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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 15) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 500000) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters. It is guaranteed that the total length of words in all problem descriptions does not exceed 500015.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence. | instruction | 0 | 97,263 | 6 | 194,526 |
Tags: bitmasks, brute force, dp
Correct Solution:
```
import itertools
def count_inversions(enumerate_seq):
tmp = list(enumerate_seq[:])
result = 0
for i in range(len(tmp)):
for j in range(len(tmp) - 1):
if tmp[j][0] > tmp[j + 1][0]:
result += 1
tmp[j], tmp[j + 1] = tmp[j + 1], tmp[j]
return result
def sub_seq(a, b):
i, j = 0, 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
i += 1
j += 1
return i == len(a)
n = int(input())
A = input().split()
B = []
m = int(input())
for _ in range(m):
B += [input().split()[1:]]
brand_new = True
best_perms = 0
best_seq = []
best_ind = -1
for i, b in enumerate(B):
for tmp in itertools.permutations(enumerate(A)):
if sub_seq([x for y, x in tmp], b):
brand_new = False
inversions = count_inversions(tmp)
similarity = n * (n - 1) // 2 - inversions + 1
if best_perms < similarity:
best_perms = similarity
best_seq = [x for y, x in tmp]
best_ind = i
if not brand_new:
print(best_ind + 1)
print('[:' + '|' * best_perms + ':]')
else:
print("Brand new problem!")
``` | output | 1 | 97,263 | 6 | 194,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 15) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 500000) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters. It is guaranteed that the total length of words in all problem descriptions does not exceed 500015.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence. | instruction | 0 | 97,264 | 6 | 194,528 |
Tags: bitmasks, brute force, dp
Correct Solution:
```
import itertools
def check(curr_words, line):
if curr_words == []:
return True
for i in range(len(line)):
if line[i] == curr_words[0]:
return check(curr_words[1:], line[i+1:])
return False
n = int(input())
words = input().split()
m = int(input())
res, idx = 0, 0
for i in range(m):
line = input().split()[1:]
for p in itertools.permutations(range(n)):
curr_words = [words[j] for j in p]
cnt = 0
for j in range(n):
cnt += len([k for k in range(j+1, n) if p[k] < p[j]])
v = n * (n-1) // 2 - cnt + 1
if check(curr_words, line[:]) and v > res:
res, idx = v, i+1
if res > 0:
print(idx)
print('[:'+str('|'*res)+':]')
else:
print('Brand new problem!')
``` | output | 1 | 97,264 | 6 | 194,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend has n cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input
The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.
Output
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Examples
Input
ee
Output
2
Input
z
Output
0
Input
0ay1
Output
2
Note
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | instruction | 0 | 97,546 | 6 | 195,092 |
Tags: brute force, implementation
Correct Solution:
```
cards = input()
flips = 0
for i in range(0, len(cards)):
if(cards[i] in '0123456789'):
if(int(cards[i])%2 != 0):
flips +=1
else:
if(cards[i] in 'aeiou'):
flips +=1
print(flips)
``` | output | 1 | 97,546 | 6 | 195,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend has n cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input
The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.
Output
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Examples
Input
ee
Output
2
Input
z
Output
0
Input
0ay1
Output
2
Note
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | instruction | 0 | 97,547 | 6 | 195,094 |
Tags: brute force, implementation
Correct Solution:
```
test_string = input()
turns = 0
for char in test_string:
if char in ['1', '3', '5', '7', '9', 'a', 'e', 'i', 'o', 'u']:
turns += 1
print(turns)
``` | output | 1 | 97,547 | 6 | 195,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend has n cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input
The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.
Output
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Examples
Input
ee
Output
2
Input
z
Output
0
Input
0ay1
Output
2
Note
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | instruction | 0 | 97,548 | 6 | 195,096 |
Tags: brute force, implementation
Correct Solution:
```
s = input("")
count = 0
for i in s:
if i in "aeiou13579":
count +=1
print (count)
``` | output | 1 | 97,548 | 6 | 195,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend has n cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input
The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.
Output
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Examples
Input
ee
Output
2
Input
z
Output
0
Input
0ay1
Output
2
Note
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | instruction | 0 | 97,549 | 6 | 195,098 |
Tags: brute force, implementation
Correct Solution:
```
s = input()
cnt = 0
for c in s:
if c in '13579':
cnt += 1
elif c in 'aeiou':
cnt += 1
print(cnt)
``` | output | 1 | 97,549 | 6 | 195,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend has n cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input
The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.
Output
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Examples
Input
ee
Output
2
Input
z
Output
0
Input
0ay1
Output
2
Note
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | instruction | 0 | 97,550 | 6 | 195,100 |
Tags: brute force, implementation
Correct Solution:
```
s = input()
c=0
for i in range(len(s)):
if s[i] == 'a' or \
s[i] == 'e' or \
s[i] == 'i' or \
s[i] == 'o' or \
s[i] == 'u' or \
s[i] == '1' or \
s[i] == '3' or \
s[i] == '5' or \
s[i] == '7' or \
s[i] == '9':
c = c + 1
print(c)
``` | output | 1 | 97,550 | 6 | 195,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend has n cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input
The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.
Output
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Examples
Input
ee
Output
2
Input
z
Output
0
Input
0ay1
Output
2
Note
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | instruction | 0 | 97,551 | 6 | 195,102 |
Tags: brute force, implementation
Correct Solution:
```
print(sum(x in 'aeiou13579' for x in input()))
``` | output | 1 | 97,551 | 6 | 195,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend has n cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input
The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.
Output
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Examples
Input
ee
Output
2
Input
z
Output
0
Input
0ay1
Output
2
Note
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | instruction | 0 | 97,552 | 6 | 195,104 |
Tags: brute force, implementation
Correct Solution:
```
s = input()
result = 0
for c in s:
if c in ['a', 'e', 'i', 'o', 'u', '1', '3', '5', '7', '9']:
result += 1
print(result)
``` | output | 1 | 97,552 | 6 | 195,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend has n cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input
The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.
