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.
Drazil is playing a math game with Varda.
Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>.
First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions:
1. x doesn't contain neither digit 0 nor digit 1.
2. <image> = <image>.
Help friends find such number.
Input
The first line contains an integer n (1 ≤ n ≤ 15) — the number of digits in a.
The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes.
Output
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
Examples
Input
4
1234
Output
33222
Input
3
555
Output
555
Note
In the first case, <image>
Submitted Solution:
```
input()
s = input()
ans = []
for i in range(len(s)):
if s[i] >= '2':
ans.append(2)
if s[i] >= '3':
ans += [3]
if s[i] >= '4':
ans += [3,2,2]
if s[i] >= '5':
ans += [5]
if s[i] >= '6':
ans += [5,3]
if s[i] >= '7':
ans += [7]
if s[i] >= '8':
ans += [7,2,2,2]
if s[i] >= '9':
ans += [7,3,3,2]
ans = reversed(sorted(ans))
print(*ans, sep='')
``` | instruction | 0 | 101,591 | 20 | 203,182 |
No | output | 1 | 101,591 | 20 | 203,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Drazil is playing a math game with Varda.
Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>.
First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions:
1. x doesn't contain neither digit 0 nor digit 1.
2. <image> = <image>.
Help friends find such number.
Input
The first line contains an integer n (1 ≤ n ≤ 15) — the number of digits in a.
The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes.
Output
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
Examples
Input
4
1234
Output
33222
Input
3
555
Output
555
Note
In the first case, <image>
Submitted Solution:
```
from math import factorial, log2
from operator import mul
from functools import reduce
def drazil():
input()
n = input()
count = [0 for _ in range(10)]
for c in n:
c = int(c)
if c == 9:
c -= 1
count[3] += 2
while True:
if c in [2,3,5,7]:
count[c] += 1
break
elif c in [4,8]:
count[2] += int(log2(c))
elif c in [0,1]:
break
elif c == 6:
count[2] += 1
count[3] += 1
elif c == 9:
count[3] += 2
else:
print('Shit', c)
input()
c -= 1
s = ''
for i, n in enumerate(count):
s = str(i)*n + s
print(s)
if __name__ == '__main__':
drazil()
``` | instruction | 0 | 101,592 | 20 | 203,184 |
No | output | 1 | 101,592 | 20 | 203,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ancient Egyptians are known to have understood difficult concepts in mathematics. The ancient Egyptian mathematician Ahmes liked to write a kind of arithmetic expressions on papyrus paper which he called as Ahmes arithmetic expression.
An Ahmes arithmetic expression can be defined as:
* "d" is an Ahmes arithmetic expression, where d is a one-digit positive integer;
* "(E1 op E2)" is an Ahmes arithmetic expression, where E1 and E2 are valid Ahmes arithmetic expressions (without spaces) and op is either plus ( + ) or minus ( - ).
For example 5, (1-1) and ((1+(2-3))-5) are valid Ahmes arithmetic expressions.
On his trip to Egypt, Fafa found a piece of papyrus paper having one of these Ahmes arithmetic expressions written on it. Being very ancient, the papyrus piece was very worn out. As a result, all the operators were erased, keeping only the numbers and the brackets. Since Fafa loves mathematics, he decided to challenge himself with the following task:
Given the number of plus and minus operators in the original expression, find out the maximum possible value for the expression on the papyrus paper after putting the plus and minus operators in the place of the original erased operators.
Input
The first line contains a string E (1 ≤ |E| ≤ 104) — a valid Ahmes arithmetic expression. All operators are erased and replaced with '?'.
The second line contains two space-separated integers P and M (0 ≤ min(P, M) ≤ 100) — the number of plus and minus operators, respectively.
It is guaranteed that P + M = the number of erased operators.
Output
Print one line containing the answer to the problem.
Examples
Input
(1?1)
1 0
Output
2
Input
(2?(1?2))
1 1
Output
1
Input
((1?(5?7))?((6?2)?7))
3 2
Output
18
Input
((1?(5?7))?((6?2)?7))
2 3
Output
16
Note
* The first sample will be (1 + 1) = 2.
* The second sample will be (2 + (1 - 2)) = 1.
* The third sample will be ((1 - (5 - 7)) + ((6 + 2) + 7)) = 18.
* The fourth sample will be ((1 + (5 + 7)) - ((6 - 2) - 7)) = 16. | instruction | 0 | 101,757 | 20 | 203,514 |
Tags: dfs and similar, dp, trees
Correct Solution:
```
from math import inf
class Node:
def __init__(self, parent = None, leftExp = None, rightExp = None, signQ = 0):
self.parent, self.leftExp, self.rightExp, self.signQ = parent, leftExp, rightExp, signQ
def __str__(self):
return "Node"
memo = {}
def Memoize(node, p, maxValue, minValue):
if not node in memo:
memo.update({node : {p : [maxValue, minValue]} })
else:
memo[node].update({p : [maxValue, minValue]})
def ExpMaxValue(root: Node, p):
m = root.signQ - p
"""if root.signQ == 1:
if p == 1:
return [root.leftExp.value + root.rightExp.value, root.leftExp.value + root.rightExp.value]
else:
return [root.leftExp.value - root.rightExp.value, root.leftExp.value - root.rightExp.value]"""
if root.signQ == 0:
return [root.value, root.value]
if root in memo:
if p in memo[root]:
return memo[root][p]
if m == 0:
value = ExpMaxValue(root.leftExp, root.leftExp.signQ)[0] + ExpMaxValue(root.rightExp, root.rightExp.signQ)[0]
Memoize(root, p, value, value)
return [value, value]
if p == 0:
value = ExpMaxValue(root.leftExp, 0)[0] - ExpMaxValue(root.rightExp, 0)[0]
Memoize(root, p, value, value)
return [value, value]
maxValue = -inf
minValue = inf
if m >= p:
for pQMid in range(2):
pQLeftMin = min(p - pQMid, root.leftExp.signQ)
for pQLeft in range(pQLeftMin + 1):
if root.leftExp.signQ - pQLeft + (1 - pQMid) > m:
continue
resLeft = ExpMaxValue(root.leftExp, pQLeft)
resRight = ExpMaxValue(root.rightExp, p - pQMid - pQLeft)
if pQMid == 1:
maxValue = max(resLeft[0] + resRight[0], maxValue)
minValue = min(resLeft[1] + resRight[1], minValue)
else:
maxValue = max(resLeft[0] - resRight[1], maxValue)
minValue = min(resLeft[1] - resRight[0], minValue)
else:
for mQMid in range(2):
mQLeftMin = min(m - mQMid, root.leftExp.signQ)
for mQLeft in range(mQLeftMin + 1):
pQLeft = root.leftExp.signQ - mQLeft
if pQLeft + (1 - mQMid) > p:
continue
resLeft = ExpMaxValue(root.leftExp, pQLeft)
resRight = ExpMaxValue(root.rightExp, p - (1 - mQMid) - pQLeft)
if mQMid == 0:
maxValue = max(resLeft[0] + resRight[0], maxValue)
minValue = min(resLeft[1] + resRight[1], minValue)
else:
maxValue = max(resLeft[0] - resRight[1], maxValue)
minValue = min(resLeft[1] - resRight[0], minValue)
Memoize(root, p, int(maxValue), int(minValue))
return [int(maxValue), int(minValue)]
def PrintNodes(root: Node):
if root.signQ != 0:
leftExp = root.leftExp if root.leftExp.signQ != 0 else root.leftExp.value
rightExp = root.rightExp if root.rightExp.signQ != 0 else root.rightExp.value
check = root.signQ == root.leftExp.signQ + root.rightExp.signQ + 1
print(root.signQ, root.parent, leftExp, rightExp, check)
PrintNodes(root.leftExp)
PrintNodes(root.rightExp)
def main():
exp = str(input())
if len(exp) == 1:
print(exp)
return
#root = Node(None, None, None)
#root.parent = root
cNode = Node()
isRight = False
for i in range(1, len(exp) - 1):
if exp[i] == '(':
if not isRight:
cNode.leftExp = Node(cNode)
cNode = cNode.leftExp
else:
cNode.rightExp = Node(cNode)
cNode = cNode.rightExp
isRight = False
elif exp[i] == '?':
isRight = True
cNode.signQ += 1
elif exp[i] == ')':
if cNode.parent != None:
cNode.parent.signQ += cNode.signQ
cNode = cNode.parent
isRight = False
else:
if not isRight:
cNode.leftExp = Node(cNode)
cNode.leftExp.value = int(exp[i])
else:
cNode.rightExp = Node(cNode)
cNode.rightExp.value = int(exp[i])
#PrintNodes(cNode)
ss = str(input()).split()
p, m = int(ss[0]), int(ss[1])
print(ExpMaxValue(cNode, p)[0])
main()
``` | output | 1 | 101,757 | 20 | 203,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ancient Egyptians are known to have understood difficult concepts in mathematics. The ancient Egyptian mathematician Ahmes liked to write a kind of arithmetic expressions on papyrus paper which he called as Ahmes arithmetic expression.
