message stringlengths 2 23.8k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum. | instruction | 0 | 61,997 | 0 | 123,994 |
Tags: constructive algorithms, graphs, implementation
Correct Solution:
```
import sys
n = int(input())
a=[]
for i in range(n):
a.append([j for j in input()])
b=[]
for i in range(n):
b.append({})
for j in a[i]:
if j in b[i]:
print('NO')
sys.exit()
b[i][j]=1
c=[-1 for i in range(26)]
d=[-1 for i in range(26)]
e=[0 for i in range(26)]
for word in a:
l=ord(word[0])-97
e[l]=1
for ch in word[1:]:
ch1 = ord(ch)-97
e[ch1]=1
if c[l]==-1:
c[l]=ch1
elif c[l]!=ch1:
print('NO')
sys.exit()
if d[ch1]==-1:
d[ch1]=l
elif d[ch1]!=l:
print('NO')
sys.exit()
l=ch1
x=[]
def abc(c,d,e,i):
e[i]=0
if c[i]==-1:
return
else:
i=c[i]
x.append(chr(i+97))
abc(c,d,e,i)
for i in range(26):
if e[i]==1 and d[i]==-1:
x.append(chr(i+97))
abc(c,d,e,i)
if sum(e)>0:
print('NO')
else:
print(''.join(x))
# Made By Mostafa_Khaled
``` | output | 1 | 61,997 | 0 | 123,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
Submitted Solution:
```
n = int(input())
A = []
for i in range (0,n):
A = A + [input()]
start = [100]*26
end = [100]*26
def num(x):
return(ord(x)-ord("a"))
def let(x):
return(chr(x+ord("a")))
def ans(x):
final = []
for i in range (0,n):
B = list(A[i])
if len(B) == 1:
if start[num(B[0])] == 100:
start[num(B[0])] = -100
if end[num(B[0])] == 100:
end[num(B[0])] = -100
else:
for j in range (0,len(B)):
if j==0:
if start[num(B[j])] == 100 or start[num(B[j])] == -100:
start[num(B[j])] = -100
if end[num(B[j])] == 100 or end[num(B[j])] == -100:
end[num(B[j])] = num(B[j+1])
elif end[num(B[j])] == num(B[j+1]): g=0
else:
return("NO")
elif 0<j<len(B)-1:
if start[num(B[j])] == 100 or start[num(B[j])] == -100:
start[num(B[j])] = num(B[j-1])
elif start[num(B[j])] != num(B[j-1]):
return("NO")
if end[num(B[j])] == 100 or end[num(B[j])] == -100:
end[num(B[j])] = num(B[j+1])
elif end[num(B[j])] != num(B[j+1]):
return("NO")
elif j == len(B)-1:
if end[num(B[j])] == 100:
end[num(B[j])] = -100
if start[num(B[j])] == 100 or start[num(B[j])] == -100:
start[num(B[j])] = num(B[j-1])
elif start[num(B[j])] == num(B[j-1]): g=0
else:
return("NO")
if len(set(start))+max(0,start.count(100)-1)+max(0,start.count(-100)-1) != 26:
return("NO")
elif len(set(end))+max(0,end.count(100)-1)+max(0,end.count(-100)-1) != 26:
return("NO")
else:
for i in range (0,26):
if start[i] != -100:
g=0
else:
final = final + [let(i)]
j = end[i]
while j != -100:
final = final + [let(j)]
j = end[j]
if len(final) != len(set(start))-min(1,start.count(100))+max(0,start.count(-100)-1):
return("NO")
else:
return("".join(final))
print(ans(A))
``` | instruction | 0 | 61,998 | 0 | 123,996 |
Yes | output | 1 | 61,998 | 0 | 123,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
Submitted Solution:
```
n = int(input())
wrong_str = False
strings = []
sets = []
for _ in range(n):
new_string = input()
new_string_set = set(new_string)
if len(new_string) != len(new_string_set):
wrong_str = True
break
strings.append(new_string)
sets.append(new_string_set)
if wrong_str:
print("NO")
exit(0)
connections = []
for _ in range(n):
connections.append((-1,-1))
changed = True
while changed:
changed = False
for i in range(len(strings)):
if strings[i] == None:
continue
for j in range(i + 1, len(strings)):
if strings[j] == None:
continue
if len(set(strings[i]).intersection(set(strings[j]))) == 0:
continue
a = strings[i]
b = strings[j]
#print(a, b)
if b in a:
strings[j] = None
changed = True
elif a in b:
strings[i] = b
strings[j] = None
changed = True
else:
is_ok = False
start_index = a.find(b[0])
if start_index != -1 and a[start_index:] in b:
strings[i] += strings[j][len(a) - start_index:]
strings[j] = None
is_ok = True
changed = True
if not is_ok:
start_index = b.find(a[0])
if start_index != -1 and b[start_index:] in a:
strings[i] = strings[j] + strings[i][len(b) - start_index:]
strings[j] = None
is_ok = True
changed = True
if not is_ok:
print("NO")
exit(0)
if wrong_str:
print("NO")
exit(0)
strings = [x for x in strings if x is not None]
whole_str = "".join(strings)
if len(whole_str) != len(set(whole_str)):
print("NO")
exit(0)
print("".join(sorted(strings)))
``` | instruction | 0 | 61,999 | 0 | 123,998 |
Yes | output | 1 | 61,999 | 0 | 123,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
Submitted Solution:
```
L = {}
R = {}
S = set()
for _ in range(int(input())):
s = input()
if len(s) == 1:
S.add(s)
for s1, s2 in zip(s, s[1:]):
if s2 != L.get(s1, s2) or s1 != R.get(s2, s1):
exit(print('NO'))
L[s1], R[s2] = s2, s1
L1, R1 = set(L), set(R)
S = S -L1 - R1
for i in L1 - R1:
while i[-1] in L:
i += L.pop(i[-1])
S.add(i)
if L:
exit(print('NO'))
print(''.join(sorted(S)))
``` | instruction | 0 | 62,000 | 0 | 124,000 |
Yes | output | 1 | 62,000 | 0 | 124,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
Submitted Solution:
```
n = int(input())
last = [-1 for i in range(26)]
left = [-1 for i in range(26)]
right = [-1 for i in range(26)]
for i in range(n):
a = list(map(lambda x: ord(x) - ord('a'), input()))
#print(a)
for x in a:
if last[x] == i:
print('NO')
exit(0)
last[x] = i
for j in range(len(a) - 1):
x = a[j]
y = a[j + 1]
if left[y] != -1 and left[y] != x or\
right[x] != -1 and right[x] != y:
print('NO')
exit(0)
left[y] = x
right[x] = y
sol = []
cnt = 0
for b in range(26):
cnt += (last[b] > -1)
if left[b] == -1 and last[b] > -1:
sol.append(b)
x = b
while right[x] != -1:
x = right[x]
sol.append(x)
if len(sol) == cnt:
for x in sol:
print(chr(x + ord('a')), end='')
print()
else:
print('NO')
``` | instruction | 0 | 62,001 | 0 | 124,002 |
Yes | output | 1 | 62,001 | 0 | 124,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
Submitted Solution:
```
u, v = {}, {}
for i in range(int(input())):
t = input()
for a, b in zip(t, t[1:]):
if u.get(a, b) != b or v.get(b, a) != a: exit(print('NO'))
u[a], v[b] = b, a
d = []
for q in set(u).difference(v):
while q[-1] in u: q += u.pop(q[-1])
d.append(q)
if u: exit(print('NO'))
print(''.join(sorted(d)))
``` | instruction | 0 | 62,002 | 0 | 124,004 |
No | output | 1 | 62,002 | 0 | 124,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
Submitted Solution:
```
StringsNumber = int(input())
FinalStrings = []
Strings = []
for i in range(StringsNumber):
Strings.append(input())
LetterGraph = {}
# Генерим граф
for i in range(len(Strings)):
if len(Strings[i]) == 1:
if Strings[i] not in LetterGraph:
LetterGraph[Strings[i]] = ""
#print("заапедил", i)
continue
for e in range(len(Strings[i]) - 1):
if Strings[i][e] not in LetterGraph:
Elements = []
for j in list(LetterGraph):
if j != Strings[i][e + 1]:
Elements.append(LetterGraph[j])
if Strings[i][e + 1] in Elements:
print("NO")
exit(0)
LetterGraph[Strings[i][e]] = Strings[i][e + 1]
continue
if LetterGraph[Strings[i][e]] == Strings[i][e + 1] or LetterGraph[Strings[i][e]] == "":
LetterGraph[Strings[i][e]] = Strings[i][e + 1]
continue
#print("Граф:", LetterGraph)
print("NO")
exit(0)
#print("Я сгенерил граф, получилось:", LetterGraph)
# Проверяем, что нету цикла
if LetterGraph:
Cycle = False
for i in LetterGraph:
if LetterGraph[i] in LetterGraph:
if LetterGraph[LetterGraph[i]] == i:
print("NO")
exit(0)
# Находим возможные первые символы
if LetterGraph:
IsIFirstSymbol = False
FirstSymbols = []
for i in LetterGraph:
IsIFirstSymbol = True
for e in LetterGraph:
if LetterGraph[e] == i:
#print(i, "не подходит, потому что", e, "указывает на него.")
IsIFirstSymbol = False
if IsIFirstSymbol:
FirstSymbols.append(i)
if not FirstSymbols:
print("NO")
exit(0)
#print("Варианты первого символа:", *FirstSymbols)
# Создаем варианты финальной строки
if LetterGraph:
Letter = ""
for i in FirstSymbols:
FinalString = i
Letter = i
for e in range(len(LetterGraph)):
if Letter in LetterGraph:
if not (LetterGraph[Letter] == ""):
FinalString += LetterGraph[Letter]
#print(Letter, "есть в графе, так что добавляем", LetterGraph[Letter], ", на которое оно указывает.")
