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109k
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float64 0
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int64 194
217k
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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 β€ m β€ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 β€ n β€ min(2^m-1, 100)) distinct binary strings a_1, a_2, β¦, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is β (k-1)/(2) β (such an element is called median), where β x β is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 β€ n β€ min(2^m-1, 100)) and m (1 β€ m β€ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, β¦, a_n β distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m β the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
for _ in range(int(input())):
mylist = []
mylist1 = []
n,m = map(int,input().split())
for _ in range(n):
mylist.append(int(input(),2))
ans = min(2**m-1,100)
for num in range(ans+1):
if num not in mylist:
mylist1.append(num)
mylist1.sort()
answer = bin(mylist1[int((len(mylist1)-1)/2)])[2:]
while len(answer)<m:
answer='0'+answer
if n == 1 and mylist == ['1']:
print('0')
else:
print(answer)
```
|
instruction
| 0
| 105,433
| 0
| 210,866
|
No
|
output
| 1
| 105,433
| 0
| 210,867
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 β€ m β€ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 β€ n β€ min(2^m-1, 100)) distinct binary strings a_1, a_2, β¦, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is β (k-1)/(2) β (such an element is called median), where β x β is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 β€ n β€ min(2^m-1, 100)) and m (1 β€ m β€ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, β¦, a_n β distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m β the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
import math
def calculate(med,count,l,dir):
if dir=='left' :
med=int(med)+1
while ( count>0 ) :
med=med-1
if (not(med in l)):
count=count-1
return med
elif dir=='right' :
med=int(med)
while ( count>0 ) :
med=med+1
if (not(med in l)):
count=count-1
return med
t=int(input())
while ( t > 0 ):
n,m=map(int,input().split())
l=[]
med=float('{:.1f}'.format(float((math.pow(2,m)-1)/2)))
#print('med start = ',med)
right=0
left=0
for i in range(0,n):
val=int(input(),2)
l.append(val)
if val > med:
right+=1
elif val <= med:
left+=1
if right>left :
count=int(((right-left)/2)+1)
#print('count',count)
med=calculate(med,count,l,'left')
elif right<left :
count=int(((left-right)+1)/2)
#print('count',count)
med=calculate(med,count,l,'right')
elif right==left :
count=1
med=calculate(med,count,l,'left')
print('{:0>{}}'.format('{:b}'.format(med),m))
#print(l)
#print('left',left,'right',right)
l.clear()
t=t-1
```
|
instruction
| 0
| 105,434
| 0
| 210,868
|
No
|
output
| 1
| 105,434
| 0
| 210,869
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 β€ m β€ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 β€ n β€ min(2^m-1, 100)) distinct binary strings a_1, a_2, β¦, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is β (k-1)/(2) β (such an element is called median), where β x β is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 β€ n β€ min(2^m-1, 100)) and m (1 β€ m β€ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, β¦, a_n β distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m β the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
from collections import Counter
T = int(input())
for _ in range(T):
n,m = map(int,input().split())
l = []
for i in range(n):
l.append(int(input(),2))
d = Counter(l)
mid = ((2**m - n)-1)//2
i = 0
j = mid
while i<=j:
if d.get(i,0) == 1:
j += 1
i += 1
r = bin(j)[2:]
r.zfill(m)
print(r)
```
|
instruction
| 0
| 105,435
| 0
| 210,870
|
No
|
output
| 1
| 105,435
| 0
| 210,871
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 β€ m β€ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 β€ n β€ min(2^m-1, 100)) distinct binary strings a_1, a_2, β¦, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is β (k-1)/(2) β (such an element is called median), where β x β is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 β€ n β€ min(2^m-1, 100)) and m (1 β€ m β€ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, β¦, a_n β distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m β the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
# data=[]
# with open('input.txt') as f:
# data = [i for i in f]
# data.reverse()
# input = data.pop
def get_ints():
return list(map(int, input().split()))
for _ in range(int(input())):
N,M = get_ints()
a = []
for i in range(N):
a.append(int(input(),2))
g = [i for i in range(1,1<<M) if not i in a]
# print(g)
if len(g)==0:
print("0"*M)
else:
d = bin(g[(len(g)-1)//2])[2:]
print("0"*(M-len(d)) + d)
```
|
instruction
| 0
| 105,436
| 0
| 210,872
|
No
|
output
| 1
| 105,436
| 0
| 210,873
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
|
instruction
| 0
| 105,437
| 0
| 210,874
|
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
class Solution:
def __init__(self, debug=False, n=None, a=None, b=None):
if n is None:
n, a, b = Solution.input()
self.n = n
self.a = a
self.b = b
self.real = [x for x in self.a]
self.inverse = [not x for x in self.a]
self.curr_zero = 0
self.curr_last = self.n - 1
self.real_idx = self.n - 1