An Ahmes arithmetic expression can be defined as:
* "d" is an Ahmes arithmetic expression, where d is a one-digit positive integer;
* "(E1 op E2)" is an Ahmes arithmetic expression, where E1 and E2 are valid Ahmes arithmetic expressions (without spaces) and op is either plus ( + ) or minus ( - ).
For example 5, (1-1) and ((1+(2-3))-5) are valid Ahmes arithmetic expressions.
On his trip to Egypt, Fafa found a piece of papyrus paper having one of these Ahmes arithmetic expressions written on it. Being very ancient, the papyrus piece was very worn out. As a result, all the operators were erased, keeping only the numbers and the brackets. Since Fafa loves mathematics, he decided to challenge himself with the following task:
Given the number of plus and minus operators in the original expression, find out the maximum possible value for the expression on the papyrus paper after putting the plus and minus operators in the place of the original erased operators.
Input
The first line contains a string E (1 ≤ |E| ≤ 104) — a valid Ahmes arithmetic expression. All operators are erased and replaced with '?'.
The second line contains two space-separated integers P and M (0 ≤ min(P, M) ≤ 100) — the number of plus and minus operators, respectively.
It is guaranteed that P + M = the number of erased operators.
Output
Print one line containing the answer to the problem.
Examples
Input
(1?1)
1 0
Output
2
Input
(2?(1?2))
1 1
Output
1
Input
((1?(5?7))?((6?2)?7))
3 2
Output
18
Input
((1?(5?7))?((6?2)?7))
2 3
Output
16
Note
* The first sample will be (1 + 1) = 2.
* The second sample will be (2 + (1 - 2)) = 1.
* The third sample will be ((1 - (5 - 7)) + ((6 + 2) + 7)) = 18.
* The fourth sample will be ((1 + (5 + 7)) - ((6 - 2) - 7)) = 16.
Submitted Solution:
```
from math import inf
class Node:
def __init__(self, parent = None, leftExp = None, rightExp = None, signQ = 0):
self.parent, self.leftExp, self.rightExp, self.signQ = parent, leftExp, rightExp, signQ
def __str__(self):
return "Node"
memo = {}
def Memoize(node, p, maxValue, minValue):
if not node in memo:
memo.update({node : {p : [maxValue, minValue]} })
else:
memo[node].update({p : [maxValue, minValue]})
def ExpMaxValue(root: Node, p, m):
"""if root.signQ == 1:
if p == 1:
return [root.leftExp.value + root.rightExp.value, root.leftExp.value + root.rightExp.value]
else:
return [root.leftExp.value - root.rightExp.value, root.leftExp.value - root.rightExp.value]"""
if root in memo:
if p in memo[root]:
return memo[root][p]
if root.signQ == 0:
return [root.value, root.value]
if m == 0:
value = ExpMaxValue(root.leftExp, root.leftExp.signQ, 0)[0] + ExpMaxValue(root.rightExp, root.rightExp.signQ, 0)[0]
Memoize(root, p, value, value)
return [value, value]
if p == 0:
value = ExpMaxValue(root.leftExp, 0, root.leftExp.signQ)[0] - ExpMaxValue(root.rightExp, 0, root.rightExp.signQ)[0]
Memoize(root, p, value, value)
return [value, value]
maxValue = -inf
minValue = inf
if m >= p:
for pQMid in range(2):
pQLeftMin = min(p - pQMid, root.leftExp.signQ)
for pQLeft in range(pQLeftMin + 1):
mQLeft = root.leftExp.signQ - pQLeft
resLeft = ExpMaxValue(root.leftExp, pQLeft, mQLeft)
resRight = ExpMaxValue(root.rightExp, p - pQMid - pQLeft, m - (1 - pQMid) - mQLeft)
if pQMid == 1:
maxValue = max(resLeft[0] + resRight[0], maxValue)
minValue = min(resLeft[1] + resRight[1], minValue)
else:
maxValue = max(resLeft[0] - resRight[1], maxValue)
minValue = min(resLeft[1] - resRight[0], minValue)
else:
for mQMid in range(2):
mQLeftMin = min(m - mQMid, root.leftExp.signQ)
for mQLeft in range(mQLeftMin + 1):
pQLeft = root.leftExp.signQ - mQLeft
resLeft = ExpMaxValue(root.leftExp, pQLeft, mQLeft)
resRight = ExpMaxValue(root.rightExp, p - (1 - mQMid) - pQLeft, m - mQMid - mQLeft)
if mQMid == 0:
maxValue = max(resLeft[0] + resRight[0], maxValue)
minValue = min(resLeft[1] + resRight[1], minValue)
else:
maxValue = max(resLeft[0] - resRight[1], maxValue)
minValue = min(resLeft[1] - resRight[0], minValue)
Memoize(root, p, maxValue, minValue)
return [maxValue, minValue]
def PrintNodes(root: Node):
if root.signQ != 0:
print(root.signQ, root.parent, root.leftExp if root.leftExp.signQ != 0 else root.leftExp.value, root.rightExp if root.rightExp.signQ != 0 else root.rightExp.value)
PrintNodes(root.leftExp)
PrintNodes(root.rightExp)
def main():
exp = str(input())
if len(exp) == 1:
print(exp)
return
#root = Node(None, None, None)
#root.parent = root
cNode = Node()
isRight = False
for i in range(1, len(exp) - 1):
if exp[i] == '(':
if not isRight:
cNode.leftExp = Node(cNode)
cNode = cNode.leftExp
else:
cNode.rightExp = Node(cNode)
cNode = cNode.rightExp
isRight = False
elif exp[i] == '?':
isRight = True
cNode.signQ += 1
elif exp[i] == ')':
if cNode.parent != None:
cNode.parent.signQ += cNode.signQ
cNode = cNode.parent
isRight = False
else:
if not isRight:
cNode.leftExp = Node(cNode)
cNode.leftExp.value = int(exp[i])
else:
cNode.rightExp = Node(cNode)
cNode.rightExp.value = int(exp[i])
#PrintNodes(cNode)
ss = str(input()).split()
p, m = int(ss[0]), int(ss[1])
print(ExpMaxValue(cNode, p, m)[0])
main()
``` | instruction | 0 | 101,758 | 20 | 203,516 |
No | output | 1 | 101,758 | 20 | 203,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ancient Egyptians are known to have understood difficult concepts in mathematics. The ancient Egyptian mathematician Ahmes liked to write a kind of arithmetic expressions on papyrus paper which he called as Ahmes arithmetic expression.
An Ahmes arithmetic expression can be defined as:
* "d" is an Ahmes arithmetic expression, where d is a one-digit positive integer;
* "(E1 op E2)" is an Ahmes arithmetic expression, where E1 and E2 are valid Ahmes arithmetic expressions (without spaces) and op is either plus ( + ) or minus ( - ).
For example 5, (1-1) and ((1+(2-3))-5) are valid Ahmes arithmetic expressions.
On his trip to Egypt, Fafa found a piece of papyrus paper having one of these Ahmes arithmetic expressions written on it. Being very ancient, the papyrus piece was very worn out. As a result, all the operators were erased, keeping only the numbers and the brackets. Since Fafa loves mathematics, he decided to challenge himself with the following task:
Given the number of plus and minus operators in the original expression, find out the maximum possible value for the expression on the papyrus paper after putting the plus and minus operators in the place of the original erased operators.
Input
The first line contains a string E (1 ≤ |E| ≤ 104) — a valid Ahmes arithmetic expression. All operators are erased and replaced with '?'.
The second line contains two space-separated integers P and M (0 ≤ min(P, M) ≤ 100) — the number of plus and minus operators, respectively.
It is guaranteed that P + M = the number of erased operators.
Output
Print one line containing the answer to the problem.
Examples
Input
(1?1)
1 0
Output
2
Input
(2?(1?2))
1 1
Output
1
Input
((1?(5?7))?((6?2)?7))
3 2
Output
18
Input
((1?(5?7))?((6?2)?7))
2 3
Output
16
Note
* The first sample will be (1 + 1) = 2.
* The second sample will be (2 + (1 - 2)) = 1.
* The third sample will be ((1 - (5 - 7)) + ((6 + 2) + 7)) = 18.
* The fourth sample will be ((1 + (5 + 7)) - ((6 - 2) - 7)) = 16.
