message stringlengths 2 23.4k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 129 108k | cluster float64 6 6 | __index_level_0__ int64 258 216k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 10), number of sentences in the chat record. Each of the next n lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.
Output
For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.
Examples
Input
5
I will go to play with you lala.
wow, welcome.
miao.lala.
miao.
miao .
Output
Freda's
OMG>.< I don't know!
OMG>.< I don't know!
Rainbow's
OMG>.< I don't know!
Submitted Solution:
```
try:
n = int(input())
for i in range(n):
s = input().split()
if (s[len(s)-1] == "lala." and s[0] == "miao.") or (not s[len(s)-1] == "lala." and not s[0] == "miao."):
print("OMG>.< I don't know!" )
elif s[0] == "miao.":
print("Rainbow's")
else:
print("Freda's")
except:
pass
``` | instruction | 0 | 64,317 | 6 | 128,634 |
No | output | 1 | 64,317 | 6 | 128,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input
Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.
The string s contains lowercase and uppercase English letters, i.e. <image>.
Output
Output a single integer, the answer to the problem.
Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | instruction | 0 | 64,482 | 6 | 128,964 |
Tags: implementation
Correct Solution:
```
s = input()
l = "Bulbasaur"
l1 = []
for i in l:
if i=="u":
l1.append(s.count("u")//2)
elif i=="a":
l1.append(s.count("a")//2)
else:
l1.append(s.count(i))
if 0 in l1:print(0)
else:print(min(l1))
``` | output | 1 | 64,482 | 6 | 128,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input
Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.
The string s contains lowercase and uppercase English letters, i.e. <image>.
Output
Output a single integer, the answer to the problem.
Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | instruction | 0 | 64,483 | 6 | 128,966 |
Tags: implementation
Correct Solution:
```
s = input()
#l = ['B', 'u', 'l', 'b', 'a', 's', 'a', 'u', 'r']
l = [0, 0, 0, 0, 0, 0, 0]
for i in s:
if i == 'B':
l[0] += 1
elif i == 'u':
l[1] += 1
elif i == 'l':
l[2] += 1
elif i == 'b':
l[3] += 1
elif i == 'a':
l[4] += 1
elif i == 's':
l[5] += 1
elif i == 'r':
l[6] += 1
l[1] //= 2
l[4] //= 2
print(min(l))
``` | output | 1 | 64,483 | 6 | 128,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input
Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.
The string s contains lowercase and uppercase English letters, i.e. <image>.
Output
Output a single integer, the answer to the problem.
Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | instruction | 0 | 64,484 | 6 | 128,968 |
Tags: implementation
Correct Solution:
```
s = input()
B = s.count("B")
u = s.count("u")//2
l = s.count("l")
b = s.count("b")
a = s.count("a")//2
s1 = s.count("s")
r = s.count("r")
print(min(B, u, l, b, a, s1, r))
``` | output | 1 | 64,484 | 6 | 128,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input
Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.
The string s contains lowercase and uppercase English letters, i.e. <image>.
Output
Output a single integer, the answer to the problem.
Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | instruction | 0 | 64,485 | 6 | 128,970 |
Tags: implementation
Correct Solution:
```
texto=input()
s=len(texto)
contador={'B':0,'u':0,'l':0,'b':0,'a':0,'s':0,'a':0,'u':0,'r':0}
for i in range(s):
letra=texto[i]
if letra in contador:
contador[letra]+=1
minimo=contador['B']
for letra in contador:
if contador[letra]<minimo:
minimo=contador[letra]
cantidad_a=contador['a']
cantidad_u=contador['u']
if (cantidad_a/2>=minimo) and (cantidad_u/2>=minimo):
print(minimo)
else:
if cantidad_a<cantidad_u:
print(cantidad_a//2)
else:
print(cantidad_u//2)
``` | output | 1 | 64,485 | 6 | 128,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input
Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.
The string s contains lowercase and uppercase English letters, i.e. <image>.
Output
Output a single integer, the answer to the problem.
Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | instruction | 0 | 64,486 | 6 | 128,972 |
Tags: implementation
Correct Solution:
```
w="Bulbasaur"
s=input()
ans=10**5+1
chk={}
for i in s:
if i in chk:
chk[i]+=1
else:
chk[i]=1
for i in w:
if i in chk:
ans=min(ans,chk[i]//[1,2][i=='a' or i=='u'])
else:
ans=0
print(ans)
``` | output | 1 | 64,486 | 6 | 128,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input
Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.
The string s contains lowercase and uppercase English letters, i.e. <image>.
Output
Output a single integer, the answer to the problem.
Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | instruction | 0 | 64,487 | 6 | 128,974 |
Tags: implementation
Correct Solution:
```
ans = 0
used = [0] * 256
a = [0] * 256
q = 'Bulbasaur'
for elem in q:
a[ord(elem)] += 1
s = input()
for elem in s:
used[ord(elem)] += 1
ans = 98148914891413
for i in range(256):
if (a[i] != 0):
ans = min(ans, used[i] // a[i])
print(ans)
``` | output | 1 | 64,487 | 6 | 128,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input
Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.
The string s contains lowercase and uppercase English letters, i.e. <image>.
Output
Output a single integer, the answer to the problem.
Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | instruction | 0 | 64,488 | 6 | 128,976 |
Tags: implementation
Correct Solution:
```
letters = set('Bulbasaur')#
string = input()
counts = []
for i in letters:
if i in ['a', 'u']:
counts.append(string.count(i)//2)
else:
counts.append(string.count(i))
print(min(counts))
``` | output | 1 | 64,488 | 6 | 128,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input
Input contains a single line containing a string s (1 ≤ |s| ≤ 105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.
The string s contains lowercase and uppercase English letters, i.e. <image>.
Output
Output a single integer, the answer to the problem.
Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | instruction | 0 | 64,489 | 6 | 128,978 |
Tags: implementation
Correct Solution:
```
s=input()
a=[]
q=s.count('B')
w=s.count('u')
e=s.count('l')
r=s.count('b')
t=s.count('a')
y=s.count('s')
u=s.count('r')
if q>=1 and w>=2 and e>=1 and r>=1 and t>=2 and y>=1 and u>=1:
w=w//2
t=t//2
a.append(q)
a.append(w)
a.append(e)
a.append(r)
a.append(t)
a.append(y)
a.append(u)
print(min(a))
else:
print(0)
``` | output | 1 | 64,489 | 6 | 128,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peterson loves to learn new languages, but his favorite hobby is making new ones. Language is a set of words, and word is a sequence of lowercase Latin letters.
Peterson makes new language every morning. It is difficult task to store the whole language, so Peterson have invented new data structure for storing his languages which is called broom. Broom is rooted tree with edges marked with letters. Initially broom is represented by the only vertex — the root of the broom. When Peterson wants to add new word to the language he stands at the root and processes the letters of new word one by one. Consider that Peterson stands at the vertex u. If there is an edge from u marked with current letter, Peterson goes through this edge. Otherwise Peterson adds new edge from u to the new vertex v, marks it with the current letter and goes through the new edge. Size of broom is the number of vertices in it.
In the evening after working day Peterson can't understand the language he made this morning. It is too difficult for bored Peterson and he tries to make it simpler. Simplification of the language is the process of erasing some letters from some words of this language. Formally, Peterson takes some positive integer p and erases p-th letter from all the words of this language having length at least p. Letters in words are indexed starting by 1. Peterson considers that simplification should change at least one word, i.e. there has to be at least one word of length at least p. Peterson tries to make his language as simple as possible, so he wants to choose p such that the size of the broom for his simplified language is as small as possible.
Peterson is pretty annoyed with this task so he asks you for help. Write a program to find the smallest possible size of the broom and integer p.
Input
The first line of input contains integer n (2 ≤ n ≤ 3·105) — the size of the broom.
Next n - 1 lines describe the broom: i-th of them contains integers ui, vi and letter xi — describing the edge from ui to vi marked with letter xi.
Vertices are numbered from 1 to n. All xi are lowercase latin letters. Vertex 1 is the root of the broom.
Edges describe correct broom which is made from Peterson's language.
Output
The first line of output should contain the minimum possible size of the broom after its simplification. The second line of output should contain integer p to choose. If there are several suitable p values, print the smallest one.
Examples
Input
5
1 2 c
2 3 a
3 4 t
2 5 t
Output
3
2
Input
16
1 2 o
2 3 f
1 4 p
4 5 i
5 6 e
6 7 c
7 8 e
4 9 r
9 10 e
10 11 t
11 12 t
12 13 y
10 14 f
14 15 i
15 16 x
Output
12
2
Note
<image>
Broom from the second sample test can be built using language "piece", "of", "pie", "pretty", "prefix". Its simplification with p = 2 obtains the language of words "pece", "o", "pe", "petty", "pefix". This language gives us the broom with minimum possible size.
Submitted Solution:
```
from collections import defaultdict
nodes = int(input())
levels = defaultdict(list)
broom = defaultdict(list)
relation = defaultdict(str)
for i in range(1, nodes):
a, b, L = [x for x in input().split()]
broom[a].append(b)
relation[a] = L
queue = ['flag','1']
flag = True
cont = 0
indice = 0
maximo = 0
maxindice = 0
lastlast = ''
while len(queue) > 0:
last = queue.pop()
# print(last)
if last != 'flag':
# print(cont)
levels[indice].append(last)
cont += 1
lista = broom[last]
for i in lista:
queue.insert(0, i)
else:
if lastlast == last:
break
queue.insert(0, 'flag')
indice += 1
# print(cont)
if cont > maximo:
maxindice = indice
maximo = cont
cont = 0
lastlast = last
a = []
for x in levels[maxindice - 1]:
a.append(relation[x])
diferencia = len(levels[maxindice]) - (len(set(a)) - 1)
# print(relation)
# print(levels)
print(nodes - maximo - diferencia)
print (maxindice - 1)
``` | instruction | 0 | 64,498 | 6 | 128,996 |
No | output | 1 | 64,498 | 6 | 128,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the Search Ultimate program that searched for strings in a text failed, Igor K. got to think: "Why on Earth does my program work so slowly?" As he double-checked his code, he said: "My code contains no errors, yet I know how we will improve Search Ultimate!" and took a large book from the shelves. The book read "Azembler. Principally New Approach".
Having carefully thumbed through the book, Igor K. realised that, as it turns out, you can multiply the numbers dozens of times faster. "Search Ultimate will be faster than it has ever been!" — the fellow shouted happily and set to work.
Let us now clarify what Igor's idea was. The thing is that the code that was generated by a compiler was far from perfect. Standard multiplying does work slower than with the trick the book mentioned.
The Azembler language operates with 26 registers (eax, ebx, ..., ezx) and two commands:
* [x] — returns the value located in the address x. For example, [eax] returns the value that was located in the address, equal to the value in the register eax.
* lea x, y — assigns to the register x, indicated as the first operand, the second operand's address. Thus, for example, the "lea ebx, [eax]" command will write in the ebx register the content of the eax register: first the [eax] operation will be fulfilled, the result of it will be some value that lies in the address written in eax. But we do not need the value — the next operation will be lea, that will take the [eax] address, i.e., the value in the eax register, and will write it in ebx.
On the first thought the second operation seems meaningless, but as it turns out, it is acceptable to write the operation as
lea ecx, [eax + ebx],
lea ecx, [k*eax]
or even
lea ecx, [ebx + k*eax],
where k = 1, 2, 4 or 8.
As a result, the register ecx will be equal to the numbers eax + ebx, k*eax and ebx + k*eax correspondingly. However, such operation is fulfilled many times, dozens of times faster that the usual multiplying of numbers. And using several such operations, one can very quickly multiply some number by some other one. Of course, instead of eax, ebx and ecx you are allowed to use any registers.
