j], s[j] = l, i[1] if i[0] == '1' else int(i[1])
F = {}
def f(i):
if not i in F:
k = bisect_left(t, i)
F[i] = s[k] if type(s[k]) == str else f((i - t[k] - 1) % s[k] + 1)
return F[i]
print(' '.join(f(i) for i in map(int, input().split())))
# Made By Mostafa_Khaled
``` | output | 1 | 50,842 | 20 | 101,685 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Right now you are to solve a very, very simple problem β to crack the safe. Four positive integers stand one by one on a circle protecting the safe. You know that to unlock this striking safe you have to make all four numbers equal to one. Operations are as follows: you may choose two adjacent numbers and increase both by one; you may choose two adjacent even numbers and divide both by two. Nothing else. Crack the safe!
Input
The single line of the input contains four space-separated integer positive numbers not greater than 109 each β four numbers on the circle in consecutive order.
Output
The output should contain "-1" (quotes for clarity) if the safe is secure, that is it's impossible to crack it. Otherwise, output should contain the sequence of operations (one operations per line) leading to unlocking the safe. You don't have to minimize the number of operations, but it should not exceed 1000. To make things clear, assume numbers stand on positions 1 through 4. Each operation is encoded by two symbols. If the following operation is dividing then first symbol is '/'; otherwise it's '+' (addition). The second symbol is the position of the first number in pair in consecutive order. (see samples for clarification).
If there are several solutions, output any of them.
Examples
Input
1 1 1 1
Output
Input
1 2 4 2
Output
/2
/3
Input
3 3 1 1
Output
+1
/1
/1
Input
2 1 2 4
Output
/3
/4 | instruction | 0 | 50,859 | 20 | 101,718 |
Tags: brute force, constructive algorithms
Correct Solution:
```
from sys import stdin
input = stdin.buffer.readline
ans = ''
def f(a, b, c):
global ans
if (a & 1) & (b & 1) and not a == b == 1:
ans += f'+{c}\n'
a += 1
b += 1
while (a & 1 ^ 1) & (b & 1 ^ 1):
ans += f'/{c}\n'
a >>= 1
b >>= 1
return a, b
a, b, c, d = map(int, input().split())
while a + b + c + d > 4:
p, q, r, s = a, b, c, d
a, b = f(a, b, 1)
b, c = f(b, c, 2)
c, d = f(c, d, 3)
d, a = f(d, a, 4)
if a == p and b == q and c == r and d == s:
if a & 1 ^ 1:
ans += '+1\n'
a += 1
b += 1
elif b & 1 ^ 1:
ans += '+2\n'
b += 1
c += 1
elif c & 1 ^ 1:
ans += '+3\n'
c += 1
d += 1
else:
ans += '+4\n'
d += 1
a += 1
print(ans)
``` | output | 1 | 50,859 | 20 | 101,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Right now you are to solve a very, very simple problem β to crack the safe. Four positive integers stand one by one on a circle protecting the safe. You know that to unlock this striking safe you have to make all four numbers equal to one. Operations are as follows: you may choose two adjacent numbers and increase both by one; you may choose two adjacent even numbers and divide both by two. Nothing else. Crack the safe!
Input
The single line of the input contains four space-separated integer positive numbers not greater than 109 each β four numbers on the circle in consecutive order.
Output
The output should contain "-1" (quotes for clarity) if the safe is secure, that is it's impossible to crack it. Otherwise, output should contain the sequence of operations (one operations per line) leading to unlocking the safe. You don't have to minimize the number of operations, but it should not exceed 1000. To make things clear, assume numbers stand on positions 1 through 4. Each operation is encoded by two symbols. If the following operation is dividing then first symbol is '/'; otherwise it's '+' (addition). The second symbol is the position of the first number in pair in consecutive order. (see samples for clarification).
If there are several solutions, output any of them.
Examples
Input
1 1 1 1
Output
Input
1 2 4 2
Output
/2
/3
Input
3 3 1 1
Output
+1
/1
/1
Input
2 1 2 4
Output
/3
/4 | instruction | 0 | 50,860 | 20 | 101,720 |
Tags: brute force, constructive algorithms
Correct Solution:
```
ring = list(map(int, input().split()))
n = len(ring)
record = []
def halve(pos):
a, b = pos % n, (pos + 1) % n
ring[a] //= 2
ring[b] //= 2
record.append('/%d' % (a + 1))
def increment(pos):
a, b = pos % n, (pos + 1) % n
ring[a] += 1
ring[b] += 1
record.append('+%d' % (a + 1))
while True:
modified = False
for a in range(n):
b = (a + 1) % n
while ring[a] + ring[b] > 3:
if ring[a] % 2 == 1 and ring[b] % 2 == 1:
increment(a)
elif ring[a] % 2 == 1:
increment(a - 1)
elif ring[b] % 2 == 1:
increment(b)
halve(a)
modified = True
if not modified:
break
while 2 in ring:
pos = ring.index(2)
increment(pos - 1)
increment(pos)
halve(pos - 1)
halve(pos)
if len(record) > 0:
print('\n'.join(record))
#print(len(record), ring)
``` | output | 1 | 50,860 | 20 | 101,721 |
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