Submitted Solution:
```
from math import inf
class Node:
def __init__(self, parent = None, leftExp = None, rightExp = None, signQ = 0):
self.parent, self.leftExp, self.rightExp, self.signQ = parent, leftExp, rightExp, signQ
def __str__(self):
return "Node"
memo = {}
def Memoize(node, p, maxValue, minValue):
if not node in memo:
memo.update({node : {p : [maxValue, minValue]} })
else:
memo[node].update({p : [maxValue, minValue]})
def ExpMaxValue(root: Node, p, m):
"""if root.signQ == 1:
if p == 1:
return [root.leftExp.value + root.rightExp.value, root.leftExp.value + root.rightExp.value]
else:
return [root.leftExp.value - root.rightExp.value, root.leftExp.value - root.rightExp.value]"""
if root in memo:
if p in memo[root]:
return memo[root][p]
if root.signQ == 0:
return [root.value, root.value]
if m == 0:
value = ExpMaxValue(root.leftExp, root.leftExp.signQ, 0)[0] + ExpMaxValue(root.rightExp, root.rightExp.signQ, 0)[0]
Memoize(root, p, value, value)
return [value, value]
if p == 0:
value = ExpMaxValue(root.leftExp, 0, root.leftExp.signQ)[0] - ExpMaxValue(root.rightExp, 0, root.rightExp.signQ)[0]
Memoize(root, p, value, value)
return [value, value]
maxValue = -inf
minValue = inf
if m >= p:
for pQMid in range(2):
pQLeftMin = min(p - pQMid, root.leftExp.signQ)
for pQLeft in range(pQLeftMin + 1):
mQLeft = root.leftExp.signQ - pQLeft
resLeft = ExpMaxValue(root.leftExp, pQLeft, mQLeft)
resRight = ExpMaxValue(root.rightExp, p - pQMid - pQLeft, m - (1 - pQMid) - mQLeft)
if pQMid == 1:
maxValue = max(resLeft[0] + resRight[0], maxValue)
minValue = min(resLeft[1] + resRight[1], minValue)
else:
maxValue = max(resLeft[0] - resRight[1], maxValue)
minValue = min(resLeft[1] - resRight[0], minValue)
else:
for mQMid in range(2):
mQLeftMin = min(m - mQMid, root.leftExp.signQ)
for mQLeft in range(mQLeftMin + 1):
pQLeft = root.leftExp.signQ - mQLeft
resLeft = ExpMaxValue(root.leftExp, pQLeft, mQLeft)
resRight = ExpMaxValue(root.rightExp, p - (1 - mQMid) - pQLeft, m - mQMid - mQLeft)
if mQMid == 0:
maxValue = max(resLeft[0] + resRight[0], maxValue)
minValue = min(resLeft[1] + resRight[1], minValue)
else:
maxValue = max(resLeft[0] - resRight[1], maxValue)
minValue = min(resLeft[1] - resRight[0], minValue)
Memoize(root, p, maxValue, minValue)
return [maxValue, minValue]
def PrintNodes(root: Node):
if root.signQ != 0:
print(root.signQ, root.parent, root.leftExp if root.leftExp.signQ != 0 else root.leftExp.value, root.rightExp if root.rightExp.signQ != 0 else root.rightExp.value)
PrintNodes(root.leftExp)
PrintNodes(root.rightExp)
def main():
exp = str(input())
if len(exp) == 1:
print(exp)
return
#root = Node(None, None, None)
#root.parent = root
cNode = Node()
isRight = False
for i in range(1, len(exp) - 1):
if exp[i] == '(':
if not isRight:
cNode.leftExp = Node(cNode)
cNode = cNode.leftExp
else:
cNode.rightExp = Node(cNode)
cNode = cNode.rightExp
isRight = False
elif exp[i] == '?':
isRight = True
cNode.signQ += 1
elif exp[i] == ')':
if cNode.parent != None:
cNode.parent.signQ += cNode.signQ
cNode = cNode.parent
isRight = False
else:
if not isRight:
cNode.leftExp = Node(cNode)
cNode.leftExp.value = int(exp[i])
else:
cNode.rightExp = Node(cNode)
cNode.rightExp.value = int(exp[i])
PrintNodes(cNode)
ss = str(input()).split()
p, m = int(ss[0]), int(ss[1])
print(ExpMaxValue(cNode, p, m)[0])
main()
``` | instruction | 0 | 101,759 | 20 | 203,518 |
No | output | 1 | 101,759 | 20 | 203,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem Statement
Dr. Suposupo developed a programming language called Shipura. Shipura supports only one binary operator ${\tt >>}$ and only one unary function ${\tt S<\ >}$.
$x {\tt >>} y$ is evaluated to $\lfloor x / 2^y \rfloor$ (that is, the greatest integer not exceeding $x / 2^y$), and ${\tt S<} x {\tt >}$ is evaluated to $x^2 \bmod 1{,}000{,}000{,}007$ (that is, the remainder when $x^2$ is divided by $1{,}000{,}000{,}007$).
The operator ${\tt >>}$ is left-associative. For example, the expression $x {\tt >>} y {\tt >>} z$ is interpreted as $(x {\tt >>} y) {\tt >>} z$, not as $x {\tt >>} (y {\tt >>} z)$. Note that these parentheses do not appear in actual Shipura expressions.
The syntax of Shipura is given (in BNF; Backus-Naur Form) as follows:
expr ::= term | expr sp ">>" sp term
term ::= number | "S" sp "<" sp expr sp ">"
sp ::= "" | sp " "
number ::= digit | number digit
digit ::= "0" | "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9"
The start symbol of this syntax is $\tt expr$ that represents an expression in Shipura. In addition, $\tt number$ is an integer between $0$ and $1{,}000{,}000{,}000$ inclusive, written without extra leading zeros.
Write a program to evaluate Shipura expressions.
Input
The input is a sequence of datasets. Each dataset is represented by a line which contains a valid expression in Shipura.
A line containing a single ${\tt \\#}$ indicates the end of the input. You can assume the number of datasets is at most $100$ and the total size of the input file does not exceed $2{,}000{,}000$ bytes.
Output
For each dataset, output a line containing the evaluated value of the expression.
Sample Input
S< S< 12 >> 2 > >
123 >> 1 >> 1
1000000000 >>129
S<S<S<S<S<2>>>>>
S <S< S<2013 >>> 11 >>> 10 >
Output for the Sample Input
81
30
0
294967268
14592400
Example
Input
S< S< 12 >> 2 > >
123 >> 1 >> 1
1000000000 >>129
S<S<S<S<S<2>>>>>
S <S< S<2013 >>> 11 >>> 10 >
#
Output
81
30
0
294967268
14592400
Submitted Solution:
```
import re
S = lambda x: x*x%int(1e9+7)
s = input()
while s != '#':
s = s.replace(' ', '')
s = re.sub(r'>>(\d+)', r'XX\1', s)
s = s.translate(str.maketrans('<>X', '()>'))
print(eval(s))
s = input()
``` | instruction | 0 | 101,999 | 20 | 203,998 |
No | output | 1 | 101,999 | 20 | 203,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number. | instruction | 0 | 102,057 | 20 | 204,114 |
Tags: brute force, probabilities
Correct Solution:
```
#!/usr/bin/env python3
import itertools
pl,pr,vl,vr,k = map(int, input().split())
l = min(pl, vl)
r = max(pr, vr)
# Generate all lucky numbers with the appropriate number of digits
# O(3**max_nr_digits) = O(3**9) < 20000
max_nr_digits = len(str(r))
lucky_numbers = []
for nr_digits in range(1,max_nr_digits+1):
lucky_numbers.extend(int(''.join(i)) for i in itertools.product("47", repeat=nr_digits))
# Filter so we only have lucky numbers that can be in the intervals
lucky_numbers = [nr for nr in lucky_numbers if (l <= nr) and (nr <= r)]
lucky_numbers.sort()
if len(lucky_numbers) < k:
print("%.12f" % 0.0)
exit()
lucky_numbers.insert(0,l-1)
lucky_numbers.append(r+1)
total_pairs = (pr-pl+1) * (vr-vl+1)
good_pairs = 0
for start_index in range(len(lucky_numbers)-k-1):
(a,b) = lucky_numbers[start_index], lucky_numbers[start_index+1]
(c,d) = lucky_numbers[start_index+k], lucky_numbers[start_index+k+1]
# A pair will have lucky_numbers[start_index+1 : start_index+1+k] in its interval when:
# The smallest number of the pair is larger than a, but at most b
# The largest number of the pair is at least c, but smaller than d
good_pairs += max((min(b,pr)-max(a+1,pl)+1),0)*max((min(d-1,vr)-max(c,vl)+1),0)
good_pairs += max((min(b,vr)-max(a+1,vl)+1),0)*max((min(d-1,pr)-max(c,pl)+1),0)
if (b == c) and (pl <= b) and (b <= pr) and (vl <=b) and (b <= vr):
# Prevent double counting of the pair (b,c) when b == c and both p and v can supply the value
good_pairs -= 1
print("%.12f" % (good_pairs / total_pairs))
``` | output | 1 | 102,057 | 20 | 204,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number. | instruction | 0 | 102,058 | 20 | 204,116 |
Tags: brute force, probabilities
Correct Solution:
```
s = list()
def rec(x):
if x > 100000000000:
return
s.append(x)
rec(x * 10 + 4)
rec(x * 10 + 7)
def f(l1, r1, l2, r2):
l1 = max(l1, l2)
r1 = min(r1, r2)
return max(r1 - l1 + 1, 0)
def main():
rec(0)
s.sort()
args = input().split()
pl, pr, vl, vr, k = int(args[0]), int(args[1]), int(args[2]), int(args[3]), int(args[4])
ans = 0
i = 1
while i + k < len(s):
l1 = s[i - 1] + 1
r1 = s[i]
l2 = s[i + k - 1]
r2 = s[i + k] - 1
a = f(l1, r1, vl, vr) * f(l2, r2, pl, pr)
b = f(l1, r1, pl, pr) * f(l2, r2, vl, vr)
ans += a + b
if k == 1 and a > 0 and b > 0:
ans -= 1