For example, let the eax register contain some number that we should multiply by 41. It takes us 2 lines:
lea ebx, [eax + 4*eax] // now ebx = 5*eax
lea eax, [eax + 8*ebx] // now eax = eax + 8*ebx = 41*eax
Igor K. got interested in the following question: what is the minimum number of lea operations needed to multiply by the given number n and how to do it? Your task is to help him.
Consider that at the initial moment of time eax contains a number that Igor K. was about to multiply by n, and the registers from ebx to ezx contain number 0. At the final moment of time the result can be located in any register.
Input
The input data contain the only integer n (1 ≤ n ≤ 255), which Igor K. is about to multiply.
Output
On the first line print number p, which represents the minimum number of lea operations, needed to do that. Then print the program consisting of p commands, performing the operations. It is guaranteed that such program exists for any n from 1 to 255.
Use precisely the following format of commands (here k is equal to 1, 2, 4 or 8, and x, y and z are any, even coinciding registers):
lea x, [y]
lea x, [y + z]
lea x, [k*y]
lea x, [y + k*z]
Please note that extra spaces at the end of a command are unacceptable.
Examples
Input
41
Output
2
lea ebx, [eax + 4*eax]
lea ecx, [eax + 8*ebx]
Input
2
Output
1
lea ebx, [eax + eax]
Input
4
Output
1
lea ebx, [4*eax] | instruction | 0 | 64,546 | 6 | 129,092 |
Tags: brute force, implementation
Correct Solution:
```
n=int(input())
mn=[float('inf')]*256
ans=""
done=set()
def gen(vals,st):
global mn,ans
if('|'.join([str(x) for x in vals]) in done):
return
done.add('|'.join([str(x) for x in vals]))
if(vals[-1]==n):
if(len(vals)<mn[n]):
mn[n]=len(vals)
ans=st
return
if(len(vals)>5):
return
md=set()
for i in range(len(vals)):
for z in [8,4,2,1]:
e=vals[i]*z
if(e>n):
continue
if(e in md):
continue
if(e>vals[-1]):
nw='e'+chr(ord('a')+len(vals))+'x'
I='e'+chr(ord('a')+i)+'x'
gen(vals+[e],st+"lea "+nw+", ["+str(z)+"*"+I+"]\n")
md.add(e)
for j in range(len(vals)):
for z in [8,4,2,1]:
e=vals[i]+z*vals[j]
if(e>n):
continue
if(e in md):
continue
if(e>vals[-1]):
nw='e'+chr(ord('a')+len(vals))+'x'
I='e'+chr(ord('a')+i)+'x'
J='e'+chr(ord('a')+j)+'x'
gen(vals+[e],st+"lea "+nw+", ["+I+" + "+str(z)+"*"+J+"]\n")
md.add(e)
gen([1],"")
print(ans.count("\n"))
print(ans,end="")
``` | output | 1 | 64,546 | 6 | 129,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the Search Ultimate program that searched for strings in a text failed, Igor K. got to think: "Why on Earth does my program work so slowly?" As he double-checked his code, he said: "My code contains no errors, yet I know how we will improve Search Ultimate!" and took a large book from the shelves. The book read "Azembler. Principally New Approach".
Having carefully thumbed through the book, Igor K. realised that, as it turns out, you can multiply the numbers dozens of times faster. "Search Ultimate will be faster than it has ever been!" — the fellow shouted happily and set to work.
Let us now clarify what Igor's idea was. The thing is that the code that was generated by a compiler was far from perfect. Standard multiplying does work slower than with the trick the book mentioned.
The Azembler language operates with 26 registers (eax, ebx, ..., ezx) and two commands:
* [x] — returns the value located in the address x. For example, [eax] returns the value that was located in the address, equal to the value in the register eax.
* lea x, y — assigns to the register x, indicated as the first operand, the second operand's address. Thus, for example, the "lea ebx, [eax]" command will write in the ebx register the content of the eax register: first the [eax] operation will be fulfilled, the result of it will be some value that lies in the address written in eax. But we do not need the value — the next operation will be lea, that will take the [eax] address, i.e., the value in the eax register, and will write it in ebx.
On the first thought the second operation seems meaningless, but as it turns out, it is acceptable to write the operation as
lea ecx, [eax + ebx],
lea ecx, [k*eax]
or even
lea ecx, [ebx + k*eax],
where k = 1, 2, 4 or 8.
As a result, the register ecx will be equal to the numbers eax + ebx, k*eax and ebx + k*eax correspondingly. However, such operation is fulfilled many times, dozens of times faster that the usual multiplying of numbers. And using several such operations, one can very quickly multiply some number by some other one. Of course, instead of eax, ebx and ecx you are allowed to use any registers.
For example, let the eax register contain some number that we should multiply by 41. It takes us 2 lines:
lea ebx, [eax + 4*eax] // now ebx = 5*eax
lea eax, [eax + 8*ebx] // now eax = eax + 8*ebx = 41*eax
Igor K. got interested in the following question: what is the minimum number of lea operations needed to multiply by the given number n and how to do it? Your task is to help him.
Consider that at the initial moment of time eax contains a number that Igor K. was about to multiply by n, and the registers from ebx to ezx contain number 0. At the final moment of time the result can be located in any register.
Input
The input data contain the only integer n (1 ≤ n ≤ 255), which Igor K. is about to multiply.
Output
On the first line print number p, which represents the minimum number of lea operations, needed to do that. Then print the program consisting of p commands, performing the operations. It is guaranteed that such program exists for any n from 1 to 255.
Use precisely the following format of commands (here k is equal to 1, 2, 4 or 8, and x, y and z are any, even coinciding registers):
lea x, [y]
lea x, [y + z]
lea x, [k*y]
lea x, [y + k*z]
Please note that extra spaces at the end of a command are unacceptable.
Examples
Input
41
Output
2
lea ebx, [eax + 4*eax]
lea ecx, [eax + 8*ebx]
Input
2
Output
1
lea ebx, [eax + eax]
Input
4
Output
1
lea ebx, [4*eax] | instruction | 0 | 64,547 | 6 | 129,094 |
Tags: brute force, implementation
Correct Solution:
```
n=int(input())
mn=[float('inf')]*256
ans=""
done=set()
def gen(vals,st):
global mn,ans
#print(*vals)
if('|'.join([str(x) for x in vals]) in done):
return
done.add('|'.join([str(x) for x in vals]))
if(vals[-1]==n):
if(len(vals)<mn[n]):
mn[n]=len(vals)
ans=st
return
if(len(vals)>5):
return
"""if(len(vals)-2>mn[vals[-1]]):
return
mn[vals[-1]]=len(vals)"""
for i in range(len(vals)):
for z in [8,4,2,1]:
e=vals[i]*z
if(e>n):
continue
if(e>vals[-1]):
nw='e'+chr(ord('a')+len(vals))+'x'
I='e'+chr(ord('a')+i)+'x'
gen(vals+[e],st+"lea "+nw+", ["+str(z)+"*"+I+"]\n")
for j in range(len(vals)):
for z in [8,4,2,1]:
e=vals[i]+z*vals[j]
if(e>n):
continue
if(e>vals[-1]):
nw='e'+chr(ord('a')+len(vals))+'x'
I='e'+chr(ord('a')+i)+'x'
J='e'+chr(ord('a')+j)+'x'
gen(vals+[e],st+"lea "+nw+", ["+I+" + "+str(z)+"*"+J+"]\n")
gen([1],"")
print(ans.count("\n"))
print(ans,end="")
``` | output | 1 | 64,547 | 6 | 129,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the Search Ultimate program that searched for strings in a text failed, Igor K. got to think: "Why on Earth does my program work so slowly?" As he double-checked his code, he said: "My code contains no errors, yet I know how we will improve Search Ultimate!" and took a large book from the shelves. The book read "Azembler. Principally New Approach".
Having carefully thumbed through the book, Igor K. realised that, as it turns out, you can multiply the numbers dozens of times faster. "Search Ultimate will be faster than it has ever been!" — the fellow shouted happily and set to work.
Let us now clarify what Igor's idea was. The thing is that the code that was generated by a compiler was far from perfect. Standard multiplying does work slower than with the trick the book mentioned.
The Azembler language operates with 26 registers (eax, ebx, ..., ezx) and two commands:
* [x] — returns the value located in the address x. For example, [eax] returns the value that was located in the address, equal to the value in the register eax.
* lea x, y — assigns to the register x, indicated as the first operand, the second operand's address. Thus, for example, the "lea ebx, [eax]" command will write in the ebx register the content of the eax register: first the [eax] operation will be fulfilled, the result of it will be some value that lies in the address written in eax. But we do not need the value — the next operation will be lea, that will take the [eax] address, i.e., the value in the eax register, and will write it in ebx.
On the first thought the second operation seems meaningless, but as it turns out, it is acceptable to write the operation as
lea ecx, [eax + ebx],
lea ecx, [k*eax]
or even
lea ecx, [ebx + k*eax],
where k = 1, 2, 4 or 8.
As a result, the register ecx will be equal to the numbers eax + ebx, k*eax and ebx + k*eax correspondingly. However, such operation is fulfilled many times, dozens of times faster that the usual multiplying of numbers. And using several such operations, one can very quickly multiply some number by some other one. Of course, instead of eax, ebx and ecx you are allowed to use any registers.
For example, let the eax register contain some number that we should multiply by 41. It takes us 2 lines:
lea ebx, [eax + 4*eax] // now ebx = 5*eax
lea eax, [eax + 8*ebx] // now eax = eax + 8*ebx = 41*eax
Igor K. got interested in the following question: what is the minimum number of lea operations needed to multiply by the given number n and how to do it? Your task is to help him.
Consider that at the initial moment of time eax contains a number that Igor K. was about to multiply by n, and the registers from ebx to ezx contain number 0. At the final moment of time the result can be located in any register.
Input
The input data contain the only integer n (1 ≤ n ≤ 255), which Igor K. is about to multiply.
Output
On the first line print number p, which represents the minimum number of lea operations, needed to do that. Then print the program consisting of p commands, performing the operations. It is guaranteed that such program exists for any n from 1 to 255.
Use precisely the following format of commands (here k is equal to 1, 2, 4 or 8, and x, y and z are any, even coinciding registers):
lea x, [y]
lea x, [y + z]
lea x, [k*y]
lea x, [y + k*z]
Please note that extra spaces at the end of a command are unacceptable.