Letter = LetterGraph[Letter]
else:
break
else:
break
FinalStrings.append(FinalString)
#print("Отдельные строки", *FinalStrings)
FinalStrings.sort()
RESULT = ""
for i in FinalStrings:
RESULT += i
print(RESULT)
``` | instruction | 0 | 62,003 | 0 | 124,006 |
No | output | 1 | 62,003 | 0 | 124,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
Submitted Solution:
```
n = int(input())
a = []
for i in range(n):
a.append(input())
used = [0 for i in range(26)]
ans = ""
s = 'qwertyuioplkjhgfdsazxcvbnm'
flag = True
answer = []
while a:
fl = False
if not flag:
break
word = a[0]
for q in range(len(word)):
let = word[q]
#print('let ',let)
if word.count(let) > 1:
flag = False
break
if fl:
break
for r in range(1, len(a)):
k = a[r]
if let in k:
# print("here ",let, k)
if (k.count(let) > 1):
flag = False
break
if k in word or word in k:
if k in word:
a.pop(r)
else:
a.pop(0)
fl = True
break
c = k.index(let)
v1 = ''
v2 = ''
if (word[:q + 1:] == k[c::]):
v1 = k + word[q + 1:]
if (word[q:] == k[:c + 1]):
v2 = word + k[c + 1:]
if (v1 and v2):
a.append(min(v1, v2))
else:
if v1:
a.append(v1)
else:
a.append(v2)
a.pop(r)
a.pop(0)
fl = True
break
if not fl:
#print(word)
answer.append(word)
a.pop(0)
answer.sort()
for i in answer:
ans = ans + i
if flag:
print(ans)
else:
print('NO')
``` | instruction | 0 | 62,004 | 0 | 124,008 |
No | output | 1 | 62,004 | 0 | 124,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.
You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).
A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.
The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.
String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.
Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.
The total length of the strings doesn't exceed 105.
Output
Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.
Examples
Input
4
mail
ai
lru
cf
Output
cfmailru
Input
3
kek
preceq
cheburek
Output
NO
Note
One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
Submitted Solution:
```
def join_st(sf, pos, ans, start, end):
for i in range(len(sf)):
gt = pos.get(sf[i], None)
if gt != None:
gt = (gt, i)
for k in range(i):
if ans[gt[0] - k - 1] == '':
ans[gt[0] - k - 1] = sf[i - k - 1]
pos[sf[i - k - 1]] = gt[0] - k - 1
start -= 1
else:
return 'NO'
for k in range(i + 1, len(sf)):
if ans[gt[0] + k - i] == '':
ans[gt[0] + k - i] = sf[k]
pos[sf[k]] = gt[0] + k - i
elif ans[gt[0] + k - i] != sf[k]:
return 'NO'
return (ans, pos, start, end)
if ans[start + 1] < sf[0]:
for i in range(len(sf)):
ans[end] = sf[i]
pos[sf[i]] = end
end += 1
else:
for i in range(1, len(sf) + 1):
ans[start] = sf[-i]
pos[sf[-i]] = start
start -= 1
return (ans, pos, start, end)
def main():
n = int(input())
strings = ['' for i in range(n)]
for i in range(n):
strings[i] = input()
pos = dict()
ans = ['' for i in range(2 * 10 ** 5 + 2)]
for i in range(len(strings[0])):
ans[10 ** 5 + 1 + i] = strings[0][i]
pos[strings[0][i]] = 10 ** 5 + 1 + i
start = 10 ** 5
end = 10 ** 5 + len(strings[0]) + 1
for i in range(1, n):
res = join_st(strings[i], pos, ans, start, end)
if res == 'NO':
print('NO')
return
else:
ans, pos, start, end = res
print(''.join(ans))
main()
``` | instruction | 0 | 62,005 | 0 | 124,010 |
No | output | 1 | 62,005 | 0 | 124,011 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15 | instruction | 0 | 62,125 | 0 | 124,250 |
"Correct Solution:
```
s = input()
K = int(input())
n = len(s)
t = s[::-1]
if 2*K+1 >= n:
print(n)
exit()
dp = [[[0 for i in range(K+1)] for j in range(n+1)] for k in range(n+1)]
if n == 2 and K == 0:
if s == t:
print(2)
exit()
else:
print(1)
exit()
ans = min(2*K+1,n)
for center in range(1,n-1):
flg = 0
sf = s[:center]
sl = t[:n-center-1]
if s[center] == s[center+1]:
flg = 1
sfl = len(sf)
sll = len(sl)
for j in range(1,sll+1):
dp[sfl][j][0] = max(dp[sfl-1][j][0],dp[sfl][j-1][0])
if sf[-1] != sl[j-1]:
for k in range(1,K+1):
dp[sfl][j][k] = max(dp[sfl-1][j-1][k-1]+1,dp[sfl-1][j][k],dp[sfl][j-1][k])
if sf[-1] == sl[j-1]:
dp[sfl][j][0] = max(dp[sfl][j][0],dp[sfl-1][j-1][0]+1)
for k in range(1,K+1):
dp[sfl][j][k] = max(dp[sfl-1][j-1][k-1]+1,dp[sfl-1][j-1][k]+1,dp[sfl-1][j][k],dp[sfl][j-1][k])
if flg:
ans= max(dp[sfl][sll][K]*2+1,dp[sfl][sll-1][K]*2+2,ans)
elif K > 0:
ans= max(dp[sfl][sll][K]*2+1,dp[sfl][sll-1][K-1]*2+2,ans)
else:
ans= max(dp[sfl][sll][K]*2+1,ans)
print(ans)
``` | output | 1 | 62,125 | 0 | 124,251 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15 | instruction | 0 | 62,126 | 0 | 124,252 |
"Correct Solution:
```
def get_ind(l, r):
mod = 3
return OFFSET * (l % mod) + r
s = input()
n = len(s)
k = int(input())
dp = [[0] * ((n + 1) * 3) for i in range(k + 1)]
OFFSET = n + 1
# 長さ1のときの初期化
for i in range(n):
dp[0][get_ind(i, i + 1)] = 1
# 長さ2~nのときのDP
for length in range(2, n + 1):
for l in range(n + 1):
for ki in range(k + 1):
r = l + length
if r > n:
break
if s[l] == s[r - 1]:
dp[ki][get_ind(l, r)] = max(2 + dp[ki][get_ind(l + 1, r - 1)], dp[ki][get_ind(l, r)])
else:
dp[ki][get_ind(l, r)] = max(dp[ki][get_ind(l + 1, r)], dp[ki][get_ind(l, r - 1)], dp[ki][get_ind(l, r)])
if ki - 1 >= 0:
dp[ki][get_ind(l, r)] = max(dp[ki - 1][get_ind(l + 1, r - 1)] + 2, dp[ki][get_ind(l, r)])
ans = 0
for i in range(k + 1):
for j in range(3 * (n + 1)):
ans = max(ans, dp[i][j])
print(ans)
``` | output | 1 | 62,126 | 0 | 124,253 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15 | instruction | 0 | 62,127 | 0 | 124,254 |
"Correct Solution:
```
"""
https://atcoder.jp/contests/agc021/tasks/agc021_d
ある部分でcutしたとする。
前半と後半のLCSを求めたとすると、交差した後のLCSの長さも同じ
→左からindex iまでと右からindex jまでのLCSを求めればいい (i < j)
あとは、普通のLCSと同様のdpでいける
dp[i][j][k] = 左からi番目の文字までと、右からindex jの文字までのk回替えた際のLCS長さ
あとは2倍すれば大体行ける？
i+1 == jでは、2倍すればおk
i+1 < j では、真ん中の文字を使えるので、2倍して1足す
"""
import sys
S = input()
S = " "+S
K = int(input())
N = len(S)-1
if N == 1:
print (1)
sys.exit()
elif N == 2:
if S[1] == S[2] or K > 0:
print (2)
else:
print (1)
sys.exit()
dp = [[[0] * (K+1) for i in range(N+2)]]
ans = 0
for i in range(1,N+1):
dp.append( [[0] * (K+1) for i in range(N+2)] )
for j in range(N,i,-1):
for k in range(K+1):
#同じ場合、どっちも進める
if S[i] == S[j]:
dp[i][j][k] = max( dp[i][j][k] , dp[i-1][j+1][k] + 1)
else:
dp[i][j][k] = max( dp[i][j][k] , dp[i-1][j][k] , dp[i][j+1][k] )
if k != 0:
dp[i][j][k] = max(dp[i][j][k] , dp[i-1][j+1][k-1] + 1)
if i+1 == j:
ans = max(ans , dp[i][j][k] * 2)
else:
ans = max(ans , dp[i][j][k] * 2 + 1)
dp[i-1] = []
print (ans)
``` | output | 1 | 62,127 | 0 | 124,255 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15 | instruction | 0 | 62,128 | 0 | 124,256 |
"Correct Solution:
```
def examA():
N = I()
ans = 0
print(ans)
return
# 参考 https://atcoder.jp/contests/agc021/submissions/8392122
def examB():
def norm2(vec):
return math.sqrt(vec[0] ** 2 + vec[1] ** 2)