i += 1
all = (pr - pl + 1) * (vr - vl + 1)
print(1.0 * ans / all)
if __name__ == '__main__':
main()
``` | output | 1 | 102,058 | 20 | 204,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number. | instruction | 0 | 102,059 | 20 | 204,118 |
Tags: brute force, probabilities
Correct Solution:
```
def gen(n, cur):
if n == 0:
global arr
arr.append(cur)
else:
gen(n - 1, cur + '4')
gen(n - 1, cur + '7')
def lseg(x1, x2):
if x2 < x1:
return 0
return x2 - x1 + 1
def inter2(x1, x2, a, b):
left = max(x1, a)
right = min(x2, b)
return (left, right)
def inter(x1, x2, a, b):
l, r = inter2(x1, x2, a, b)
return lseg(l, r)
def minus(a, b, c, d):
d1 = (-1, c - 1)
d2 = (d + 1, 10 ** 10)
d1 = inter(a, b, *d1)
d2 = inter(a, b, *d2)
return d1 + d2
arr = []
for i in range(1, 11):
gen(i, '')
arr = [0] + sorted(list(map(int, arr)))
pl, pr, vl, vr, k = map(int, input().split())
ans = 0
for start in range(1, len(arr) - k + 1):
if arr[start + k - 1] > max(vr, pr):
break
if arr[start] < min(pl, vl):
continue
goleft = inter(pl, pr, arr[start - 1] + 1, arr[start])
goright = inter(vl, vr, arr[start + k - 1], arr[start + k] - 1)
left, right = inter2(pl, pr, arr[start - 1] + 1, arr[start]), \
inter2(vl, vr, arr[start + k - 1], arr[start + k] - 1)
ans += goleft * goright
left1 = inter2(pl, pr, arr[start + k - 1], arr[start + k] - 1)
right1 = inter2(vl, vr, arr[start - 1] + 1, arr[start])
ans += minus(right1[0], right1[1], right[0], right[1]) * lseg(*left1)
ans += (lseg(*right1) - minus(right1[0], right1[1], right[0], right[1])) \
* minus(left1[0], left1[1], left[0], left[1])
ans /= (pr - pl + 1) * (vr - vl + 1)
print(ans)
``` | output | 1 | 102,059 | 20 | 204,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number. | instruction | 0 | 102,060 | 20 | 204,120 |
Tags: brute force, probabilities
Correct Solution:
```
#!/usr/bin/env python3
vl, vr, pl, pr, k = map(int, input().split())
lucky = [4, 7]
sz = 0
for i in range(1, 9):
base = 10 ** i
psz, sz = sz, len(lucky)
for j in [4 * base, 7 * base]:
for pos in range(psz, sz):
lucky.append(j + lucky[pos])
ans = 0
for i in range(0, len(lucky)-k+1):
ll, lr = 1 if i == 0 else lucky[i-1] + 1, lucky[i]
rl, rr = lucky[i+k-1], 1_000_000_000 if i+k == len(lucky) else lucky[i+k] - 1
vc = max(0, min(vr, lr) - max(vl, ll) + 1)
pc = max(0, min(pr, rr) - max(pl, rl) + 1)
cnt1 = vc * pc
vc = max(0, min(pr, lr) - max(pl, ll) + 1)
pc = max(0, min(vr, rr) - max(vl, rl) + 1)
cnt2 = vc * pc
ans += cnt1 + cnt2
if k == 1 and cnt1 > 0 and cnt2 > 0:
ans -= 1
print(ans / ((vr - vl + 1) * (pr - pl + 1)))
``` | output | 1 | 102,060 | 20 | 204,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number. | instruction | 0 | 102,061 | 20 | 204,122 |
Tags: brute force, probabilities
Correct Solution:
```
import itertools
def generate_happy():
happy_digits = '47'
happies = []
for num_len in range(1, 10):
happies.extend(itertools.product(happy_digits, repeat=num_len))
return [int(''.join(num)) for num in happies]
def clamp_segment(start, end, min_b, max_b):
if start > end or min_b > max_b:
raise ValueError(f"{start} {end} {min_b} {max_b}")
if start > max_b or end < min_b:
return None, None
new_start = start
new_end = end
if start < min_b:
new_start = min_b
if end > max_b:
new_end = max_b
return new_start, new_end
def get_window_bounds(happies, happy_count, happy_window_ind):
return (happies[happy_window_ind - 1] if happy_window_ind > 0 else 0,
happies[happy_window_ind],
happies[happy_window_ind + happy_count - 1],
(happies[happy_window_ind + happy_count]
if happy_window_ind + happy_count < len(happies)
else 10 ** 9 + 1
)
)
def get_good_segm_count_ordered(
lower_start, lower_end, upper_start, upper_end,
prev_happy, this_happy, last_happy, next_happy):
(lower_bound_start, lower_bound_end) = clamp_segment(
lower_start, lower_end,
prev_happy + 1, this_happy
)
(upper_bound_start, upper_bound_end) = clamp_segment(
upper_start, upper_end,
last_happy, next_happy - 1
)
if lower_bound_start is None or upper_bound_start is None:
return 0
return (lower_bound_end - lower_bound_start + 1) *\
(upper_bound_end - upper_bound_start + 1)
def get_good_segm_count(happies, happy_count, happy_window_ind,
start_1, end_1, start_2, end_2):
prev_happy, this_happy, last_happy, next_happy = get_window_bounds(
happies, happy_count, happy_window_ind
)
first_is_lower_count = get_good_segm_count_ordered(
start_1, end_1, start_2, end_2,
prev_happy, this_happy, last_happy, next_happy
)
second_is_lower_count = get_good_segm_count_ordered(
start_2, end_2, start_1, end_1,
prev_happy, this_happy, last_happy, next_happy
)
this_happy = happies[happy_window_ind]
if (happy_count == 1
and start_1 <= this_happy <= end_2
and start_2 <= this_happy <= end_2):
second_is_lower_count -= 1
return first_is_lower_count + second_is_lower_count
def main(args=None):
if args is None:
p_start, p_end, v_start, v_end, happy_count = map(int, input().split())
else:
p_start, p_end, v_start, v_end, happy_count = args
happies = generate_happy()
good_segments_count = 0
for happy_window_ind in range(len(happies) - happy_count + 1):
good_segments_count += get_good_segm_count(
happies, happy_count, happy_window_ind,
p_start, p_end, v_start, v_end
)
# print(get_good_segm_count(
# happies, happy_count, happy_window_ind,
# p_start, p_end, v_start, v_end
# ), happy_window_ind)
all_segments_count = (p_end - p_start + 1) * (v_end - v_start + 1)
return good_segments_count / all_segments_count
print(main())
``` | output | 1 | 102,061 | 20 | 204,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number. | instruction | 0 | 102,062 | 20 | 204,124 |
Tags: brute force, probabilities
Correct Solution:
```
import itertools as it
all_lucky = []
for length in range(1, 10):
for comb in it.product(['7', '4'], repeat=length):
all_lucky += [int(''.join(comb))]
all_lucky.sort()
# print(len(all_lucky))
pl, pr, vl, vr, k = map(int, input().split())
result = 0
def inters_len(a, b, c, d):
a, b = sorted([a, b])
c, d = sorted([c, d])
return (max([a, c]), min([b, d]))
def check_for_intervals(pl, pr, vl, vr):
global result
for i in range(1, len(all_lucky) - k):
le, re = i, i + k - 1
a, b = inters_len(all_lucky[le - 1] + 1, all_lucky[le], pl, pr)
left_len = max([0, b - a + 1])
c, d = inters_len(all_lucky[re], all_lucky[re + 1] - 1, vl, vr)
right_len = max([0, d - c + 1])
result += (left_len / (pr - pl + 1)) * (right_len / (vr - vl + 1))
if b == c and left_len * right_len > 1e-6:
result -= (1 / (pr - pl + 1) / (vr - vl + 1)) / 2
a, b = inters_len(0, all_lucky[0], pl, pr)
left_len = max([0, b - a + 1])
c, d = inters_len(all_lucky[k - 1], all_lucky[k] - 1, vl, vr)
right_len = max([0, d - c + 1])
#print((left_len / (pr - pl + 1)) * (right_len / (vr - vl + 1)))
#print(left_len, right_len)
result += (left_len / (pr - pl + 1)) * (right_len / (vr - vl + 1))
if b == c and left_len * right_len > 1e-6:
result -= (1 / (pr - pl + 1) / (vr - vl + 1)) / 2
a, b = inters_len(all_lucky[-(k + 1)] + 1, all_lucky[-k], pl, pr)
left_len = max([0, b - a + 1])
c, d = inters_len(all_lucky[-1], 10**9, vl, vr)
right_len = max([0, d - c + 1])
#print((left_len / (pr - pl + 1)) * (right_len / (vr - vl + 1)))
result += (left_len / (pr - pl + 1)) * (right_len / (vr - vl + 1))
if b == c and left_len * right_len > 1e-6:
result -= (1 / (pr - pl + 1) / (vr - vl + 1)) / 2
#print(all_lucky[:5])
if k != len(all_lucky):
check_for_intervals(pl, pr, vl, vr)
check_for_intervals(vl, vr, pl, pr)
else:
a, b = inters_len(0, all_lucky[0], pl, pr)
left_len = max([0, b - a + 1])
c, d = inters_len(all_lucky[-1], 10**9, vl, vr)
right_len = max([0, d - c + 1])
result += (left_len / (pr - pl + 1)) * (right_len / (vr - vl + 1))
if b == c and left_len * right_len > 1e-6:
result -= (1 / (pr - pl + 1) / (vr - vl + 1)) / 2
a, b = inters_len(0, all_lucky[0], vl, vr)
left_len = max([0, b - a + 1])
c, d = inters_len(all_lucky[-1], 10**9, pl, pr)
right_len = max([0, d - c + 1])
result += (left_len / (vr - vl + 1)) * (right_len / (pr - pl + 1))
if b == c and left_len * right_len > 1e-6:
result -= (1 / (pr - pl + 1) / (vr - vl + 1)) / 2
print(result)
# Made By Mostafa_Khaled
``` | output | 1 | 102,062 | 20 | 204,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number.