Examples
Input
41
Output
2
lea ebx, [eax + 4*eax]
lea ecx, [eax + 8*ebx]
Input
2
Output
1
lea ebx, [eax + eax]
Input
4
Output
1
lea ebx, [4*eax]
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
if n == 1:
print(0)
exit()
inf = 10**9
dp = [inf] * (n + 1)
dp[1] = 0
prev = [[-1, -1, -1, -1] for _ in range(n + 1)]
def ok():
path = []
used_num = {n}
stack = [n]
while stack:
z = stack.pop()
if z == 1:
continue
x, mx, y, my = prev[z]
path.append([x, mx, y, my, z])
if x not in used_num:
used_num.add(x)
stack.append(x)
if y != -1 and y not in used_num:
used_num.add(y)
stack.append(y)
register_name = {x: f'e{chr(97+i)}x' for i, x in enumerate(sorted(used_num))}
ans = []
used = [0] * len(path)
for ub in sorted(used_num):
for i, (x, mx, y, my, z) in enumerate(path):
if not used[i] and max(x, y, z) <= ub:
if y == -1:
ans.append(f'lea {register_name[z]}, [{str(mx)+"*" if mx > 1 else ""}{register_name[x]}]')
else:
ans.append(f'lea {register_name[z]}, [{str(mx)+"*" if mx > 1 else ""}{register_name[x]} + {str(my)+"*" if my > 1 else ""}{register_name[y]}]')
used[i] = 1
print(len(ans))
print('\n'.join(ans))
exit()
def calc(x, y):
visited = set()
stack = [x, y]
res = 0
while stack:
num = stack.pop()
if num == 1:
continue
res += 1
if prev[num][0] != -1 and prev[num][0] not in visited:
visited.add(prev[num][0])
stack.append(prev[num][0])
if prev[num][2] != -1 and prev[num][2] not in visited:
visited.add(prev[num][2])
stack.append(prev[num][2])
return res
while True:
flag = 0
next_dp = dp[:]
for x in range(1, n):
if dp[x] == inf:
continue
for m in (2, 4, 8):
if x * m <= n and next_dp[x * m] > dp[x] + 1:
flag = 1
next_dp[x * m] = dp[x] + 1
prev[x * m] = [x, m, -1, -1]
for y in range(1, n):
if dp[y] == inf:
continue
for my in (1, 2, 4, 8):
if x + y * my <= n and next_dp[x + y * my] > calc(x, y) + 1:
flag = 1
next_dp[x + y * my] = calc(x, y) + 1
prev[x + y * my] = [x, 1, y, my]
dp = next_dp
if not flag:
ok()
``` | instruction | 0 | 64,548 | 6 | 129,096 |
No | output | 1 | 64,548 | 6 | 129,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the Search Ultimate program that searched for strings in a text failed, Igor K. got to think: "Why on Earth does my program work so slowly?" As he double-checked his code, he said: "My code contains no errors, yet I know how we will improve Search Ultimate!" and took a large book from the shelves. The book read "Azembler. Principally New Approach".
Having carefully thumbed through the book, Igor K. realised that, as it turns out, you can multiply the numbers dozens of times faster. "Search Ultimate will be faster than it has ever been!" — the fellow shouted happily and set to work.
Let us now clarify what Igor's idea was. The thing is that the code that was generated by a compiler was far from perfect. Standard multiplying does work slower than with the trick the book mentioned.
The Azembler language operates with 26 registers (eax, ebx, ..., ezx) and two commands:
* [x] — returns the value located in the address x. For example, [eax] returns the value that was located in the address, equal to the value in the register eax.
* lea x, y — assigns to the register x, indicated as the first operand, the second operand's address. Thus, for example, the "lea ebx, [eax]" command will write in the ebx register the content of the eax register: first the [eax] operation will be fulfilled, the result of it will be some value that lies in the address written in eax. But we do not need the value — the next operation will be lea, that will take the [eax] address, i.e., the value in the eax register, and will write it in ebx.
On the first thought the second operation seems meaningless, but as it turns out, it is acceptable to write the operation as
lea ecx, [eax + ebx],
lea ecx, [k*eax]
or even
lea ecx, [ebx + k*eax],
where k = 1, 2, 4 or 8.
As a result, the register ecx will be equal to the numbers eax + ebx, k*eax and ebx + k*eax correspondingly. However, such operation is fulfilled many times, dozens of times faster that the usual multiplying of numbers. And using several such operations, one can very quickly multiply some number by some other one. Of course, instead of eax, ebx and ecx you are allowed to use any registers.
For example, let the eax register contain some number that we should multiply by 41. It takes us 2 lines:
lea ebx, [eax + 4*eax] // now ebx = 5*eax
lea eax, [eax + 8*ebx] // now eax = eax + 8*ebx = 41*eax
Igor K. got interested in the following question: what is the minimum number of lea operations needed to multiply by the given number n and how to do it? Your task is to help him.
Consider that at the initial moment of time eax contains a number that Igor K. was about to multiply by n, and the registers from ebx to ezx contain number 0. At the final moment of time the result can be located in any register.
Input
The input data contain the only integer n (1 ≤ n ≤ 255), which Igor K. is about to multiply.
Output
On the first line print number p, which represents the minimum number of lea operations, needed to do that. Then print the program consisting of p commands, performing the operations. It is guaranteed that such program exists for any n from 1 to 255.
Use precisely the following format of commands (here k is equal to 1, 2, 4 or 8, and x, y and z are any, even coinciding registers):
lea x, [y]
lea x, [y + z]
lea x, [k*y]
lea x, [y + k*z]
Please note that extra spaces at the end of a command are unacceptable.
Examples
Input
41
Output
2
lea ebx, [eax + 4*eax]
lea ecx, [eax + 8*ebx]
Input
2
Output
1
lea ebx, [eax + eax]
Input
4
Output
1
lea ebx, [4*eax]
Submitted Solution:
```
import sys
from array import array # noqa: F401
from itertools import product
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
if n == 1:
print(0)
exit()
inf = 10**9
dp = [inf] * (n + 1)
dp[1] = 0
prev = [[-1, -1, -1, -1] for _ in range(n + 1)]
def ok():
path = []
used_num = {n}
stack = [n]
while stack:
z = stack.pop()
if z == 1:
continue
x, mx, y, my = prev[z]
path.append([x, mx, y, my, z])
if x not in used_num:
used_num.add(x)
stack.append(x)
if y != -1 and y not in used_num:
used_num.add(y)
stack.append(y)
register_name = {x: f'e{chr(97+i)}x' for i, x in enumerate(sorted(used_num))}
ans = []
used = [0] * len(path)
for ub in sorted(used_num):
for i, (x, mx, y, my, z) in enumerate(path):
if not used[i] and max(x, y, z) <= ub:
if y == -1:
ans.append(f'lea {register_name[z]}, [{str(mx)+"*" if mx > 1 else ""}{register_name[x]}]')
else:
ans.append(f'lea {register_name[z]}, [{str(mx)+"*" if mx > 1 else ""}{register_name[x]} + {str(my)+"*" if my > 1 else ""}{register_name[y]}]')
used[i] = 1
print(len(ans))
print('\n'.join(ans))
exit()
while True:
next_dp = dp[:]
for x in range(1, n):
if dp[x] == inf:
continue
for m in (2, 4, 8):
if x * m <= n and next_dp[x * m] > dp[x] + 1:
next_dp[x * m] = dp[x] + 1
prev[x * m] = [x, m, -1, -1]
for y in range(1, n):
if dp[y] == inf:
continue
for mx, my in product((1, 2, 4, 8), repeat=2):
if x * mx + y * my <= n and next_dp[x * mx + y * my] > dp[x] + dp[y] + 1:
next_dp[x * mx + y * my] = dp[x] + dp[y] + 1
prev[x * mx + y * my] = [x, mx, y, my]
dp = next_dp
if dp[n] != inf:
ok()
``` | instruction | 0 | 64,549 | 6 | 129,098 |
No | output | 1 | 64,549 | 6 | 129,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the Search Ultimate program that searched for strings in a text failed, Igor K. got to think: "Why on Earth does my program work so slowly?" As he double-checked his code, he said: "My code contains no errors, yet I know how we will improve Search Ultimate!" and took a large book from the shelves. The book read "Azembler. Principally New Approach".
Having carefully thumbed through the book, Igor K. realised that, as it turns out, you can multiply the numbers dozens of times faster. "Search Ultimate will be faster than it has ever been!" — the fellow shouted happily and set to work.
Let us now clarify what Igor's idea was. The thing is that the code that was generated by a compiler was far from perfect. Standard multiplying does work slower than with the trick the book mentioned.
The Azembler language operates with 26 registers (eax, ebx, ..., ezx) and two commands:
* [x] — returns the value located in the address x. For example, [eax] returns the value that was located in the address, equal to the value in the register eax.
* lea x, y — assigns to the register x, indicated as the first operand, the second operand's address. Thus, for example, the "lea ebx, [eax]" command will write in the ebx register the content of the eax register: first the [eax] operation will be fulfilled, the result of it will be some value that lies in the address written in eax. But we do not need the value — the next operation will be lea, that will take the [eax] address, i.e., the value in the eax register, and will write it in ebx.
On the first thought the second operation seems meaningless, but as it turns out, it is acceptable to write the operation as
lea ecx, [eax + ebx],
lea ecx, [k*eax]
or even
lea ecx, [ebx + k*eax],
where k = 1, 2, 4 or 8.
As a result, the register ecx will be equal to the numbers eax + ebx, k*eax and ebx + k*eax correspondingly. However, such operation is fulfilled many times, dozens of times faster that the usual multiplying of numbers. And using several such operations, one can very quickly multiply some number by some other one. Of course, instead of eax, ebx and ecx you are allowed to use any registers.
For example, let the eax register contain some number that we should multiply by 41. It takes us 2 lines:
lea ebx, [eax + 4*eax] // now ebx = 5*eax
lea eax, [eax + 8*ebx] // now eax = eax + 8*ebx = 41*eax
Igor K. got interested in the following question: what is the minimum number of lea operations needed to multiply by the given number n and how to do it? Your task is to help him.
Consider that at the initial moment of time eax contains a number that Igor K. was about to multiply by n, and the registers from ebx to ezx contain number 0. At the final moment of time the result can be located in any register.
Input
The input data contain the only integer n (1 ≤ n ≤ 255), which Igor K. is about to multiply.
Output
On the first line print number p, which represents the minimum number of lea operations, needed to do that. Then print the program consisting of p commands, performing the operations. It is guaranteed that such program exists for any n from 1 to 255.
Use precisely the following format of commands (here k is equal to 1, 2, 4 or 8, and x, y and z are any, even coinciding registers):
lea x, [y]
lea x, [y + z]
lea x, [k*y]
lea x, [y + k*z]
Please note that extra spaces at the end of a command are unacceptable.
Examples
Input
41
Output
2
lea ebx, [eax + 4*eax]
lea ecx, [eax + 8*ebx]
Input
2
Output
1
lea ebx, [eax + eax]
Input
4
Output
1
lea ebx, [4*eax]
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
if n == 1:
print(0)
exit()
inf = 10**9
dp = [inf] * (n + 1)
dp[1] = 0
prev = [[-1, -1, -1, -1] for _ in range(n + 1)]
def ok():
path = []
used_num = {n}
stack = [n]
while stack:
z = stack.pop()
if z == 1:
continue
x, mx, y, my = prev[z]
path.append([x, mx, y, my, z])
if x not in used_num:
used_num.add(x)
stack.append(x)
if y != -1 and y not in used_num:
used_num.add(y)
stack.append(y)
register_name = {x: f'e{chr(97+i)}x' for i, x in enumerate(sorted(used_num))}
ans = []
flag = [0] * (n + 1)
used = [0] * len(path)
for ub in sorted(used_num):
flag[ub] = 1
for i, (x, mx, y, my, z) in enumerate(path):
if not used[i] and (flag[x] and (y == -1 or flag[y]) and flag[z]):
if y == -1:
ans.append(f'lea {register_name[z]}, [{str(mx)+"*" if mx > 1 else ""}{register_name[x]}]')
else:
ans.append(f'lea {register_name[z]}, [{str(mx)+"*" if mx > 1 else ""}{register_name[x]} + {str(my)+"*" if my > 1 else ""}{register_name[y]}]')
used[i] = 1
print(len(ans))
print('\n'.join(ans))
exit()
while True:
flag = 0
next_dp = dp[:]
for x in range(1, n):
if dp[x] == inf:
continue
for m in (2, 4, 8):
if x * m <= n and next_dp[x * m] > dp[x] + 1:
flag = 1
next_dp[x * m] = dp[x] + 1
prev[x * m] = [x, m, -1, -1]
for y in range(1, n):
if dp[y] == inf:
continue
for my in (1, 2, 4, 8):
if x + y * my <= n and next_dp[x + y * my] > dp[x] + dp[y] + 1:
flag = 1
next_dp[x + y * my] = dp[x] + dp[y] + 1
prev[x + y * my] = [x, 1, y, my]
dp = next_dp
if not flag:
ok()
``` | instruction | 0 | 64,550 | 6 | 129,100 |
No | output | 1 | 64,550 | 6 | 129,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the Search Ultimate program that searched for strings in a text failed, Igor K. got to think: "Why on Earth does my program work so slowly?" As he double-checked his code, he said: "My code contains no errors, yet I know how we will improve Search Ultimate!" and took a large book from the shelves. The book read "Azembler. Principally New Approach".
