message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N and another string T of length M. These strings consist of lowercase English letters.
A string X is called a good string when the following conditions are all met:
* Let L be the length of X. L is divisible by both N and M.
* Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.
* Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.
Determine if there exists a good string. If it exists, find the length of the shortest such string.
Constraints
* 1 \leq N,M \leq 10^5
* S and T consist of lowercase English letters.
* |S|=N
* |T|=M
Input
Input is given from Standard Input in the following format:
N M
S
T
Output
If a good string does not exist, print `-1`; if it exists, print the length of the shortest such string.
Examples
Input
3 2
acp
ae
Output
6
Input
6 3
abcdef
abc
Output
-1
Input
15 9
dnsusrayukuaiia
dujrunuma
Output
45
Submitted Solution:
```
from fractions import gcd
n, m = map(int, input().split())
s = list(input())
t = list(input())
i = 1
ans = n * m // gcd(n, m)
g = gcd(n, m)
for i in range(g):
if s[i * n // g] != t[i * m // g]:
ans = -1
break
print(ans)
``` | instruction | 0 | 94,219 | 0 | 188,438 |
Yes | output | 1 | 94,219 | 0 | 188,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N and another string T of length M. These strings consist of lowercase English letters.
A string X is called a good string when the following conditions are all met:
* Let L be the length of X. L is divisible by both N and M.
* Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.
* Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.
Determine if there exists a good string. If it exists, find the length of the shortest such string.
Constraints
* 1 \leq N,M \leq 10^5
* S and T consist of lowercase English letters.
* |S|=N
* |T|=M
Input
Input is given from Standard Input in the following format:
N M
S
T
Output
If a good string does not exist, print `-1`; if it exists, print the length of the shortest such string.
Examples
Input
3 2
acp
ae
Output
6
Input
6 3
abcdef
abc
Output
-1
Input
15 9
dnsusrayukuaiia
dujrunuma
Output
45
Submitted Solution:
```
n,m=map(int,input().split());g=m-max([a*n%m for a in range(m)]);print(-1if input()[::n//g]!=input()[::m//g]else n*m//g)
``` | instruction | 0 | 94,220 | 0 | 188,440 |
Yes | output | 1 | 94,220 | 0 | 188,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N and another string T of length M. These strings consist of lowercase English letters.
A string X is called a good string when the following conditions are all met:
* Let L be the length of X. L is divisible by both N and M.
* Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.
* Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.
Determine if there exists a good string. If it exists, find the length of the shortest such string.
Constraints
* 1 \leq N,M \leq 10^5
* S and T consist of lowercase English letters.
* |S|=N
* |T|=M
Input
Input is given from Standard Input in the following format:
N M
S
T
Output
If a good string does not exist, print `-1`; if it exists, print the length of the shortest such string.
Examples
Input
3 2
acp
ae
Output
6
Input
6 3
abcdef
abc
Output
-1
Input
15 9
dnsusrayukuaiia
dujrunuma
Output
45
Submitted Solution:
```
import fractions
n, m = map(int,input().split())
s = input()
t = input()
q = fractions.gcd(n, m)
l = n * m // q #lcm(n, m)
for i in range(0, q):
if s[i * n//q] != t[i * m//q]:
print('-1')
exit()
print(l)
``` | instruction | 0 | 94,221 | 0 | 188,442 |
Yes | output | 1 | 94,221 | 0 | 188,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N and another string T of length M. These strings consist of lowercase English letters.
A string X is called a good string when the following conditions are all met:
* Let L be the length of X. L is divisible by both N and M.
* Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.
* Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.
Determine if there exists a good string. If it exists, find the length of the shortest such string.
Constraints
* 1 \leq N,M \leq 10^5
* S and T consist of lowercase English letters.
* |S|=N
* |T|=M
Input
Input is given from Standard Input in the following format:
N M
S
T
Output
If a good string does not exist, print `-1`; if it exists, print the length of the shortest such string.