Submitted Solution:
```
import itertools
def generate_happy():
happy_digits = '47'
happies = []
for num_len in range(1, 10):
happies.extend(itertools.product(happy_digits, repeat=num_len))
return [int(''.join(num)) for num in happies]
def clamp_segment(start, end, min_b, max_b):
if start > end or min_b > max_b:
raise ValueError(f"{start} {end} {min_b} {max_b}")
if start > max_b or end < min_b:
return None, None
new_start = start
new_end = end
if start < min_b:
new_start = min_b
if end > max_b:
new_end = max_b
return new_start, new_end
def get_window_bounds(happies, happy_count, happy_window_ind):
return (happies[happy_window_ind - 1] if happy_window_ind > 0 else 1,
happies[happy_window_ind],
happies[happy_window_ind + happy_count - 1],
(happies[happy_window_ind + happy_count]
if happy_window_ind + happy_count < len(happies)
else 10 ** 9
)
)
def get_good_segm_count_ordered(
lower_start, lower_end, upper_start, upper_end,
prev_happy, this_happy, last_happy, next_happy):
(lower_bound_start, lower_bound_end) = clamp_segment(
lower_start, lower_end,
prev_happy, this_happy
)
(upper_bound_start, upper_bound_end) = clamp_segment(
upper_start, upper_end,
last_happy, next_happy
)
if lower_bound_start is None or upper_bound_start is None:
return 0
return (lower_bound_end - lower_bound_start + 1) *\
(upper_bound_end - upper_bound_start + 1)
def get_good_segm_count(happies, happy_count, happy_window_ind,
start_1, end_1, start_2, end_2):
prev_happy, this_happy, last_happy, next_happy = get_window_bounds(
happies, happy_count, happy_window_ind
)
first_is_lower_count = get_good_segm_count_ordered(
start_1, end_1, start_2, end_2,
prev_happy, this_happy, last_happy, next_happy
)
second_is_lower_count = get_good_segm_count_ordered(
start_2, end_2, start_1, end_2,
prev_happy, this_happy, last_happy, next_happy
)
return first_is_lower_count + second_is_lower_count
def main(args=None):
if args is None:
p_start, p_end, v_start, v_end, happy_count = map(int, input().split())
else:
p_start, p_end, v_start, v_end, happy_count = args
happies = generate_happy()
good_segments_count = 0
for happy_window_ind in range(len(happies) - happy_count + 1):
good_segments_count += get_good_segm_count(
happies, happy_count, happy_window_ind,
p_start, p_end, v_start, v_end
)
all_segments_count = (p_end - p_start + 1) * (v_end - v_start + 1)
return good_segments_count / all_segments_count
main()
``` | instruction | 0 | 102,063 | 20 | 204,126 |
No | output | 1 | 102,063 | 20 | 204,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number.
Submitted Solution:
```
import itertools
def generate_happy():
happy_digits = '47'
happies = []
for num_len in range(1, 10):
happies.extend(itertools.product(happy_digits, repeat=num_len))
return [int(''.join(num)) for num in happies]
def clamp_segment(start, end, min_b, max_b):
if start > end or min_b > max_b:
raise ValueError(f"{start} {end} {min_b} {max_b}")
if start > max_b or end < min_b:
return None, None
new_start = start
new_end = end
if start < min_b:
new_start = min_b
if end > max_b:
new_end = max_b
return new_start, new_end
def get_window_bounds(happies, happy_count, happy_window_ind):
return (happies[happy_window_ind - 1] if happy_window_ind > 0 else 0,
happies[happy_window_ind],
happies[happy_window_ind + happy_count - 1],
(happies[happy_window_ind + happy_count]
if happy_window_ind + happy_count < len(happies)
else 10 ** 9 + 1
)
)
def get_good_segm_count_ordered(
lower_start, lower_end, upper_start, upper_end,
prev_happy, this_happy, last_happy, next_happy):
(lower_bound_start, lower_bound_end) = clamp_segment(
lower_start, lower_end,
prev_happy + 1, this_happy
)
(upper_bound_start, upper_bound_end) = clamp_segment(
upper_start, upper_end,
last_happy, next_happy - 1
)
if lower_bound_start is None or upper_bound_start is None:
return 0
return (lower_bound_end - lower_bound_start + 1) *\
(upper_bound_end - upper_bound_start + 1)
def get_good_segm_count(happies, happy_count, happy_window_ind,
start_1, end_1, start_2, end_2):
prev_happy, this_happy, last_happy, next_happy = get_window_bounds(
happies, happy_count, happy_window_ind
)
first_is_lower_count = get_good_segm_count_ordered(
start_1, end_1, start_2, end_2,
prev_happy, this_happy, last_happy, next_happy
)
second_is_lower_count = get_good_segm_count_ordered(
start_2, end_2, start_1, end_1,
prev_happy, this_happy, last_happy, next_happy
)
return first_is_lower_count + second_is_lower_count
def main(args=None):
if args is None:
p_start, p_end, v_start, v_end, happy_count = map(int, input().split())
else:
p_start, p_end, v_start, v_end, happy_count = args
happies = generate_happy()
good_segments_count = 0
for happy_window_ind in range(len(happies) - happy_count + 1):
good_segments_count += get_good_segm_count(
happies, happy_count, happy_window_ind,
p_start, p_end, v_start, v_end
)
# print(get_good_segm_count(
# happies, happy_count, happy_window_ind,
# p_start, p_end, v_start, v_end
# ), happy_window_ind)
all_segments_count = (p_end - p_start + 1) * (v_end - v_start + 1)
return good_segments_count / all_segments_count
main()
``` | instruction | 0 | 102,064 | 20 | 204,128 |
No | output | 1 | 102,064 | 20 | 204,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number.
Submitted Solution:
```
def gen(n, cur):
if n == 0:
global arr
arr.append(cur)
else:
gen(n - 1, cur + '4')
gen(n - 1, cur + '7')
def inter(x1, x2, a, b):
left = max(x1, a)
right = min(x2, b)
if right < left:
return 0
return right - left + 1
arr = []
for i in range(1, 11):
gen(i, '')
arr = [0] + sorted(list(map(int, arr)))
pl, pr, vl, vr, k = map(int, input().split())
ans = 0
for start in range(1, len(arr) - k + 1):
if arr[start + k - 1] > max(vr, pr):
break
if arr[start] < min(pl, vl):
continue
goleft = inter(pl, pr, arr[start - 1] + 1, arr[start])
goright = inter(vl, vr, arr[start + k - 1], arr[start + k] - 1)
ans += goleft * goright
goright = inter(vl, vr, arr[start - 1] + 1, arr[start])
goleft = inter(pl, pr, arr[start + k - 1], arr[start + k] - 1)
ans += goleft * goright
ans /= (pr - pl + 1) * (vr - vl + 1)
print(ans)
``` | instruction | 0 | 102,065 | 20 | 204,130 |
No | output | 1 | 102,065 | 20 | 204,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.