Output
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Examples
Input
ee
Output
2
Input
z
Output
0
Input
0ay1
Output
2
Note
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | instruction | 0 | 97,553 | 6 | 195,106 |
Tags: brute force, implementation
Correct Solution:
```
s = input()
l = len(s)
match = ['a','e','i','u','o']
match_num = ['0','2','4','6','8']
count = 0
for i in range(l):
if s[i].isdigit():
if not (s[i] in match_num):
count += 1
else:
if s[i] in match:
count += 1
print(count)
``` | output | 1 | 97,553 | 6 | 195,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not. | instruction | 0 | 97,562 | 6 | 195,124 |
Tags: *special, strings, trees
Correct Solution:
```
# python3
# utf-8
class Trie:
def __init__(self):
self.letter___node = {}
self.words_nr = 0
def add_word(self, word):
word = word + '$'
curr_node = self
for letter in word:
if letter not in curr_node.letter___node:
curr_node.letter___node[letter] = Trie()
curr_node.words_nr += 1
curr_node = curr_node.letter___node[letter]
def check_word(self, word):
word = word + '$'
curr_node = self
for letter in word:
if letter not in curr_node.letter___node:
return False
curr_node = curr_node.letter___node[letter]
return True
def count_word(self, word):
word = word + '$'
curr_node = self
curr_state = 0
presses_saved = 0
for letter in word:
if letter not in curr_node.letter___node:
if curr_state == 1:
# print(presses_saved)
if '$' in curr_node.letter___node:
return min(len(word) - 1,
len(word) - 1 - presses_saved + 1
)
else:
return len(word) - 1
if curr_state == 0:
return len(word) - 1
if curr_node.words_nr > 1:
curr_node = curr_node.letter___node[letter]
elif curr_node.words_nr == 1:
# print(letter, presses_saved)
if curr_state == 0:
curr_state = 1
presses_saved += 1
curr_node = curr_node.letter___node[letter]
elif curr_node.words_nr == 0:
if curr_state == 1:
return min(len(word) - 1,
len(word) - 1 - presses_saved + 1
)
elif curr_state == 0:
return len(word) - 1
if curr_node.words_nr == 0:
presses_saved -= 1
if curr_state == 1:
return min(len(word) - 1,
len(word) - 1 - presses_saved + 1
)
elif curr_state == 0:
return len(word) - 1
text = ''
while(1):
try:
line = input()
if line == '':
raise Exception('e')
text += line + '\n'
except:
break
# print(text)
ans = 0
syms = ['\n', '.', ',', '?', '!', "'", '-']
for sym in syms:
text = text.replace(sym, ' ')
ans += text.count(' ')
idx___word = text.split(' ')
root = Trie()
root.add_word('$')
root.add_word('$$')
for word in idx___word:
if word == '':
continue
count = root.count_word(word)
check = root.check_word(word)
# print(word, check, count)
ans += count
if not check:
root.add_word(word)
print(ans)
``` | output | 1 | 97,562 | 6 | 195,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not. | instruction | 0 | 97,563 | 6 | 195,126 |
Tags: *special, strings, trees
Correct Solution:
```
class Ddict:
def __init__(self):
self.dicts={}
def add(self,key):
d=self.dicts
for i in key:
if i not in d:
d[i]={}
d=d[i]
d[' ']=''
def find(self,key):
if key=='':
return '',''
d=self.dicts
q=[]
h=[key[0]]
for i in key:
if i not in d:
if ' ' in d and len(d)==1:
return ''.join(q),''.join(h)
return '',''
q.append(i)
if len(d)!=1:
h=q[:]
d=d[i]
if ' ' in d and len(d)==1:
return ''.join(q),''.join(h)
return '',''
words = Ddict()
ans=0
while True:
try:
x=input()
if not x:
break
except:
break
ans+=len(x)+1
ws=[[]]
for i in x:
if i in '.,?!\'- ':
if ws[-1]:
ws.append([])
else:
ws[-1].append(i)
ws=list(map(lambda e:''.join(e),ws))
for w in ws:
next_word,helped_word = words.find(w)
if next_word and next_word!=helped_word:
ans-=len(next_word)-len(helped_word)-1
words.add(w)
print(ans)
``` | output | 1 | 97,563 | 6 | 195,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not. | instruction | 0 | 97,564 | 6 | 195,128 |
Tags: *special, strings, trees
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
class Trie:
class Node:
def __init__(self, char: int = 1):
self.char = char
self.children = []
self.counter = 1
def __init__(self):
self.root = Trie.Node()
self.ans = 0
def add(self, word):
node = self.root
for char in word:
found_in_child = False
for child in node.children:
if child.char == char:
child.counter += 1
node = child
found_in_child = True
break
if not found_in_child:
new_node = Trie.Node(char)
node.children.append(new_node)
node = new_node
def query(self, prefix, root=None):
if not root: root = self.root
node = root
if not root.children:
return 0
prefix = [prefix]
for char in prefix:
char_not_found = True
for child in node.children:
if child.char == char:
char_not_found = False
node = child
break
if char_not_found:
return 0
return node
a = []
tr = Trie()
while True:
try:
s = input()
if not s:break
except EOFError:
break
a += [s]
def calc(s):
node = tr.root
found = False
ind = -1
its = 0
for char in s:
if not found:
ind += 1
next = tr.query(char, node)
if not next:
break
if next.counter == 1 and not found:
found = True
node = next
its += 1
if found and node.counter == 1:
if not node.children:
tr.ans += min(len(s), ind + 2 + len(s)-its)
elif node.children:
tr.ans += len(s)
else:
tr.ans += min(len(s), ind + 2)
else:
tr.ans += len(s)
if not (its == len(s) and not node.children):
tr.add(s)
for s in a:
tr.ans += 1
cur = []
if 1:
for k in s:
if not k.isalpha():
tr.ans += 1
if cur:
calc(cur)
cur = []
else:
cur += [ord(k)]
if cur:
calc(cur)
print(tr.ans)
``` | output | 1 | 97,564 | 6 | 195,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not. | instruction | 0 | 97,565 | 6 | 195,130 |
Tags: *special, strings, trees
Correct Solution:
```
import sys
import re
SEPARATORS = "[.,? !'-]"
class TrieNode(object):
def __init__(self):
self.terminal = False
self.go = {}
self.count = 0
def insert(node, s):
nodes = [node]
unique, auto = 0, 0
for c in s:
if c not in node.go:
node.go[c] = TrieNode()
node = node.go[c]
nodes.append(node)
if node.count == 1:
unique += 1
if node.terminal:
auto = max(unique - 2, 0)
if not node.terminal:
node.terminal = True
for node in nodes:
node.count += 1
return auto
root = TrieNode()
answer = 0
for line in sys.stdin:
answer += len(line)
for word in filter(None, re.split(SEPARATORS, line.strip())):
answer -= insert(root, word)
print(answer)
``` | output | 1 | 97,565 | 6 | 195,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not. | instruction | 0 | 97,566 | 6 | 195,132 |
Tags: *special, strings, trees
Correct Solution:
```
import sys
def read_text():
text = ''
for line in sys.stdin:
text += line
return text
# def read_text():
# text = ''
# next_line = input()
# while len(next_line) > 0:
# text += next_line + '\n'
# next_line = input()
# return text
def matching_word(word, bor, symbol, search):
if symbol in search:
search = search[symbol]
word_found = 'size' in search and search['size'] == 1
return get_match(word, bor) if word_found else 0, search
return 0, ()
def contains_word(word, bor):
search = bor
for symbol in word:
if symbol in search:
search = search[symbol]
else:
return False
words_left = 0
for key, value in search.items():
if key == 'size':
continue
words_left += value['size']
return search['size'] - words_left == 1
def get_match(word, bor):
search = bor
match = ''
for symbol in word:
match += symbol
search = search[symbol]
while len(search.keys()) == 2:
symbol = next(filter(lambda item: item != 'size', search.keys()))
match += symbol
search = search[symbol]
return match
def add_to_bor(word, bor):
search = bor
for symbol in word:
if symbol not in search:
search[symbol] = dict([('size', 0)])
search = search[symbol]
search['size'] += 1
def replace_index(text, start, match):
index = start
for symbol in match:
if symbol == text[index]:
index += 1
else:
return -1
return start + len(match) - 1
def get_updated_search(word, bor):
search = bor
for symbol in word:
search = search[symbol]
return search
text = list(read_text())
text_size = len(text)
bor = dict()
separators = ['.', ',', '?', '!', '\'', '-', ' ', '\n']
typed = 0
index = 0
word = ''
search = bor
match = 0
while index < text_size:
symbol = text[index]
typed += 1
if symbol not in separators:
word += symbol
if match != -1:
match, search = matching_word(word, bor, symbol, search)
if match != 0 and match != word:
new_index = replace_index(text, index + 1 - len(word), match)
if new_index != -1:
typed += 1
index = new_index
word = match
search = get_updated_search(word, bor)
else:
match = -1
elif len(word) > 0:
if not contains_word(word, bor):
add_to_bor(word, bor)
search = bor
word = ''
match = 0
index += 1
print(typed)
``` | output | 1 | 97,566 | 6 | 195,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not.
Submitted Solution:
```
def main(file = None):
text = ""
part = input()
while part:
text += part +'\n'
try:
part = input()
except EOFError:
part = ""
total = len(text)
word = ""
for l in text:
if l.isalpha():
word += l
elif word:
short = shortcut(word)
total -= short
update(word)
word = ""
return total
prev = set()
def shortcut(word):
for i in range(1, len(word)-2):
occ = 0
part = word[:i]
for auto in prev:
if auto.startswith(part) and auto != part:
occ += 1
latest = auto
if occ == 1 and word.startswith(latest):
return len(latest)-len(part)-1
return 0
def update(word):
prev.add(word)
if __name__ == '__main__':
print(main())
``` | instruction | 0 | 97,567 | 6 | 195,134 |
No | output | 1 | 97,567 | 6 | 195,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not.
Submitted Solution:
```
import fileinput
words_stack = dict()
words = list()
def get_similar(word):
was_new = False
end = False
count = 0
need_write = 0
new_word_count = 0
for i in range(1,len(word) + 1):
little_word = word[0:i]
if not words_stack.get(little_word):
if i == len(word):
words_stack[little_word] = 1
else:
words_stack[little_word] = 2
new_word_count += 1
if not was_new:
for j in range(1, i):
little_word = word[0:j]
words_stack[little_word] = 3
was_new = True
else:
number_word = words_stack[little_word]
if number_word == 1:
end = True
count += 1
elif number_word == 2:
count += 1
else:
need_write += 1
if end:
return (new_word_count + 2 + need_write) if count >= 2 else (new_word_count + count + need_write)
else:
return new_word_count + count + need_write
answer = 0
word = str("")
prep = str("")
for line in fileinput.input():
for char in list(line):
if char in ["'", ',', ' ', '\n', '.', '-', '!', '?']:
plus = get_similar(word)
answer += plus
word = ""
prep += char
else:
plus = get_similar(prep)
answer += plus
prep = ""
word += str(char)
answer += get_similar(prep)
answer += get_similar(word)
print(answer)
``` | instruction | 0 | 97,568 | 6 | 195,136 |
No | output | 1 | 97,568 | 6 | 195,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not.
Submitted Solution:
```
import sys
def next_word_start(str, beg):
for i in range(beg, len(str)):
if str[i] in punctuation:
return i
return len(str)
def was_semi_typed(words, lookup):
max_size = 0
res = ""
for word in words:
similarity = 0
if len(lookup) < len(word):
continue
for i in range(len(word)):
if word[i] == lookup[i]:
similarity += 1
if max_size < similarity and i == len(word) - 1:
max_size = similarity
res = word
else:
break
return res
prefixes = set()
punctuation = ['.', ',', '?', '-', '!', '\'', ' ', '\n']
inp = sys.stdin.readlines()
data = ""
for lines in inp:
data += lines.title().lower()
words = set()
non_unique_prefixes = set()
result = 0
i = 0
last_word_end = -1
while i < len(data):
word_ended = data[i] in punctuation
if word_ended:
words.add(data[last_word_end + 1 : i])
last_word_end = i
# we have printed a punt mark
result += 1
i += 1
else:
current_prefix = data[last_word_end + 1 : i + 1]
if current_prefix in non_unique_prefixes:
# if it's in non unique, we need to continue typing letters
result += 1
i += 1
continue
if not (current_prefix in prefixes):
prefixes.add(current_prefix)
result += 1
i += 1
else:
# seems to be we have this word. lookup front to know what the word is...
next_space = next_word_start(data, last_word_end + 1)
lookup = data[last_word_end + 1 : next_space]