self.inv = False
self.debug = debug
if debug:
self.current_state = [x for x in self.a]
self.operations = [self.n + 1000]
@staticmethod
def input():
n = int(input())
a = [x == '1' for x in input()]
b = [x == '1' for x in input()]
return n, a, b
def iteration(self):
if self.inv:
if self.inverse[self.curr_last] != self.b[self.real_idx]:
if self.inverse[self.curr_zero] != self.inverse[self.curr_last]:
self.operation(0)
self.real[self.curr_zero], self.inverse[self.curr_zero] = self.inverse[self.curr_zero], self.real[
self.curr_zero]
self.operation(self.real_idx)
self.inv = False
self.curr_zero, self.curr_last = self.curr_last, self.curr_zero
else:
if self.real[self.curr_last] != self.b[self.real_idx]:
if self.real[self.curr_zero] != self.real[self.curr_last]:
self.operation(0)
self.real[self.curr_zero], self.inverse[self.curr_zero] = self.inverse[self.curr_zero], self.real[
self.curr_zero]
self.operation(self.real_idx)
self.inv = True
self.curr_zero, self.curr_last = self.curr_last, self.curr_zero
if self.inv:
self.curr_last += 1
else:
self.curr_last -= 1
self.real_idx -= 1
def solve(self):
while self.real_idx >= 0:
self.iteration()
return self.return_answer()
def special_cases(self):
if self.n == 1:
if self.a[0] != self.b[0]:
self.operation(0)
return self.return_answer
# if self.n == 2:
# if self.a[1] != self.b[1]:
# if self.a[0] != self.a[1]:
# self.operation(0)
# self.a[0] = not self.a[0]
# self.operation(1)
# self.a = [not x for x in self.a][::-1]
# if self.debug:
# assert self.current_state[1] == self.b[1]
# if self.a[0] != self.b[0]:
# self.operation(1)
# return self.return_answer
def operation(self, i):
self.operations.append(i + 1)
if self.debug:
self.current_state[:i + 1] = [not x for x in self.current_state[:i + 1]][::-1]
if self.n < 2:
return
def return_answer(self):
if self.debug:
assert all(x == y for x, y in zip(self.b, self.current_state))
ans = f'{len(self.operations) - 1} {" ".join(str(x) for x in self.operations[1:])}'
print(ans)
return ans
def solve():
n = int(input())
a = [int(x) for x in input()]
b = [int(x) for x in input()]
# state
real = [x for x in a]
inverse = [(x + 1) % 2 for x in a]
inv_idx = 0
real_idx = n - 1
inv = False
operations = []
# fix all
seq = []
while real_idx > 0:
if not inv:
if real[real_idx] == b[real_idx]:
real_idx -= 1
inv_idx += 1
seq.append(real[real_idx])
continue
if real[0] != real[real_idx]:
operations.append(1)
real[0], inverse[-1] = inverse[-1], real[0]
operations.append(real_idx + 1)
real[real_idx] = b[real_idx]
seq.append(real[real_idx])
inverse[inv_idx] = (real[real_idx] + 1) % 2
inv = True
else:
if inverse[inv_idx] == b[real_idx]:
real_idx -= 1
inv_idx += 1
seq.append(inverse[inv_idx])
continue
if inverse[-1] != inverse[inv_idx]:
operations.append(1)
real[0], inverse[-1] = inverse[-1], real[0]
operations.append(real_idx + 1)
inverse[inv_idx] = b[real_idx]
real[real_idx] = (b[real_idx] + 1) % 2
seq.append(inverse[inv_idx])
inv = False
real_idx -= 1
inv_idx += 1
# fix 0
if (inv and inverse[-1] != b[0]) or (real[0] != b[0] and not inv):
operations.append(1)
seq.append(b[0])
print(seq, b)
print(len(operations), *operations)
if __name__ == '__main__':
for _ in range(int(input())):
n, a, b = Solution.input()
Solution(False, n, a, b).solve()
```
|
output
| 1
| 105,437
| 0
| 210,875
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
|
instruction
| 0
| 105,438
| 0
| 210,876
|
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
from sys import stdin
t = int(stdin.readline().strip())
for _ in range(t):
n = int(stdin.readline().strip())
a = stdin.readline().strip()
b = stdin.readline().strip()
#n,m = list(map(int, stdin.readline().strip().split(' ')))
out = []
for i in range(n-1):
if a[i] != a[i+1]:
out.append(i+1)
current = a[-1]
for i in range(n-1, -1, -1):
if b[i] != current:
out.append(i+1)
current = '0' if current == '1' else '1'
print(len(out), end=" ")
print(*out)
```
|
output
| 1
| 105,438
| 0
| 210,877
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
|
instruction
| 0
| 105,439
| 0
| 210,878
|
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools
from collections import deque,defaultdict,OrderedDict
import collections
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
#Solving Area Starts-->
#Solving Area Starts-->
for _ in range(ri()):
n=ri()
a=rs()
b=rs()
s=""
if a==b:
print(0)
continue
if n==1:
print(1,1)
continue
ans1=[]
ans2=[]
z=0
c=0
for j in range(n):
if a[j]=='0':
break
else:
c+=1
if c>0:
ans1.append(c)
i=c
while i<n:
if a[i]=='1':
if i-1>=0:
ans1.append(i)
s='1'*(i)
if a[i:]=='1'*(n-len(s)):
z=1
break
for k in range(i,n):
if a[k]=='0':
break
ans1.append(k)
i=k+1
else:
i+=1
if z==1:
ans1.append(n)
z=0
c=0
for j in range(n):
if b[j]=='0':
break
else:
c+=1
if c>0:
ans2.append(c)
i=c
while i<n:
if b[i]=='1':
if i-1>=0:
ans2.append(i)
s='1'*(i)
if b[i:]=='1'*(n-len(s)):
z=1
break
for k in range(i,n):
if b[k]=='0':
break
ans2.append(k)
i=k+1
else:
i+=1
if z==1:
ans2.append(n)
# print(ans1)
# print(ans2)
print(len(ans1)+len(ans2),*(ans1+ans2[::-1]))
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
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, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
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()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
```
|
output
| 1