Input
The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).
Output
On the single line print the result with an absolute error of no more than 10 - 9.
Examples
Input
1 10 1 10 2
Output
0.320000000000
Input
5 6 8 10 1
Output
1.000000000000
Note
Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number.
Submitted Solution:
```
#!/usr/bin/env python3
vl, vr, pl, pr, k = map(int, input().split())
lucky = [4, 7]
sz = 0
for i in range(1, 9):
base = 10 ** i
psz, sz = sz, len(lucky)
for j in [4 * base, 7 * base]:
for pos in range(psz, sz):
lucky.append(j + lucky[pos])
ans = 0
for i in range(0, len(lucky)-k+1):
ll, lr = 1 if i == 0 else lucky[i-1] + 1, lucky[i]
rl, rr = lucky[i+k-1], 1_000_000_000 if i+k == len(lucky) else lucky[i+k]
vc = max(0, min(vr, lr) - max(vl, ll) + 1)
pc = max(0, min(pr, rr) - max(pl, rl) + 1)
ans += vc * pc
vc = max(0, min(pr, lr) - max(pl, ll) + 1)
pc = max(0, min(vr, rr) - max(vl, rl) + 1)
ans += vc * pc
print(ans / ((vr-vl+1) * (pr - pl + 1)))
``` | instruction | 0 | 102,066 | 20 | 204,132 |
No | output | 1 | 102,066 | 20 | 204,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1 | instruction | 0 | 102,253 | 20 | 204,506 |
Tags: brute force, greedy, math
Correct Solution:
```
resList = []
testCount = int(input())
for test in range(testCount):
num = int(input())
sum = 0
numStr=""
if(num>45):
resList.append(str(-1))
elif(len(str(num)) == 1):
resList.append(str(num))
else:
for i in reversed(range(1,10)):
if(num == 0):
break
elif(num < i):
sum = sum + num
numStr += str(num)
break
sum = sum + i
num = num - i
numStr += str(i)
resList.append(numStr[::-1])
for res in resList:
print(res)
``` | output | 1 | 102,253 | 20 | 204,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1 | instruction | 0 | 102,254 | 20 | 204,508 |
Tags: brute force, greedy, math
Correct Solution:
```
def suma(n):
s=0
while n!=0 :
s=s+n%10
n=n//10
return s
te=int(input())
for i in range(te):
num=int(input())
res="987654321"
if num>45:
print("-1")
else:
for i in range(100):
if res[-1]=="0":
res=res[:-1]
if suma(int(res))==num:
print(res[::-1])
break
res=str(int(res)-1)
``` | output | 1 | 102,254 | 20 | 204,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1 | instruction | 0 | 102,255 | 20 | 204,510 |
Tags: brute force, greedy, math
Correct Solution:
```
for _ in range (int(input())):
n=int(input())
if n==43:
print("13456789")
continue
elif n==44:
print("23456789")
continue
elif n==45:
print("123456789")
continue
s=set()
curr=9
num=0
copy=n
ch=1
while n>9:
if curr<1:
ch=0
break
num=num*10+curr
s.add(curr)
n-=curr
curr-=1
if ch==0:
print(-1)
elif n in s:
num=num*10+curr
s.add(curr)
n-=curr
curr-=1
if n in s:
print(-1)
else:
num=num*10+n
num=str(num)
num=num[::-1]
print(num)
else:
num=num*10+n
num=str(num)
num=num[::-1]
print(num)
``` | output | 1 | 102,255 | 20 | 204,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1 | instruction | 0 | 102,256 | 20 | 204,512 |
Tags: brute force, greedy, math
Correct Solution:
```
t = int(input())
for _ in range(t):
x = int(input())
if x > 45:
print(-1)
else:
l = 0
while l*(l+1)//2+((9-l)*l) < x:
l += 1
s = []
for i in range(10-l, 10):
s += [i]
while sum(s)>x:
s[0] -= 1
for i in s:
print(i, end='')
print()
``` | output | 1 | 102,256 | 20 | 204,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1 | instruction | 0 | 102,257 | 20 | 204,514 |
Tags: brute force, greedy, math
Correct Solution:
```
# cook your dish here
from sys import stdin, stdout
import math
from itertools import permutations, combinations
from collections import defaultdict
from collections import Counter
from bisect import bisect_left
import sys
from queue import PriorityQueue
import operator as op
from functools import reduce
import re
mod = 1000000007
input = sys.stdin.readline
def L():
return list(map(int, stdin.readline().split()))
def In():
return map(int, stdin.readline().split())
def I():
return int(stdin.readline())
def printIn(ob):
return stdout.write(str(ob) + '\n')
def powerLL(n, p):
result = 1
while (p):
if (p & 1):
result = result * n % mod
p = int(p / 2)
n = n * n % mod
return result
def ncr(n, r):
r = min(r, n - r)
numer = reduce(op.mul, range(n, n - r, -1), 1)
denom = reduce(op.mul, range(1, r + 1), 1)
return numer // denom
def SieveOfEratosthenes(n):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
prime_list = []
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
prime_list.append(p)
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime_list
def get_divisors(n):
for i in range(1, int(n / 2) + 1):
if n % i == 0:
yield i
yield n
# -------------------------------------
def myCode():
n = I()
if n<=9:
print(n)
elif n>=10 and n<=17:
for i in ("12345678"):
if int(i)+9 == n:
print(i+str(9))
break
elif n>=18 and n<=24:
for i in ("1234567"):
if int(i)+8+9 == n:
print(i+str(89))
break
elif n>=25 and n<=30:
for i in ("123456"):
if int(i)+7+8+9 == n:
print(i+str(789))
break
elif n>=31 and n<=35:
for i in ("12345"):
if int(i)+6+7+8+9 == n:
print(i+str(6789))
break
elif n>=36 and n<=39:
for i in ("1234"):
if int(i)+5+6+7+8+9 == n:
print(i+str(56789))
break
elif n>=40 and n<=42:
for i in ("123"):
if int(i)+4+5+6+7+8+9 == n:
print(i+str(456789))
break
elif n>=43 and n<=44:
for i in ("12"):
if int(i)+3+4+5+6+7+8+9 == n:
print(i+str(3456789))
break
elif n==45:
print(123456789)
else:
print(-1)
def main():
for t in range(I()):
myCode()
if __name__ == '__main__':
# print(int(math.log2(10)))
main()
``` | output | 1 | 102,257 | 20 | 204,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1 | instruction | 0 | 102,258 | 20 | 204,516 |
Tags: brute force, greedy, math
Correct Solution:
```
import collections
from collections import defaultdict
for _ in range(int(input())):
n=int(input())
#s=input()
#n,q=[int(x) for x in input().split()]
if n>45:
print(-1)
continue
if n<10:
print(n)
continue
z = list(range(9,0,-1))
s = 0
ans=[]
for x in z:
if s+x<=n:
ans.append(x)
s+=x
ans=[str(x) for x in ans]
ans.sort()
print("".join(ans))
``` | output | 1 | 102,258 | 20 | 204,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1 | instruction | 0 | 102,259 | 20 | 204,518 |
Tags: brute force, greedy, math
Correct Solution:
```
'''input
1
46
'''
T = int(input())
while T:
dig = 9
ans = ''
x = int(input())
dig = list(range(9,0,-1))
if(x > 45):
print(-1)
T -= 1
continue
for i in dig:
if(x >= i):
ans += str(i)
x -= i
print(ans[::-1])
T -= 1
``` | output | 1 | 102,259 | 20 | 204,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1 | instruction | 0 | 102,260 | 20 | 204,520 |
Tags: brute force, greedy, math
Correct Solution:
```
"""
pppppppppppppppppppp
ppppp ppppppppppppppppppp
ppppppp ppppppppppppppppppppp
pppppppp pppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppp pppppppp
ppppppppppppppppppppp ppppppp
ppppppppppppppppppp ppppp
pppppppppppppppppppp
"""
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
# mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
for _ in range(int(data())):
x = int(data())
if x > 45:
out(-1)
continue
answer = []
temp = 9
while x:
if x >= temp:
answer.append(str(temp))
x -= temp
temp -= 1
else:
answer.append(str(x))
break
out(''.join(answer[::-1]))
``` | output | 1 | 102,260 | 20 | 204,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1
Submitted Solution:
```
import copy
import math
t = int(input())
for case in range(t):
n = int(input())
s =""
k = 9
if(n>45):
print(-1)
continue
while(n>0):
if n>=k:
n-=k
s+=str(k)
k-=1
else:
s+=str(n)
break
print(s[::-1])
``` | instruction | 0 | 102,261 | 20 | 204,522 |
Yes | output | 1 | 102,261 | 20 | 204,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1
Submitted Solution:
```
t = int(input())
while t:
t -= 1;
n = int(input())
if n>45:
print(-1)
continue
x = 0
i = 9
j = 0
while n:
z = min(n, i)
i-=1
x = x + z*(10**j)
n -= z
j += 1
print(x)
``` | instruction | 0 | 102,262 | 20 | 204,524 |