Having carefully thumbed through the book, Igor K. realised that, as it turns out, you can multiply the numbers dozens of times faster. "Search Ultimate will be faster than it has ever been!" — the fellow shouted happily and set to work.
Let us now clarify what Igor's idea was. The thing is that the code that was generated by a compiler was far from perfect. Standard multiplying does work slower than with the trick the book mentioned.
The Azembler language operates with 26 registers (eax, ebx, ..., ezx) and two commands:
* [x] — returns the value located in the address x. For example, [eax] returns the value that was located in the address, equal to the value in the register eax.
* lea x, y — assigns to the register x, indicated as the first operand, the second operand's address. Thus, for example, the "lea ebx, [eax]" command will write in the ebx register the content of the eax register: first the [eax] operation will be fulfilled, the result of it will be some value that lies in the address written in eax. But we do not need the value — the next operation will be lea, that will take the [eax] address, i.e., the value in the eax register, and will write it in ebx.
On the first thought the second operation seems meaningless, but as it turns out, it is acceptable to write the operation as
lea ecx, [eax + ebx],
lea ecx, [k*eax]
or even
lea ecx, [ebx + k*eax],
where k = 1, 2, 4 or 8.
As a result, the register ecx will be equal to the numbers eax + ebx, k*eax and ebx + k*eax correspondingly. However, such operation is fulfilled many times, dozens of times faster that the usual multiplying of numbers. And using several such operations, one can very quickly multiply some number by some other one. Of course, instead of eax, ebx and ecx you are allowed to use any registers.
For example, let the eax register contain some number that we should multiply by 41. It takes us 2 lines:
lea ebx, [eax + 4*eax] // now ebx = 5*eax
lea eax, [eax + 8*ebx] // now eax = eax + 8*ebx = 41*eax
Igor K. got interested in the following question: what is the minimum number of lea operations needed to multiply by the given number n and how to do it? Your task is to help him.
Consider that at the initial moment of time eax contains a number that Igor K. was about to multiply by n, and the registers from ebx to ezx contain number 0. At the final moment of time the result can be located in any register.
Input
The input data contain the only integer n (1 ≤ n ≤ 255), which Igor K. is about to multiply.
Output
On the first line print number p, which represents the minimum number of lea operations, needed to do that. Then print the program consisting of p commands, performing the operations. It is guaranteed that such program exists for any n from 1 to 255.
Use precisely the following format of commands (here k is equal to 1, 2, 4 or 8, and x, y and z are any, even coinciding registers):
lea x, [y]
lea x, [y + z]
lea x, [k*y]
lea x, [y + k*z]
Please note that extra spaces at the end of a command are unacceptable.
Examples
Input
41
Output
2
lea ebx, [eax + 4*eax]
lea ecx, [eax + 8*ebx]
Input
2
Output
1
lea ebx, [eax + eax]
Input
4
Output
1
lea ebx, [4*eax]
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
if n == 1:
print(0)
exit()
inf = 10**9
dp = [inf] * (n + 1)
dp[1] = 0
prev = [[-1, -1, -1, -1] for _ in range(n + 1)]
def ok():
path = []
used_num = {n}
stack = [n]
while stack:
z = stack.pop()
if z == 1:
continue
x, mx, y, my = prev[z]
path.append([x, mx, y, my, z])
if x not in used_num:
used_num.add(x)
stack.append(x)
if y != -1 and y not in used_num:
used_num.add(y)
stack.append(y)
register_name = {x: f'e{chr(97+i)}x' for i, x in enumerate(sorted(used_num))}
ans = []
used = [0] * len(path)
for ub in sorted(used_num):
for i, (x, mx, y, my, z) in enumerate(path):
if not used[i] and max(x, y, z) <= ub:
if y == -1:
ans.append(f'lea {register_name[z]}, [{str(mx)+"*" if mx > 1 else ""}{register_name[x]}]')
else:
ans.append(f'lea {register_name[z]}, [{str(mx)+"*" if mx > 1 else ""}{register_name[x]} + {str(my)+"*" if my > 1 else ""}{register_name[y]}]')
used[i] = 1
print(len(ans))
print('\n'.join(ans))
exit()
def calc(x, y):
visited = {x, y, 1}
stack = [x, y]
res = 0
while stack:
num = stack.pop()
if num == 1:
continue
res += 1
if prev[num][0] != -1 and prev[num][0] not in visited:
visited.add(prev[num][0])
stack.append(prev[num][0])
if prev[num][2] != -1 and prev[num][2] not in visited:
visited.add(prev[num][2])
stack.append(prev[num][2])
return res
while True:
flag = 0
next_dp = dp[:]
for x in range(1, n):
if dp[x] == inf:
continue
for m in (2, 4, 8):
if x * m <= n and next_dp[x * m] > dp[x] + 1:
flag = 1
next_dp[x * m] = dp[x] + 1
prev[x * m] = [x, m, -1, -1]
for y in range(1, n):
if dp[y] == inf:
continue
for my in (1, 2, 4, 8):
if x + y * my <= n and next_dp[x + y * my] > calc(x, y) + 1:
flag = 1
next_dp[x + y * my] = calc(x, y) + 1
prev[x + y * my] = [x, 1, y, my]
dp = next_dp
if not flag:
for i in range(0, n + 1, 10):
print(f'{i:3d}:', *dp[i:i + 10], file=sys.stderr)
ok()
``` | instruction | 0 | 64,551 | 6 | 129,102 |
No | output | 1 | 64,551 | 6 | 129,103 |
Provide a correct Python 3 solution for this coding contest problem.
Digits Are Not Just Characters
Mr. Manuel Majorana Minore made a number of files with numbers in their names. He wants to have a list of the files, but the file listing command commonly used lists them in an order different from what he prefers, interpreting digit sequences in them as ASCII code sequences, not as numbers. For example, the files file10, file20 and file3 are listed in this order.
Write a program which decides the orders of file names interpreting digit sequences as numeric values.
Each file name consists of uppercase letters (from 'A' to 'Z'), lowercase letters (from 'a' to 'z'), and digits (from '0' to '9').
A file name is looked upon as a sequence of items, each being either a letter or a number. Each single uppercase or lowercase letter forms a letter item. Each consecutive sequence of digits forms a number item.
Two item are ordered as follows.
* Number items come before letter items.
* Two letter items are ordered by their ASCII codes.
* Two number items are ordered by their values when interpreted as decimal numbers.
Two file names are compared item by item, starting from the top, and the order of the first different corresponding items decides the order of the file names. If one of them, say $A$, has more items than the other, $B$, and all the items of $B$ are the same as the corresponding items of $A$, $B$ should come before.
For example, three file names in Sample Input 1, file10, file20, and file3 all start with the same sequence of four letter items f, i, l, and e, followed by a number item, 10, 20, and 3, respectively. Comparing numeric values of these number items, they are ordered as file3 $<$ file10 $<$ file20.
Input
The input consists of a single test case of the following format.
$n$
$s_0$
$s_1$
:
$s_n$
The integer $n$ in the first line gives the number of file names ($s_1$ through $s_n$) to be compared with the file name given in the next line ($s_0$). Here, $n$ satisfies $1 \leq n \leq 1000$.
The following $n + 1$ lines are file names, $s_0$ through $s_n$, one in each line. They have at least one and no more than nine characters. Each of the characters is either an uppercase letter, a lowercase letter, or a digit.
Sequences of digits in the file names never start with a digit zero (0).
Output
For each of the file names, $s_1$ through $s_n$, output one line with a character indicating whether it should come before $s_0$ or not. The character should be "-" if it is to be listed before $s_0$; otherwise, it should be "+", including cases where two names are identical.
Sample Input 1
2
file10
file20
file3
Sample Output 1
+
-
Sample Input 2
11
X52Y
X
X5
X52
X52Y
X52Y6
32
ABC
XYZ
x51y
X8Y
X222
Sample Output 2
-
-
-
+
+
-
-
+
+
-
+
Example
Input
2
file10
file20
file3
Output
+
- | instruction | 0 | 64,777 | 6 | 129,554 |
"Correct Solution:
```
N = int(input())
S = [input() for _ in range(N + 1)]
def convert(s):
l = []
for i in range(len(s)):
if s[i].isdigit():
if len(l) != 0 and isinstance(l[-1], list):
l[-1].append(s[i])
else:
l.append([s[i]])
else:
l.append(s[i])
return [int(''.join(e)) if isinstance(e, list) else e for e in l]
S = [convert(e) for e in S]
t = S[0]
def compare(s):
for i in range(min([len(s), len(t)])):
if type(s[i]) == type(t[i]):
if s[i] < t[i]:
return '-'
elif s[i] > t[i]:
return '+'
else:
if type(s[i]) == int:
return '-'
else:
return '+'
if len(t) <= len(s):
return '+'
else:
return '-'
for s in S[1:]:
print(compare(s))
``` | output | 1 | 64,777 | 6 | 129,555 |
Provide a correct Python 3 solution for this coding contest problem.
Digits Are Not Just Characters
Mr. Manuel Majorana Minore made a number of files with numbers in their names. He wants to have a list of the files, but the file listing command commonly used lists them in an order different from what he prefers, interpreting digit sequences in them as ASCII code sequences, not as numbers. For example, the files file10, file20 and file3 are listed in this order.
Write a program which decides the orders of file names interpreting digit sequences as numeric values.
Each file name consists of uppercase letters (from 'A' to 'Z'), lowercase letters (from 'a' to 'z'), and digits (from '0' to '9').
A file name is looked upon as a sequence of items, each being either a letter or a number. Each single uppercase or lowercase letter forms a letter item. Each consecutive sequence of digits forms a number item.
Two item are ordered as follows.
* Number items come before letter items.
* Two letter items are ordered by their ASCII codes.
* Two number items are ordered by their values when interpreted as decimal numbers.
Two file names are compared item by item, starting from the top, and the order of the first different corresponding items decides the order of the file names. If one of them, say $A$, has more items than the other, $B$, and all the items of $B$ are the same as the corresponding items of $A$, $B$ should come before.
For example, three file names in Sample Input 1, file10, file20, and file3 all start with the same sequence of four letter items f, i, l, and e, followed by a number item, 10, 20, and 3, respectively. Comparing numeric values of these number items, they are ordered as file3 $<$ file10 $<$ file20.
Input
The input consists of a single test case of the following format.
$n$
$s_0$
$s_1$
:
$s_n$
The integer $n$ in the first line gives the number of file names ($s_1$ through $s_n$) to be compared with the file name given in the next line ($s_0$). Here, $n$ satisfies $1 \leq n \leq 1000$.
The following $n + 1$ lines are file names, $s_0$ through $s_n$, one in each line. They have at least one and no more than nine characters. Each of the characters is either an uppercase letter, a lowercase letter, or a digit.
Sequences of digits in the file names never start with a digit zero (0).
Output
For each of the file names, $s_1$ through $s_n$, output one line with a character indicating whether it should come before $s_0$ or not. The character should be "-" if it is to be listed before $s_0$; otherwise, it should be "+", including cases where two names are identical.