| 105,439
| 0
| 210,879
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
|
instruction
| 0
| 105,440
| 0
| 210,880
|
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
#e=[int(x) for x in input().split()]
#a=list(input())
#b=list(input())
a=[int(x) for x in input()]
b=[int(x) for x in input()]
a.append(0)
b.append(0)
ans=[]
for i in range(n):
if a[i]!=a[i+1]:
ans.append(i+1)
#a= [i-a[j] for j in range(i,-1,-1)]+ a[i+1:]
for i in range(n-1,-1,-1):
if b[i]!=b[i+1]:
ans.append(i+1)
print(len(ans),end=" ")
for x in ans:
print(x,end=" ")
print()
```
|
output
| 1
| 105,440
| 0
| 210,881
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
|
instruction
| 0
| 105,441
| 0
| 210,882
|
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
# import sys
# import math
# input = sys.stdin.readline
from collections import deque
from queue import LifoQueue
for _ in range(int(input())):
n = int(input())
a = input()
b = input()
a1 = []
a2 = []
# if n==1:
# if a[0]==b[0]:
# print(0)
# else:
# print(1,1)
for i in range(0,n,2):
if a[i:i+2]=='01':
a1.append(i+1)
a1.append(i+2)
elif a[i:i+2]=='10':
if i!=0:
a1.append(i)
a1.append(i+1)
elif a[i:i+2]=='11':
if i!=0:
a1.append(i)
a1.append(i+2)
elif a[i:i+2]=='1':
if n-1>0:
a1.append(n-1)
a1.append(n)
for i in range(0,n,2):
if b[i:i+2]=='01':
a2.append(i+1)
a2.append(i+2)
elif b[i:i+2]=='10':
if i!=0:
a2.append(i)
a2.append(i+1)
elif b[i:i+2]=='11':
if i!=0:
a2.append(i)
a2.append(i+2)
elif b[i:i+2]=='1':
if n-1>0:
a2.append(n-1)
a2.append(n)
a2.reverse()
c = a1+a2
# print(*c)
for i in range(len(c)-1):
if c[i]==c[i+1]:
c.pop(i)
c.pop(i)
break
print(len(c),*c)
```
|
output
| 1
| 105,441
| 0
| 210,883
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
|
instruction
| 0
| 105,442
| 0
| 210,884
|
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
a=input()
b=input()
aToAllOnesOrZeros=[]
for i in range(n-1):
if a[i]!=a[i+1]: #flip up to a[i]. Then a[0,1,...i+1] will be uniform. a will become all = to a[-1]
aToAllOnesOrZeros.append(i+1)
bToAllOnesOrZeros=[]
for i in range(n-1):
if b[i]!=b[i+1]: #b will become all = to b[-1]
bToAllOnesOrZeros.append(i+1)
res=aToAllOnesOrZeros
if a[-1]!=b[-1]: #flip all aTransformed so that aTransformed==bTransformed
res.append(n)
for i in range(len(bToAllOnesOrZeros)-1,-1,-1):
res.append(bToAllOnesOrZeros[i]) #reverse the process
l=len(res)
res.insert(0,l)
print(' '.join([str(x) for x in res]))
#def flip(s,l):
# S=list(s[:l])
# for i in range(l):
# S[i]='0' if S[i]=='1' else '1'
# return ''.join(S)+s[l:]
#
#y=a
#for i in range(n-1):
# if y[i]!=y[i+1]:
# y=flip(y,i+1)
#print(y) #y will be all a[-1]
#
#z=b
#for i in range(n-1):
# if z[i]!=z[i+1]:
# z=flip(z,i+1)
#print(z) #z will be all b[-1]
```
|
output
| 1
| 105,442
| 0
| 210,885
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
|
instruction
| 0
| 105,443
| 0
| 210,886
|
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import collections
t1=int(input())
for _ in range(t1):
flipflag=0
bitflag=0
n=int(input())
a2=input()
a=collections.deque([])
for i in range(n):
a.append(a2[i])
b=input()
ans=[]
for i in range(n-1,-1,-1):
if flipflag==0:
p=a[-1]
else:
p=a[0]
if bitflag==1:
if p=='0':
p='1'
else:
p='0'
if flipflag==0:
q=a[0]
else:
q=a[-1]
if bitflag==1:
if q=='0':
q='1'
else:
q='0'
if p!=b[i]:
if q!=b[i]:
ans.append(str(i+1))
if flipflag==0:
a.popleft()
else:
a.pop()
flipflag=1-flipflag
bitflag=1-bitflag
else:
ans.append('1')
ans.append(str(i+1))
if flipflag==0:
a.popleft()
else:
a.pop()
flipflag=1-flipflag
bitflag=1-bitflag
else:
if flipflag==0:
a.pop()
else:
a.popleft()
print(len(ans))
if len(ans)>0:
print(' '.join(ans))
```
|
output
| 1
| 105,443
| 0
| 210,887
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
|
instruction
| 0
| 105,444
| 0
| 210,888
|
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import deque
t = int(input())
for _ in range(t):
n = int(input())
a = input()
b = input()
q = deque([i for i in range(n)])
rev = False
ans = []
for i in range(n)[::-1]:
if rev:
a_ind = q[0]
if a[a_ind] != b[i]:
q.popleft()
continue
else:
if a[q[-1]] == b[i]:
ans.append(i + 1)
else:
ans.append(1)
ans.append(i + 1)
q.pop()
rev = False
else:
a_ind = q[-1]
if a[a_ind] == b[i]:
q.pop()
continue
else:
if a[q[0]] != b[i]:
ans.append(i + 1)
else:
ans.append(1)
ans.append(i + 1)
q.popleft()
rev = True
print(len(ans), *ans)
```
|
output
| 1
| 105,444
| 0
| 210,889
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
import os
t1=int(input())
while t1:
t1-=1
n=int(input())
s=input()
t=input()
s+="0"
t+="0"
ans=[]
for i in range(n):
if s[i]!=s[i+1]:
ans.append(i+1)
for i in range(n,0,-1):
if t[i]!=t[i-1]:
ans.append(i)
print(len(ans),end=' ')
for x in ans:
print(x,end=' ')
print()
```
|
instruction
| 0
| 105,445
| 0
| 210,890
|
Yes
|
output
| 1
| 105,445
| 0
| 210,891
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
def convert_to_1(s):
L=[]
for i in range(len(s)-1):
if s[i]!=s[i+1]:L.append(i+1)
if s[-1]=='0':L.append(len(s))
return L
for i in ' '*(int(input())):
n=int(input())
s1=input()
s2=input()
L1=convert_to_1(s1)
L2=convert_to_1(s2)
L=L1+L2[::-1]
print(len(L),end=' ')
for i in L:print(i,end=' ')
print()
```
|
instruction
| 0
| 105,446
| 0
| 210,892
|
Yes
|
output
| 1
| 105,446
| 0
| 210,893
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from sys import stdin, stdout