# any 2 points must have different position.
def ConvexHull(point_list):
pos2idx = {point_list[i]: i for i in range(len(point_list))}
y_val = defaultdict(list)
x_list = sorted(list(set([p[0] for p in point_list])))
for x, y in point_list:
y_val[x].append(y)
upper = [(x_list[0], max(y_val[x_list[0]]))]
lower = [(x_list[0], min(y_val[x_list[0]]))]
prev = float('inf')
for xi in x_list[1:]:
x0, y0 = upper[-1]
x1, y1 = xi, max(y_val[xi])
if (y1 - y0) / (x1 - x0) < prev:
upper.append((x1, y1))
prev = (y1 - y0) / (x1 - x0)
else:
while True:
x0, y0 = upper[-1]
if len(upper) == 1:
upper.append((x1, y1))
break
x00, y00 = upper[-2]
if (y1 - y0) / (x1 - x0) > (y1 - y00) / (x1 - x00):
upper.pop()
else:
prev = (y1 - y0) / (x1 - x0)
upper.append((x1, y1))
break
prev = -float('inf')
for xi in x_list[1:]:
x0, y0 = lower[-1]
x1, y1 = xi, min(y_val[xi])
if (y1 - y0) / (x1 - x0) > prev:
lower.append((x1, y1))
prev = (y1 - y0) / (x1 - x0)
else:
while True:
x0, y0 = lower[-1]
if len(lower) == 1:
lower.append((x1, y1))
break
x00, y00 = lower[-2]
if (y1 - y0) / (x1 - x0) < (y1 - y00) / (x1 - x00):
lower.pop()
else:
prev = (y1 - y0) / (x1 - x0)
lower.append((x1, y1))
break
# return upper, lower
# return [pos2idx[xy] for xy in upper], [pos2idx[xy] for xy in lower]
upper_idx, lower_idx = [pos2idx[xy] for xy in upper], [pos2idx[xy] for xy in lower]
if upper_idx[-1] == lower_idx[-1]:
upper_idx.pop()
CH_idx = upper_idx
CH_idx.extend(reversed(lower_idx))
if CH_idx[0] == CH_idx[-1] and len(CH_idx) > 1:
CH_idx.pop()
return CH_idx
N = I()
P = [[]for _ in range(N)]
D = defaultdict(int)
for i in range(N):
x,y = LI()
P[i] = (x,y)
D[(x,y)] = i
C = ConvexHull(P)
ans = [0]*N
if len(C)==2:
for c in C:
ans[c] = 0.5
for v in ans:
print(v)
return
#print(C)
for i,c in enumerate(C):
s, t, u = C[i - 1], C[i], C[(i + 1) % len(C)]
x0, y0 = P[s]
x1, y1 = P[t]
x2, y2 = P[u]
vec0 = (y0 - y1, x1 - x0)
vec1 = (y1 - y2, x2 - x1)
ans[t] = math.acos((vec0[0] * vec1[0] + vec0[1] * vec1[1]) / (norm2(vec0) * norm2(vec1))) / (2 * math.pi)
for v in ans:
print(v)
return
def examC():
ans = 0
print(ans)
return
def examD():
S = SI()
N = len(S)
K = I()
if K>150:
K = 150
dp = [[[0]*(K+1) for _ in range(N)] for i in range(N)]
for i in range(N):
for j in range(K+1):
dp[i][i][j] = 1
for i in range(2,N+1):
for k in range(K+1):
for l,r in enumerate(range(i-1,N)):
if S[l]==S[r]:
dp[l][r][k] = dp[l+1][r-1][k] + 2
else:
if k>0:
dp[l][r][k] = max(dp[l+1][r][k],dp[l][r-1][k],dp[l+1][r-1][k-1]+2)
else:
dp[l][r][k] = max(dp[l + 1][r][k], dp[l][r - 1][k])
ans = max(dp[0][-1])
#print(dp)
print(ans)
return
def examE():
ans = 0
print(ans)
return
def examF():
ans = 0
print(ans)
return
from decimal import Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]
sys.setrecursionlimit(10**7)
if __name__ == '__main__':
examD()
"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""
``` | output | 1 | 62,128 | 0 | 124,257 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15 | instruction | 0 | 62,129 | 0 | 124,258 |
"Correct Solution:
```
import sys
readline = sys.stdin.buffer.readline
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
def solve():
s = list(ns())
ls = len(s)
lk = ni()
dp = [[[0]*ls for _ in range(ls)] for i in range(2)]
ans = 0
for i in range(ls):
dp[0][i][i] = 1
for k in range(lk+1):
v = k & 1
for dis in range(1, ls):
for i in range(ls - dis):
j = i + dis
dp[v][i][j] = 0
if s[i] == s[j]:
dp[v][i][j] = dp[v][i+1][j-1] + 2
elif k:
dp[v][i][j] = dp[v^1][i+1][j-1] + 2
dp[v][i][j] = max(dp[v][i][j], dp[v][i+1][j], dp[v][i][j-1])
ans = max(ans, dp[v][0][-1])
print(ans)
return
solve()
``` | output | 1 | 62,129 | 0 | 124,259 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15 | instruction | 0 | 62,130 | 0 | 124,260 |
"Correct Solution:
```
def main():
s = input()
n = len(s)
k = int(input())
# dp[使った回数][左端からの距離][左端]
dp = [[[1]*(n-i) for i in range(n)] for _ in range(k+1)]
for i in range(n-1):
if s[i] == s[i+1]:
dp[0][1][i] = 2
if k > 0:
for cnt in range(1, k+1):
for i in range(n-1):
dp[cnt][1][i] = 2
for cnt in range(k):
for d in range(2, n):
for left in range(n-d):
right = left+d
plus = (s[left] == s[right])*2
dp[cnt][d][left] = max(
dp[cnt][d][left], dp[cnt][d-1][left],
dp[cnt][d-1][left+1], dp[cnt][d-2][left+1]+plus)
dp[cnt+1][d][left] = max(
dp[cnt+1][d][left], dp[cnt][d][left], dp[cnt][d-2][left+1]+2)
for d in range(2, n):
for left in range(n-d):
right = left+d
plus = (s[left] == s[right])*2
dp[k][d][left] = max(
dp[k][d][left], dp[k][d-1][left], dp[k][d-1][left+1], dp[k][d-2][left+1]+plus)
print(dp[-1][-1][0])
main()
``` | output | 1 | 62,130 | 0 | 124,261 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15 | instruction | 0 | 62,131 | 0 | 124,262 |
"Correct Solution:
```
def main():
s = input()
n = len(s)
k = int(input())
# dp[使った回数][左端からの距離][左端]
dp = [[[1]*(n-i) for i in range(n)] for _ in range(k+1)]
if n == 1:
print(1)
return
dp2 = dp[0][1]
for i in range(n-1):
if s[i] == s[i+1]:
dp2[i] = 2
if k > 0:
for cnt in range(1, k+1):
dp2 = dp[cnt][1]
for i in range(n-1):
dp2[i] = 2
for cnt in range(k):
dp2 = dp[cnt]
for d in range(2, n):
dp3 = dp2[d]
for left in range(n-d):
right = left+d
plus = (s[left] == s[right])*2
dp3[left] = max(
dp3[left], dp2[d-1][left],
dp2[d-1][left+1], dp2[d-2][left+1]+plus)
dp[cnt+1][d][left] = max(
dp[cnt+1][d][left], dp3[left], dp2[d-2][left+1]+2)
dp2 = dp[k]
for d in range(2, n):
dp3 = dp2[d]
for left in range(n-d):
right = left+d
plus = (s[left] == s[right])*2
dp3[left] = max(
dp3[left], dp2[d-1][left], dp2[d-1][left+1], dp2[d-2][left+1]+plus)
print(dp[-1][-1][0])
main()
``` | output | 1 | 62,131 | 0 | 124,263 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15 | instruction | 0 | 62,132 | 0 | 124,264 |
"Correct Solution:
```
def solve(s, k):
n = len(s)
pdp = [[0] * (k + 1) for _ in range(n + 1)]
for l in range(n - 1, -1, -1):
dp = [[0] * (k + 1) for _ in range(n + 1)]
for r in range(l, n):
dpr, dpr1 = dp[r], pdp[r - 1]
inc = 1 if l == r else 2
is_same = s[l] == s[r]
for p in range(k + 1):
if is_same:
dpr[p] = dpr1[p] + inc
else:
if p:
dpr[p] = max(pdp[r][p], dp[r - 1][p], dpr1[p - 1] + inc)
else:
dpr[p] = max(pdp[r][p], dp[r - 1][p])
pdp = dp
return pdp[-2][k]
print(solve(input(), int(input())))
``` | output | 1 | 62,132 | 0 | 124,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15
Submitted Solution:
```
import sys
readline = sys.stdin.readline
S = list(map(ord, readline().strip()))
N = len(S)
S = [None] + S + [None]
K = int(readline())
if N == 1:
ans = 1
elif N == 2:
ans = 1
if S[1] == S[2]:
ans = 2
if K:
ans = 2
else:
dp1 = [[0]*(K+1) for _ in range(N+2)]
dp2 = [[0]*(K+1) for _ in range(N+2)]
for i in range(1, N+1):
dp1[i][0] = 1
for i in range(1, N):
dp2[i][0] = 1
if S[i] == S[i+1]:
dp2[i][0] = 2
elif K:
dp2[i][1] = 2
for s in range(3, N+1):
dp3 = [[0]*(K+1) for _ in range(N+2)]
for i in range(1, N-s+2):
j = i+s-1
if S[i] == S[j]:
for k in range(K+1):
dp3[i][k] = max(dp2[i+1][k], dp2[i][k], 2+dp1[i+1][k])
else:
dp3[i][0] = max(dp2[i+1][0], dp2[i][0], dp2[i+1][0])
for k in range(1, K+1):
dp3[i][k] = max(dp2[i+1][k], dp2[i][k], dp1[i+1][k], 2+dp1[i+1][k-1])
dp1 = [k[:] for k in dp2]
dp2 = [k[:] for k in dp3]
ans = max(dp3[1][k] for k in range(K+1))
print(ans)
'''
MLEした
dp = [[[0]*(K+1) for _ in range(N+2)] for _ in range(N+2)]
for i in range(1, N+1):
dp[i][i][0] = 1
for i in range(1, N):
dp[i][i+1][0] = 1
if S[i] == S[i+1]:
dp[i][i+1][0] = 2
elif K:
dp[i][i+1][1] = 2
for s in range(3, N+1):
for i in range(1, N-s+2):
j = i+s-1
if S[i] == S[j]:
for k in range(K+1):
dp[i][j][k] = max(dp[i+1][j][k], dp[i][j-1][k], 2+dp[i+1][j-1][k])
else:
dp[i][j][0] = max(dp[i+1][j][0], dp[i][j-1][0], dp[i+1][j-1][0])
for k in range(1, K+1):
dp[i][j][k] = max(dp[i+1][j][k], dp[i][j-1][k], dp[i+1][j-1][k], 2+dp[i+1][j-1][k-1], dp[i+1][j-1][k])
print(max(dp[1][N][k] for k in range(K+1)))
'''
``` | instruction | 0 | 62,133 | 0 | 124,266 |
Yes | output | 1 | 62,133 | 0 | 124,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15
Submitted Solution:
```
import sys
readline = sys.stdin.readline
S = list(map(ord, readline().strip()))
N = len(S)
S = [None] + S + [None]
K = int(readline())
if N == 1:
ans = 1
elif N == 2:
ans = 1
if S[1] == S[2]:
ans = 2
if K:
ans = 2
else:
dp1 = [[0]*(K+1) for _ in range(N+2)]
dp2 = [[0]*(K+1) for _ in range(N+2)]
for i in range(1, N+1):
dp1[i][0] = 1
for i in range(1, N):
dp2[i][0] = 1
if S[i] == S[i+1]:
dp2[i][0] = 2
elif K:
dp2[i][1] = 2
for s in range(3, N+1):
dp3 = [[0]*(K+1) for _ in range(N+2)]
for i in range(1, N-s+2):
j = i+s-1
if S[i] == S[j]:
for k in range(K+1):
dp3[i][k] = max(dp2[i+1][k], dp2[i][k], 2+dp1[i+1][k])
else:
dp3[i][0] = max(dp2[i+1][0], dp2[i][0], dp2[i+1][0])
for k in range(1, K+1):
dp3[i][k] = max(dp2[i+1][k], dp2[i][k], dp1[i+1][k], 2+dp1[i+1][k-1])
'''
dp1 = [k[:] for k in dp2]
dp2 = [k[:] for k in dp3]
'''
dp1 = dp2[:]
dp2 = dp3[:]
ans = max(dp3[1][k] for k in range(K+1))
print(ans)
'''
MLEした
dp = [[[0]*(K+1) for _ in range(N+2)] for _ in range(N+2)]
for i in range(1, N+1):
dp[i][i][0] = 1
for i in range(1, N):
dp[i][i+1][0] = 1
if S[i] == S[i+1]:
dp[i][i+1][0] = 2
elif K:
dp[i][i+1][1] = 2
for s in range(3, N+1):
for i in range(1, N-s+2):
j = i+s-1
if S[i] == S[j]:
for k in range(K+1):
dp[i][j][k] = max(dp[i+1][j][k], dp[i][j-1][k], 2+dp[i+1][j-1][k])
else:
dp[i][j][0] = max(dp[i+1][j][0], dp[i][j-1][0], dp[i+1][j-1][0])
for k in range(1, K+1):
dp[i][j][k] = max(dp[i+1][j][k], dp[i][j-1][k], dp[i+1][j-1][k], 2+dp[i+1][j-1][k-1], dp[i+1][j-1][k])
print(max(dp[1][N][k] for k in range(K+1)))
'''
``` | instruction | 0 | 62,134 | 0 | 124,268 |
Yes | output | 1 | 62,134 | 0 | 124,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15
Submitted Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
s = input()
K = int(input())
n = len(s)
dp = [[[0]*(K+1) for _ in range(n+2-i)] for i in range(n+1)]
for i in range(n):
for j in range(K+1):
dp[i][1][j] = 1
for l in range(2, n+1):
for i in range(n-l+1):
for k in range(K+1):
# print(l, i, i+l, j)
if s[i]==s[i+l-1]:
dp[i][l][k] = dp[i+1][l-2][k] + 2
else:
dp[i][l][k] = max(dp[i+1][l-1][k], dp[i][l-1][k])
if k>=1:
dp[i][l][k] = max(dp[i][l][k], dp[i+1][l-2][k-1]+2)
ans = max(dp[0][i][j] for i in range(n+2) for j in range(K+1))
print(ans)
``` | instruction | 0 | 62,135 | 0 | 124,270 |
Yes | output | 1 | 62,135 | 0 | 124,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15
Submitted Solution:
```
S = input()
K = int(input())
N = len(S)
# MLE対策
if K * 2 >= N-1:
print(N)
exit()
dp = [[[0 for _ in range(K+1)] for _ in range(N)] for _ in range(N)]
for i in range(N):
for k in range(K+1):
dp[i][i][k] = 1
for _j in range(1, N):
for i in range(N):
j = i + _j
if j < N:
for k in range(K+1):
dp[i][j][k] = max(dp[i][j][k], dp[i+1][j][k], dp[i][j-1][k])
if _j == 1:
if S[i] == S[j]:
dp[i][j][k] = 2
else:
if k+1 <= K:
dp[i][j][k+1] = 2
else:
if S[i] == S[j]:
dp[i][j][k] = max(dp[i][j][k], dp[i+1][j-1][k] + 2)
else:
if k+1 <= K:
dp[i][j][k+1] = max(dp[i][j][k], dp[i+1][j-1][k] + 2)
print(max(dp[0][N-1]))
``` | instruction | 0 | 62,136 | 0 | 124,272 |
Yes | output | 1 | 62,136 | 0 | 124,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15
Submitted Solution:
```
def examA():
N = I()
ans = 0
print(ans)
return
# 参考 https://atcoder.jp/contests/agc021/submissions/8392122
def examB():
def norm2(vec):
return math.sqrt(vec[0] ** 2 + vec[1] ** 2)