Sample Input 1
2
file10
file20
file3
Sample Output 1
+
-
Sample Input 2
11
X52Y
X
X5
X52
X52Y
X52Y6
32
ABC
XYZ
x51y
X8Y
X222
Sample Output 2
-
-
-
+
+
-
-
+
+
-
+
Example
Input
2
file10
file20
file3
Output
+
- | instruction | 0 | 64,778 | 6 | 129,556 |
"Correct Solution:
```
import re
n=int(input())
s=[ ''.join(map(lambda x : '{0:09d}'.format(int(x)) if '0' <= x[0] <= '9' else x, re.sub(r'\d+', r' \g<0> ',input()).split()))for _ in range(n+1) ]
for i in range(1,n+1):
print('-' if s[i] < s[0] else '+')
``` | output | 1 | 64,778 | 6 | 129,557 |
Provide a correct Python 3 solution for this coding contest problem.
Digits Are Not Just Characters
Mr. Manuel Majorana Minore made a number of files with numbers in their names. He wants to have a list of the files, but the file listing command commonly used lists them in an order different from what he prefers, interpreting digit sequences in them as ASCII code sequences, not as numbers. For example, the files file10, file20 and file3 are listed in this order.
Write a program which decides the orders of file names interpreting digit sequences as numeric values.
Each file name consists of uppercase letters (from 'A' to 'Z'), lowercase letters (from 'a' to 'z'), and digits (from '0' to '9').
A file name is looked upon as a sequence of items, each being either a letter or a number. Each single uppercase or lowercase letter forms a letter item. Each consecutive sequence of digits forms a number item.
Two item are ordered as follows.
* Number items come before letter items.
* Two letter items are ordered by their ASCII codes.
* Two number items are ordered by their values when interpreted as decimal numbers.
Two file names are compared item by item, starting from the top, and the order of the first different corresponding items decides the order of the file names. If one of them, say $A$, has more items than the other, $B$, and all the items of $B$ are the same as the corresponding items of $A$, $B$ should come before.
For example, three file names in Sample Input 1, file10, file20, and file3 all start with the same sequence of four letter items f, i, l, and e, followed by a number item, 10, 20, and 3, respectively. Comparing numeric values of these number items, they are ordered as file3 $<$ file10 $<$ file20.
Input
The input consists of a single test case of the following format.
$n$
$s_0$
$s_1$
:
$s_n$
The integer $n$ in the first line gives the number of file names ($s_1$ through $s_n$) to be compared with the file name given in the next line ($s_0$). Here, $n$ satisfies $1 \leq n \leq 1000$.
The following $n + 1$ lines are file names, $s_0$ through $s_n$, one in each line. They have at least one and no more than nine characters. Each of the characters is either an uppercase letter, a lowercase letter, or a digit.
Sequences of digits in the file names never start with a digit zero (0).
Output
For each of the file names, $s_1$ through $s_n$, output one line with a character indicating whether it should come before $s_0$ or not. The character should be "-" if it is to be listed before $s_0$; otherwise, it should be "+", including cases where two names are identical.
Sample Input 1
2
file10
file20
file3
Sample Output 1
+
-
Sample Input 2
11
X52Y
X
X5
X52
X52Y
X52Y6
32
ABC
XYZ
x51y
X8Y
X222
Sample Output 2
-
-
-
+
+
-
-
+
+
-
+
Example
Input
2
file10
file20
file3
Output
+
- | instruction | 0 | 64,779 | 6 | 129,558 |
"Correct Solution:
```
n = int(input())
ss = [input() for _ in range(n + 1)]
sp = [[s[0]] for s in ss]
for i, s in enumerate(ss):
for j in range(1, len(s)):
if s[j - 1].isdigit() != s[j].isdigit():
sp[i].append('')
sp[i][-1] += s[j]
s0 = sp[0]
for s in sp[1:]:
p = 0
m = min(len(s0), len(s))
while p < m and s0[p] == s[p]:
p += 1
if p == m:
print('-+'[len(s0) <= len(s)])
continue
a = s0[p].isdigit()
b = s[p].isdigit()
if a and b:
print('-+'[int(s0[p]) <= int(s[p])])
elif a or b:
print('-+'[a])
else:
print('-+'[s0[p] <= s[p]])
``` | output | 1 | 64,779 | 6 | 129,559 |
Provide a correct Python 3 solution for this coding contest problem.
Digits Are Not Just Characters
Mr. Manuel Majorana Minore made a number of files with numbers in their names. He wants to have a list of the files, but the file listing command commonly used lists them in an order different from what he prefers, interpreting digit sequences in them as ASCII code sequences, not as numbers. For example, the files file10, file20 and file3 are listed in this order.
Write a program which decides the orders of file names interpreting digit sequences as numeric values.
Each file name consists of uppercase letters (from 'A' to 'Z'), lowercase letters (from 'a' to 'z'), and digits (from '0' to '9').
A file name is looked upon as a sequence of items, each being either a letter or a number. Each single uppercase or lowercase letter forms a letter item. Each consecutive sequence of digits forms a number item.
Two item are ordered as follows.
* Number items come before letter items.
* Two letter items are ordered by their ASCII codes.
* Two number items are ordered by their values when interpreted as decimal numbers.
Two file names are compared item by item, starting from the top, and the order of the first different corresponding items decides the order of the file names. If one of them, say $A$, has more items than the other, $B$, and all the items of $B$ are the same as the corresponding items of $A$, $B$ should come before.
For example, three file names in Sample Input 1, file10, file20, and file3 all start with the same sequence of four letter items f, i, l, and e, followed by a number item, 10, 20, and 3, respectively. Comparing numeric values of these number items, they are ordered as file3 $<$ file10 $<$ file20.
Input
The input consists of a single test case of the following format.
$n$
$s_0$
$s_1$
:
$s_n$
The integer $n$ in the first line gives the number of file names ($s_1$ through $s_n$) to be compared with the file name given in the next line ($s_0$). Here, $n$ satisfies $1 \leq n \leq 1000$.
The following $n + 1$ lines are file names, $s_0$ through $s_n$, one in each line. They have at least one and no more than nine characters. Each of the characters is either an uppercase letter, a lowercase letter, or a digit.
Sequences of digits in the file names never start with a digit zero (0).
Output
For each of the file names, $s_1$ through $s_n$, output one line with a character indicating whether it should come before $s_0$ or not. The character should be "-" if it is to be listed before $s_0$; otherwise, it should be "+", including cases where two names are identical.
Sample Input 1
2
file10
file20
file3
Sample Output 1
+
-
Sample Input 2
11
X52Y
X
X5
X52
X52Y
X52Y6
32
ABC
XYZ
x51y
X8Y
X222
Sample Output 2
-
-
-
+
+
-
-
+
+
-
+
Example
Input
2
file10
file20
file3
Output
+
- | instruction | 0 | 64,780 | 6 | 129,560 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
def ssplit(s):
tp=0
tmp=""
ret=[]
for c in s:
if c.isdigit():
tmp+=c
else:
if tmp!="":
ret.append((0,int(tmp)))
tmp=""
ret.append((1,ord(c)))
if tmp!="":
ret.append((0,int(tmp)))
return ret
n=int(input())
s=ssplit(input())
for i in range(n):
z=ssplit(input())
if z>=s:
print('+')
else:
print('-')
``` | output | 1 | 64,780 | 6 | 129,561 |
Provide a correct Python 3 solution for this coding contest problem.
Digits Are Not Just Characters
Mr. Manuel Majorana Minore made a number of files with numbers in their names. He wants to have a list of the files, but the file listing command commonly used lists them in an order different from what he prefers, interpreting digit sequences in them as ASCII code sequences, not as numbers. For example, the files file10, file20 and file3 are listed in this order.
Write a program which decides the orders of file names interpreting digit sequences as numeric values.
Each file name consists of uppercase letters (from 'A' to 'Z'), lowercase letters (from 'a' to 'z'), and digits (from '0' to '9').
A file name is looked upon as a sequence of items, each being either a letter or a number. Each single uppercase or lowercase letter forms a letter item. Each consecutive sequence of digits forms a number item.
Two item are ordered as follows.
* Number items come before letter items.
* Two letter items are ordered by their ASCII codes.
* Two number items are ordered by their values when interpreted as decimal numbers.
Two file names are compared item by item, starting from the top, and the order of the first different corresponding items decides the order of the file names. If one of them, say $A$, has more items than the other, $B$, and all the items of $B$ are the same as the corresponding items of $A$, $B$ should come before.
For example, three file names in Sample Input 1, file10, file20, and file3 all start with the same sequence of four letter items f, i, l, and e, followed by a number item, 10, 20, and 3, respectively. Comparing numeric values of these number items, they are ordered as file3 $<$ file10 $<$ file20.
Input
The input consists of a single test case of the following format.
$n$
$s_0$
$s_1$
:
$s_n$
The integer $n$ in the first line gives the number of file names ($s_1$ through $s_n$) to be compared with the file name given in the next line ($s_0$). Here, $n$ satisfies $1 \leq n \leq 1000$.
The following $n + 1$ lines are file names, $s_0$ through $s_n$, one in each line. They have at least one and no more than nine characters. Each of the characters is either an uppercase letter, a lowercase letter, or a digit.
Sequences of digits in the file names never start with a digit zero (0).
Output
For each of the file names, $s_1$ through $s_n$, output one line with a character indicating whether it should come before $s_0$ or not. The character should be "-" if it is to be listed before $s_0$; otherwise, it should be "+", including cases where two names are identical.
Sample Input 1
2
file10
file20
file3
Sample Output 1
+
-
Sample Input 2
11
X52Y
X
X5
X52
X52Y
X52Y6
32
ABC
XYZ
x51y
X8Y
X222
Sample Output 2
-
-
-
+
+
-
-
+
+
-
+
Example
Input
2
file10
file20
file3
Output
+
- | instruction | 0 | 64,781 | 6 | 129,562 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**6)
def tl(s):
pre = False
ref = []
for i in s:
if ord(i) < 58:
if pre:
ref[-1] = [True, ref[-1][1]*10 + int(i)]
else:
pre = True
ref += [[True, int(i)]]
else:
ref += [[False, i]]
pre=False
return ref
def main():
n = int(input())
pibot = tl(input())
for _ in range(n):
temp = tl(input())
ans = "-"
for i in range(len(temp)):
if i >= len(pibot):
ans = "+"
break
if pibot[i] == temp[i]:
if pibot == temp:
ans = "+"
break
continue
if pibot[i][0] and not(temp[i][0]):
ans = "+"
elif not(pibot[i][0]) and temp[i][0]:
ans = "-"
elif pibot[i][0]:
if pibot[i][1] < temp[i][1]:
ans = "+"
else:
ans = "-"
else:
if ord(pibot[i][1]) < ord(temp[i][1]):
ans = "+"
else:
ans = "-"
break
print(ans)
return
main()
``` | output | 1 | 64,781 | 6 | 129,563 |
Provide a correct Python 3 solution for this coding contest problem.
Digits Are Not Just Characters
Mr. Manuel Majorana Minore made a number of files with numbers in their names. He wants to have a list of the files, but the file listing command commonly used lists them in an order different from what he prefers, interpreting digit sequences in them as ASCII code sequences, not as numbers. For example, the files file10, file20 and file3 are listed in this order.
Write a program which decides the orders of file names interpreting digit sequences as numeric values.
Each file name consists of uppercase letters (from 'A' to 'Z'), lowercase letters (from 'a' to 'z'), and digits (from '0' to '9').
A file name is looked upon as a sequence of items, each being either a letter or a number. Each single uppercase or lowercase letter forms a letter item. Each consecutive sequence of digits forms a number item.
Two item are ordered as follows.
* Number items come before letter items.
* Two letter items are ordered by their ASCII codes.
* Two number items are ordered by their values when interpreted as decimal numbers.
Two file names are compared item by item, starting from the top, and the order of the first different corresponding items decides the order of the file names. If one of them, say $A$, has more items than the other, $B$, and all the items of $B$ are the same as the corresponding items of $A$, $B$ should come before.