# 0 1 2 3 4 5 6 7 8 9
# 9 8 7 6 5 4 3 2 1 (0) 9
# 1 2 3 4 5 6 7 8 (9) 8
# 8 7 6 5 4 3 2 (1) 7
# 2 3 4 5 6 7 (8) 6
# 7 6 5 4 3 (2) 5
# ..........
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
a = stdin.readline().strip()
b = stdin.readline().strip()
ans = []
idx = 0
flip = False
for i in range(n-1, -1, -1):
if (not flip and a[idx] == b[i]) or (flip and a[idx] != b[i]):
ans.append(1)
ans.append(i+1)
if flip:
idx -= i
else:
idx += i
flip = not flip
stdout.write(str(len(ans)) + ' ')
if len(ans) > 0:
stdout.write(' '.join(map(str, ans)) + '\n')
```
|
instruction
| 0
| 105,447
| 0
| 210,894
|
Yes
|
output
| 1
| 105,447
| 0
| 210,895
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
#!/usr/bin/env python3
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
a=list(input())
b=list(input())
ans=[]
l=0
r=n-1
for i in range(n-1,-1,-1):
if a[i]!=b[i]:
r=i
break
else:
r=0
if l==r:
if a[0]==b[0]:
print(0)
else:
print(1,1)
continue
else:
flip=False
count=r
while l!=r:
if flip==False:
if a[l]==b[count]:
if a[l]=='0':
a[l]='1'
else:
a[l]='0'
ans.append(1)
ans.append(count+1)
else:
ans.append(count+1)
flip=True
else:
if a[r]!=b[count]:
if a[r]=='0':
a[r]='1'
else:
a[r]='0'
ans.append(1)
ans.append(count+1)
else:
ans.append(count+1)
flip=False
for i in range(count,-1,-1):
if flip==True:
if a[l+count-i]==b[i]:
l=l+count-i
count=i
break
else:
if a[r-count+i]!=b[i]:
r=r-count+i
count=i
break
else:
if flip==False:
l=r
else:
r=l
if flip==False:
if a[l]!=b[0]:
ans.append(1)
else:
if a[r]==b[0]:
ans.append(1)
ans=[len(ans)]+ans
print(*ans)
```
|
instruction
| 0
| 105,448
| 0
| 210,896
|
Yes
|
output
| 1
| 105,448
| 0
| 210,897
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
import sys
#comment these out later
#sys.stdin = open("in.in", "r")
#sys.stdout = open("out.out", "w")
inp = sys.stdin.read().split(); ii = 0
t = int(inp[ii]); ii += 1
for _ in range(t):
toprint = []
n = int(inp[ii]); ii += 1
a = list(map(int, inp[ii])); ii += 1
b = list(map(int, inp[ii])); ii += 1
lead = a[0]
for i in range(n-1, -1, -1):
x = a[i]
y = b[i]
if lead == y:
toprint.append(1)
toprint.append(i + 1)
lead = y
else:
toprint.append(i + 1)
lead = y
if toprint == []:
print(0)
else:
print(" ".join(list(map(str, toprint))))
```
|
instruction
| 0
| 105,449
| 0
| 210,898
|
No
|
output
| 1
| 105,449
| 0
| 210,899
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from sys import stdin
tt = int(input())
for loop in range(tt):
n = int(input())
a = input()
b = input()
ans = []
x = 0
for i in range(n-1,-1,-1):
if x == 0 and a[i] != b[i]:
x ^= 1
ans.append(i+1)
elif x == 1 and a[i] == b[i]:
x ^= 1
ans.append(i+1)
print (len(ans),*ans)
```
|
instruction
| 0
| 105,450
| 0
| 210,900
|
No
|
output
| 1
| 105,450
| 0
| 210,901
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
t=input()
ans=0
A=[]
for i in range(n-1,-1,-1):
if ans%2==1 and t[i]!=s[i]:
continue
elif ans%2==0 and t[i]==s[i]:
continue
else:
ans+=1
A.append(i+1)
t=s
print(ans,end=" ")
for i in A:
print(i,end=" ")
print()
```
|
instruction
| 0
| 105,451
| 0
| 210,902
|
No
|
output
| 1
| 105,451
| 0
| 210,903
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 10^5) β the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0β€ kβ€ 2n), followed by k integers p_1,β¦,p_k (1β€ p_iβ€ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01β 11β 00β 10.
In the second test case, we have 01011β 00101β 11101β 01000β 10100β 00100β 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=1
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
//If you Know me , Then you probably don't know me
"""
###########################---START-CODING---##############################
num=int(z())
for _ in range( num ):
n=int(z())
T1=fast()
T2=fast()
ans=[]
k=1
for i in T1:
if i=='1':
ans.append(k)
k+=1
k=1
for i in T2:
if i=='1':
ans.append(k)
k+=1
print(len(ans),*ans)
```
|
instruction
| 0
| 105,452
| 0
| 210,904
|
No
|
output
| 1
| 105,452
| 0
| 210,905
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+β¦ +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+β¦+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1β€ nβ€ 100, 0β€ kβ€ β n/2 β) β the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,β¦,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
|
instruction
| 0
| 105,478
| 0
| 210,956
|
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
def solve():
n,k=map(int,input().split())
s=input()
if k>=n/2:
return "NO"
for i in range(k):
if s[i]!=s[-(i+1)]:
return "NO"
else:
return "YES"
for _ in range(int(input())):
print(solve())
```
|
output
| 1
| 105,478
| 0
| 210,957
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+β¦ +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+β¦+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1β€ nβ€ 100, 0β€ kβ€ β n/2 β) β the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,β¦,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
|
instruction
| 0
| 105,479
| 0
| 210,958
|
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
def solve(n, k, s):
if k == 0:
return "YES"
count = 0
if n%2:
for i in range(n//2):
if s[i] == s[-i-1]:
count += 1
else:
break
if count >= k:
return "YES"
else:
return "NO"
else:
for i in range(n//2 - 1):
if s[i] == s[-i-1]:
count += 1
else:
break
if count>=k:
return "YES"
else:
return "NO"