# any 2 points must have different position.
def ConvexHull(point_list):
pos2idx = {point_list[i]: i for i in range(len(point_list))}
y_val = defaultdict(list)
x_list = sorted(list(set([p[0] for p in point_list])))
for x, y in point_list:
y_val[x].append(y)
upper = [(x_list[0], max(y_val[x_list[0]]))]
lower = [(x_list[0], min(y_val[x_list[0]]))]
prev = float('inf')
for xi in x_list[1:]:
x0, y0 = upper[-1]
x1, y1 = xi, max(y_val[xi])
if (y1 - y0) / (x1 - x0) < prev:
upper.append((x1, y1))
prev = (y1 - y0) / (x1 - x0)
else:
while True:
x0, y0 = upper[-1]
if len(upper) == 1:
upper.append((x1, y1))
break
x00, y00 = upper[-2]
if (y1 - y0) / (x1 - x0) > (y1 - y00) / (x1 - x00):
upper.pop()
else:
prev = (y1 - y0) / (x1 - x0)
upper.append((x1, y1))
break
prev = -float('inf')
for xi in x_list[1:]:
x0, y0 = lower[-1]
x1, y1 = xi, min(y_val[xi])
if (y1 - y0) / (x1 - x0) > prev:
lower.append((x1, y1))
prev = (y1 - y0) / (x1 - x0)
else:
while True:
x0, y0 = lower[-1]
if len(lower) == 1:
lower.append((x1, y1))
break
x00, y00 = lower[-2]
if (y1 - y0) / (x1 - x0) < (y1 - y00) / (x1 - x00):
lower.pop()
else:
prev = (y1 - y0) / (x1 - x0)
lower.append((x1, y1))
break
# return upper, lower
# return [pos2idx[xy] for xy in upper], [pos2idx[xy] for xy in lower]
upper_idx, lower_idx = [pos2idx[xy] for xy in upper], [pos2idx[xy] for xy in lower]
if upper_idx[-1] == lower_idx[-1]:
upper_idx.pop()
CH_idx = upper_idx
CH_idx.extend(reversed(lower_idx))
if CH_idx[0] == CH_idx[-1] and len(CH_idx) > 1:
CH_idx.pop()
return CH_idx
N = I()
P = [[]for _ in range(N)]
D = defaultdict(int)
for i in range(N):
x,y = LI()
P[i] = (x,y)
D[(x,y)] = i
C = ConvexHull(P)
ans = [0]*N
if len(C)==2:
for c in C:
ans[c] = 0.5
for v in ans:
print(v)
return
#print(C)
for i,c in enumerate(C):
s, t, u = C[i - 1], C[i], C[(i + 1) % len(C)]
x0, y0 = P[s]
x1, y1 = P[t]
x2, y2 = P[u]
vec0 = (y0 - y1, x1 - x0)
vec1 = (y1 - y2, x2 - x1)
ans[t] = math.acos((vec0[0] * vec1[0] + vec0[1] * vec1[1]) / (norm2(vec0) * norm2(vec1))) / (2 * math.pi)
for v in ans:
print(v)
return
def examC():
ans = 0
print(ans)
return
def examD():
S = SI()
N = len(S)
K = I()
dp = defaultdict(lambda:[0 for _ in range(K+1)])
for i in range(N):
for j in range(K+1):
dp[(i,i)][j] = 1
for i in range(2,N+1):
for k in range(K+1):
for l,r in enumerate(range(i-1,N)):
if S[l]==S[r]:
dp[(l,r)][k] = dp[(l+1,r-1)][k] + 2
else:
if k>0:
dp[(l,r)][k] = max(dp[(l+1,r)][k],dp[(l,r-1)][k],dp[(l+1,r-1)][k-1]+2)
else:
dp[(l,r)][k] = max(dp[(l + 1,r)][k], dp[(l,r - 1)][k])
ans = max(dp[(0,N-1)])
#print(dp)
print(ans)
return
def examE():
ans = 0
print(ans)
return
def examF():
ans = 0
print(ans)
return
from decimal import Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]
sys.setrecursionlimit(10**7)
if __name__ == '__main__':
examD()
"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""
``` | instruction | 0 | 62,137 | 0 | 124,274 |
No | output | 1 | 62,137 | 0 | 124,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
S = list(input().rstrip())
K = int(input())
L = len(S)
def main():
dp = [[[0]*(K+1) for _ in range(L+1)] for _ in range(L+1)]
ans = 0
for ind in reversed(range(L)):
# [0, ind), [ind, L-1]
s = S[ind]
for i in range(ind):
for k in reversed(range(K+1)):
if S[i] == s:
dp[i+1][L-ind][k] = dp[i][L-ind-1][k] + 1
else:
dp[i+1][L-ind][k] = max(dp[i][L-ind][k], dp[i+1][L-ind-1][k])
if k < K:
dp[i+1][L-ind][k+1] = max(dp[i+1][L-ind][k+1], dp[i+1][L-ind][k]+1)
for k in range(K+1):
ans = max(ans, 2*dp[ind][L-ind][k])
ans = max(ans, 2*dp[ind][L-ind-1][k] + 1)
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 62,138 | 0 | 124,276 |
No | output | 1 | 62,138 | 0 | 124,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15
Submitted Solution:
```
def main():
s = input()
n = len(s)
k = int(input())
# dp[左端][左端からの右端の距離][使った回数]
dp = [[[1]*(k+1) for _ in [0]*(n-i)] for i in range(n)]
for i in range(n-1):
if s[i] == s[i+1]:
dp[i][1][0] = 2
for cnt in range(k):
for d in range(2, n):
for left in range(n-d):
right = left+d
plus = (s[left] == s[right])*2
dp[left][d][cnt] = max(
dp[left][d][cnt], dp[left][d-1][cnt],
dp[left+1][d-1][cnt], dp[left+1][d-2][cnt]+plus)
dp[left][d][cnt+1] = max(
max(dp[left][d][cnt:cnt+2]), dp[left+1][d-2][cnt]+2)
for d in range(2, n):
for left in range(n-d):
right = left+d
plus = (s[left] == s[right])*2
dp[left][d][k] = max(
dp[left][d][k], dp[left][d-1][k], dp[left+1][d-1][k], dp[left+1][d-2][k]+plus)
print(dp[0][-1][-1])
main()
``` | instruction | 0 | 62,139 | 0 | 124,278 |
No | output | 1 | 62,139 | 0 | 124,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has decided to give a string to his mother.
The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T. That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).
Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value. Find the highest possible value achievable.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq |S|
* S consists of lowercase English letters.
* K is an integer.
Input
Input is given from Standard Input in the following format:
S
K
Output
Print the highest possible value achievable.
Examples
Input
abcabcabc
1
Output
7
Input
atcodergrandcontest
3
Output
15
Submitted Solution:
```
def lcs(str1, str2):
'''文字列str1, str2の最長共通部分列(Lowest Common Subsequence, LCS)を求める
計算量O(len(str1)*len(str2))
'''
dp = [[0]*(len(str2) + 1) for _ in range(len(str1) + 1)]
for i in range(len(str1)):
for j in range(len(str2)):
if str1[i] == str2[j]:
dp[i+1][j+1] = dp[i][j] + 1
else:
dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
return dp[len(str1)][len(str2)]
s = input()
n = len(s)
k = int(input())
ans = 0
for i in range(n):
t = s[0:i]
rev_t = s[i+1:][::-1]
min_len = min(len(t), len(rev_t))
ans = max(min(lcs(t, rev_t) + k, min_len) * 2 + 1, ans)
for i in range(n + 1):
t = s[0:i]
rev_t = s[i:][::-1]
min_len = min(len(t), len(rev_t))
ans = max(min(lcs(t, rev_t) + k, min_len) * 2, ans)
print(ans)
``` | instruction | 0 | 62,140 | 0 | 124,280 |
No | output | 1 | 62,140 | 0 | 124,281 |
Provide a correct Python 3 solution for this coding contest problem.