For example, three file names in Sample Input 1, file10, file20, and file3 all start with the same sequence of four letter items f, i, l, and e, followed by a number item, 10, 20, and 3, respectively. Comparing numeric values of these number items, they are ordered as file3 $<$ file10 $<$ file20.
Input
The input consists of a single test case of the following format.
$n$
$s_0$
$s_1$
:
$s_n$
The integer $n$ in the first line gives the number of file names ($s_1$ through $s_n$) to be compared with the file name given in the next line ($s_0$). Here, $n$ satisfies $1 \leq n \leq 1000$.
The following $n + 1$ lines are file names, $s_0$ through $s_n$, one in each line. They have at least one and no more than nine characters. Each of the characters is either an uppercase letter, a lowercase letter, or a digit.
Sequences of digits in the file names never start with a digit zero (0).
Output
For each of the file names, $s_1$ through $s_n$, output one line with a character indicating whether it should come before $s_0$ or not. The character should be "-" if it is to be listed before $s_0$; otherwise, it should be "+", including cases where two names are identical.
Sample Input 1
2
file10
file20
file3
Sample Output 1
+
-
Sample Input 2
11
X52Y
X
X5
X52
X52Y
X52Y6
32
ABC
XYZ
x51y
X8Y
X222
Sample Output 2
-
-
-
+
+
-
-
+
+
-
+
Example
Input
2
file10
file20
file3
Output
+
- | instruction | 0 | 64,782 | 6 | 129,564 |
"Correct Solution:
```
#!/usr/bin/python3
import array
from fractions import Fraction
import functools
import itertools
import math
import os
import sys
def main():
N = read_int()
A = inp()
for _ in range(N):
print('-' if solve(inp(), A) < 0 else '+')
def solve(A, B):
i = 0
j = 0
N = len(A)
M = len(B)
while i < N and j < M:
c = A[i]
d = B[i]
if '0' <= c <= '9':
if '0' <= d <= '9':
ii = i + 1
while ii < N and '0' <= A[ii] <= '9':
ii += 1
jj = j + 1
while jj < M and '0' <= B[jj] <= '9':
jj += 1
a = int(A[i : ii])
b = int(B[j : jj])
if a != b:
return a - b
i = ii
j = jj
else:
return -1
else:
if '0' <= d <= '9':
return 1
if c != d:
return ord(c) - ord(d)
i += 1
j += 1
if i < N:
return 1
if j < M:
return -1
return 0
###############################################################################
# AUXILIARY FUNCTIONS
DEBUG = 'DEBUG' in os.environ
def inp():
return sys.stdin.readline().rstrip()
def read_int():
return int(inp())
def read_ints():
return [int(e) for e in inp().split()]
def dprint(*value, sep=' ', end='\n'):
if DEBUG:
print(*value, sep=sep, end=end)
if __name__ == '__main__':
main()
``` | output | 1 | 64,782 | 6 | 129,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}. | instruction | 0 | 65,207 | 6 | 130,414 |
Tags: brute force, implementation
Correct Solution:
```
from string import ascii_lowercase
n = int(input())
hashed_words = {}
word_buffer = []
for i in range(n):
data = [i for i in input().strip()]
if len(set(data)) <= 2:
if ''.join(set(data)) in hashed_words.keys():
hashed_words[''.join(set(data))] += len(data)
else:
hashed_words[''.join(set(data))] = len(data)
if len(hashed_words) > 0:
my_max = 0
for i in ascii_lowercase:
for j in ascii_lowercase:
if i != j:
current = 0
if i in hashed_words:
current += hashed_words[i]
if j in hashed_words:
current += hashed_words[j]
if i+j in hashed_words:
current += hashed_words[i+j]
if j+i in hashed_words:
current += hashed_words[j+i]
if my_max < current:
my_max = current
print(my_max)
else:
print(0)
``` | output | 1 | 65,207 | 6 | 130,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}. | instruction | 0 | 65,208 | 6 | 130,416 |
Tags: brute force, implementation
Correct Solution:
```
def main():
n = int(input())
a = [input() for i in range(n)]
result = 0
for i in range(26):
for j in range(i + 1, 26):
t = 0
ci = chr(i + ord('a'))
cj = chr(j + ord('a'))
for s in a:
if s.count(ci) + s.count(cj) == len(s):
t += len(s)
result = max(result, t)
print(result)
main()
``` | output | 1 | 65,208 | 6 | 130,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}. | instruction | 0 | 65,209 | 6 | 130,418 |
Tags: brute force, implementation
Correct Solution:
```
import string
n = int(input())
a = []
for i in range(n):
a.append(input())
ans = 0
small = string.ascii_lowercase
for i in small:
for j in small:
tmp = 0
for s in a:
cnt = 0
for c in s:
if c == i or c == j:
cnt += 1
if cnt == len(s):
tmp += len(s)
ans = max(ans, tmp)
print(ans)
``` | output | 1 | 65,209 | 6 | 130,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}. | instruction | 0 | 65,210 | 6 | 130,420 |
Tags: brute force, implementation
Correct Solution:
```
import itertools
n = int(input())
words = [input() for x in range(n)]
letterList = []
for i in words:
for character in i:
if character not in letterList:
letterList.append(character)
maxLength = 0
if len(letterList) == 1:
for word in words:
maxLength += len(word)
print(maxLength)
exit()
allCombinations = itertools.combinations(letterList,2)
for c in allCombinations:
a,b = list(c)[0],list(c)[1]
length = 0
for word in words:
j = 0
for character in word:
j += 1
if character != a and character != b:
break
if j == len(word):
length += len(word)
if length > maxLength:
maxLength = length
print(maxLength)
``` | output | 1 | 65,210 | 6 | 130,421 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}. | instruction | 0 | 65,211 | 6 | 130,422 |
Tags: brute force, implementation
Correct Solution:
```
words = [input() for _ in range(int(input()))]
alphabet = 'abcdefghijklmnopqrstuvwxyz'
Max = 0
for c in alphabet:
for cc in alphabet:
s = 0
for w in words:
if set([ i for i in w]) <= {c, cc}:
s += len(w)
Max = max(Max, s)
print(Max)
``` | output | 1 | 65,211 | 6 | 130,423 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}. | instruction | 0 | 65,212 | 6 | 130,424 |
Tags: brute force, implementation
Correct Solution:
```
d={}
for _ in range(int(input())):
s=input()
S=set(s)
if len(S)<=2:
k="".join(sorted(S))
if d.__contains__(k):
d[k]+=len(s)
else:
d[k]=len(s)
l=[[len(set(i)),set(i),d[i]] for i in d.keys()]
l1=[]
l=sorted(l,reverse=True)
for i in range(len(l)):
if l[i][0]==1:
l1=l[i:]
del l[i:]
break
if len(l)==0 and len(l1)==0:
print(0)
elif len(l)!=0 and len(l1)!=0:
for i in l:
for j in l1:
if len(i[1]|j[1])==2:
i[2]+=j[2]
d={}
for i in l1:
k=i[1].pop()
if d.__contains__(k):
d[k]+=i[2]
else:
d[k]=i[2]
if len(d)>1:
k=list(sorted(d.values(),reverse=True))
print(max(k[0]+k[1],max(i[2] for i in l)))
else:
print(max(max(d.values()),max(i[2] for i in l)))
elif len(l)==0 and len(l1)!=0:
d={}
for i in l1:
k=i[1].pop()
if d.__contains__(k):
d[k]+=i[2]
else:
d[k]=i[2]
if len(d)>1:
k=list(sorted(d.values(),reverse=True))
print(k[0]+k[1])
else:
print(max(d.values()))
else:
print(max(i[2] for i in l))
``` | output | 1 | 65,212 | 6 | 130,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}. | instruction | 0 | 65,213 | 6 | 130,426 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
one = dict()
two = dict()
for i in range(n):
f = input()
q = set()
for j in f:
if j not in q:
q.add(j)
if len(q) < 3:
if len(q) == 1:
if ''.join(sorted(list(q))) in one:
one[''.join(sorted(list(q)))] += len(f)
else:
one[''.join(sorted(list(q)))] = len(f)
else:
if ''.join(sorted(list(q))) in two:
two[''.join(sorted(list(q)))] += len(f)
else:
two[''.join(sorted(list(q)))] = len(f)
ans = 0
for i in two:
if i[0] in one:
two[i] += one[i[0]]
if i[1] in one:
two[i] += one[i[1]]
if ans < two[i]:
ans = two[i]
for i in one:
if one[i] > ans:
ans = one[i]
for j in one:
if i != j and one[i] + one[j] > ans:
ans = one[i] + one[j]
print(ans)
``` | output | 1 | 65,213 | 6 | 130,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}. | instruction | 0 | 65,214 | 6 | 130,428 |
Tags: brute force, implementation
Correct Solution:
```
import itertools
import functools
N = int(input())
words = [ input() for i in range(N) ]
words_bit = [[functools.reduce(lambda x, y : x | y, [0] + [ 1 << ord(ch) - ord('a') for ch in words[i]])][0] for i in range(N)]
print(max([sum([ len(words[k]) for k in range(N) if not words_bit[k] & ~(1 << i | 1 << j)]) for i in range(26) for j in range(26)]))
``` | output | 1 | 65,214 | 6 | 130,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
Submitted Solution:
```
def f(c):
if type(c) == int:
return chr(c + 97)
if type(c) == str:
return ord(c) - 97
def g(s):
r = []
for i in s:
if i not in r:
r.append(i)
if len(r) > 2:
return r
return r
m = [[0] * 26 for i in range(26)]
n = int(input())
for i in range(n):
s = input()
a = g(s)
if len(a) == 1:
m[f(a[0])][f(a[0])] += len(s)
elif len(a) == 2:
m[f(a[0])][f(a[1])] += len(s)
m[f(a[1])][f(a[0])] += len(s)
cnt = 0
for i in range(26):
for j in range(26):
if i == j:
cnt = max(cnt, m[i][j])
else:
cnt = max(cnt, m[i][i] + m[i][j] + m[j][j])
print(cnt)
``` | instruction | 0 | 65,215 | 6 | 130,430 |
Yes | output | 1 | 65,215 | 6 | 130,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
Submitted Solution:
```
n = int(input())
a = []
b = []
for i in range(n):
s = input()
a.append(sorted(list(set(s))))
b.append(len(s))
v = 0
for i in range(97,97+25):
for j in range(98,97+26):
u = 0
for k in range(n):
for c in a[k]:
if ord(c) != i and ord(c) != j:
break
else:
u += b[k]
v = max(u, v)
print(v)
``` | instruction | 0 | 65,216 | 6 | 130,432 |
Yes | output | 1 | 65,216 | 6 | 130,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
Submitted Solution:
```
n=int(input())
i=0
x=[]
while i<n:
x.append(str(input()))
i+=1
i=25
r=0
while i:
j=i
while j:
j-=1
t=0
k=n
while k:
k-=1
y=x[k]
z=len(y)
ok=1
while z:
z-=1
if ord(y[z])-ord('a')!=i:
if ord(y[z])-ord('a')!=j:ok=0
if ok:t+=len(y)
if r<t:r=t
i-=1
print(r)
``` | instruction | 0 | 65,217 | 6 | 130,434 |
Yes | output | 1 | 65,217 | 6 | 130,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
Submitted Solution:
```
n = int(input())
s = []
for i in range(n):
k = input()
s.append(k)
maxans = -1
for i in range(26):
for j in range(26):
ans = 0
for k in s:
for p in range(len(k)):
if k[p] != chr(ord('a') + i) and k[p] != chr(ord('a') + j):
break
if k[p] == chr(ord('a') + i) or k[p] == chr(ord('a') + j):
ans += len(k)
if ans > maxans:
maxans = ans
print(maxans)
``` | instruction | 0 | 65,218 | 6 | 130,436 |
Yes | output | 1 | 65,218 | 6 | 130,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
Submitted Solution:
```
a='abcdefghijklmnopqrstuvwxyz'
n=int(input())
l=[]
ma=0
for i in range(n) :
l.append(input())
for i in range(len(a)) :
for j in range(len(a)) :
s=0
for x in l :
l1=set(x)
if len(l1)==2 :
if a[i] in l1 and a[j] in l1 :
s+=len(x)
if len(l1)==1 :
if a[i] in l1 or a[j] in l1 :
s+=len(x)
ma=max(ma,s)
print(ma)
``` | instruction | 0 | 65,219 | 6 | 130,438 |
No | output | 1 | 65,219 | 6 | 130,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
Submitted Solution:
```
N = int(input())
words = []
import string
for _ in range(N):
word = input()
o = { letter: False for letter in string.ascii_lowercase}
letters = []
count = 0
for letter in word:
if o[letter] == False:
o[letter] = True
letters.append(letter)
count += 1
if count < 3:
words.append([word, letters])
max_total = 0
for i in range(len(words)):
letters = words[i][1]
total_length = len(words[i][0])
for j in range(len(words)):
if i == j:
continue
if len(letters) == 1 and letters[0] in words[j][1]:
letters = words[j][1]
total_length += len(words[j][0])
else:
if letters == words[j][1]:
total_length += len(words[j][0])
elif len(words[j][1]) == 1 and words[j][1][0] in letters:
total_length += len(words[j][0])
if total_length > max_total:
max_total = total_length
print(max_total)
``` | instruction | 0 | 65,220 | 6 | 130,440 |
No | output | 1 | 65,220 | 6 | 130,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
Submitted Solution:
```
n = int(input())
words = []
letters = []
for i in range(n):
word = input()
if len(word) > 1:
a = word[0]
b = None
for j in word:
if j == a:
continue
else:
if b == None:
b = j
if j == b:
continue
break
else:
words.append(word)
if a not in letters:
letters.append(a)
if b not in letters:
letters.append(b)
else:
words.append(word)
if word not in letters:
letters.append(word)
len_let = len(letters)
maximum = 0
for i in range(len_let):
for j in range(i + 1, len_let):
a = letters[i]
b = letters[j]
counter = 0
for word in words:
for z in word:
if a != z and b != z:
break
else:
counter += len(word)
if counter > maximum:
maximum = counter
print(maximum)
``` | instruction | 0 | 65,221 | 6 | 130,442 |
No | output | 1 | 65,221 | 6 | 130,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.
Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.
Input
The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.
Output
Print a single integer — the maximum possible total length of words in Andrew's article.
Examples
Input
4
abb
cacc
aaa
bbb
Output
9
Input
5
a
a
bcbcb
cdecdecdecdecdecde
aaaa
Output
6
Note
In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.
In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.
Submitted Solution:
```
import sys
stuff = sys.stdin.readlines()
maximum_length = int(stuff[0])
stuff.pop(0)
stuff = [n.rstrip('\n') for n in stuff]
result = [len(n) for n in stuff if len(n) < maximum_length]
print(sum(result))
``` | instruction | 0 | 65,222 | 6 | 130,444 |
No | output | 1 | 65,222 | 6 | 130,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course. | instruction | 0 | 65,313 | 6 | 130,626 |
Tags: brute force, constructive algorithms, dp, implementation, strings
Correct Solution:
```
import sys
import re
def mentioned_usernames(line):
return {x for x in re.split(r'[^A-Za-z0-9]+', line)}
t = int(input())
for ti in range(t):
possible_users = []
messages = []
n = int(input())
usernames = set(input().split())
# print("usernames =", usernames, file=sys.stderr)
m = int(input())
for i in range(m):
user, text = input().split(':')
messages.append(text)
if user == '?':
mu = mentioned_usernames(text)
possible_users.append(usernames - mentioned_usernames(text))
else:
possible_users.append({user})
# print(possible_users, file=sys.stderr)
for i in range(m-1):
if len(possible_users[i]) == 1:
possible_users[i+1].difference_update(possible_users[i])
for i in range(m-1, 0, -1):
if len(possible_users[i]) == 1:
possible_users[i-1].difference_update(possible_users[i])
# print(possible_users, file=sys.stderr)
res = []
is_possible = True
prev_user = '$'
for i in range(m):
if possible_users[i]:
pusers = possible_users[i] - {prev_user}
resx = next(iter(pusers))
res.append(resx)
prev_user = resx
else:
is_possible = False
break
if is_possible:
for i in range(m):
print('{}:{}'.format(res[i], messages[i]))
else:
print("Impossible")
``` | output | 1 | 65,313 | 6 | 130,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course. | instruction | 0 | 65,314 | 6 | 130,628 |
Tags: brute force, constructive algorithms, dp, implementation, strings
Correct Solution:
```
import sys
import re
def mentioned_usernames(line):
return {x for x in re.split(r'[^A-Za-z0-9]+', line)}
t = int(input())
for ti in range(t):
possible_users = []
messages = []
n = int(input())
usernames = set(input().split())
# print("usernames =", usernames, file=sys.stderr)
m = int(input())
for i in range(m):
user, text = input().split(':')
messages.append(text)
if user == '?':
mu = mentioned_usernames(text)
possible_users.append(usernames - mentioned_usernames(text))
else:
possible_users.append({user})
# print(possible_users, file=sys.stderr)
is_fixed = [False] * m
for i in range(m-1):
if len(possible_users[i]) == 1:
possible_users[i+1].difference_update(possible_users[i])
is_fixed = True
for i in range(m-1, 0, -1):
if len(possible_users[i]) == 1:
possible_users[i-1].difference_update(possible_users[i])
is_fixed = True
# print(possible_users, file=sys.stderr)
res = []
is_possible = True
prev_user = '$'
for i in range(m):
if possible_users[i]:
pusers = possible_users[i] - {prev_user}
resx = next(iter(pusers))
res.append(resx)
prev_user = resx
else:
is_possible = False
break
if is_possible:
for i in range(m):
print('{}:{}'.format(res[i], messages[i]))
else:
print("Impossible")
``` | output | 1 | 65,314 | 6 | 130,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course. | instruction | 0 | 65,315 | 6 | 130,630 |
Tags: brute force, constructive algorithms, dp, implementation, strings
Correct Solution:
```
from math import *
from sys import *
from decimal import *
def gcd(a,b):
if b:
return gcd(b,a%b)
return a
t=int(stdin.readline())
for sssss in range(t):
n=int(stdin.readline())
names=stdin.readline().split()
dn=dict()
for i in range(len(names)):
dn[names[i]]=i
text=[]
ps=[]
m=int(stdin.readline())
for i in range(m):
nxt=stdin.readline()
getname=nxt.split(":")[0]
nxt=nxt.split(":")[1]
text.append(nxt)
if getname=="?":
pos=set([z for z in range(n)])
hell=""
shit=set([".",",","!","?"])
for y in nxt:
if y in shit:
hell+=" "
else:
hell+=y
nhere=hell.split()
for j in nhere:
pos.discard(dn.get(j))
if i and len(ps[-1])==1:
for dd in ps[-1]:
pos.discard(dd)
ps.append(pos)
else:
if i:
ps[-1].discard(dn[getname])
ps.append(set([dn[getname]]))
#print(ps)
dp=[[False]*n]
for i in ps[0]:
dp[0][i]=True
for i in range(1,m):
#print(dp)
dp.append([False]*n)
for j in ps[i]:
for k in range(n):
if k!=j and dp[-2][k]==True:
dp[-1][j]=True
break
if True not in dp[-1]:
stdout.write("Impossible\n")
continue
#print(dp)
yy=[0]*m
ban=-1
for i in range(m-1,-1,-1):
for j in range(n):
if j!=ban and dp[i][j]:
ban=j
yy[i]=j
break
#print(yy,names)
for i in range(m):
stdout.write(names[yy[i]]+":"+text[i])
``` | output | 1 | 65,315 | 6 | 130,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course. | instruction | 0 | 65,316 | 6 | 130,632 |
Tags: brute force, constructive algorithms, dp, implementation, strings
Correct Solution:
```
#!/usr/bin/python3
# BEGIN template
import sys
import re
import pprint
def dbg(x,y=''):
if len(y) > 0: y += ' = '
sys.stderr.write('\n>>> '+y+pprint.pformat(x)+'\n')
oo = 0x3f3f3f3f3f3f3f3f
# END template
def main():
for t in range(int(input())):
impo = False
# input
n = int(input())
users = input().split()
users_set = set(users)
users.insert(0,'0')
m = int(input())
msg = [None]*(m+5)
for i in range(1,m+1):
user, text = input().split(':')
alts = set()
if user != '?': user = users.index(user)
else:
# this shit is pretty fucked up, dude
for alt in users_set - {x for x in re.split(r'[^A-Za-z0-9]+',text)}:
alts.add(users.index(alt))
msg[i] = dict(user=user, text=text, users=alts)
# remove before and after
for i in range(1,m+1):
if 1 <= i-1: msg[i]['users'].discard(msg[i-1]['user'])
if i+1 <= m: msg[i]['users'].discard(msg[i+1]['user'])
if msg[i]['user'] == '?' and len(msg[i]['users']) == 0:
impo = True
break
if impo:
print('Impossible')
continue
# compute answer
dp = [[0 for j in range(n+5)] for i in range(m+5)]
for i in range(n+5):
dp[m+1][i] = oo
for i in range(m,0,-1):
u = msg[i]['user']
for j in range(n+1):
if u != '?':
if u != j and dp[i+1][u]: dp[i][j] = u
else:
for alt in msg[i]['users']:
if alt != j and dp[i+1][alt]:
dp[i][j] = alt
break
# output
if not dp[1][0]:
print('Impossible')
continue
j = 0
for i in range(1,m+1):
print(users[dp[i][j]]+':'+msg[i]['text'])
j = dp[i][j]
main()
``` | output | 1 | 65,316 | 6 | 130,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course. | instruction | 0 | 65,317 | 6 | 130,634 |
Tags: brute force, constructive algorithms, dp, implementation, strings
Correct Solution:
```
import re
t = int(input())
delimiters = "?", ".", " ", ",", "!", ":"
regexPattern = '|'.join(map(re.escape, delimiters))
for i in range(t):
n = int(input())
usernames = {x for x in str.split(input(), ' ')}
m = int(input())
possibilities = []
for j in range(m):
possibilities.append({x for x in usernames})
messages = []
for j in range(m):
messages.append(input())
for j in range(m):
if (messages[j][0] != '?'):
messageSplit = re.split(':', messages[j])
possibilities[j] = {messageSplit[0]}
else:
messageSplit = re.split(regexPattern, messages[j])
for token in messageSplit:
if token in usernames:
possibilities[j] = possibilities[j] - {token}
changed = True
while changed:
changed = False
for j in range(m):
if len(possibilities[j]) == 1:
(poss,) = possibilities[j]
if j < m-1 and poss in possibilities[j+1]:
changed = True
possibilities[j+1] = possibilities[j+1] - possibilities[j]
if j > 0 and poss in possibilities[j-1]:
changed = True
possibilities[j-1] = possibilities[j-1] - possibilities[j]
worked = True
for j in range(m):
if len(possibilities[j]) == 0:
worked = False
if not worked:
print("Impossible")
else :
for j in range(m):
poss = next(iter(possibilities[j]))
if (messages[j][0] == '?'):
print (poss + messages[j][1:])
else:
print(messages[j])
if (j < m-1):
possibilities[j+1] = possibilities[j+1] - {poss}
``` | output | 1 | 65,317 | 6 | 130,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course. | instruction | 0 | 65,318 | 6 | 130,636 |
Tags: brute force, constructive algorithms, dp, implementation, strings
Correct Solution:
```
from sys import *
t=int(stdin.readline())
for sssss in range(t):
text,ps,dn,n,names,m=[],[],dict(),int(stdin.readline()),stdin.readline().split(),int(stdin.readline())
for i in range(len(names)):
dn[names[i]]=i
for i in range(m):
nxt=stdin.readline()
getname=nxt.split(":")[0]
nxt=nxt.split(":")[1]
text.append(nxt)
if getname=="?":
pos,hell,shit=set([z for z in range(n)]),"",set([".",",","!","?"])