# need to know if this word is part of smth. E.g: we can suggest "test", but user printed "tested".
semi_typed = was_semi_typed(words, lookup)
suggestion_size = len(was_semi_typed(words, lookup))
if semi_typed in words and semi_typed != '' and current_prefix != semi_typed:
result += 2 # +1 because we printed a new letter, +1 because we accept hint
append = len(lookup) - suggestion_size
# внутри надо удалить префиксы слова semi_typed и добавить новый
# например. Мы написали слова thun, затем по букве t предлагаем ввести "thun". Если пользователь принял
# подсказку и дополнил её чем-то, например, "thunder", то нельзя реагировать на t, th, thu, thun
# (добавить их в список non-unique и удалить из unique(prefixes). При этом в unique добавить thund,
# как новый префикс для реагирования
if append > 0:
for j in range(len(semi_typed)):
if semi_typed[:j + 1] in prefixes:
prefixes.remove(semi_typed[:j + 1])
non_unique_prefixes.add(semi_typed[:j + 1])
# добавляем все уникальные префиксы нашего слова
for k in range(len(lookup)):
if not (lookup[:len(semi_typed) + k] in non_unique_prefixes):
prefixes.add(lookup[:len(semi_typed) + k])
# добавляем наше новое слово в словарь:)
words.add(lookup)
result += append # problem is here. Hopefully fixed
i = next_space # todo
last_word_end = i
else:
prefixes.remove(current_prefix)
non_unique_prefixes.add(current_prefix)
i += 1
result += 1
print(result)
``` | instruction | 0 | 97,569 | 6 | 195,138 |
No | output | 1 | 97,569 | 6 | 195,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arcady is a copywriter. His today's task is to type up an already well-designed story using his favorite text editor.
Arcady types words, punctuation signs and spaces one after another. Each letter and each sign (including line feed) requires one keyboard click in order to be printed. Moreover, when Arcady has a non-empty prefix of some word on the screen, the editor proposes a possible autocompletion for this word, more precisely one of the already printed words such that its prefix matches the currently printed prefix if this word is unique. For example, if Arcady has already printed «codeforces», «coding» and «codeforces» once again, then there will be no autocompletion attempt for «cod», but if he proceeds with «code», the editor will propose «codeforces».
With a single click Arcady can follow the editor's proposal, i.e. to transform the current prefix to it. Note that no additional symbols are printed after the autocompletion (no spaces, line feeds, etc). What is the minimum number of keyboard clicks Arcady has to perform to print the entire text, if he is not allowed to move the cursor or erase the already printed symbols?
A word here is a contiguous sequence of latin letters bordered by spaces, punctuation signs and line/text beginnings/ends. Arcady uses only lowercase letters. For example, there are 20 words in «it's well-known that tic-tac-toe is a paper-and-pencil game for two players, x and o.».
Input
The only line contains Arcady's text, consisting only of lowercase latin letters, spaces, line feeds and the following punctuation signs: «.», «,», «?», «!», «'» and «-». The total amount of symbols doesn't exceed 3·105. It's guaranteed that all lines are non-empty.
Output
Print a single integer — the minimum number of clicks.
Examples
Input
snow affects sports such as skiing, snowboarding, and snowmachine travel.
snowboarding is a recreational activity and olympic and paralympic sport.
Output
141
Input
'co-co-co, codeforces?!'
Output
25
Input
thun-thun-thunder, thunder, thunder
thunder, thun-, thunder
thun-thun-thunder, thunder
thunder, feel the thunder
lightning then the thunder
thunder, feel the thunder
lightning then the thunder
thunder, thunder
Output
183
Note
In sample case one it's optimal to use autocompletion for the first instance of «snowboarding» after typing up «sn» and for the second instance of «snowboarding» after typing up «snowb». This will save 7 clicks.
In sample case two it doesn't matter whether to use autocompletion or not.
Submitted Solution:
```
import sys
def read_text():
text = ''
for line in sys.stdin:
text += line
return text
def matching_word(word, bor, symbol, search):
if symbol in search:
search = search[symbol]
word_found = 'size' in search and search['size'] == 1
return get_match(word, bor) if word_found else 0, search
return 0, ()
def contains_word(word, bor):
search = bor
for symbol in word:
if symbol in search:
search = search[symbol]
else:
return False
return len(search.keys()) > 0
def get_match(word, bor):
search = bor
match = ''
for symbol in word:
match += symbol
search = search[symbol]
while len(search.keys()) == 2:
symbol = next(filter(lambda item: item != 'size', search.keys()))
match += symbol
search = search[symbol]
return match
def add_to_bor(word, bor):
search = bor
for symbol in word:
if symbol not in search:
search[symbol] = dict([('size', 0)])
search = search[symbol]
search['size'] += 1
def replace_index(text, start, match):
index = start
for symbol in match:
if symbol == text[index]:
index += 1
else:
return -1
return start + len(match) - 1
text = list(read_text())
text_size = len(text)
bor = dict()
separators = ['.', ',', '?', '!', '\'', '-', ' ', '\n']
typed = 0
index = 0
word = ''
search = bor
while index < text_size:
symbol = text[index]
typed += 1
if symbol not in separators:
word += symbol
match, search = matching_word(word, bor, symbol, search)
if match != 0 and match != word:
new_index = replace_index(text, index + 1 - len(word), match)
if new_index != -1:
typed += 1
index = new_index
word = match
elif len(word) > 0:
if not contains_word(word, bor):
add_to_bor(word, bor)
search = bor
word = ''
index += 1
print(typed)