Almost periodic string
Given the string S. Answer Q queries for this string S. In the i-th query, determine if S [l_i, \ r_i] can be a string of period t_i when you can change from S [l_i, \ r_i] to one character. S [l, \ r] represents the substring of the string S from the lth to the rth.
If the character string W is a character string with period t, then for i \ = \ 1, \ 2, \ ..., \ | W | − t, W_ {i} = W_ {i + t}. I will do it.
Constraints
* 1 ≤ | S | ≤ 10 ^ 5
* 1 ≤ Q ≤ 10 ^ 5
* 1 ≤ l_i ≤ r_i ≤ | S |
* 1 ≤ t_i ≤ r_i − l_i + 1
* S consists of only lowercase letters
Input Format
Input is given from standard input in the following format.
S
Q
l_1 r_1 t_1
...
l_Q r_Q t_Q
Output Format
Output over Q lines. On line i, print the answer to the i-th query with `Yes` or` No`.
Sample Input 1
abcabcaxcabc
Four
1 9 3
8 12 3
1 4 2
2 3 2
Sample Output 1
Yes
Yes
No
Yes
Sample Input 2
isuruu
Four
3 6 1
3 6 2
3 6 3
2 4 1
Sample Output 2
Yes
Yes
Yes
No
Example
Input
abcabcaxcabc
4
1 9 3
8 12 3
1 4 2
2 3 2
Output
Yes
Yes
No
Yes | instruction | 0 | 62,227 | 0 | 124,454 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
base = 37; MOD = 10**9 + 9
S = readline().strip()
L = len(S)
H = [0]*(L+1)
pw = [1]*(L+1)
ca = ord('a')
v = 0
for i in range(L):
H[i+1] = v = (v * base + (ord(S[i]) - ca)) % MOD
v = 1
for i in range(1, L+1):
pw[i] = v = v * base % MOD
Q = int(readline())
for i in range(Q):
l, r, t = map(int, readline().split()); l -= 1
m = r-l
if ((H[r]-H[r-t]) - (H[l+t]-H[l]) * pw[m-t]) % MOD == 0:
write("Yes\n")
continue
left = 0; right = m-t+1
base = H[l] - H[l+t]
while left+1 < right:
mid = (left + right) >> 1
if ((H[l+mid]-H[l+t+mid]) - base * pw[mid]) % MOD == 0:
left = mid
else:
right = mid
l1 = left
left = 0; right = m-t-l1+1
base = (H[r-t]-H[r]) % MOD
while left+1 < right:
mid = (left + right) >> 1
if (H[r-t-mid]-H[r-mid]) * pw[mid] % MOD == base:
left = mid
else:
right = mid
l2 = left
if l1+l2+1 == m-t:
if l1 <= t or l2 <= t:
write("Yes\n")
else:
write("No\n")
else:
if ((H[l+m-t-l2-1]-H[l+m-l2-1]) - (H[l+l1+1]-H[l+t+l1+1]) * pw[m-t-l2-l1-2]) % MOD != 0:
write("No\n")
else:
p1 = l1; p2 = m-t-l2-1
if p2 - p1 == t and S[l+p2-t] == S[l+p2+t]:
write("Yes\n")
else:
write("No\n")
solve()
``` | output | 1 | 62,227 | 0 | 124,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4]. | instruction | 0 | 62,309 | 0 | 124,618 |
Tags: combinatorics, dp, implementation
Correct Solution:
```
s = input()
A = []
a = 0
b = 0
for i in range(len(s)):
if s[i] == 'a':
a += 1
b = 0
if s[i] == 'b':
b += 1
if a > 0:
A.append(a);
a = 0
if a > 0:
A.append(a)
if len(A) > 0:
dp = [0] * (len(A) + 1)
dp[0] = A[0]
for i in range(1, len(A)):
dp[i] = dp[i - 1] + A[i] * (1 + dp[i - 1])
print(dp[len(A) - 1] % 1000000007)
else:
print(0)
``` | output | 1 | 62,309 | 0 | 124,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4]. | instruction | 0 | 62,310 | 0 | 124,620 |
Tags: combinatorics, dp, implementation
Correct Solution:
```
r = t = 0
s = input()
p = list(s)
for k in s:
if k == 'b':
t = r
if k == 'a':
r = (r + t + 1) % (10**9 + 7)
print(r)
``` | output | 1 | 62,310 | 0 | 124,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4]. | instruction | 0 | 62,311 | 0 | 124,622 |
Tags: combinatorics, dp, implementation
Correct Solution:
```
s = input()
mod = 10**9 + 7
L = []
c = 0
f = 0
for i in s:
if i == 'a':
f = 0
c += 1
elif i == 'b' and f == 0:
L.append(c)
c = 0
f = 1
if c > 0 and f == 0:
L.append(c)
res = 1
for i in range(len(L)):
res *= (L[i]+1)%mod
print((res-1)%mod)
``` | output | 1 | 62,311 | 0 | 124,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4]. | instruction | 0 | 62,312 | 0 | 124,624 |
Tags: combinatorics, dp, implementation
Correct Solution:
```
mod = int(1e9) + 7
s = input()
n = len(s)
ans = 1
kol = 0
tmp = 0
for c in s:
if c == 'b':
ans *= (kol + 1)
ans = ans % mod
kol = 0
if c == 'a':
kol += 1
tmp += 1
print((ans * (kol + 1) - 1) % mod)
``` | output | 1 | 62,312 | 0 | 124,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4]. | instruction | 0 | 62,313 | 0 | 124,626 |
Tags: combinatorics, dp, implementation
Correct Solution:
```
import re
s = input()
s = re.sub(r'[^a-b]', '', s)
s = re.sub(r'b+', 'b', s)
next_b = [len(s)-1 for _ in range(len(s))]
last_b_index = -1
i = len(s) - 1
while i>=0:
if last_b_index !=-1:
next_b[i] = last_b_index
if s[i] == 'b':
last_b_index = i
i -= 1
dp = [-1 for _ in range(len(s)+1)]
dp[len(s)] = 0
i = len(s) - 1
while i>=0:
dp[i] = dp[i + 1]
if s[i] == 'a':
dp[i] += 1 + (dp[next_b[i]+1])
dp[i] %= 1000000007
i -= 1
print(dp[0])
``` | output | 1 | 62,313 | 0 | 124,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4]. | instruction | 0 | 62,314 | 0 | 124,628 |
Tags: combinatorics, dp, implementation
Correct Solution:
```
import sys
if __name__ == "__main__":
# single variables
s = [str(val) for val in input().split()][0]
singles = 0
for i in s:
if(i == 'a'):
singles +=1
count = 0
groups = []
for i in s:
if(i == 'a'):
count += 1
if(i == 'b' and count != 0):
groups.append(count)
count = 0
if(count != 0):
groups.append(count)
modval = int(1e9+7)
tots = 1
for i in groups:
tots = ((tots % modval) * ((i+1) % modval)) % modval
print(tots-1)
``` | output | 1 | 62,314 | 0 | 124,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4]. | instruction | 0 | 62,315 | 0 | 124,630 |
Tags: combinatorics, dp, implementation
Correct Solution:
```
line = list(input()) + ['b']
a = []
count = 0
for i in line:
if i == 'b' and count > 0:
a.append(count)
count = 0
if i == 'a':
count += 1
ans = 1
for i in a:
ans *= i + 1
ans %= 1000000007
print(ans - 1)
``` | output | 1 | 62,315 | 0 | 124,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4]. | instruction | 0 | 62,316 | 0 | 124,632 |
Tags: combinatorics, dp, implementation
Correct Solution:
```
st, count, ans = input(), 0, 1
for char in st:
if (char == 'a'): count+=1
elif(char == 'b' and count > 0):
ans*=(count+1)
count = 0
ans = (ans*(count+1)-1) if count > 0 else ans-1
print(ans%(10**9+7))
``` | output | 1 | 62,316 | 0 | 124,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4].
Submitted Solution:
```
s = input()
m = 10**9+7
a = []
j = 0
for i in range(len(s)):
if s[i] == 'b':
a.append(j)
j = 0
elif s[i] == 'a':
j += 1
if j != 0:
a.append(j)
if len(a) == 0:
print(0)
exit()
ans = a[0]
for i in range(1, len(a)):
ans = (ans+ans*a[i]+a[i])%m
print(ans)
``` | instruction | 0 | 62,317 | 0 | 124,634 |
Yes | output | 1 | 62,317 | 0 | 124,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4].
Submitted Solution:
```
old_palavra = input()
a_palavra = []
for letra in old_palavra:
if letra == "a":
a_palavra.append(letra)
if letra == "b" and len(a_palavra) > 0 and a_palavra[-1] == "a":
a_palavra.append(letra)
matrix_palavra = [[]]
arrayinterno = 0
for i in a_palavra:
if i == "a":
matrix_palavra[arrayinterno].append(i)
else:
matrix_palavra.append([])
arrayinterno += 1
tamanhos_de_arrays = []
for i in matrix_palavra:
tamanhos_de_arrays.append(len(i) +1)
resultado = tamanhos_de_arrays[0]
for i in range(1, len(tamanhos_de_arrays)):
resultado *= tamanhos_de_arrays[i]
print((resultado -1) % 1000000007)
``` | instruction | 0 | 62,318 | 0 | 124,636 |
Yes | output | 1 | 62,318 | 0 | 124,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4].