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = input()
print(solve(n, k, s))
```
|
output
| 1
| 105,479
| 0
| 210,959
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+β¦ +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+β¦+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1β€ nβ€ 100, 0β€ kβ€ β n/2 β) β the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,β¦,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
|
instruction
| 0
| 105,480
| 0
| 210,960
|
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
t = int(input())
for _ in range(t):
n,k = map(int,input().split())
s = input()
if 2*k==n:
print("NO")
continue
if k==0:
print("YES")
continue
if s[:k]==s[-k:][::-1]:
print("YES")
else:
print("NO")
```
|
output
| 1
| 105,480
| 0
| 210,961
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+β¦ +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+β¦+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1β€ nβ€ 100, 0β€ kβ€ β n/2 β) β the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,β¦,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
|
instruction
| 0
| 105,481
| 0
| 210,962
|
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
S = input()
print()
if k == 0:
print("YES")
continue
elif S[:k] == (S[::-1])[:k]:
if n//2 >= k + 1 - n%2:
print("YES")
continue
print("NO")
```
|
output
| 1
| 105,481
| 0
| 210,963
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+β¦ +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+β¦+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1β€ nβ€ 100, 0β€ kβ€ β n/2 β) β the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,β¦,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
|
instruction
| 0
| 105,482
| 0
| 210,964
|
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
# from math import *
# from bisect import *
# from heapq import *
# from collections import defaultdict as dd
# from collections import OrderedDict as odict
# from collections import Counter as cc
# from collections import deque
# sys.setrecursionlimit(5*10**5+100) #this is must for dfs
# ______________________________________________________________________________________________________
# segment tree for range minimum query
# n = int(input())
# a = list(map(int,input().split()))
# st = [float('inf') for i in range(4*len(a))]
# def build(a,ind,start,end):
# if start == end:
# st[ind] = a[start]
# else:
# mid = (start+end)//2
# build(a,2*ind+1,start,mid)
# build(a,2*ind+2,mid+1,end)
# st[ind] = min(st[2*ind+1],st[2*ind+2])
# build(a,0,0,n-1)
# def query(ind,l,r,start,end):
# if start>r or end<l:
# return float('inf')
# if l<=start<=end<=r:
# return st[ind]
# mid = (start+end)//2
# return min(query(2*ind+1,l,r,start,mid),query(2*ind+2,l,r,mid+1,end))
# ______________________________________________________________________________________________________
# Checking prime in O(root(N))
# def isprime(n):
# if (n % 2 == 0 and n > 2) or n == 1: return 0
# else:
# s = int(n**(0.5)) + 1
# for i in range(3, s, 2):
# if n % i == 0:
# return 0
# return 1
# def lcm(a,b):
# return (a*b)//gcd(a,b)
# ______________________________________________________________________________________________________
# nCr under mod
# def C(n,r,mod = 10**9+7):
# if r>n:
# return 0
# num = den = 1
# for i in range(r):
# num = (num*(n-i))%mod
# den = (den*(i+1))%mod
# return (num*pow(den,mod-2,mod))%mod
# ______________________________________________________________________________________________________
# For smallest prime factor of a number
# M = 2*(10**5)+10
# pfc = [i for i in range(M)]
# def pfcs(M):
# for i in range(2,M):
# if pfc[i]==i:
# for j in range(i+i,M,i):
# if pfc[j]==j:
# pfc[j] = i
# return
# pfcs(M)
# ______________________________________________________________________________________________________
tc = 1
tc = int(input())
for _ in range(tc):
n,k = inp()
s = str(input())
f = True
i = 0
j = n-1
while(k>0):
if i+1>=j:
f = False
break
# print(i,j)
if s[i]==s[j]:
pass
else:
f = False
break
i+=1
j-=1
k-=1
print("YES" if f else "NO")
```
|
output
| 1
| 105,482
| 0
| 210,965
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+β¦ +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+β¦+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1β€ nβ€ 100, 0β€ kβ€ β n/2 β) β the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,β¦,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
|
instruction
| 0
| 105,483
| 0
| 210,966
|
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
# by the authority of GOD author: Kritarth Sharma #
import math
from collections import defaultdict,Counter
from itertools import permutations
from decimal import Decimal, localcontext
from collections import defaultdict
ii = lambda : int(input())
li = lambda:list(map(int,input().split()))
def main():
for _ in range(ii()):
n,k=li()
s=input()
c=0
if k==0:
print("YES")
else:
a1=s[:n//2]
a2=s[n//2+1:][::-1]
c=0
for i in range(min(len(a1),len(a2))):
if a1[i]==a2[i]:
c+=1
else:
break
if 2*k+1<=n and c>=k:
print("YES")
else:
print("NO")
import os,sys
from io import BytesIO,IOBase
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from math import gcd
def find(l):
c=0
for i in range(len(l)):
if l[i]>1:
c=i
return c
else:
return None
def checkDivisibility(n, digit) :