Examples
Input
3 2
acp
ae
Output
6
Input
6 3
abcdef
abc
Output
-1
Input
15 9
dnsusrayukuaiia
dujrunuma
Output
45
Submitted Solution:
```
def gcd(a, b):
while b:
a, b = b, a % b
return a
N,M = map(int,input().split())
s = input()
t = input()
g = gcd(N,M)
l = N*M//g
n = N//g
m = M//g
ans = l
for i in range(g):
if s[i*n]!=t[i*m]:
ans = -1
break
print(ans)
``` | instruction | 0 | 94,222 | 0 | 188,444 |
Yes | output | 1 | 94,222 | 0 | 188,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N and another string T of length M. These strings consist of lowercase English letters.
A string X is called a good string when the following conditions are all met:
* Let L be the length of X. L is divisible by both N and M.
* Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.
* Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.
Determine if there exists a good string. If it exists, find the length of the shortest such string.
Constraints
* 1 \leq N,M \leq 10^5
* S and T consist of lowercase English letters.
* |S|=N
* |T|=M
Input
Input is given from Standard Input in the following format:
N M
S
T
Output
If a good string does not exist, print `-1`; if it exists, print the length of the shortest such string.
Examples
Input
3 2
acp
ae
Output
6
Input
6 3
abcdef
abc
Output
-1
Input
15 9
dnsusrayukuaiia
dujrunuma
Output
45
Submitted Solution:
```
# A
def gcd(a, b):
while b:
a, b = b, a % b
return a
N, M = list(map(int, input().split()))
S = input()
T = input()
re = -1
# 1文字目が必ず一緒であること
if S[0] == T[0]:
if N == M:
if S == T:
re = N
else:
min_num = (N * M) // gcd(N, M) # 最小公倍数を求める（基本はこれが答え）
re = min_num
# 重なる部分が同じ文字かを調べる
s_d = min_num // N
t_d = min_num // M
# 最小公倍数みる
st_min = (s_d * t_d) // gcd(s_d, t_d)
st = st_min
i = 1
while st // s_d < N and st // t_d < M:
if S[st // s_d] != T[st // t_d]:
re = -1
break
else:
i += 1
st = st * i
print(re)
``` | instruction | 0 | 94,223 | 0 | 188,446 |
No | output | 1 | 94,223 | 0 | 188,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N and another string T of length M. These strings consist of lowercase English letters.
A string X is called a good string when the following conditions are all met:
* Let L be the length of X. L is divisible by both N and M.
* Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.
* Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.
Determine if there exists a good string. If it exists, find the length of the shortest such string.
Constraints
* 1 \leq N,M \leq 10^5
* S and T consist of lowercase English letters.
* |S|=N
* |T|=M
Input
Input is given from Standard Input in the following format:
N M
S
T
Output
If a good string does not exist, print `-1`; if it exists, print the length of the shortest such string.
Examples
Input
3 2
acp
ae
Output
6
Input
6 3
abcdef
abc
Output
-1
Input
15 9
dnsusrayukuaiia
dujrunuma
Output
45
Submitted Solution:
```
n,m=map(int,input().split())
s=input()
t=input()
def gcd(x,y):
r=x%y
return gcd(y,r) if r else y
#最小公倍数
def lcm(x, y):
return (x * y) // gcd(x, y)
L=lcm(n,m)
if s[0]==t[0]:
X=[""]*(L+1)
else:
print(-1)
exit()
for i in range(1,n):
X[int(i*L/n)+1]=s[i]
for i in range(1,m):
if X[int(i*L/m)+1] !="":
if X[int(i*L/m)+1] !=t[i]:
print(-1)
exit()
print(L)
``` | instruction | 0 | 94,224 | 0 | 188,448 |
No | output | 1 | 94,224 | 0 | 188,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N and another string T of length M. These strings consist of lowercase English letters.
A string X is called a good string when the following conditions are all met:
* Let L be the length of X. L is divisible by both N and M.
* Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.
* Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.
Determine if there exists a good string. If it exists, find the length of the shortest such string.
Constraints
* 1 \leq N,M \leq 10^5
* S and T consist of lowercase English letters.
* |S|=N
* |T|=M
Input
Input is given from Standard Input in the following format:
N M
S
T
Output
If a good string does not exist, print `-1`; if it exists, print the length of the shortest such string.