Submitted Solution:
```
import sys
import os
from io import BytesIO, IOBase
#########################
# imgur.com/Pkt7iIf.png #
#########################
# returns the list of prime numbers less than or equal to n:
'''def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r'''
# returns all the divisors of a number n(takes an additional parameter start):
'''def divs(n, start=1):
divisors = []
for i in range(start, int(math.sqrt(n) + 1)):
if n % i == 0:
if n / i == i:
divisors.append(i)
else:
divisors.extend([i, n // i])
return divisors'''
# returns the number of factors of a given number if a primes list is given:
'''def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
return divs_number'''
# returns the leftmost and rightmost positions of x in a given list d(if x isnot present then returns (-1,-1)):
'''def flin(d, x, default=-1):
left = right = -1
for i in range(len(d)):
if d[i] == x:
if left == -1: left = i
right = i
if left == -1:
return (default, default)
else:
return (left, right)'''
# returns (b**p)%m
'''def modpow(b,p,m):
res=1
b=b%m
while(p):
if p&1:
res=(res*b)%m
b=(b*b)%m
p>>=1
return res'''
# if m is a prime this can be used to determine (1//a)%m or modular multiplicative inverse of a number a
'''def mod_inv(a,m):
return modpow(a,m-2,m)'''
# returns the ncr%m for (if m is a prime number) for very large n and r
'''def ncr(n,r,m):
res=1
if r==0:
return 1
if n-r<r:
r=n-r
p,k=1,1
while(r):
res=((res%m)*(((n%m)*mod_inv(r,m))%m))%m
n-=1
r-=1
return res'''
# returns ncr%m (if m is a prime number and there should be a list fact which stores the factorial values upto n):
'''def ncrlis(n,r,m):
return (fact[n]*(mod_inv(fact[r],m)*mod_inv(fact[n-r],m))%m)%m'''
#factorial list which stores factorial of values upto n:
'''mx_lmt=10**5+10
fact=[1 for i in range(mx_lmt)]
for i in range(1,mx_lmt):
fact[i]=(i*fact[i-1])%mod'''
#count xor of numbers from 1 to n:
'''def xor1_n(n):
d={0:n,1:1,2:n+1,3:0}
return d[n&3]'''
def ceil(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return defaultdict(int)
def ddl(): return defaultdict(list)
def ddd(): return defaultdict(defaultdict(int))
# region fastio
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")
from collections import defaultdict
#from collections import deque
#from collections import OrderedDict
#from math import gcd
#import time
#import itertools
#import timeit
#import random
#from bisect import bisect_left as bl
#from bisect import bisect_right as br
#from bisect import insort_left as il
#from bisect import insort_right as ir
#from heapq import *
#mod=998244353
mod=10**9+7
for _ in range(1):
s=input()
s+='b'
f=0
cnt=0
res=1
for i in range(len(s)):
if s[i]=='a':
cnt+=1
if s[i]=='b':
res*=(1+cnt)
res%=mod
cnt=0
print(res-1)
``` | instruction | 0 | 62,319 | 0 | 124,638 |
Yes | output | 1 | 62,319 | 0 | 124,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4].
Submitted Solution:
```
from functools import reduce
s = [len(z)+1 for z in ''.join(y for y in [x for x in input() if x == 'a' or x == 'b']).split('b') if len(z) > 0]
print((reduce(lambda x, y: (x*y) % int(1e9+7), s) - 1) % int(1e9+7) if len(s) > 0 else 0)
``` | instruction | 0 | 62,320 | 0 | 124,640 |
Yes | output | 1 | 62,320 | 0 | 124,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4].
Submitted Solution:
```
#------------------------------warmup----------------------------
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")
#-------------------game starts now-----------------------------------------------------
s=input()
n=len(s)
c=0
t=0
ans=0
r=0
for i in range(n):
if s[i]=='a':
t=1
r=(r+1)%1000000007
ans=(ans+c%1000000007)%1000000007
if s[i]=='b' and t==1:
c=(c+1)%1000000007
t=0
print((ans+r)%1000000007)
``` | instruction | 0 | 62,321 | 0 | 124,642 |
No | output | 1 | 62,321 | 0 | 124,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4].
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
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")
#-------------------game starts now-----------------------------------------------------
mod=10**9+7
s=input()
n=len(s)
count=s.count('a')
b=1
curr=0
a=list()
for i in s:
if i=='a':
curr+=1
else:
if b>0 and curr>0:
a.append(curr)
curr=0
b=0
if i=='b':
b=1
if b>0 and curr>0:
a.append(curr)
post=[0]*len(a)
if len(a)>0:
post[-1]=a[-1]
for i in range (len(a)-2,-1,-1):
post[i]=post[i+1]+a[i]
c=0
for i in range (len(a)-1):
c+=(a[i]*post[i+1])
c%=mod
#print(count,c)
count+=c
print(count)
``` | instruction | 0 | 62,322 | 0 | 124,644 |
No | output | 1 | 62,322 | 0 | 124,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4].
Submitted Solution:
```
# call the main method
def main():
s = input()
a = 1
out = 1
for i in s:
if i == 'b':
out *= a
a = 1
if i == 'a':
a = a + 1
out *= a
out = out -1
out %= 1000000000 + 7
print(out - 1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 62,323 | 0 | 124,646 |
No | output | 1 | 62,323 | 0 | 124,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut found a string s. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p_1, p_2, …, p_k, such that:
1. For each i (1 ≤ i ≤ k), s_{p_i} = 'a'.
2. For each i (1 ≤ i < k), there is such j that p_i < j < p_{i + 1} and s_j = 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 10^9 + 7.
Input
The first line contains the string s (1 ≤ |s| ≤ 10^5) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p_1, p_2, …, p_k modulo 10^9 + 7.
Examples
Input
abbaa
Output
5
Input
baaaa
Output
4
Input
agaa
Output
3
Note
In the first example, there are 5 possible sequences. [1], [4], [5], [1, 4], [1, 5].
In the second example, there are 4 possible sequences. [2], [3], [4], [5].
In the third example, there are 3 possible sequences. [1], [3], [4].
Submitted Solution:
```
s=list(input())
p=s.count('a')
n=len(s)
b=[]
mod=10**9+7
for i in range(n):
if s[i]=='a':
b.append(i)
m=len(b)
if m<=1:
print(p)
else:
k=0
j=1
#print(b)
y=1
for i in range(b[0]+1,b[m-1]):
if s[i]=='b':
#print(k,j)
if b[k]<i<b[j]:
#print(m-j)
p=(p+m-j)%mod
k+=1
y=0
#print(p)
if s[i]=='a':
j+=1
y=1
#print(j)
if k<j and y==0:
#print(j)
p=(p+m-j)%mod
print(p)
``` | instruction | 0 | 62,324 | 0 | 124,648 |
No | output | 1 | 62,324 | 0 | 124,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and hence decided to came up with something. Roman invented a pizza's recipe, while Denis invented a string multiplication. According to Denis, the result of multiplication (product) of strings s of length m and t is a string t + s_1 + t + s_2 + … + t + s_m + t, where s_i denotes the i-th symbol of the string s, and "+" denotes string concatenation. For example, the product of strings "abc" and "de" is a string "deadebdecde", while the product of the strings "ab" and "z" is a string "zazbz". Note, that unlike the numbers multiplication, the product of strings s and t is not necessarily equal to product of t and s.
Roman was jealous of Denis, since he invented such a cool operation, and hence decided to invent something string-related too. Since Roman is beauty-lover, he decided to define the beauty of the string as the length of the longest substring, consisting of only one letter. For example, the beauty of the string "xayyaaabca" is equal to 3, since there is a substring "aaa", while the beauty of the string "qwerqwer" is equal to 1, since all neighboring symbols in it are different.
In order to entertain Roman, Denis wrote down n strings p_1, p_2, p_3, …, p_n on the paper and asked him to calculate the beauty of the string ( … (((p_1 ⋅ p_2) ⋅ p_3) ⋅ … ) ⋅ p_n, where s ⋅ t denotes a multiplication of strings s and t. Roman hasn't fully realized how Denis's multiplication works, so he asked you for a help. Denis knows, that Roman is very impressionable, he guarantees, that the beauty of the resulting string is at most 10^9.
Input
The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings, wroted by Denis.
Next n lines contain non-empty strings p_1, p_2, …, p_n, consisting of lowercase english letters.
It's guaranteed, that the total length of the strings p_i is at most 100 000, and that's the beauty of the resulting product is at most 10^9.
Output
Print exactly one integer — the beauty of the product of the strings.
Examples
Input
3
a
b
a
Output
3
Input
2
bnn
a
Output
1
Note
In the first example, the product of strings is equal to "abaaaba".
In the second example, the product of strings is equal to "abanana".
Submitted Solution:
```
ALPH = 'abcdefghijklmnopqrstuvwxyz'
MAX = 10 ** 9
def cnt(s):
c = {ch : 0 for ch in ALPH}
i = 0
while i < len(s):
j = i + 1
while j < len(s) and s[i] == s[j]:
j += 1
c[s[i]] = max(c[s[i]], j - i)
i = j
return c
def nxt(c, t):
nc = cnt(t)
for ch in ALPH:
if c[ch] and not nc[ch]:
nc[ch] = 1
f = 0
while f < len(t) and t[f] == t[0]:
f += 1
r = 0
while r < len(t) and t[-1 - r] == t[-1]:
r += 1
if t[0] == t[-1]:
if f == len(t):
nc[t[0]] = max(nc[t[0]], c[t[0]] + (c[t[0]] + 1) * len(t))
elif c[t[0]]:
nc[t[0]] = max(nc[t[0]], f + 1 + r)
else:
nc[t[0]] = max(nc[t[0]], f + (c[t[0]] > 0))
nc[t[-1]] = max(nc[t[-1]], r + (c[t[-1]] > 0))
return {x : min(MAX, y) for x, y in nc.items()}
n = int(input())
c = cnt(input())
for i in range(n - 1):
c = nxt(c, input())
print(max(c.values()))
``` | instruction | 0 | 62,333 | 0 | 124,666 |
Yes | output | 1 | 62,333 | 0 | 124,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and hence decided to came up with something. Roman invented a pizza's recipe, while Denis invented a string multiplication. According to Denis, the result of multiplication (product) of strings s of length m and t is a string t + s_1 + t + s_2 + … + t + s_m + t, where s_i denotes the i-th symbol of the string s, and "+" denotes string concatenation. For example, the product of strings "abc" and "de" is a string "deadebdecde", while the product of the strings "ab" and "z" is a string "zazbz". Note, that unlike the numbers multiplication, the product of strings s and t is not necessarily equal to product of t and s.
Roman was jealous of Denis, since he invented such a cool operation, and hence decided to invent something string-related too. Since Roman is beauty-lover, he decided to define the beauty of the string as the length of the longest substring, consisting of only one letter. For example, the beauty of the string "xayyaaabca" is equal to 3, since there is a substring "aaa", while the beauty of the string "qwerqwer" is equal to 1, since all neighboring symbols in it are different.
In order to entertain Roman, Denis wrote down n strings p_1, p_2, p_3, …, p_n on the paper and asked him to calculate the beauty of the string ( … (((p_1 ⋅ p_2) ⋅ p_3) ⋅ … ) ⋅ p_n, where s ⋅ t denotes a multiplication of strings s and t. Roman hasn't fully realized how Denis's multiplication works, so he asked you for a help. Denis knows, that Roman is very impressionable, he guarantees, that the beauty of the resulting string is at most 10^9.
Input
The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings, wroted by Denis.
Next n lines contain non-empty strings p_1, p_2, …, p_n, consisting of lowercase english letters.
It's guaranteed, that the total length of the strings p_i is at most 100 000, and that's the beauty of the resulting product is at most 10^9.
Output
Print exactly one integer — the beauty of the product of the strings.