# If the digit divides the
# number then return true
# else return false.
return (digit != 0 and n % digit == 0)
# Function to check if
# all digits of n divide
# it or not
def allDigitsDivide( n) :
nlist = list(map(int, set(str(n))))
for digit in nlist :
if digit!=0 and not (checkDivisibility(n, digit)) :
return False
return True
def lcm(s):
a=[int(d) for d in str(s)]
lc = a[0]
for i in a[1:]:
if i!=0:
lc = lc*i//gcd(lc, i)
return(lc)
if __name__ == "__main__":
main()
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# If prime[p] is not
# changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
l=[1,]
# Print all prime numbers
for p in range(2, n+1):
if prime[p]:
l.append(p)
return l
def fact(n):
return 1 if (n == 1 or n == 0) else n * fact(n - 1)
def prime(n) :
if (n <= 1) :
return False
if (n <= 3) :
return True
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
```
|
output
| 1
| 105,483
| 0
| 210,967
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+β¦ +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+β¦+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1β€ nβ€ 100, 0β€ kβ€ β n/2 β) β the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,β¦,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
|
instruction
| 0
| 105,484
| 0
| 210,968
|
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
import sys
import math
import heapq
import bisect
from collections import Counter
from collections import defaultdict
from io import BytesIO, IOBase
import string
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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
self.BUFSIZE = 8192
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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 = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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:
self.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")
def get_int():
return int(input())
def get_ints():
return list(map(int, input().split(' ')))
def get_int_grid(n):
return [get_ints() for _ in range(n)]
def get_str():
return input().split(' ')
def yes_no(b):
if b:
return "YES"
else:
return "NO"
def binary_search(good, left, right, delta=1, right_true=False):
"""
Performs binary search
----------
Parameters
----------
:param good: Function used to perform the binary search
:param left: Starting value of left limit
:param right: Starting value of the right limit
:param delta: Margin of error, defaults value of 1 for integer binary search
:param right_true: Boolean, for whether the right limit is the true invariant
:return: Returns the most extremal value interval [left, right] which is good function evaluates to True,
alternatively returns False if no such value found
"""
limits = [left, right]
while limits[1] - limits[0] > delta:
if delta == 1:
mid = sum(limits) // 2
else:
mid = sum(limits) / 2
if good(mid):
limits[int(right_true)] = mid
else:
limits[int(~right_true)] = mid
if good(limits[int(right_true)]):
return limits[int(right_true)]
else:
return False
def prefix_sums(a, drop_zero=False):
p = [0]
for x in a:
p.append(p[-1] + x)
if drop_zero:
return p[1:]
else:
return p
def prefix_mins(a, drop_zero=False):
p = [float('inf')]
for x in a:
p.append(min(p[-1], x))
if drop_zero:
return p[1:]
else:
return p
def solve_a():
n, k = get_ints()
s = input().strip()
j = 0
while j <= (n - 3) // 2 and s[j] == s[-(j + 1)]:
j += 1
return yes_no(k <= j)
t = get_int()
for _ in range(t):
print(solve_a())
```
|
output
| 1
| 105,484
| 0
| 210,969
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+β¦ +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+β¦+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1β€ nβ€ 100, 0β€ kβ€ β n/2 β) β the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,β¦,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
|
instruction
| 0
| 105,485
| 0
| 210,970
|
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
t=int(input())
for z in range(t):
n,k=[int(q)for q in input().split()]
s=input()
a=s[:k]+s[n-k:]
if k==n/2:
print("NO")
elif k==0 or a==a[::-1]:
print("YES")
else:
print("NO")
```
|
output
| 1
| 105,485
| 0
| 210,971
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce the designation <image>, where x is a string, n is a positive integer and operation " + " is the string concatenation operation. For example, [abc, 2] = abcabc.
We'll say that string s can be obtained from string t, if we can remove some characters from string t and obtain string s. For example, strings ab and aΡba can be obtained from string xacbac, and strings bx and aaa cannot be obtained from it.
Sereja has two strings, w = [a, b] and q = [c, d]. He wants to find such maximum integer p (p > 0), that [q, p] can be obtained from string w.
Input
The first line contains two integers b, d (1 β€ b, d β€ 107). The second line contains string a. The third line contains string c. The given strings are not empty and consist of lowercase English letters. Their lengths do not exceed 100.
Output
In a single line print an integer β the largest number p. If the required value of p doesn't exist, print 0.
Examples
Input
10 3
abab
bab
Output
3
|
instruction
| 0
| 105,560
| 0
| 211,120
|
Tags: binary search, dfs and similar, strings
Correct Solution:
```
b, p = map(int, input().split())
s1 = input()
s2 = input()
cnt = [0] * len(s2)
nxt = [0] * len(s2)
for i in range(len(s2)):
pos = i
for j in range(len(s1)):
if s1[j] == s2[pos]:
pos += 1
if pos == len(s2):
cnt[i] += 1
pos = 0
nxt[i] = pos
ans = 0
poss = 0
for i in range(b):
ans += cnt[poss]
poss = nxt[poss]
print(ans // p)
```
|
output
| 1
| 105,560
| 0
| 211,121
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce the designation <image>, where x is a string, n is a positive integer and operation " + " is the string concatenation operation. For example, [abc, 2] = abcabc.
We'll say that string s can be obtained from string t, if we can remove some characters from string t and obtain string s. For example, strings ab and aΡba can be obtained from string xacbac, and strings bx and aaa cannot be obtained from it.
Sereja has two strings, w = [a, b] and q = [c, d]. He wants to find such maximum integer p (p > 0), that [q, p] can be obtained from string w.
Input
The first line contains two integers b, d (1 β€ b, d β€ 107). The second line contains string a. The third line contains string c. The given strings are not empty and consist of lowercase English letters. Their lengths do not exceed 100.
Output
In a single line print an integer β the largest number p. If the required value of p doesn't exist, print 0.
Examples
Input
10 3
abab
bab
Output
3
Submitted Solution:
```
b, p = map(int, input().split())
s1 = input()
s2 = input()
cnt = 0
pos = 0
flag = True
while flag:
for i in range(len(s1)):
if s1[i] == s2[pos]:
pos += 1
if i == len(s1) - 1:
cnt += 1
if pos == len(s2):
flag = False
break
ans = b // cnt // p
print(ans)
```
|
instruction
| 0
| 105,561
| 0
| 211,122
|
No
|
output
| 1
| 105,561
| 0
| 211,123
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce the designation <image>, where x is a string, n is a positive integer and operation " + " is the string concatenation operation. For example, [abc, 2] = abcabc.
We'll say that string s can be obtained from string t, if we can remove some characters from string t and obtain string s. For example, strings ab and aΡba can be obtained from string xacbac, and strings bx and aaa cannot be obtained from it.
Sereja has two strings, w = [a, b] and q = [c, d]. He wants to find such maximum integer p (p > 0), that [q, p] can be obtained from string w.
Input
The first line contains two integers b, d (1 β€ b, d β€ 107). The second line contains string a. The third line contains string c. The given strings are not empty and consist of lowercase English letters. Their lengths do not exceed 100.
Output
In a single line print an integer β the largest number p. If the required value of p doesn't exist, print 0.
Examples
Input
10 3
abab
bab
Output
3
Submitted Solution:
```
import math
def a_can_from_b(a, b):
pt = 0
for chr in a:
try:
while b[pt] != chr:
pt += 1
except IndexError:
return False
return True
class CodeforcesTask314BSolution:
def __init__(self):
self.result = ''
self.b_d = []
self.a = ''
self.c = ''
def read_input(self):
self.b_d = [int(x) for x in input().split(" ")]
self.a = input()
self.c = input()
def process_task(self):
left = 1
right = math.ceil(self.b_d[0] / self.b_d[1])
while right - left > 1:
mid = left + (right - left) // 2
#print(left, right, mid)
if a_can_from_b(self.c * (self.b_d[1] * mid), self.a * self.b_d[0]):
left = mid
else:
right = mid
if not a_can_from_b(self.c * (self.b_d[1] * left), self.a * self.b_d[0]):
left -= 1
self.result = str(left)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask314BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
```
|
instruction
| 0
| 105,562
| 0
| 211,124
|
No
|
output
| 1
| 105,562
| 0
| 211,125
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce the designation <image>, where x is a string, n is a positive integer and operation " + " is the string concatenation operation. For example, [abc, 2] = abcabc.