Examples
Input
3 2
acp
ae
Output
6
Input
6 3
abcdef
abc
Output
-1
Input
15 9
dnsusrayukuaiia
dujrunuma
Output
45
Submitted Solution:
```
n, m = list(map(int, input().split()))
s = input()
t = input()
flag = False
def gcd(a, b):
while b:
a, b = b, a%b
return a
l = int(n*m / gcd(n, m))
if max(n, m) % min(n, m) != 0:
print(l if s[0] == t[0] else -1)
exit()
if len(s) < len(t):
tmp = s
s = t
t = tmp
if len(s) == len(t):
if s == t:
print(len(s))
exit()
else:
print(-1)
exit()
for j in [1, 2, 3, 5]:
for i in range(1, m):
if s[j*i*(n//m)] != t[i]:
print(-1)
exit()
print(l)
``` | instruction | 0 | 94,225 | 0 | 188,450 |
No | output | 1 | 94,225 | 0 | 188,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N and another string T of length M. These strings consist of lowercase English letters.
A string X is called a good string when the following conditions are all met:
* Let L be the length of X. L is divisible by both N and M.
* Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.
* Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.
Determine if there exists a good string. If it exists, find the length of the shortest such string.
Constraints
* 1 \leq N,M \leq 10^5
* S and T consist of lowercase English letters.
* |S|=N
* |T|=M
Input
Input is given from Standard Input in the following format:
N M
S
T
Output
If a good string does not exist, print `-1`; if it exists, print the length of the shortest such string.
Examples
Input
3 2
acp
ae
Output
6
Input
6 3
abcdef
abc
Output
-1
Input
15 9
dnsusrayukuaiia
dujrunuma
Output
45
Submitted Solution:
```
from fractions import gcd
N.M=(int(i) for i in input.split())
S=input()
T=input()
gcd_NM=gcd(N,M)
lcm_NM=(N*M//gcd_NM)
A=True
for i in range(gcd_NM):
if S[(N//gcd_NM)*i]!=T[(M//gcd_NM)*i]:
A=False
break
if A:
print(lcm_NM)
else:
print(-1)
``` | instruction | 0 | 94,226 | 0 | 188,452 |
No | output | 1 | 94,226 | 0 | 188,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer value x and a string s consisting of digits from 1 to 9 inclusive.
A substring of a string is a contiguous subsequence of that string.
Let f(l, r) be the sum of digits of a substring s[l..r].
Let's call substring s[l_1..r_1] x-prime if
* f(l_1, r_1) = x;
* there are no values l_2, r_2 such that
* l_1 ≤ l_2 ≤ r_2 ≤ r_1;
* f(l_2, r_2) ≠ x;
* x is divisible by f(l_2, r_2).
You are allowed to erase some characters from the string. If you erase a character, the two resulting parts of the string are concatenated without changing their order.
What is the minimum number of characters you should erase from the string so that there are no x-prime substrings in it? If there are no x-prime substrings in the given string s, then print 0.
Input
The first line contains a string s (1 ≤ |s| ≤ 1000). s contains only digits from 1 to 9 inclusive.
The second line contains an integer x (1 ≤ x ≤ 20).
Output
Print a single integer — the minimum number of characters you should erase from the string so that there are no x-prime substrings in it. If there are no x-prime substrings in the given string s, then print 0.
Examples
Input
116285317
8
Output
2
Input
314159265359
1
Output
2
Input
13
13
Output
0
Input
3434343434
7
Output
5
Note
In the first example there are two 8-prime substrings "8" and "53". You can erase these characters to get rid of both: "116285317". The resulting string "1162317" contains no 8-prime substrings. Removing these characters is also a valid answer: "116285317".
In the second example you just have to erase both ones.
In the third example there are no 13-prime substrings. There are no substrings with the sum of digits equal to 13 at all.
In the fourth example you can have neither "34", nor "43" in a string. Thus, you have to erase either all threes or all fours. There are 5 of each of them, so it doesn't matter which.
Submitted Solution:
```
s=input()
x=input()
c=0
if s.count(x)>0 :
c+=s.count(x)
print(c)
``` | instruction | 0 | 94,575 | 0 | 189,150 |
No | output | 1 | 94,575 | 0 | 189,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer value x and a string s consisting of digits from 1 to 9 inclusive.