``` | instruction | 0 | 97,570 | 6 | 195,140 |
No | output | 1 | 97,570 | 6 | 195,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb". | instruction | 0 | 97,821 | 6 | 195,642 |
Tags: dp, hashing, strings
Correct Solution:
```
import math
from collections import defaultdict
def main():
p = 31
m = 1e9 + 9
def string_hash(s):
hash_value = 0
p_pow = 1
for c in s:
hash_value = (hash_value + (ord(c)-ord('a')+1) * p_pow) % m
p_pow = (p_pow * p) % m
return hash_value
n = int(input())
words = input().split()
eq = [[False] * n for _ in range(n)]
for i in range(n):
for j in range(n):
if i == j:
eq[i][j] = eq[j][i] = True
else:
eq[i][j] = eq[i][j] = words[i] == words[j]
dp = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n-1, -1, -1):
for j in range(n-1, -1, -1):
if not eq[i][j]:
dp[i][j] = 0
else:
if i < n-1 and j < n-1:
dp[i][j] = dp[i+1][j+1] + 1
else:
dp[i][j] = 1
all = sum(len(w) for w in words) + n-1
ans = all
for start in range(n):
size = 1
while start + size < n:
num = 1
pos = start + size
while pos + size <= n:
if dp[start][pos] >= size:
num += 1
pos += size
else:
pos += 1
if num > 1:
segsize = sum(len(words[i]) for i in range(start, start+size)) + size-1
newsize = size
ans = min(ans, all - num * segsize + num * newsize)
size += 1
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 97,821 | 6 | 195,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb". | instruction | 0 | 97,822 | 6 | 195,644 |
Tags: dp, hashing, strings
Correct Solution:
```
n = int(input())
arr = input()
final = len(arr)
arr = arr.split()
lens = [0 for x in range(n)]
visit = [0 for x in range(n)]
cnt = 0
ans = 0
for i in range(n):
if visit[i]:
continue
lens[cnt] = len(arr[i])
for j in range(i+1,n):
if arr[j]==arr[i]:
arr[j] = cnt
visit[j] = 1
arr[i] = cnt
cnt += 1
for i in range(n):
for j in range(i,n):
temp = arr[i:j+1]
ind = 1
found = 0
len2 = j-i+1
cur = 0
kmp = [0 for x in range(len2)]
while ind < len2:
if temp[ind] == temp[cur]:
cur += 1
kmp[ind] = cur
ind += 1
else:
if cur != 0:
cur -= 1
else:
kmp[ind] = 0
ind += 1
ind = 0
cur = 0
while ind < n:
if arr[ind] == temp[cur]:
ind += 1
cur += 1
if cur == len2:
found += 1
cur = 0
elif ind < n and temp[cur] != arr[ind]:
if cur != 0:
cur = kmp[cur-1]
else:
ind += 1
if found>1:
res = 0
for k in temp:
res += (lens[k]-1)*(found)
res += (len(temp)-1)*(found)
ans = max(ans,res)
print(final-ans)
``` | output | 1 | 97,822 | 6 | 195,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb". | instruction | 0 | 97,823 | 6 | 195,646 |
Tags: dp, hashing, strings
Correct Solution:
```
n = int(input())
s = input()
a = list(s.split())
eq = [[0 for i in range(n)] for j in range(n)]
dp = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
eq[i][i] = 1
for j in range(0, i):
if a[i] == a[j]:
eq[i][j] += 1
eq[j][i] += 1
for i in range(n - 1, -1, -1):
for j in range(n - 1, -1, -1):
if eq[i][j] == 1:
if i < n - 1 and j < n - 1:
dp[i][j] = dp[i + 1][j + 1] + 1
else:
dp[i][j] = 1
allsum = n - 1
for k in a:
allsum += len(k)
ans = allsum
for i in range(n):
sx = 0
j = 0
while i + j < n:
sx += len(a[i + j])
cnt = 1
pos = i + j + 1
while pos < n:
if dp[i][pos] > j:
cnt += 1
pos += j
pos += 1
cur = allsum - sx*cnt + (j + 1)*cnt - j*cnt
if cnt > 1 and ans > cur:
ans = cur
j += 1
print(ans)
``` | output | 1 | 97,823 | 6 | 195,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb". | instruction | 0 | 97,824 | 6 | 195,648 |
Tags: dp, hashing, strings
Correct Solution:
```
# import time
N = 303
eq = []
dp = []
for i in range(N):
eq.append([False] * N)
for i in range(N):
dp.append([0] * N)
n = int(input())
s = input()
# t = time.time()
allsum = len(s)
s = s.split()
for i in range(n):
eq[i][i] = True
for j in range(i):
eq[i][j] = eq[j][i] = s[i] == s[j]
for i in range(n - 1, -1, -1):
for j in range(n - 1, -1, -1):
if eq[i][j]:
if i < n - 1 and j < n - 1:
dp[i][j] = dp[i + 1][j + 1] + 1
else:
dp[i][j] = 1
ans = allsum
for i in range(n):
su = 0
for j in range(1, n - i + 1):
su += len(s[i + j - 1])
cnt = 1
pos = i + j
while pos < n:
if dp[i][pos] >= j:
cnt += 1
pos += j - 1
pos += 1
cur = allsum - su * cnt + cnt
if cnt > 1 and ans > cur:
# print(allsum, su, cnt, j)
ans = cur
print(ans)
# print(time.time() - t)
``` | output | 1 | 97,824 | 6 | 195,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb". | instruction | 0 | 97,825 | 6 | 195,650 |
Tags: dp, hashing, strings
Correct Solution:
```
# import time
N = 303
eq = []
dp = []
for i in range(N):
eq.append([False] * N)
for i in range(N):
dp.append([0] * N)
n = int(input())
s = input()
# t = time.time()
allsum = len(s)
s = s.split()
for i in range(n):
eq[i][i] = True
for j in range(i):
eq[i][j] = eq[j][i] = s[i] == s[j]
for i in range(n - 1, -1, -1):
for j in range(n - 1, -1, -1):
if eq[i][j]:
if i < n - 1 and j < n - 1:
dp[i][j] = dp[i + 1][j + 1] + 1
else:
dp[i][j] = 1
ans = allsum
for i in range(n):
su = 0
for j in range(n - i):
su += len(s[i + j])
cnt = 1
pos = i + j + 1
while pos < n:
if dp[i][pos] > j:
cnt += 1
pos += j
pos += 1
cur = allsum - su * cnt + (j + 1) * cnt - j * cnt
if cnt > 1 and ans > cur:
# print(allsum, su, cnt, j)
ans = cur
print(ans)
# print(time.time() - t)
``` | output | 1 | 97,825 | 6 | 195,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb". | instruction | 0 | 97,826 | 6 | 195,652 |
Tags: dp, hashing, strings
Correct Solution:
```
from sys import stdin
def kmp(pat, txt):
leng = 0;i = 1
ans=0
M = len(pat) ;N = len(txt) ;lps = [0]*M ;j = 0
#Calculo de lps, prefifo propio mas largo que tambien es sufijo de pat[0:i]
while i < M:
if pat[i]== pat[leng]:
leng += 1
lps[i] = leng
i += 1
elif leng != 0:
leng = lps[leng-1]
else:
lps[i] = 0
i += 1
i = 0
while i < N:
if pat[j] == txt[i]:
i += 1;j += 1
if j == M:
if ((i-j)==0 or txt[i-j-1]==" ") and ((i-j+len(pat))==(len(txt)) or txt[i-j+len(pat)]==" "):