for y in nxt:
if y in shit:
hell+=" "
else:
hell+=y
nhere=hell.split()
for j in nhere:
pos.discard(dn.get(j))
if i and len(ps[-1])==1:
for dd in ps[-1]:
pos.discard(dd)
ps.append(pos)
else:
if i:
ps[-1].discard(dn[getname])
ps.append(set([dn[getname]]))
dp,yy,ban=[[False]*n],[0]*m,-1
for i in ps[0]:
dp[0][i]=True
for i in range(1,m):
dp.append([False]*n)
for j in ps[i]:
for k in range(n):
if k!=j and dp[-2][k]==True:
dp[-1][j]=True
break
if True not in dp[-1]:
stdout.write("Impossible\n")
continue
for i in range(m-1,-1,-1):
for j in range(n):
if j!=ban and dp[i][j]:
ban=j
yy[i]=j
break
for i in range(m):
stdout.write(names[yy[i]]+":"+text[i])
``` | output | 1 | 65,318 | 6 | 130,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course. | instruction | 0 | 65,319 | 6 | 130,638 |
Tags: brute force, constructive algorithms, dp, implementation, strings
Correct Solution:
```
import re
def proc(msgs):
chg = False
for i, msg in enumerate(msgs):
if msg[0] != '?': continue
ppn = msg[1]
if i > 0: ppn.discard(msgs[i-1][0])
if i != len(msgs)-1: ppn.discard(msgs[i+1][0])
if len(ppn) == 1:
msg[0] = ppn.pop()
chg = True
return chg
def is_valid(msgs):
for i, msg in enumerate(msgs):
if msg[0] == '?' or (i > 0 and msg[0] == msgs[i-1][0]) or (i != len(msgs)-1 and msg[0] == msgs[i+1][0]):
return False
return True
TC = int(input())
for tc in range(0, TC):
N = int(input())
persons = set(input().split())
M = int(input())
msgs = []
for m in range(M):
line = input()
m1 = re.match(r'(.*)\:(.*)', line)
ppn = set(persons)
for un in re.findall(r'[a-zA-Z0-9]+', m1.group(2)):
ppn.discard(un)
msgs.append([m1.group(1), ppn, m1.group(2)])
while True:
if proc(msgs): continue
chgt = False
for msg in msgs:
if msg[0] == '?' and len(msg[1]) > 0:
msg[0] = msg[1].pop()
chgt = True
break
if not chgt: break
if is_valid(msgs):
for msg in msgs:
print('%s:%s' % (msg[0], msg[2]))
else:
print('Impossible')
``` | output | 1 | 65,319 | 6 | 130,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course. | instruction | 0 | 65,320 | 6 | 130,640 |
Tags: brute force, constructive algorithms, dp, implementation, strings
Correct Solution:
```
import re
def foo():
n = int(input())
who = input().split()
m = int(input())
msg = []
l = []
for num in range(m):
a, t = input().split(':')
msg.append(t)
st = set()
if a != '?':
st.add(a)
else:
names = re.split('\W+', t)
for w in who:
if not w in names:
st.add(w)
l.append(st)
d2 = []
for num in range(1, m):
d = {}
for w1 in l[num]:
for w2 in l[num - 1]:
if w1 != w2:
d[w1] = w2
break
l[num] = [x for x in d]
d2.append(d)
curr = None
for w in l[m - 1]:
curr = w
res = [curr]
for num in list(reversed(range(1, m))):
curr = d2[num - 1].get(curr, None) # d2[num - 1][curr]
res.append(curr)
res = list(reversed(res))
if None in res:
print("Impossible")
else:
for num in range(m):
print(res[num] + ':' + msg[num])
t = int(input())
for _ in range(t):
foo()
``` | output | 1 | 65,320 | 6 | 130,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course.
Submitted Solution:
```
import re
def add_used(id, s, used):
used[id-1].add(s)
used[id].add(s)
used[id+1].add(s)
tcase = int(input())
for cas in range(tcase):
n = int(input())
names = input().split()
m = int(input())
res = True
ans = [''] * (m+1)
msg = [''] * (m+1)
used = [set() for i in range(m+1)]
for i in range(m):
ans[i], msg[i] = input().split(':')
if ans[i] != '?':
add_used(i, ans[i], used)
mentioned = re.split('\W+', msg[i])
for s in mentioned:
if s in names:
used[i].add(s)
for i in range(m):
if ans[i] == '?' and len(used[i]) == n-1:
ans[i] = list(set(names) - used[i])[0]
add_used(i, ans[i], used)
for i in range(m):
if ans[i] == '?':
if len(used[i]) == n:
res = False
else:
ans[i] = list(set(names) - used[i])[0]
add_used(i, ans[i], used)
for i in range(m-1):
if ans[i] == ans[i+1]:
res = False
if res:
for i in range(m):
print(ans[i] + ':' + msg[i])
else:
print('Impossible')
``` | instruction | 0 | 65,321 | 6 | 130,642 |
Yes | output | 1 | 65,321 | 6 | 130,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course.
Submitted Solution:
```
import re
def proc(msgs):
chg = False
for i, msg in enumerate(msgs):
if msg[0] != '?': continue
ppn = msg[1]
if i > 0: ppn.discard(msgs[i-1][0])
if i != len(msgs)-1: ppn.discard(msgs[i+1][0])
if len(ppn) == 1:
msg[0] = ppn.pop()
chg = True
return chg
def is_valid(msgs):
for i, msg in enumerate(msgs):
if msg[0] == '?' or (i > 0 and msg[0] == msgs[i-1][0]) or (i != len(msgs)-1 and msg[0] == msgs[i+1][0]):
return False
return True
TC = int(input())
for tc in range(0, TC):
N = int(input())
persons = set(input().split())
M = int(input())
msgs = []
for m in range(M):
line = input()
m1 = re.match(r'(.*)\:(.*)', line)
ppn = set(persons)
for un in re.findall(r'[a-zA-Z0-9]+', m1.group(2)):
ppn.discard(un)
msgs.append([m1.group(1), ppn, m1.group(2)])
while True:
if proc(msgs): continue
chgt = False
for msg in msgs:
if msg[0] == '?' and len(msg[1]) > 0:
msg[0] = msg[1].pop()
chgt = True
break
if not chgt: break
if is_valid(msgs):
for msg in msgs:
print('%s:%s' % (msg[0], msg[2]))
else:
print('Impossible')
# Made By Mostafa_Khaled
``` | instruction | 0 | 65,322 | 6 | 130,644 |
Yes | output | 1 | 65,322 | 6 | 130,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course.
Submitted Solution:
```
#!/usr/bin/python3
# BEGIN template
import sys
import re
import pprint
def dbg(x,y=''):
if len(y) > 0: y += ' = '
sys.stderr.write('\n>>> '+y+pprint.pformat(x)+'\n')
oo = 0x3f3f3f3f3f3f3f3f
# END template
def main():
t = int(input())
for t in range(t):
# input
n = int(input())
users = set(input().split())
m = int(input())
msg = []
for i in range(m):
user, text = input().split(':')
alts = set()
if user != '?':
alts.add(user)
else:
# this shit is pretty fucked up, dude
alts = users - {x for x in re.split(r'[^A-Za-z0-9]+',text)}
msg.append(dict(user=user, text=text, users=alts))
# remove before and after
for i in range(m-1):
if len(msg[i]['users']) == 1:
msg[i+1]['users'].difference_update(msg[i]['users'])
for i in range(m-1,0,-1):
if len(msg[i]['users']) == 1:
msg[i-1]['users'].difference_update(msg[i]['users'])
# compute answer
last = ''
impo = False
for i in range(m):
msg[i]['users'].discard(last)
if len(msg[i]['users']) == 0:
impo = True
break
last = next(iter(msg[i]['users']))
msg[i]['user'] = last
if impo:
print('Impossible')
continue
for i in range(m):
print(msg[i]['user']+':'+msg[i]['text'])
'''
dp = [[0 for j in range(n+5)] for i in range(m+5)]
for i in range(1,n+5):
dp[m+1][i] = oo
for i in range(m,0,-1):
for j in range(n+1):
for k in msg[i]['users']:
if k != j and dp[i+1][k]:
dp[i][j] = k
break
# output
if not dp[1][0]:
print('Impossible')
continue
j = 0
for i in range(1,m+1):
print(users[dp[i][j]]+':'+msg[i]['text'])
j = dp[i][j]'''
main()
``` | instruction | 0 | 65,323 | 6 | 130,646 |
Yes | output | 1 | 65,323 | 6 | 130,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input
The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of messages in the chat. The next m line contain the messages in the following formats, one per line:
* <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat.
* <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output
Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Examples
Input
1
2
Vladik netman
2
?: Hello, Vladik!
?: Hi
Output
netman: Hello, Vladik!
Vladik: Hi
Input
1
2
netman vladik
3
netman:how are you?
?:wrong message
vladik:im fine
Output
Impossible
Input
2
3
netman vladik Fedosik
2
?: users are netman, vladik, Fedosik
vladik: something wrong with this chat
4
netman tigerrrrr banany2001 klinchuh
4
?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
?: yes, netman
banany2001: yes of course.
Output
Impossible
netman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?
klinchuh: yes, coach!
tigerrrrr: yes, netman
banany2001: yes of course.
Submitted Solution:
```
#!/usr/bin/python3
# BEGIN template
import sys
import re
import pprint
def dbg(x,y=''):
if len(y) > 0: y += ' = '
sys.stderr.write('\n>>> '+y+pprint.pformat(x)+'\n')
oo = 0x3f3f3f3f3f3f3f3f
# END template
def main():
for t in range(int(input())):
# input
n = int(input())
users = input().split()
users_set = set(users)
users.insert(0,'0')
m = int(input())
msg = [None]*(m+5)
for i in range(1,m+1):
user, text = input().split(':')
alts = set()
if user != '?':
user = users.index(user)
alts.add(user)
else:
# this shit is pretty fucked up, dude
for alt in users_set - set(re.split('[^A-Za-z0-9]+',text)):
alts.add(users.index(alt))
msg[i] = dict(user=user, text=text, users=alts)
# remove before and after
for i in range(1,m):
if len(msg[i]['users']) == 1:
msg[i+1]['users'].difference_update(msg[i]['users'])
for i in range(m,1,-1):
if len(msg[i]['users']) == 1:
msg[i-1]['users'].difference_update(msg[i]['users'])
'''removed = True
while removed:
removed = False
for i in range(1,m+1):
if 1 <= i-1:
bef = len(msg[i]['users'])
msg[i]['users'].discard(msg[i-1]['user'])
if len(msg[i]['users']) < bef: removed = True
if i+1 <= m:
bef = len(msg[i]['users'])
msg[i]['users'].discard(msg[i+1]['user'])
if len(msg[i]['users']) < bef: removed = True'''
# compute answer
impo = False
last = 0
for i in range(1,m+1):
msg[i]['users'].discard(last)
if len(msg[i]['users']) == 0:
impo = True
break
last = next(iter(msg[i]['users']))
msg[i]['user'] = last
# output
if impo:
print('Impossible')
continue
for i in range(1,m+1):
print(users[msg[i]['user']]+':'+msg[i]['text'])
main()
``` | instruction | 0 | 65,324 | 6 | 130,648 |
Yes | output | 1 | 65,324 | 6 | 130,649 |
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