A substring of a string is a contiguous subsequence of that string.
Let f(l, r) be the sum of digits of a substring s[l..r].
Let's call substring s[l_1..r_1] x-prime if
* f(l_1, r_1) = x;
* there are no values l_2, r_2 such that
* l_1 ≤ l_2 ≤ r_2 ≤ r_1;
* f(l_2, r_2) ≠ x;
* x is divisible by f(l_2, r_2).
You are allowed to erase some characters from the string. If you erase a character, the two resulting parts of the string are concatenated without changing their order.
What is the minimum number of characters you should erase from the string so that there are no x-prime substrings in it? If there are no x-prime substrings in the given string s, then print 0.
Input
The first line contains a string s (1 ≤ |s| ≤ 1000). s contains only digits from 1 to 9 inclusive.
The second line contains an integer x (1 ≤ x ≤ 20).
Output
Print a single integer — the minimum number of characters you should erase from the string so that there are no x-prime substrings in it. If there are no x-prime substrings in the given string s, then print 0.
Examples
Input
116285317
8
Output
2
Input
314159265359
1
Output
2
Input
13
13
Output
0
Input
3434343434
7
Output
5
Note
In the first example there are two 8-prime substrings "8" and "53". You can erase these characters to get rid of both: "116285317". The resulting string "1162317" contains no 8-prime substrings. Removing these characters is also a valid answer: "116285317".
In the second example you just have to erase both ones.
In the third example there are no 13-prime substrings. There are no substrings with the sum of digits equal to 13 at all.
In the fourth example you can have neither "34", nor "43" in a string. Thus, you have to erase either all threes or all fours. There are 5 of each of them, so it doesn't matter which.
Submitted Solution:
```
s=input()
l=list(s)
x=eval(input())
z=0
if str(x)==s:
print(0)
else:
if str(x) in s:
z=l.count(str(x))
else:
for i in range(len(l)-1,1):
for j in range(i+1,i+2):
if (int(l[i])+int(l[j]))==x:
z=z+1
print(z)
``` | instruction | 0 | 94,576 | 0 | 189,152 |
No | output | 1 | 94,576 | 0 | 189,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer value x and a string s consisting of digits from 1 to 9 inclusive.
A substring of a string is a contiguous subsequence of that string.
Let f(l, r) be the sum of digits of a substring s[l..r].
Let's call substring s[l_1..r_1] x-prime if
* f(l_1, r_1) = x;
* there are no values l_2, r_2 such that
* l_1 ≤ l_2 ≤ r_2 ≤ r_1;
* f(l_2, r_2) ≠ x;
* x is divisible by f(l_2, r_2).
You are allowed to erase some characters from the string. If you erase a character, the two resulting parts of the string are concatenated without changing their order.
What is the minimum number of characters you should erase from the string so that there are no x-prime substrings in it? If there are no x-prime substrings in the given string s, then print 0.
Input
The first line contains a string s (1 ≤ |s| ≤ 1000). s contains only digits from 1 to 9 inclusive.
The second line contains an integer x (1 ≤ x ≤ 20).
Output
Print a single integer — the minimum number of characters you should erase from the string so that there are no x-prime substrings in it. If there are no x-prime substrings in the given string s, then print 0.
Examples
Input
116285317
8
Output
2
Input
314159265359
1
Output
2
Input
13
13
Output
0
Input
3434343434
7
Output
5
Note
In the first example there are two 8-prime substrings "8" and "53". You can erase these characters to get rid of both: "116285317". The resulting string "1162317" contains no 8-prime substrings. Removing these characters is also a valid answer: "116285317".
In the second example you just have to erase both ones.
In the third example there are no 13-prime substrings. There are no substrings with the sum of digits equal to 13 at all.
In the fourth example you can have neither "34", nor "43" in a string. Thus, you have to erase either all threes or all fours. There are 5 of each of them, so it doesn't matter which.