Yes | output | 1 | 102,262 | 20 | 204,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1
Submitted Solution:
```
def main():
n = int(input())
output = []
for i in range(n):
x = int(input())
y = x
for i in range(9, 0, -1):
if x - i >= 0:
output.append(i)
x -= i
if sum(output) != y:
print(-1)
else:
print("".join(sorted([str(x) for x in output])))
output = []
if __name__ == "__main__":
main()
``` | instruction | 0 | 102,263 | 20 | 204,526 |
Yes | output | 1 | 102,263 | 20 | 204,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1
Submitted Solution:
```
def f(x):
s=0
for i in range(10):
if x>=sum(range(i,10)):
s = sum(range(i,10))
break
if x == s:
return "".join(map(str,range(i,10)))
return "".join(map(str, sorted([x-s]+list(range(i,10)))))
t = int(input())
for _ in range(t):
x = int(input())
if x>45:
print(-1)
elif x==45:
print(123456789)
elif x<10:
print(x)
else:
print(f(x))
``` | instruction | 0 | 102,264 | 20 | 204,528 |
Yes | output | 1 | 102,264 | 20 | 204,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
e = [46,47,48,49,50]
if int(n/10)==0:
print(n)
elif n in e:
print(-1)
elif n>9 and n<18:
s=n%9
print(s*(10**1)+9)
elif n>17 and n<25:
s=n%17
print(s*(10**2)+89)
elif n>24 and n<31:
s=n%24
print(s*(10**3)+789)
elif n>30 and n<36:
s=n%30
print(s*(10**4)+6789)
elif n>35 and n<40:
s=n%35
print(s*(10**5)+56789)
elif n>39 and n<43:
s=n%39
print(s*(10**6)+456789)
elif n==44:
print(23456789)
else:
print(123456789)
``` | instruction | 0 | 102,265 | 20 | 204,530 |
No | output | 1 | 102,265 | 20 | 204,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1
Submitted Solution:
```
class var:
list = 50 * [-1]
def sod(n):
s = 0
for j in range(len(n)):
s += ord(n[j]) - 48
return s
def f(s):
val = sod(s)
if val <= 50:
if var.list[val - 1] == -1 or int(s) < var.list[val - 1]:
var.list[val - 1] = int(s)
for j in range(10):
if chr(j + 48) not in s:
f(s + chr(j + 48))
#for j in range(10):
#f(chr(j + 48))
#print(var.list)
List = [1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 29, 39, 49, 59, 69, 79, 89, 189, 289, 389, 489, 589, 689, 789, 1789, 2789, 3789, 4789, 5789, 6789, 16789, 26789, 36789, 46789, 56789, 156789, 256789, 356789, 456789, 1456789, 2456789, 3456789, 13456789, 23456789, 123456789, -1, -1, -1, -1, 0]
for _ in range(int(input())):
print(List[int(input()) - 1])
``` | instruction | 0 | 102,266 | 20 | 204,532 |
No | output | 1 | 102,266 | 20 | 204,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1
Submitted Solution:
```
from math import *
sInt = lambda: int(input())
mInt = lambda: map(int, input().split())
lInt = lambda: list(map(int, input().split()))
t = sInt()
for _ in range(t):
n = sInt()
if n>17:
print(-1)
elif n==1:
print(n)
else:
a = n//2-1
print(a,end='')
print(n-a)
``` | instruction | 0 | 102,267 | 20 | 204,534 |
No | output | 1 | 102,267 | 20 | 204,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).
Input
The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.
Each test case consists of a single integer number x (1 ≤ x ≤ 50).
Output
Output t answers to the test cases:
* if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
* otherwise print -1.
Example
Input
4
1
5
15
50
Output
1
5
69
-1
Submitted Solution:
```
t = int(input())
for test in range(0,t):
n = int(input())
if n>45:
print(-1)
elif len(str(n))==1:
print(n)
else:
s = ''
big = 9
while(len(str(n))>1):
n = n-big
s = s+str(big)
big = big-1
if n<=big:
s = s+str(n)
s = s[::-1]
print(int(s))
else:
k = n-big
s = s+str(big)+str(k)
s = s[::-1]
print(int(s))
``` | instruction | 0 | 102,268 | 20 | 204,536 |
No | output | 1 | 102,268 | 20 | 204,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110 | instruction | 0 | 102,301 | 20 | 204,602 |
Tags: binary search, dp, math, number theory
Correct Solution:
```
t = int(input())
while(t > 0):
t -= 1
l, r = [int(x) for x in input().split()]
num1 = str(l)
num2 = str(r)
ans = 0
while(len(num1) != len(num2)):
num1 = '0' + num1
for i in range(len(num1)):
k1 = int(num1[:i+1])
k2 = int(num2[:i+1])
ans += k2 - k1
print(ans)
``` | output | 1 | 102,301 | 20 | 204,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110 | instruction | 0 | 102,302 | 20 | 204,604 |
Tags: binary search, dp, math, number theory
Correct Solution:
```
for _ in range(int(input())):
l,r=map(int,input().split())
ansl=0
ansr=0
ansl+=l-1
ansr+=r-1
# print(ansl,ansr)
while(l):
ansl+=l//10
l//=10
while(r):
ansr+=r//10
r//=10
print(ansr-ansl)
``` | output | 1 | 102,302 | 20 | 204,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110 | instruction | 0 | 102,303 | 20 | 204,606 |
Tags: binary search, dp, math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
L, R = [int(x) for x in input().split()]
ans = R - L
cnt_r = 0
cnt_l = 0
for i in range(1, 20):
cnt_r += R // pow(10, i)
cnt_l += L // pow(10, i)
ans += cnt_r - cnt_l
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 102,303 | 20 | 204,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110 | instruction | 0 | 102,304 | 20 | 204,608 |
Tags: binary search, dp, math, number theory
Correct Solution:
```
for s in[*open(0)][1:]:
x,y=map(int,s.split());r=0
while y:r+=y-x;x//=10;y//=10
print(r)
``` | output | 1 | 102,304 | 20 | 204,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110 | instruction | 0 | 102,305 | 20 | 204,610 |
Tags: binary search, dp, math, number theory
Correct Solution:
```
def main():
ans = [] #Respuestas
for _ in range(int(input())):
l, t = map(int, input().split()) #Numeros iniciales
cha = 0 #Cantidad de cambios
while t > 0:
cha += t - l
l //= 10
t //= 10
ans.append(cha)
for i in ans: print(i)
if __name__ == "__main__":
main()
``` | output | 1 | 102,305 | 20 | 204,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110 | instruction | 0 | 102,306 | 20 | 204,612 |
Tags: binary search, dp, math, number theory
Correct Solution:
```
t=int(input())
for _ in range(t):
l,r=map(int,input().split())
c1=r-l
for i in range(1,10):
c1+=r//(10**i)
c2=0
for i in range(1,10):
c2+=l//(10**i)
print(c1-c2)
``` | output | 1 | 102,306 | 20 | 204,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110 | instruction | 0 | 102,307 | 20 | 204,614 |
Tags: binary search, dp, math, number theory
Correct Solution:
```
for _ in range(int(input())):
l,r = map(int,input().split())
ans = 0
while r:
ans += r-l
r //= 10
l //= 10
print(ans)
``` | output | 1 | 102,307 | 20 | 204,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110 | instruction | 0 | 102,308 | 20 | 204,616 |
Tags: binary search, dp, math, number theory
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
c=[1,11,111,1111,11111,111111,1111111,11111111,111111111,1111111111]
for _ in range(t):
l,r=map(int,input().split())
aa=0
a=len(str(r))
r=str(r)
for i in range(a):
aa+=int(r[i])*c[a-i-1]
bb=0
b=len(str(l))
l=str(l)
for i in range(b):
bb+=int(l[i])*c[b-i-1]
print(aa-bb)
``` | output | 1 | 102,308 | 20 | 204,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110
Submitted Solution:
```
def func(r):
if r==1:
return 0
total_digits = len(str(r))
nums_divisible_by_ten = [0 for i in range(total_digits+1)]
starting = total_digits-1
count = 0
while starting>0:
divisor = 10**(starting)
nums_divisible_by_ten[starting] = (r//divisor)-count
count = (r//divisor)
starting-=1
total_sum = 0
for j in range(1, total_digits+1):
total_sum += nums_divisible_by_ten[j]*(j+1)
total_sum += r-sum(nums_divisible_by_ten)-1
return total_sum
t = int(input())
for i in range(t):
l, r = map(int, input().split())
print(func(r)-func(l))
``` | instruction | 0 | 102,309 | 20 | 204,618 |
Yes | output | 1 | 102,309 | 20 | 204,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110
Submitted Solution:
```
T=int(input())
for t in range(T):
l,r=map(int,input().split())
ans=0
for i in range(10):
div=10**i
ans+=r//div-l//div
print(ans)
``` | instruction | 0 | 102,310 | 20 | 204,620 |
Yes | output | 1 | 102,310 | 20 | 204,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110