ans+=1
i=i-j+len(pat)
j =0
else:
j = lps[j-1]
elif i < N and pat[j] != txt[i]:
if j != 0:
j = lps[j-1]
else:
i += 1
return ans
n=int(stdin.readline().strip())
s1=stdin.readline().strip().split()
s=[]
x=1
d=dict()
d1=dict()
st=set()
ans=n-1
s2=""
for i in s1:
ans+=len(i)
if i not in st:
d.update({i:x})
d1.update({x:len(i)})
x+=1
st.add(i)
s.append(d[i])
s2+=str(d[i])
if len(s)<n:
s2+=" "
acum=0
s3=s2.split()
tot=ans
for i in range(n):
x=0
y=0
for j in range(i,n):
x+=(d1[s[j]]-1)
y+=len(s3[j])
z=kmp(s2[acum:acum+y+j-i],s2)
if z>1:
aux=tot-(z*(x+j-i))
#print(s2[acum:acum+y+j-i],"/ ",aux,acum,y,j-i)
if aux<ans:
#print(s2[acum:acum+y+j-i+1],"/ ",aux)
ans=aux
acum+=len(s3[i])+1
print(ans)
``` | output | 1 | 97,826 | 6 | 195,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb". | instruction | 0 | 97,827 | 6 | 195,654 |
Tags: dp, hashing, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
s = input()
a = list(s.split())
eq = [[0 for i in range(n)] for j in range(n)]
dp = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
eq[i][i] = 1
for j in range(0, i):
if a[i] == a[j]:
eq[i][j] += 1
eq[j][i] += 1
for i in range(n - 1, -1, -1):
for j in range(n - 1, -1, -1):
if eq[i][j] == 1:
if i < n - 1 and j < n - 1:
dp[i][j] = dp[i + 1][j + 1] + 1
else:
dp[i][j] = 1
allsum = n - 1
for k in a:
allsum += len(k)
ans = allsum
for i in range(n):
sx = 0
j = 0
while i + j < n:
sx += len(a[i + j])
cnt = 1
pos = i + j + 1
while pos < n:
if dp[i][pos] > j:
cnt += 1
pos += j
pos += 1
cur = allsum - sx*cnt + (j + 1)*cnt - j*cnt
if cnt > 1 and ans > cur:
ans = cur
j += 1
print(ans)
``` | output | 1 | 97,827 | 6 | 195,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb".
Submitted Solution:
```
n = int(input())
arr = input()
final = len(arr)
arr = arr.split()
lens = [0 for x in range(n)]
visit = [0 for x in range(n)]
cnt = 0
ans = 0
for i in range(n):
if visit[i]:
continue
lens[cnt] = len(arr[i])
for j in range(i+1,n):
if arr[j]==arr[i]:
arr[j] = cnt
visit[j] = 1
arr[i] = cnt
cnt += 1
for i in range(n):
for j in range(i,n):
temp = arr[i:j+1]
ind = j+1
need = len(temp)
found = 0
while ind < n:
if arr[ind]==temp[0]:
cnt = 0
while ind+cnt < n:
if arr[ind+cnt]==temp[cnt]:
cnt += 1
else:
break
if cnt==need:
found += 1
break
ind += cnt
continue
ind += 1
if found:
res = 0
for k in temp:
res += (lens[k]-1)*(found+1)
res += (len(temp)-1)*(found+1)
ans = max(ans,res)
print(final-ans)
``` | instruction | 0 | 97,828 | 6 | 195,656 |
No | output | 1 | 97,828 | 6 | 195,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb".
Submitted Solution:
```
n=int(input())
s=input().split()
flag=[[s[i]!=s[j] for j in range(n)]for i in range(n)]
b=[0]+[len(i)-1 for i in s]
for i in range(n):b[i+1]+=b[i]
ans=0
for i in range(n-1):
for j in range(i+1,n):
t=min(j-i,n-j)
for k in range(t):
if flag[i+k][j+k]:
break
if k:
ans=max(ans,b[i+k+1]-b[i]+k)
print(((b[n]+n)+(n-1))-ans*2)
``` | instruction | 0 | 97,829 | 6 | 195,658 |
No | output | 1 | 97,829 | 6 | 195,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb".
Submitted Solution:
```
#!/usr/bin/env python3
import math
import sys
import re
def get_abbreviation(s):
abbrev = ""
for w in s:
abbrev += w[0].upper()
return abbrev
results = {}
def calculate_savings(o):
ss_word_count = len(o[0].split(" ")) # This will be the size of abbrev (ex. a ab abbbb abbbb = 4 (AAAA))
ss_length = len(o[0]) # This will be the size saved for each word (ex. a ab abbb = 9
# To get the savings we take how much is saved per occurrence ss_len - ss_word_count
# and multiple by the total number of occurrences len(o)
return (ss_length - ss_word_count) * len(o)
def substring_length(start, end, testWords, testString):
mid = math.ceil((start + end)/2)
substring = " ".join(testWords[start:end])
if substring not in results:
r = '\\b{}\\b'.format(substring)
o = re.findall(r, testString)
if len(o) > 1:
savings = calculate_savings(o)
#print("testString: {} substring: {} o: {} savings: {}".format(testString, substring, o, savings))
results[substring] = savings
if start < mid and end > mid:
substring_length(start, mid, testWords, testString)
substring_length(mid, end, testWords, testString)
input_count = sys.stdin.readline()
input_word = sys.stdin.readline()
testString = input_word.rstrip()
testWords = testString.split(" ")
substring_length(0, len(testWords), testWords, testString)
#print(results)
sorted_by_value = sorted(results.items(), key=lambda kv: kv[1], reverse=True)
to_replace = sorted_by_value[0][0]
#print(sorted_by_value)
abbreviation = get_abbreviation(to_replace.split(" "))
#print("Converting {} to {}".format(to_replace, abbreviation))
r = '\\b{}\\b'.format(to_replace)
finalString = re.sub(r, abbreviation, testString)
#print("Converted original {} to final {} length: {}".format(testString, finalString, len(finalString)))
print(len(finalString))
``` | instruction | 0 | 97,830 | 6 | 195,660 |
No | output | 1 | 97,830 | 6 | 195,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a text consisting of n space-separated words. There is exactly one space character between any pair of adjacent words. There are no spaces before the first word and no spaces after the last word. The length of text is the number of letters and spaces in it. w_i is the i-th word of text. All words consist only of lowercase Latin letters.
Let's denote a segment of words w[i..j] as a sequence of words w_i, w_{i + 1}, ..., w_j. Two segments of words w[i_1 .. j_1] and w[i_2 .. j_2] are considered equal if j_1 - i_1 = j_2 - i_2, j_1 ≥ i_1, j_2 ≥ i_2, and for every t ∈ [0, j_1 - i_1] w_{i_1 + t} = w_{i_2 + t}. For example, for the text "to be or not to be" the segments w[1..2] and w[5..6] are equal, they correspond to the words "to be".