Submitted Solution:
```
s=input()
l=list(s)
x=eval(input())
z=0
if str(x)==s:
print(0)
else:
if str(x) in s:
z=l.count(str(x))
else:
for i in range(len(l)-1,l):
for j in range(i+1,i+2):
if (int(l[i])+int(l[j]))==x:
z=z+1
print(z)
``` | instruction | 0 | 94,577 | 0 | 189,154 |
No | output | 1 | 94,577 | 0 | 189,155 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb | instruction | 0 | 94,663 | 0 | 189,326 |
Tags: constructive algorithms, greedy, math, matrices
Correct Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
BA = h[1][2] - BB
AB = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB
Now34 = AB + NowB + BB
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB - ABB)] + ["b" for j in range(BBB)] + ["a" for j in range(BB - BBB)] + ["b" for j in range(BAB)] + ["a" for j in range(BA - BAB)] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
``` | output | 1 | 94,663 | 0 | 189,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb | instruction | 0 | 94,664 | 0 | 189,328 |
Tags: constructive algorithms, greedy, math, matrices
Correct Solution:
```
def get_input():
a, b, d = map(int, input().split())
c, e = map(int, input().split())
f = int(input())
return [a, b, c, d, e, f]
def check_condition(a, b, c, d, e, f):
condition1 = (a + b + c) % 2 == 0
condition2 = (d + e + a) % 2 == 0
condition3 = (e + f + c) % 2 == 0
condition4 = (d + f + b) % 2 == 0
condition = condition1 and condition2 and condition3 and condition4
return condition
def find_t(a, b, c, d, e, f):
t_min1 = round((d + f - a - c) / 2)
t_min2 = round((e + f - a - b) / 2)
t_min3 = round((d + e - b - c) / 2)
t_min4 = 0
t_min = max(t_min1, t_min2, t_min3, t_min4)
t_max1 = round((d + e - a) / 2)
t_max2 = round((e + f - c) / 2)
t_max3 = round((d + f - b) / 2)
t_max = min(t_max1, t_max2, t_max3)
if t_min <= t_max:
return t_min
else:
return -1
def find_all(a, b, c, d, e, f, t):
x1 = round((a + c - d - f) / 2 + t)
x2 = round((d + f - b) / 2 - t)
y1 = round((a + b - e - f) / 2 + t)
y2 = round((e + f - c) / 2 - t)
z1 = round((b + c - d - e) / 2 + t)
z2 = round((d + e - a) / 2 - t)
return [x1, x2, y1, y2, z1, z2]
def generate_string(x1, x2, y1, y2, z1, z2, t):
n = x1 + x2 + y1 + y2 + z1 + z2 + t
s1 = ''.join(['a'] * n)
s2 = ''.join(['a'] * (z1 + z2 + t)) + ''.join(['b'] * (x1 + x2 + y1 + y2))
s3 = ''.join(['a'] * t) + ''.join(['b'] * (y1 + y2 + z1 + z2)) + ''.join(['a'] * (x1 + x2))
s4 = ''.join(['b'] * (t + z2)) + ''.join(['a'] * (z1 + y2)) + ''.join(['b'] * (y1 + x2)) + ''.join(['a'] * x1)
return [s1, s2, s3, s4]
def __main__():
fail_output = "-1"
a, b, c, d, e, f = map(int, get_input())
if not(check_condition(a, b, c, d, e, f)):
print(fail_output)
return False
t = find_t(a, b, c, d, e, f)
if t < 0:
print(fail_output)
return False
x1, x2, y1, y2, z1, z2 = map(int, find_all(a, b, c, d, e, f, t))
s1, s2, s3, s4 = map(str, generate_string(x1, x2, y1, y2, z1, z2, t))
print(str(x1 + x2 + y1 + y2 + z1 + z2 + t) + '\n')
print(s1 + '\n')
print(s2 + '\n')
print(s3 + '\n')
print(s4 + '\n')
__main__()
# Made By Mostafa_Khaled
``` | output | 1 | 94,664 | 0 | 189,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB;
Now34 = AB + NowB + BB;
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += Now
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB)[ABB:]] + ["b" for j in range(BBB)] + ["a" for j in range(BB)[BBB:]] + ["b" for j in range(BAB)] + ["a" for j in range(BA)[BAB:]] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
``` | instruction | 0 | 94,665 | 0 | 189,330 |