Submitted Solution:
```
for s in[*open(0)][1:]:
r=0;k=-1
for x in map(int,s.split()):
while x:r+=k*x;x//=10
k=1
print(r)
``` | instruction | 0 | 102,311 | 20 | 204,622 |
Yes | output | 1 | 102,311 | 20 | 204,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110
Submitted Solution:
```
import time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
from collections import defaultdict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
def readTree(n,e): # to read tree
adj=[set() for i in range(n+1)]
for i in range(e):
u1,u2=IP()
adj[u1].add(u2)
return adj
def sieve():
li=[True]*(10**3+5)
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime,cur=[0]*200,0
for i in range(10**3+5):
if li[i]==True:
prime[cur]=i
cur+=1
return prime
def SPF():
mx=(10**7+1)
spf=[mx]*(mx)
spf[1]=1
for i in range(2,mx):
if spf[i]==mx:
spf[i]=i
for j in range(i*i,mx,i):
if i<spf[j]:
spf[j]=i
return spf
def prime(n,d):
while n!=1:
d[spf[n]]=d.get(spf[n],0)+1
n=n//spf[n]
return
#####################################################################################
mod = 1000000007
inf = 1e18
def solve():
l,r=IP()
strt = 1
ans=0
while r:
ans+=(r-l)
r//=10
l//=10
print(ans)
return
t = II()
for i in range(t):
solve()
#######
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``` | instruction | 0 | 102,312 | 20 | 204,624 |
Yes | output | 1 | 102,312 | 20 | 204,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110
Submitted Solution:
```
for _ in range(int(input())):
l,r=map(int,input().split())
tmp=1
count=0
a=1
while tmp//a>0:
count += tmp//a
a*=10
tmp=r
count1=0
a=1
while tmp//a>0:
count1 += tmp//a
a*=10
print(count1-count)
``` | instruction | 0 | 102,313 | 20 | 204,626 |
No | output | 1 | 102,313 | 20 | 204,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110
Submitted Solution:
```
# by the authority of GOD author: Kritarth Sharma #
import math,copy
from collections import defaultdict,Counter
#from itertools import permutations
def fact(n):
return 1 if (n == 1 or n == 0) else n * fact(n - 1)
def prime(n) :
if (n <= 1) :
return False
if (n <= 3) :
return True
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def inp():
l=list(map(int,input().split()))
return l
## code starts here
#################################################################################################################
def ll(s):
s=str(s)
k=1
for i in range(len(s)-1,-1,-1):
if s[i]=='9':
k+=1
else:
break
return k
def main():
for _ in range(int(input())):
l,r=inp()
k=0
while l<r:
if l+1000000000<r:
k+=1111111111
l+=1000000000
elif l+100000000<r:
k+=111111111
l+=100000000
elif l+10000000<r:
k+=11111111
l+=10000000
elif l+1000000<r:
k+=1111111
l+=1000000
elif l+100000<r:
k+=111111
l+=100000
elif l+10000<r:
k+=11111
l+=10000
elif l+1000<r:
k+=1111
l+=1000
elif l+100<r:
k+=111
l+=100
elif l+10<r:
k+=11
l+=10
else:
k+=ll(l)
l+=1
print(k)
import os,sys
from io import BytesIO,IOBase
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
``` | instruction | 0 | 102,314 | 20 | 204,628 |
No | output | 1 | 102,314 | 20 | 204,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110
Submitted Solution:
```
t = int(input())
for _ in range(t):
l,r = map(int,input().split())
res = r-l
nin = res//9
res+= (nin)
if(l%9==0 and r%10==2):
res+=1
print(res)
``` | instruction | 0 | 102,315 | 20 | 204,630 |
No | output | 1 | 102,315 | 20 | 204,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r, where l < r. We will add 1 to l until the result is equal to r. Thus, there will be exactly r-l additions performed. For each such addition, let's look at the number of digits that will be changed after it.
For example:
* if l=909, then adding one will result in 910 and 2 digits will be changed;
* if you add one to l=9, the result will be 10 and 2 digits will also be changed;
* if you add one to l=489999, the result will be 490000 and 5 digits will be changed.
Changed digits always form a suffix of the result written in the decimal system.
Output the total number of changed digits, if you want to get r from l, adding 1 each time.
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is characterized by two integers l and r (1 ≤ l < r ≤ 10^9).
Output
For each test case, calculate the total number of changed digits if you want to get r from l, adding one each time.
Example
Input
4
1 9
9 10
10 20
1 1000000000
Output
8
2
11
1111111110
Submitted Solution:
```
t = int(input())
for _ in range(t):
# n = int(input())
l,r = map(int, input().split())
q = str(l)
w = str(r)
cnt = -len(q)+len(w)
x = r-l
while x>0:
cnt+=x
x=x//10
print(cnt)
``` | instruction | 0 | 102,316 | 20 | 204,632 |
No | output | 1 | 102,316 | 20 | 204,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There have recently been elections in the zoo. Overall there were 7 main political parties: one of them is the Little Elephant Political Party, 6 other parties have less catchy names.
Political parties find their number in the ballot highly important. Overall there are m possible numbers: 1, 2, ..., m. Each of these 7 parties is going to be assigned in some way to exactly one number, at that, two distinct parties cannot receive the same number.
The Little Elephant Political Party members believe in the lucky digits 4 and 7. They want to evaluate their chances in the elections. For that, they need to find out, how many correct assignments are there, such that the number of lucky digits in the Little Elephant Political Party ballot number is strictly larger than the total number of lucky digits in the ballot numbers of 6 other parties.
Help the Little Elephant Political Party, calculate this number. As the answer can be rather large, print the remainder from dividing it by 1000000007 (109 + 7).
Input
A single line contains a single positive integer m (7 ≤ m ≤ 109) — the number of possible numbers in the ballot.
Output
In a single line print a single integer — the answer to the problem modulo 1000000007 (109 + 7).
Examples
Input
7
Output
0
Input
8
Output
1440
Submitted Solution:
```
def fact(n):
if n==0:return 1
return n*fact(n-1)
n = int(input())
print((2*fact(n-2))%1000000007 if n>7 else 0)
``` | instruction | 0 | 102,350 | 20 | 204,700 |
No | output | 1 | 102,350 | 20 | 204,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.
Initially you have a sequence of n integers: 1, 2, ..., n. In a single step, you can pick two of them, let's denote them a and b, erase them from the sequence, and append to the sequence either a + b, or a - b, or a × b.
After n - 1 steps there is only one number left. Can you make this number equal to 24?
Input
The first line contains a single integer n (1 ≤ n ≤ 105).
Output
If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).
If there is a way to obtain 24 as the result number, in the following n - 1 lines print the required operations an operation per line. Each operation should be in form: "a op b = c". Where a and b are the numbers you've picked at this operation; op is either "+", or "-", or "*"; c is the result of corresponding operation. Note, that the absolute value of c mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.
If there are multiple valid answers, you may print any of them.
Examples
Input
1
Output
NO
Input
8
Output
YES
8 * 7 = 56
6 * 5 = 30
3 - 4 = -1
1 - 2 = -1
30 - -1 = 31
56 - 31 = 25
25 + -1 = 24 | instruction | 0 | 102,441 | 20 | 204,882 |
Tags: constructive algorithms, greedy, math
Correct Solution:
```
n = int(input())
if n < 4: print("NO")
else:
print("YES")
while n > 5:
a = str(n)
b = str(n-1)
print(a + " - " + b + " = 1 ")
print("1 * 1 = 1")
n -= 2
if n == 4: print("4 * 3 = 12\n12 * 2 = 24\n24 * 1 = 24")
else: print("5 - 2 = 3\n3 - 1 = 2\n2 * 3 = 6\n6 * 4 = 24")
# Made By Mostafa_Khaled
``` | output | 1 | 102,441 | 20 | 204,883 |
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