An abbreviation is a replacement of some segments of words with their first uppercase letters. In order to perform an abbreviation, you have to choose at least two non-intersecting equal segments of words, and replace each chosen segment with the string consisting of first letters of the words in the segment (written in uppercase). For example, for the text "a ab a a b ab a a b c" you can replace segments of words w[2..4] and w[6..8] with an abbreviation "AAA" and obtain the text "a AAA b AAA b c", or you can replace segments of words w[2..5] and w[6..9] with an abbreviation "AAAB" and obtain the text "a AAAB AAAB c".
What is the minimum length of the text after at most one abbreviation?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 300) — the number of words in the text.
The next line contains n space-separated words of the text w_1, w_2, ..., w_n. Each word consists only of lowercase Latin letters.
It is guaranteed that the length of text does not exceed 10^5.
Output
Print one integer — the minimum length of the text after at most one abbreviation.
Examples
Input
6
to be or not to be
Output
12
Input
10
a ab a a b ab a a b c
Output
13
Input
6
aa bb aa aa bb bb
Output
11
Note
In the first example you can obtain the text "TB or not TB".
In the second example you can obtain the text "a AAAB AAAB c".
In the third example you can obtain the text "AB aa AB bb".
Submitted Solution:
```
wNum = int(input())
text = str(input())
oldSize = len(text)
tData = text.split(' ')
diff = 0
for s in range(1, wNum // 2 + 1):
for i in range(wNum - s):
words = ' '.join(tData[i:i + s])
cnt = text.count(words)
if cnt != 1:
score = (len(words) - s) * text.count(words)
diff = score if score > diff else diff
print(oldSize - diff)
``` | instruction | 0 | 97,831 | 6 | 195,662 |
No | output | 1 | 97,831 | 6 | 195,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a> | instruction | 0 | 97,970 | 6 | 195,940 |
Tags: implementation
Correct Solution:
```
spacecnt = 0;
for tag in (input().split('>'))[:-1]:
if tag.find('/') != -1:
spacecnt -= 2
print(' '*spacecnt+tag+'>')
else:
print(' '*spacecnt+tag+'>')
spacecnt += 2
``` | output | 1 | 97,970 | 6 | 195,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a> | instruction | 0 | 97,971 | 6 | 195,942 |
Tags: implementation
Correct Solution:
```
from string import ascii_lowercase
def print_end_tag(level, character):
print(" " * level + f"</{character}>" if level else f"</{character}>")
def print_start_tag(level, character):
print(" " * level + f"<{character}>" if level else f"<{character}>")
def main():
xml_line = input()
counter = 0
last_character = ''
for character in xml_line:
if character == '<' or character == '>':
continue
if character in ascii_lowercase:
end_tag = True if last_character == '/' else False
if not end_tag:
print_start_tag(counter, character)
counter += 1
else:
counter -= 1
print_end_tag(counter, character)
last_character = character
main()
``` | output | 1 | 97,971 | 6 | 195,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a> | instruction | 0 | 97,972 | 6 | 195,944 |
Tags: implementation
Correct Solution:
```
xml = input()
tags = xml.split('>')[:-1]
for i in range(len(tags)):
tags[i] += '>'
h = 0
for tag in tags:
if '</' not in tag:
print(' ' * 2 * h + tag)
h += 1
else:
h -= 1
print(' ' * 2 * h + tag)
``` | output | 1 | 97,972 | 6 | 195,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a> | instruction | 0 | 97,973 | 6 | 195,946 |
Tags: implementation
Correct Solution:
```
str_input = input()
queue = []
indentation = 0
while(str_input != ""):
if(str_input[1] == '/'):
print("%s%s"%(indentation * " ", str_input[0:4]))
queue.pop()
str_input = str_input[4:]
if(str_input != "" and str_input[1] == "/"):
indentation -= 2
else:
print("%s%s"%(indentation * " ", str_input[0:3]))
queue.append(str_input[1:2])
str_input = str_input[3:]
if(str_input != "" and str_input[0:3] != "</%s"%(queue[-1])):
indentation += 2
``` | output | 1 | 97,973 | 6 | 195,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a> | instruction | 0 | 97,974 | 6 | 195,948 |
Tags: implementation
Correct Solution:
```
s = input().split(">")
del s[-1]
s = [i+">" for i in s]
h = 0
for i in s:
if "/" in i:
h-=1
print(h*2*" "+i)
else:
print(h*2*" "+i)
h+=1
``` | output | 1 | 97,974 | 6 | 195,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a> | instruction | 0 | 97,975 | 6 | 195,950 |
Tags: implementation
Correct Solution:
```
h = 0
for tag in input().replace('><', '> <').split():
if '/' in tag:
h -= 2
print(h*' ' + tag)
if '/' not in tag:
h += 2
``` | output | 1 | 97,975 | 6 | 195,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a> | instruction | 0 | 97,976 | 6 | 195,952 |
Tags: implementation
Correct Solution:
```
a = input().split('>')
#print(*a)
bal = 0
for s in a:
if len(s) == 0:
continue
if (s[1] == '/' ):
bal -= 1
print(bal * " " + s+ ">")
else:
print(bal * " " + s + ">")
bal +=1
``` | output | 1 | 97,976 | 6 | 195,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a> | instruction | 0 | 97,977 | 6 | 195,954 |
Tags: implementation
Correct Solution:
```
s=input()
s1=s.split('<')
del s1[0]
c=0
for i in s1:
if '/' in i:
print(' '*(c-1),'<',i,sep='')
c-=1
else:
print(' '*c,'<',i,sep='')
c+=1
``` | output | 1 | 97,977 | 6 | 195,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a>
Submitted Solution:
```
import math
for m in range(1,2):
#t = int(input())
s = str(input())
curr=[]
i=0
#print(len(s))
j=0
for j in range(len(s)*2):
#print(i)
if i>=len(s):
break
if "/" in s[i:i+3+1]:
#print(s[i:i+4])
if s[i+2] in curr:
# print(curr)
curr.reverse()
index1= curr.index(s[i+2])
# print(index1)
index1= len(curr) - index1-1
# print(curr)
print(index1* " "+ s[i:i+3+1])
curr.remove(s[i+2])
curr.reverse()
i+=4
else:
#print(1)
#if i>=37:
# index1 = curr.index(s[i+1])
#print(s[i+1] not in curr, curr)
if s[i+1] in curr:
curr.append(s[i+1])
curr.reverse()
index1 = curr.index(s[i+1])
index1=len(curr) - index1-1
print(index1* " "+ s[i:i+3])
curr.reverse()
i+=3
continue
curr.append(s[i+1])
index1 = curr.index(s[i+1])
print(index1*" "+s[i:i+3])
i+=3
``` | instruction | 0 | 97,978 | 6 | 195,956 |
Yes | output | 1 | 97,978 | 6 | 195,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a>
Submitted Solution:
```
text = input().strip()
visited = []
spacing_level = 0
for i in range(len(text)):
letter = text[i]
if letter in "</>":
continue
if text[i-1] == '/':
spacing_level -= 1
print(" " * spacing_level + "</" + letter + ">")
else:
print(" " * spacing_level + "<" + letter + ">")
spacing_level += 1
``` | instruction | 0 | 97,979 | 6 | 195,958 |
Yes | output | 1 | 97,979 | 6 | 195,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a>
Submitted Solution:
```
s,k=input().split('<')[1:],0
for c in s:
if c[0]=='/':
print(' '*(k-1),'<',c,sep='')
k-=1
else:
print(' '*k,'<',c,sep='')
k+=1
``` | instruction | 0 | 97,980 | 6 | 195,960 |
Yes | output | 1 | 97,980 | 6 | 195,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a>
Submitted Solution:
```
s = input().split('<')[1:]
level = -1
arr = set()
f = False
for i in s:
f = False
if i[0]!='/':
f = True
if f: level+=1
print(" "*(2*level) + '<' + i)
if not f: level-=1
# print(2*level)
``` | instruction | 0 | 97,981 | 6 | 195,962 |
Yes | output | 1 | 97,981 | 6 | 195,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a>
Submitted Solution:
```
a = input().split('>')
print(*a)
bal = 0
for s in a:
if len(s) == 0:
continue
if (s[1] == '/' ):
bal -= 1
print(bal * " " + s+ ">")
else:
print(bal * " " + s + ">")
bal +=1
``` | instruction | 0 | 97,982 | 6 | 195,964 |
No | output | 1 | 97,982 | 6 | 195,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a>
Submitted Solution:
```
from string import ascii_lowercase
def print_end_tag(level, character):
print(" " * level, f"</{character}>")
def print_start_tag(level, character):
print(" " * level, f"<{character}>")
def main():
xml_line = input()
counter = 0
last_character = ''
for character in xml_line:
if character == '<' or character == '>':
continue
if character in ascii_lowercase:
end_tag = True if last_character == '/' else False
if not end_tag:
print_start_tag(counter, character)
counter += 1
else:
counter -= 1
print_end_tag(counter, character)
last_character = character
main()
``` | instruction | 0 | 97,983 | 6 | 195,966 |
No | output | 1 | 97,983 | 6 | 195,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a>
Submitted Solution:
```
s = input().split('<')[1:]
level = 1
arr = set()
for i in s:
print(" "*level + '<' + i)
if i[0]!='/':
level += 1
else:
level -= 1
``` | instruction | 0 | 97,984 | 6 | 195,968 |
No | output | 1 | 97,984 | 6 | 195,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a string <x> as an opening tag, where x is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type </x>, where x is the same letter.
Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair.
Let's define the notion of a XML-text:
* an empty string is a XML-text
* if s is a XML-text, then s'=<a>+s+</a> also is a XML-text, where a is any small Latin letter
* if s1, s2 are XML-texts, then s1+s2 also is a XML-text
You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form:
* each tag (opening and closing) is located on a single line
* print before the tag 2 * h spaces, where h is the level of the tag's nestedness.
Input
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Output
Print the given XML-text according to the above-given rules.
Examples
Input
<a><b><c></c></b></a>
Output
<a>
<b>
<c>
</c>
</b>
</a>
Input
<a><b></b><d><c></c></d></a>
Output
<a>
<b>
</b>
<d>
<c>
</c>
</d>
</a>
Submitted Solution:
```
n=input()
ls=[]
i=0
while (i<len(n)):
if(n[i]=='<'):
p=''
j=i
while(n[j]!='>'):
p+=n[j]
j+=1
p+='>'
ls.append(p)
i+=1
k=0
i=0
while i<len(ls):
flag=0
if (ls[i].replace(' ','')[1]=='/'):
i+=1
continue
else:
j=i+1
while (j<len(ls) and ls[j].replace(' ','').replace('/','')!=ls[i].replace(' ','')):
ls[j]=' '*2+ls[j]
j+=1
i+=1
for i in ls:
print(i)
``` | instruction | 0 | 97,985 | 6 | 195,970 |
No | output | 1 | 97,985 | 6 | 195,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES | instruction | 0 | 99,223 | 6 | 198,446 |
Tags: implementation, math
Correct Solution:
```
# coding=utf-8
s = str(input())
links = 0
for i in range(len(s)):
if s[i] == '-':
links += 1
pearls = len(s) - links
if pearls == 0:
print('YES')
elif links%pearls == 0:
print('YES')
else:
print('NO')
``` | output | 1 | 99,223 | 6 | 198,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES | instruction | 0 | 99,224 | 6 | 198,448 |
Tags: implementation, math
Correct Solution:
```
s = input()
ls = len(s)
ac = 0
bc = 0
for i in range(ls):
if s[i] == 'o':
ac += 1
elif s[i] == '-':
bc += 1
if ac == 0:
print('YES')
elif bc % ac == 0:
print('YES')
else:
print('NO')
``` | output | 1 | 99,224 | 6 | 198,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES | instruction | 0 | 99,225 | 6 | 198,450 |
Tags: implementation, math
Correct Solution:
```
a = input()
t = a.count("o")
r = len(a) - t
if t == 0:
print("YES")
elif r%(t) == 0:
print("YES")
else:
print("NO")
``` | output | 1 | 99,225 | 6 | 198,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES | instruction | 0 | 99,226 | 6 | 198,452 |
Tags: implementation, math
Correct Solution:
```
x=input()
# print(x.count('-')%x.count('o'))
if 'o' not in x:
print('YES')
elif '-' not in x:
print('YES')
elif x.count('o')==1:
print('YES')
elif x.count('-')%x.count('o')==0:
print('YES')
else:
print('NO')
``` | output | 1 | 99,226 | 6 | 198,453 |
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