No | output | 1 | 94,665 | 0 | 189,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB
Now34 = AB + NowB + BB
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB - ABB)] + ["b" for j in range(BBB)] + ["a" for j in range(BB - BBB)] + ["b" for j in range(BAB)] + ["a" for j in range(BA - BAB)] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
``` | instruction | 0 | 94,666 | 0 | 189,332 |
No | output | 1 | 94,666 | 0 | 189,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB;
Now34 = AB + NowB + BB;
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB)[ABB:]] + ["b" for j in range(BBB)] + ["a" for j in range(BB)[BBB:]] + ["b" for j in range(BAB)] + ["a" for j in range(BA)[BAB:]] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
``` | instruction | 0 | 94,667 | 0 | 189,334 |
No | output | 1 | 94,667 | 0 | 189,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | instruction | 0 | 94,959 | 0 | 189,918 |
Tags: dp, implementation
Correct Solution:
```
from sys import stdin, stdout
n = int(stdin.readline())
s = stdin.readline().strip()
cnt = [[0, 0] for i in range(n)]
cnt[0][1] = int(s[0])
cnt[0][0] = 1 - int(s[0])
for i in range(1, n):
cnt[i][1] += cnt[i - 1][1] + int(s[i])
cnt[i][0] = i + 1 - cnt[i][1]
cnt.append([0, 0])
ans = 0
for i in range(n):
value = int(s[i])
current = i + ans
while current < n:
if max(cnt[current][1] - cnt[i - 1][1], cnt[current][0] - cnt[i - 1][0]) != min(cnt[current][1] - cnt[i - 1][1], cnt[current][0] - cnt[i - 1][0]):
current += (max(cnt[current][1] - cnt[i - 1][1], cnt[current][0] - cnt[i - 1][0]) - min(cnt[current][1] - cnt[i - 1][1], cnt[current][0] - cnt[i - 1][0]))
else:
ans = max(ans, current - i + 1)
current += 1
stdout.write(str(ans))
``` | output | 1 | 94,959 | 0 | 189,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | instruction | 0 | 94,960 | 0 | 189,920 |
Tags: dp, implementation
Correct Solution:
```
n = int(input()); res = []
s = input(); heights = {0 : [-1]}; current = 0
for i in range(n):
current += (-1)**int(s[i])
if current not in heights:
heights[current] = [i]
else: heights[current].append(i)
for i in heights.values():
res.append(max(i) - min(i))
print(max(res))
``` | output | 1 | 94,960 | 0 | 189,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | instruction | 0 | 94,961 | 0 | 189,922 |
Tags: dp, implementation
Correct Solution:
```
"""
Code of Ayush Tiwari
Codeforces: servermonk
Codechef: ayush572000
"""
def solution():
n=int(input())
s=input()
cnt1=[0]*(n+1)
cnt0=[0]*(n+1)
balance=[0]
for i in range(n):
if s[i]=='1':
cnt1[i+1]+=cnt1[i]+1
cnt0[i+1]=cnt0[i]
else:
cnt1[i+1]=cnt1[i]
cnt0[i+1]+=cnt0[i]+1
x=cnt1[i+1]-cnt0[i+1]
balance.append(x)
d={}
ans=0
for i in range(n+1):
if d.get(balance[i],-1)==-1:
d[balance[i]]=i
else:
ans=max(ans,i-d[balance[i]])
print(ans)
solution()
``` | output | 1 | 94,961 | 0 | 189,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | instruction | 0 | 94,962 | 0 | 189,924 |
Tags: dp, implementation
Correct Solution:
```
import traceback
import math
from collections import defaultdict
from functools import lru_cache
def main():
N = int(input())
nums = input()
dp = [0] * N
tot = 0
for i in range(N):
tot += 1 if nums[i] == '1' else -1
dp[i] = tot
# if tot == 0:
# return N
loc = {}
loc[0] = -1
ans = 0
for i in range(N):
loc.setdefault(dp[i], i)
ans = max(ans, i - loc[dp[i]])
return ans
try:
ans = main()
print(ans)
except Exception as e:
# print(e)
traceback.print_exc()
``` | output | 1 | 94,962 | 0 | 189,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | instruction | 0 | 94,963 | 0 | 189,926 |