Examples
Input
3
a
b
a
Output
3
Input
2
bnn
a
Output
1
Note
In the first example, the product of strings is equal to "abaaaba".
In the second example, the product of strings is equal to "abanana".
Submitted Solution:
```
t = [0] * 26
def get_k(c):
return ord(c) - ord('a')
def analyze(s):
length_of_str = len(s)
pos = 0
beauty_table = [0] * 26
for i in range(1, length_of_str):
if s[i] != s[pos]:
k = get_k(s[pos])
beauty_table[k] = max(beauty_table[k], i - pos)
pos = i
k = get_k(s[pos])
beauty_table[k] = max(beauty_table[k], length_of_str - pos)
pos = 0
while pos < length_of_str and s[pos] == s[0]:
pos += 1
left_beauty = pos
pos = length_of_str - 1
while pos > 0 and s[pos] == s[length_of_str - 1]:
pos -= 1
right_beauty = length_of_str - pos - 1
return beauty_table, left_beauty, right_beauty
r = []
for _ in range(int(input())):
p = input()
if all(x == p[0] for x in p): # pure
k = get_k(p[0])
for i in range(26):
if i == k:
t[i] = len(p) * (t[i] + 1) + t[i]
else:
t[i] = min(1, t[i])
else:
for i in range(26):
t[i] = min(1, t[i])
bt, lb, rb = analyze(p)
lk, rk = get_k(p[0]), get_k(p[-1])
if lk == rk:
t[lk] = lb + rb + t[lk]
else:
t[lk], t[rk] = t[lk] + lb, t[rk] + rb
for i in range(26):
t[i] = max(t[i], bt[i])
# r.append(max(t))
# print('\ntableInfo: ', end= ' ')
# for i in range(26):
# print('{}:{}/'.format(chr(i + ord('a')), t[i]), end=' ')
# print('')
# print(' '.join(map(str, r)))
print(max(t))
``` | instruction | 0 | 62,334 | 0 | 124,668 |
Yes | output | 1 | 62,334 | 0 | 124,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and hence decided to came up with something. Roman invented a pizza's recipe, while Denis invented a string multiplication. According to Denis, the result of multiplication (product) of strings s of length m and t is a string t + s_1 + t + s_2 + … + t + s_m + t, where s_i denotes the i-th symbol of the string s, and "+" denotes string concatenation. For example, the product of strings "abc" and "de" is a string "deadebdecde", while the product of the strings "ab" and "z" is a string "zazbz". Note, that unlike the numbers multiplication, the product of strings s and t is not necessarily equal to product of t and s.
Roman was jealous of Denis, since he invented such a cool operation, and hence decided to invent something string-related too. Since Roman is beauty-lover, he decided to define the beauty of the string as the length of the longest substring, consisting of only one letter. For example, the beauty of the string "xayyaaabca" is equal to 3, since there is a substring "aaa", while the beauty of the string "qwerqwer" is equal to 1, since all neighboring symbols in it are different.
In order to entertain Roman, Denis wrote down n strings p_1, p_2, p_3, …, p_n on the paper and asked him to calculate the beauty of the string ( … (((p_1 ⋅ p_2) ⋅ p_3) ⋅ … ) ⋅ p_n, where s ⋅ t denotes a multiplication of strings s and t. Roman hasn't fully realized how Denis's multiplication works, so he asked you for a help. Denis knows, that Roman is very impressionable, he guarantees, that the beauty of the resulting string is at most 10^9.
Input
The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings, wroted by Denis.
Next n lines contain non-empty strings p_1, p_2, …, p_n, consisting of lowercase english letters.
It's guaranteed, that the total length of the strings p_i is at most 100 000, and that's the beauty of the resulting product is at most 10^9.
Output
Print exactly one integer — the beauty of the product of the strings.
Examples
Input
3
a
b
a
Output
3
Input
2
bnn
a
Output
1
Note
In the first example, the product of strings is equal to "abaaaba".
In the second example, the product of strings is equal to "abanana".
Submitted Solution:
```
t = [0] * 26
def get_k(c):
return ord(c) - ord('a')
def analyze(s):
length_of_str = len(s)
pos = 0
beauty_table = [0] * 26
for i in range(1, length_of_str):
if s[i] != s[pos]:
k = get_k(s[pos])
beauty_table[k] = max(beauty_table[k], i - pos)
pos = i
k = get_k(s[pos])
beauty_table[k] = max(beauty_table[k], length_of_str - pos)
pos = 0
while pos < length_of_str and s[pos] == s[0]:
pos += 1
left_beauty = pos
pos = length_of_str - 1
while pos > 0 and s[pos] == s[length_of_str - 1]:
pos -= 1
right_beauty = length_of_str - pos - 1
return beauty_table, left_beauty, right_beauty
r = []
for _ in range(int(input())):
p = input()
if all(x == p[0] for x in p): # pure
k = get_k(p[0])
for i in range(26):
if i == k:
t[i] = len(p) * (t[i] + 1) + t[i]
else:
t[i] = min(1, t[i])
else:
bt, lb, rb = analyze(p)
lk, rk = get_k(p[0]), get_k(p[-1])
t[lk] = max(min(1, t[lk]) + lb, t[lk])
t[rk] = max(min(1, t[rk]) + rb, t[rk])
for i in range(26):
t[i] = max(t[i], bt[i])
r.append(max(t))
# print('\ntableInfo: ', end= ' ')
# for i in range(26):
# print('{}:{}/'.format(chr(i + ord('a')), t[i]), end=' ')
# print('')
# print(' '.join(map(str, r)))
print(max(t))
``` | instruction | 0 | 62,335 | 0 | 124,670 |
No | output | 1 | 62,335 | 0 | 124,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and hence decided to came up with something. Roman invented a pizza's recipe, while Denis invented a string multiplication. According to Denis, the result of multiplication (product) of strings s of length m and t is a string t + s_1 + t + s_2 + … + t + s_m + t, where s_i denotes the i-th symbol of the string s, and "+" denotes string concatenation. For example, the product of strings "abc" and "de" is a string "deadebdecde", while the product of the strings "ab" and "z" is a string "zazbz". Note, that unlike the numbers multiplication, the product of strings s and t is not necessarily equal to product of t and s.
Roman was jealous of Denis, since he invented such a cool operation, and hence decided to invent something string-related too. Since Roman is beauty-lover, he decided to define the beauty of the string as the length of the longest substring, consisting of only one letter. For example, the beauty of the string "xayyaaabca" is equal to 3, since there is a substring "aaa", while the beauty of the string "qwerqwer" is equal to 1, since all neighboring symbols in it are different.
In order to entertain Roman, Denis wrote down n strings p_1, p_2, p_3, …, p_n on the paper and asked him to calculate the beauty of the string ( … (((p_1 ⋅ p_2) ⋅ p_3) ⋅ … ) ⋅ p_n, where s ⋅ t denotes a multiplication of strings s and t. Roman hasn't fully realized how Denis's multiplication works, so he asked you for a help. Denis knows, that Roman is very impressionable, he guarantees, that the beauty of the resulting string is at most 10^9.
Input
The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings, wroted by Denis.
Next n lines contain non-empty strings p_1, p_2, …, p_n, consisting of lowercase english letters.
It's guaranteed, that the total length of the strings p_i is at most 100 000, and that's the beauty of the resulting product is at most 10^9.
Output
Print exactly one integer — the beauty of the product of the strings.
Examples
Input
3
a
b
a
Output
3
Input
2
bnn
a
Output
1
Note
In the first example, the product of strings is equal to "abaaaba".
In the second example, the product of strings is equal to "abanana".
Submitted Solution:
```
n = int(input())
s = []
p = ""
for i in range(n):
s.append(input())
def pd(a,b):
s = ""
a = list(a)
for i in a:
s = s + b + i
s = s + b
return(s)
p = pd(s[0],s[1])
for i in range(2,n):
p = pd(p,s[i])
s = list(p)
c = 1
m = [1]
for i in range(len(s) - 1):
if s[i] == s[i + 1]:
c = c + 1
else:
c = 0
m.append(c)
m.append(c)
print(max(m))
``` | instruction | 0 | 62,336 | 0 | 124,672 |
No | output | 1 | 62,336 | 0 | 124,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and hence decided to came up with something. Roman invented a pizza's recipe, while Denis invented a string multiplication. According to Denis, the result of multiplication (product) of strings s of length m and t is a string t + s_1 + t + s_2 + … + t + s_m + t, where s_i denotes the i-th symbol of the string s, and "+" denotes string concatenation. For example, the product of strings "abc" and "de" is a string "deadebdecde", while the product of the strings "ab" and "z" is a string "zazbz". Note, that unlike the numbers multiplication, the product of strings s and t is not necessarily equal to product of t and s.
Roman was jealous of Denis, since he invented such a cool operation, and hence decided to invent something string-related too. Since Roman is beauty-lover, he decided to define the beauty of the string as the length of the longest substring, consisting of only one letter. For example, the beauty of the string "xayyaaabca" is equal to 3, since there is a substring "aaa", while the beauty of the string "qwerqwer" is equal to 1, since all neighboring symbols in it are different.
In order to entertain Roman, Denis wrote down n strings p_1, p_2, p_3, …, p_n on the paper and asked him to calculate the beauty of the string ( … (((p_1 ⋅ p_2) ⋅ p_3) ⋅ … ) ⋅ p_n, where s ⋅ t denotes a multiplication of strings s and t. Roman hasn't fully realized how Denis's multiplication works, so he asked you for a help. Denis knows, that Roman is very impressionable, he guarantees, that the beauty of the resulting string is at most 10^9.
Input
The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings, wroted by Denis.
Next n lines contain non-empty strings p_1, p_2, …, p_n, consisting of lowercase english letters.
It's guaranteed, that the total length of the strings p_i is at most 100 000, and that's the beauty of the resulting product is at most 10^9.
Output
Print exactly one integer — the beauty of the product of the strings.
Examples
Input
3
a
b
a
Output
3
Input
2
bnn
a
Output
1
Note
In the first example, the product of strings is equal to "abaaaba".
In the second example, the product of strings is equal to "abanana".