We'll say that string s can be obtained from string t, if we can remove some characters from string t and obtain string s. For example, strings ab and aΡba can be obtained from string xacbac, and strings bx and aaa cannot be obtained from it.
Sereja has two strings, w = [a, b] and q = [c, d]. He wants to find such maximum integer p (p > 0), that [q, p] can be obtained from string w.
Input
The first line contains two integers b, d (1 β€ b, d β€ 107). The second line contains string a. The third line contains string c. The given strings are not empty and consist of lowercase English letters. Their lengths do not exceed 100.
Output
In a single line print an integer β the largest number p. If the required value of p doesn't exist, print 0.
Examples
Input
10 3
abab
bab
Output
3
Submitted Solution:
```
b, p = map(int, input().split())
s1 = input()
s2 = input()
cnt = 0
pos = 0
flag = True
while flag and cnt < b:
for i in range(len(s1)):
if i == 0:
cnt += 1
if s1[i] == s2[pos]:
pos += 1
if pos == len(s2):
flag = False
break
if flag == False:
ans = b // cnt // p
else:
ans = 0
print(ans)
```
|
instruction
| 0
| 105,563
| 0
| 211,126
|
No
|
output
| 1
| 105,563
| 0
| 211,127
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce the designation <image>, where x is a string, n is a positive integer and operation " + " is the string concatenation operation. For example, [abc, 2] = abcabc.
We'll say that string s can be obtained from string t, if we can remove some characters from string t and obtain string s. For example, strings ab and aΡba can be obtained from string xacbac, and strings bx and aaa cannot be obtained from it.
Sereja has two strings, w = [a, b] and q = [c, d]. He wants to find such maximum integer p (p > 0), that [q, p] can be obtained from string w.
Input
The first line contains two integers b, d (1 β€ b, d β€ 107). The second line contains string a. The third line contains string c. The given strings are not empty and consist of lowercase English letters. Their lengths do not exceed 100.
Output
In a single line print an integer β the largest number p. If the required value of p doesn't exist, print 0.
Examples
Input
10 3
abab
bab
Output
3
Submitted Solution:
```
b, p = map(int, input().split())
s1 = input()
s2 = input()
cnt = 0
pos = 0
flag = True
while flag and cnt < b:
for i in range(len(s1)):
if i == 0:
cnt += 1
if s1[i] == s2[pos]:
pos += 1
if pos == len(s2):
flag = False
break
if flag:
ans = b // cnt // p
else:
ans = 0
print(ans)
```
|
instruction
| 0
| 105,564
| 0
| 211,128
|
No
|
output
| 1
| 105,564
| 0
| 211,129
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
|
instruction
| 0
| 105,788
| 0
| 211,576
|
Tags: combinatorics, greedy, implementation, math
Correct Solution:
```
MOD = int(1e9)+7
def solve():
s = input()
r=c=0
for i in range(len(s)-1, -1, -1):
if s[i] == 'b': c+=1
else:
r+=c
c= (c*2)%MOD
return r%MOD
print(solve())
```
|
output
| 1
| 105,788
| 0
| 211,577
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
|
instruction
| 0
| 105,789
| 0
| 211,578
|
Tags: combinatorics, greedy, implementation, math
Correct Solution:
```
MOD = 10**9 + 7
s = input()
bcount, count = 0, 0
for c in reversed(s):
if c == 'b':
bcount += 1
else:
count += bcount
bcount *= 2
if bcount > 2**62:
bcount %= MOD
print(count % MOD)
```
|
output
| 1
| 105,789
| 0
| 211,579
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
|
instruction
| 0
| 105,790
| 0
| 211,580
|
Tags: combinatorics, greedy, implementation, math
Correct Solution:
```
x = input()
str0 = x
x=x[::-1]
# print(x)
mod = 10**9 + 7
ans = 0
sum0 = 0
for i in x:
if i == 'b':
ans += 1
if i == 'a':
sum0 += ans
sum0 = sum0 % mod
ans *=2
ans = ans % mod
print(sum0)
```
|
output
| 1
| 105,790
| 0
| 211,581
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
|
instruction
| 0
| 105,791
| 0
| 211,582
|
Tags: combinatorics, greedy, implementation, math
Correct Solution:
```
s = input()
l = [x for x in range(len(s)) if s[x] == 'a']
# print(l)
i, step = 0, 0
while l:
step = (len(s) + 2*step - l.pop() - 1 - i) % (10 ** 9 + 7)
i += 1
print(step)
```
|
output
| 1
| 105,791
| 0
| 211,583
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
|
instruction
| 0
| 105,792
| 0
| 211,584
|
Tags: combinatorics, greedy, implementation, math
Correct Solution:
```
def main():
s=input()
r=0
c=0
m=(10**9)+7
for i in range(len(s)-1,-1,-1):
if s[i]=='b':
c=(c%m)+1
else:
r=((r%m)+(c%m))%m
c=(c*2)%m
print(r)
if __name__=='__main__':
main()
```
|
output
| 1
| 105,792
| 0
| 211,585
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
|
instruction
| 0
| 105,793
| 0
| 211,586
|
Tags: combinatorics, greedy, implementation, math
Correct Solution:
```
mod = 1000000007
qtdb = 0
jogadas = 0
s = input()
for i in range(len(s)-1, -1, -1):
if(s[i] == 'b'):
qtdb = (qtdb + 1) % mod
else:
jogadas = (jogadas + qtdb) % mod
qtdb = (2 * qtdb) % mod
print(jogadas)
```
|
output
| 1
| 105,793
| 0
| 211,587
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
|
instruction
| 0
| 105,794
| 0
| 211,588
|
Tags: combinatorics, greedy, implementation, math
Correct Solution:
```
s = input()
ans, cnt = 0, 0
for i in range(1, len(s) + 1):
if s[-i] == 'a':
ans = (2 * ans + i - 1 - cnt) % (10 ** 9 + 7)
cnt += 1
print(ans)
```
|
output
| 1
| 105,794
| 0
| 211,589
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
|
instruction
| 0
| 105,795
| 0
| 211,590
|
Tags: combinatorics, greedy, implementation, math
Correct Solution:
```
s = input()[::-1]
a, b = 0, 0
mod = 10 ** 9 + 7
for i in s:
if i == 'b':
b += 1
else:
a += b
a %= mod
b <<= 1
b %= mod
print(a)
```
|
output
| 1
| 105,795
| 0
| 211,591
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
Submitted Solution:
```
inp = input()
cnt_b = 0
ans = 0
mod = 1000000000 + 7
for i in inp[::-1]:
if i == 'b':
cnt_b += 1
else:
ans = (ans + cnt_b) % mod
cnt_b = (cnt_b * 2) % mod
print(ans)
```
|
instruction
| 0
| 105,796
| 0
| 211,592
|
Yes
|
output
| 1
| 105,796
| 0
| 211,593
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
Submitted Solution:
```
s = input()
a = 0
m = 10 ** 9 + 7
t = 0
for i in range(len(s)):
if s[~i] == 'a':
a = (a + t) % m
t = (t * 2) % m
else:
t += 1
print(a)
```
|
instruction
| 0
| 105,797
| 0
| 211,594
|
Yes
|
output
| 1
| 105,797
| 0
| 211,595
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
Submitted Solution:
```
s = input().rstrip()
mod = 10 ** 9 + 7
a_count = 0
rs = 0
cur_mod = (2 ** a_count) % mod
for c in s:
if c == 'a':
a_count += 1
cur_mod = (cur_mod * 2) % mod
elif c == 'b':
rs = (rs + cur_mod - 1) % mod
print(rs)
```
|
instruction
| 0
| 105,798
| 0
| 211,596
|
Yes
|
output
| 1
| 105,798
| 0
| 211,597
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
Submitted Solution:
```
MOD = 1e9 + 7
s = input()
ans = 0
b = 0
for c in s[::-1]:
if c == 'a':
ans = (ans + b) % MOD
b = (2 * b) % MOD
else:
b = (b + 1) % MOD
print(int(ans))
```
|
instruction
| 0
| 105,799
| 0
| 211,598
|
Yes
|
output
| 1
| 105,799
| 0
| 211,599
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
Submitted Solution:
```
n=input()
i=0
while "ab" in n:
n=n.replace("ab","bba")
i+=1
print(i%(10**6+7))
```
|
instruction
| 0
| 105,800
| 0
| 211,600
|
No
|
output
| 1
| 105,800
| 0
| 211,601
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
Submitted Solution:
```
n = input()
r = 0
c = 0
t = []
for i in range(len(n)):
if n[i]=="a":
c+=1
else:
t.append(c)
c = 0
for i in t:
if i==1:
r+=1
elif i>1:
r+=i+1
print(r%(10**9+7))
```
|
instruction
| 0
| 105,801
| 0
| 211,602
|
No
|
output
| 1
| 105,801
| 0
| 211,603
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
Submitted Solution:
```
s=input()
ans=0
j=0
for i in range(0,len(s)):
if i>=j:
cnt1=0
while i<len(s) and s[i]=='a':
cnt1=cnt1+1
i=i+1
cnt2=0
while i<len(s) and s[i]=='b':
cnt2=cnt2+1
i=i+1
j=i;
if cnt1>0 and cnt2>0:
ans=ans + cnt2
bb= cnt2*2
cnt1=cnt1-1
ans = ans + bb*(pow(2,cnt1,1000000007)-1)
#print(ans)
print(ans%1000000007)
```
|
instruction
| 0
| 105,802
| 0
| 211,604
|
No
|
output
| 1
| 105,802
| 0
| 211,605
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" β "bba".
The second example: "aab" β "abba" β "bbaba" β "bbbbaa".
Submitted Solution:
```
s = input()
r = 0
na = 0
for i in range(len(s)):
if s[i] == 'a':
na += 1
else:
r += (2**na - 1)%1000000007
print(r)
```
|
instruction
| 0
| 105,803
| 0
| 211,606
|
No
|
output
| 1
| 105,803
| 0
| 211,607
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to take the entrance examination of Kyoto University tomorrow and have decided to memorize a set of strings S that is expected to appear in the examination. Since it is really tough to memorize S as it is, you have decided to memorize a single string T that efficiently contains all the strings in S.
You have already confirmed the following conditions for S and T are satisfied.
* Every string in S is a consecutive subsequence(1) in T .
* For every pair of strings x, y (x \neq y) in S , x is not a subsequence(2) of y.
Note that (1) is ''consecutive subsequence'', while (2) is ''subsequence''.
The next day, you opened the problem booklet at the examination and found that you can get a full score only if you remember S. However, You have forgot how to reconstruct S from T . Since all you remember is that T satisfies the above conditions, first you have decided to find the maximum possible number of elements in S .
Constraints
* 1 \leq |T| \leq 10^5
* T consists of only lowercase letters.
Partial points
* 30 points will be awarded for passing the test set satisfying the condition: 1 \leq |T| \leq 50 .
* Another 30 points will be awarded for passing the test set satisfying the condition: 1 \leq |T| \leq 10^3.
Input
The input is given from Standard Input in the following format:
T
The input only consists of T on one line.
Output
Print the maximum number of elements inS .
Examples
Input
abcabc
Output
3
Input
abracadabra
Output
7
Input
abcbabbcabbc
Output
8
Input
bbcacbcbcabbabacccbbcacbaaababbacabaaccbccabcaabba
Output
44
Submitted Solution:
```
from collections import Counter
T = input()
M = 0
for i in range(1, 1001):
c = Counter(T[j:j+i] for j in range(0, len(T)-i+1))
M = max(len(c), M)
print(M)
```
|
instruction
| 0
| 106,007
| 0
| 212,014
|
No
|
output
| 1
| 106,007
| 0
| 212,015
|
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