Submitted Solution:
```
import sys
from collections import Counter
readline = sys.stdin.readline
N = int(readline())
S = [readline().strip() for i in range(N)]
def calc(s):
res = []
last = s[0]
cnt = 0
for c in s:
if c == last:
cnt += 1
else:
res.append((last, cnt))
last = c
cnt = 1
if cnt > 0:
res.append((last, cnt))
return res
rr = calc(S[N-1])
if len(rr) > 1:
ans = max(x for c, x in rr)
c0, x0 = rr[0]
c1, x1 = rr[-1]
m0 = m1 = 0
for i in range(N-1):
s = S[i]
for c in s:
if c1 == c == c0:
ans = max(ans, x0+x1+1)
elif c == c0:
ans = max(ans, x0+1)
elif c == c1:
ans = max(ans, x1+1)
else:
c0, x0 = rr[0]
rr0 = calc(S[0])
r = 0
for c, x in rr0:
if c0 == c:
r = max(r, x)
for i in range(1, N-1):
rr1 = calc(S[i])
if len(rr1) == 1:
c1, x1 = rr1[0]
if c0 == c1:
r = (r+1)*x1 + r
else:
r = +(r > 0)
else:
d0, y0 = rr1[0]
d1, y1 = rr1[-1]
if c0 != d0:
y0 = 0
if c0 != d1:
y1 = 0
if r > 0:
r = y0 + y1 + 1
else:
r = max(y0, y1)
ans = (r+1)*x0 + r
print(ans)
``` | instruction | 0 | 62,337 | 0 | 124,674 |
No | output | 1 | 62,337 | 0 | 124,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman and Denis are on the trip to the programming competition. Since the trip was long, they soon got bored, and hence decided to came up with something. Roman invented a pizza's recipe, while Denis invented a string multiplication. According to Denis, the result of multiplication (product) of strings s of length m and t is a string t + s_1 + t + s_2 + … + t + s_m + t, where s_i denotes the i-th symbol of the string s, and "+" denotes string concatenation. For example, the product of strings "abc" and "de" is a string "deadebdecde", while the product of the strings "ab" and "z" is a string "zazbz". Note, that unlike the numbers multiplication, the product of strings s and t is not necessarily equal to product of t and s.
Roman was jealous of Denis, since he invented such a cool operation, and hence decided to invent something string-related too. Since Roman is beauty-lover, he decided to define the beauty of the string as the length of the longest substring, consisting of only one letter. For example, the beauty of the string "xayyaaabca" is equal to 3, since there is a substring "aaa", while the beauty of the string "qwerqwer" is equal to 1, since all neighboring symbols in it are different.
In order to entertain Roman, Denis wrote down n strings p_1, p_2, p_3, …, p_n on the paper and asked him to calculate the beauty of the string ( … (((p_1 ⋅ p_2) ⋅ p_3) ⋅ … ) ⋅ p_n, where s ⋅ t denotes a multiplication of strings s and t. Roman hasn't fully realized how Denis's multiplication works, so he asked you for a help. Denis knows, that Roman is very impressionable, he guarantees, that the beauty of the resulting string is at most 10^9.
Input
The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings, wroted by Denis.
Next n lines contain non-empty strings p_1, p_2, …, p_n, consisting of lowercase english letters.
It's guaranteed, that the total length of the strings p_i is at most 100 000, and that's the beauty of the resulting product is at most 10^9.
Output
Print exactly one integer — the beauty of the product of the strings.
Examples
Input
3
a
b
a
Output
3
Input
2
bnn
a
Output
1
Note
In the first example, the product of strings is equal to "abaaaba".
In the second example, the product of strings is equal to "abanana".
Submitted Solution:
```
def mult(s, t):
f1, l1, cntf1, cntl1, beaut1, n1 = s
f2, l2, cntf2, cntl2, beaut2, n2, p = t
f3, l3, cntf3, cntl3, beaut3, n3 = 0, 0, 0, 0, beaut1, 0
f3 = f1
l3 = l1
n3 = n2 * (n1 + 1) + n1
if cntf1 == n1 and f1 == f2:
cntf3 = n1 * (cntf2 + 1) + cntf2
else:
cntf3 = cntf1
if l1 == n1 and l1 == l2:
cntl3 = n1 * (cntl2 + 1) + cntl2
else:
cntl3 = cntl1
if f1 != l1:
if f1 in p:
beaut3 = max(beaut3, cntf1 + 1)
if l1 in p:
beaut3 = max(beaut3, cntl1 + 1)
elif cntf1 == n1:
beaut3 = max(n1, beaut3)
ans = 0
h = 0
for d in p:
if d == f1:
h += 1
ans = max(ans, h)
else:
h = 0
ans = max(ans, h)
beaut3 = max(beaut3, n1 * (ans + 1) + ans)
else:
if f1 in p:
beaut3 = max(beaut3, 1 + cntf1 + cntl1)
else:
beaut3 = max(beaut3, cntf1, cntl1)
return [f3, l3, cntf3, cntl3, beaut3, n3]
n = int(input())
p = []
for i in range(n):
p.append(input())
pp = []
for s in p:
f = s[0]
l = s[-1]
cntf = 0
cntl = 0
beaut = 1
hep = 1
for i in s:
if i == f:
cntf += 1
else:
break
for i in s[::-1]:
if i == l:
cntl += 1
else:
break
for i in range(1, len(s)):
if s[i] == s[i - 1]:
hep += 1
else:
beaut = max(beaut, hep)
hep = 1
beaut = max(beaut, hep)
pp.append([f, l, cntf, cntl, beaut, len(s), s])
p = pp[::-1]
lasts = p[0][:-1]
for i in range(1, len(p)):
lasts = mult(lasts, p[i])
print(lasts[-2])
``` | instruction | 0 | 62,338 | 0 | 124,676 |
No | output | 1 | 62,338 | 0 | 124,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase English letters and a number k. Let's call a string consisting of lowercase English letters beautiful if the number of occurrences of each letter in that string is divisible by k. You are asked to find the lexicographically smallest beautiful string of length n, which is lexicographically greater or equal to string s. If such a string does not exist, output -1.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of the description contains two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and number k respectively.
The second line contains string s consisting of lowercase English letters.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case output in a separate line lexicographically smallest beautiful string of length n, which is greater or equal to string s, or -1 if such a string does not exist.
Example
Input
4
4 2
abcd
3 1
abc
4 3
aaaa
9 3
abaabaaaa
Output
acac
abc
-1
abaabaaab
Note
In the first test case "acac" is greater than or equal to s, and each letter appears 2 or 0 times in it, so it is beautiful.
In the second test case each letter appears 0 or 1 times in s, so s itself is the answer.
We can show that there is no suitable string in the third test case.
In the fourth test case each letter appears 0, 3, or 6 times in "abaabaaab". All these integers are divisible by 3. | instruction | 0 | 62,511 | 0 | 125,022 |
Tags: binary search, brute force, constructive algorithms, greedy, strings
Correct Solution:
```
gans = []
for _ in range(int(input())):
n, k = map(int, input().split())
u = list(input())
for i in range(n):
u[i] = ord(u[i]) - ord('a')
z = []
for i in range(26):
z.append([])
z[-1].append(0)
z[u[0]][0] += 1
z[u[0]][0] %= k
for i in range(1, n):
for j in range(26):
z[j].append(z[j][i - 1])
z[u[i]][i] += 1
z[u[i]][i] %= k
for i in range(n - 1, -1, -1):
sm = 0
for j in range(26):
sm += (k - z[j][i]) % k
if sm % k == (n - i - 1) % k and sm <= n - i - 1:
if i == n - 1:
ans = []
for r in range(n):
ans.append(chr(u[r] + ord('a')))
gans.append(''.join(ans))
break
ans = u[:i + 1]
if sm + k > n - i - 1 or u[i + 1] == ord('z') - ord('a'):
for j in range(u[i + 1] + 1, 26):
if z[j][i] != 0:
ans.append(j)
z[j][i] += 1
z[j][i] %= k
break
else:
continue
else:
ans.append(u[i + 1] + 1)
z[u[i + 1] + 1][i] += 1
z[u[i + 1] + 1][i] %= k
s = []
for j in range(26):
for r in range((k - z[j][i]) % k):
s.append(j)
for j in range(n - len(ans) - len(s)):
ans.append(0)
ans += s
for j in range(len(ans)):
ans[j] = chr(ans[j] + ord('a'))
gans.append(''.join(ans))
break
else:
if n % k == 0 and u[0] != z:
ans = [u[0] + 1]
for i in range(n - k):
ans.append(0)
for i in range(k - 1):
ans.append(u[0] + 1)
for i in range(n):
ans[i] = chr(ans[i] + ord('a'))
gans.append(''.join(ans))
else:
gans.append(str(-1))
print('\n'.join(gans))
``` | output | 1 | 62,511 | 0 | 125,023 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase English letters and a number k. Let's call a string consisting of lowercase English letters beautiful if the number of occurrences of each letter in that string is divisible by k. You are asked to find the lexicographically smallest beautiful string of length n, which is lexicographically greater or equal to string s. If such a string does not exist, output -1.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of the description contains two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and number k respectively.
The second line contains string s consisting of lowercase English letters.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case output in a separate line lexicographically smallest beautiful string of length n, which is greater or equal to string s, or -1 if such a string does not exist.
Example
Input
4
4 2
abcd
3 1
abc
4 3
aaaa
9 3
abaabaaaa
Output
acac
abc
-1
abaabaaab
Note
In the first test case "acac" is greater than or equal to s, and each letter appears 2 or 0 times in it, so it is beautiful.
In the second test case each letter appears 0 or 1 times in s, so s itself is the answer.
We can show that there is no suitable string in the third test case.
In the fourth test case each letter appears 0, 3, or 6 times in "abaabaaab". All these integers are divisible by 3. | instruction | 0 | 62,512 | 0 | 125,024 |
Tags: binary search, brute force, constructive algorithms, greedy, strings
Correct Solution:
```
import sys, math, itertools, random, bisect
from collections import defaultdict
INF = sys.maxsize
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()
mod = 10**9 + 7
def get(x,k):
return (k - x%k) % k
for _ in range(int(input())):
n,k = get_ints()
s = input()
if n%k!=0: print(-1)
else:
cnt = [0]*26
for i in s:
cnt[ord(i)-ord('a')] += 1
tot = 0
flag = 1
for i in range(26):
tot += get(cnt[i],k)
if tot==0:
print(s)
flag = 0
for i in range(n-1,-1,-1):
if not flag: break
tot -= get(cnt[ord(s[i])-ord('a')], k)
cnt[ord(s[i])-ord('a')] -= 1
tot += get(cnt[ord(s[i])-ord('a')], k)
for j in range(ord(s[i])-ord('a')+1, 26):
lst_tot = tot
tot -= get(cnt[j],k)
cnt[j] += 1
tot += get(cnt[j],k)
if tot <= n-i-1:
print(s[:i],end="")
print(chr(ord('a')+j),end="")
rem = ""
rem += 'a'*(n-i-1 - tot)
for w in range(26):
f = get(cnt[w],k)
rem += chr(ord('a')+w)*f
# rem = "".join(sorted(rem))
print(rem)
flag = 0
break
cnt[j] -= 1
tot = lst_tot
``` | output | 1 | 62,512 | 0 | 125,025 |
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