Tags: dp, implementation
Correct Solution:
```
from itertools import accumulate
n=int(input())
m={0:-1}
a=0
for i,b in enumerate(accumulate(map(lambda c:2*int(c)-1,input()))):
x=m.get(b)
if x is None:
m[b]=i
else:
a=max(a,i-x)
print(a)
``` | output | 1 | 94,963 | 0 | 189,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | instruction | 0 | 94,964 | 0 | 189,928 |
Tags: dp, implementation
Correct Solution:
```
n=int(input())
l=[int(i) for i in input()]
d={}
for i in range(n):
if l[i]==0:
l[i]=-1
else:
l[i]=1
from collections import defaultdict
d=defaultdict(int)
#d[0]=1
sm=0
ans=0
maxi=0
d={}
d[0]=-1
for i in range(n):
sm+=l[i]
# print(sm)
if sm in d:
maxi=max(maxi,i-d[sm])
if sm not in d:
d[sm]=i
print(maxi)
exit()
ans=0
for i in d:
c=d[i]
ans+=c*(c-1)//2
print(ans)
``` | output | 1 | 94,964 | 0 | 189,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | instruction | 0 | 94,965 | 0 | 189,930 |
Tags: dp, implementation
Correct Solution:
```
import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = int(input())
s = [int(x) for x in list(input())]
l = []
z = 0
o = 0
d = {}
d[0] = -1
for i in range(n):
if(s[i] == 0):
z+=1
else:
o+=1
l.append(z-o)
ans = 0
for i in range(n):
if(l[i] not in d):
d[l[i]] = i
else:
ans = max(ans, i-d[l[i]])
# print(l)
print(ans)
``` | output | 1 | 94,965 | 0 | 189,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | instruction | 0 | 94,966 | 0 | 189,932 |
Tags: dp, implementation
Correct Solution:
```
n = int(input())
zo = input()
dr = [0]
ml = 0
mini = {0: 0}
for i in range(n):
if zo[i]=='0':
dr.append(dr[i]-1)
else:
dr.append(dr[i]+1)
if not(dr[i+1] in mini):
mini[dr[i+1]] = i+1
else:
ml = max(ml, i+1 - mini[dr[i+1]])
print(ml)
``` | output | 1 | 94,966 | 0 | 189,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
Submitted Solution:
```
n = int(input())
l = [-2] * (n * 2 + 1)
l[n], r = -1, 0
for i, c in enumerate(input()):
n += {'0': -1, '1': 1}[c]
if l[n] == -2:
l[n] = i
else:
i -= l[n]
if r < i:
r = i
print(r)
``` | instruction | 0 | 94,967 | 0 | 189,934 |
Yes | output | 1 | 94,967 | 0 | 189,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
Submitted Solution:
```
from collections import defaultdict
n = int(input())
s = [int(i) for i in input().strip()]
balance = 0
ans = 0
seen = defaultdict(lambda: -2)
seen[0] = -1
for j, i in enumerate(s):
if i:
balance += 1
else:
balance -= 1
if seen[balance] == -2:
seen[balance] = j
ans = max(ans, j - seen[balance])
print(ans)
``` | instruction | 0 | 94,968 | 0 | 189,936 |
Yes | output | 1 | 94,968 | 0 | 189,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
Submitted Solution:
```
n=int(input())
T=input()
d={0:-1}
s=0
mix=0
for i in range(n):
if(T[i]=='0'):
s+=-1
else :
s+=1
if s not in d:
d[s]=i
else:
if(i-d[s])> mix:
mix=i-d[s]
print(mix)
``` | instruction | 0 | 94,969 | 0 | 189,938 |
Yes | output | 1 | 94,969 | 0 | 189,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
Submitted Solution:
```
n = input()
s = input()
res = 0
ans = 0
dp = {}
dp[0] = -1
for i in range(int(n)):
if s[i] == '1':
res += 1
else:
res -= 1
if res not in dp:
dp[res] = i
else:
ans = max(ans, i - dp[res])
print(ans)
``` | instruction | 0 | 94,970 | 0 | 189,940 |
Yes | output | 1 | 94,970 | 0 | 189,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
Submitted Solution:
```
n = int(input())
s = str(input())
count0 = s.count('0')
count1 = s.count('1')
if count0 == 0 or count1 == 0:
print('0')
elif count0 == count1:
print(n)
elif count0>count1:
print(2*count1)
else:
print(2*count0)
``` | instruction | 0 | 94,971 | 0 | 189,942 |
No | output | 1 | 94,971 | 0 | 189,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
Submitted Solution:
```
##a = list(map(int, input().split()))
##print(' '.join(map(str, res)))
n = int(input())
s = input()
ps = [0 for i in range(n+1)]
for i in range(0, n):
ps[i+1] = ps[i]+(0 if s[i] == '0' else 1)
l = 0
r = n//2
while l < r:
x = (l+r+1)//2
flag = False
for i in range(n-2*x+1):
if (ps[i+2*x]-ps[i]) == x:
flag = True
break
if flag:
l = x
else:
r = x-1
print(2*l)
``` | instruction | 0 | 94,972 | 0 | 189,944 |
No | output | 1 | 94,972 | 0 | 189,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
Submitted Solution:
```
def ch(s):
if s.count('0')==s.count('1'):
return True
else:
return False
n=int(input())
s=input()
c=0
for i in range(n):
for j in range(i+1,n):
x=s[i:j]
if ch(x):
if len(x)>c:
c=len(x)
print(c)
``` | instruction | 0 | 94,973 | 0 | 189,946 |
No | output | 1 | 94,973 | 0 | 189,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of s.
Input
The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.
The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.
Output
If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.
Examples
Input
8
11010111
Output
4
Input
3
111
Output
0
Note
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring.
Submitted Solution:
```
import sys
placeholder = sys.stdin.readline().strip().split(" ")
string = sys.stdin.readline().strip().split(" ")
"""
string = "111"
"""
num_zeros = 0
num_ones = 0
for x in string:
if(x == '0'):
num_zeros+=1
elif(x == '1'):
num_ones+=1
print(2 * min(num_zeros, num_ones))
``` | instruction | 0 | 94,974 | 0 | 189,948 |
No | output | 1 | 94,974 | 0 | 189,949 |
Provide a correct Python 3 solution for this coding contest problem.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob | instruction | 0 | 95,115 | 0 | 190,230 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
class Node:
def __init__(self, sigma, depth):
self.end = False
self.child = [None] * sigma
self.depth = depth
def __setitem__(self, i, x):
self.child[i] = x
def __getitem__(self, i):
return self.child[i]
sigma = 2
root = Node(sigma, 0)
def add_trie(S):
vn = root
for cs in S:
if vn[cs] is None:
vn[cs] = Node(sigma, vn.depth + 1)
vn = vn[cs]
vn.end = True
ans = 0
N, L = map(int, readline().split())
for _ in range(N):
S = list(map(int, readline().strip()))
add_trie(S)
cnt = 0
stack = [root]
while stack:
vn = stack.pop()
for i in range(2):
if vn[i] is None:
r = L - vn.depth
cnt ^= -r&r
else:
stack.append(vn[i])
print('Alice' if cnt else 'Bob')
``` | output | 1 | 95,115 | 0 | 190,231 |
Provide a correct Python 3 solution for this coding contest problem.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob | instruction | 0 | 95,116 | 0 | 190,232 |
"Correct Solution:
```
N, L = map(int, input().split())
make = lambda:[None, None, 0]
root = make()
def construct(s):
n = root
for i in s:
if n[i] is None:
n[i] = n = make()
else:
n = n[i]
n[2] = 1
for i in range(N):
s = map(int, input())
construct(s)
caps = {}
st = [(root, 0, 0)]
while st:
n, i, l = st.pop()
if i:
if n[1] is None:
caps[L - l] = caps.get(L - l, 0) + 1
else:
if not n[1][2]:
st.append((n[1], 0, l+1))
else:
st.append((n, 1, l))
if n[0] is None:
caps[L - l] = caps.get(L - l, 0) + 1
else:
if not n[0][2]:
st.append((n[0], 0, l+1))
ans = 0
for v in caps:
k = caps[v]
if k % 2 == 0:
continue
v -= 1
r = 1
while v % 4 == 3:
v //= 4
r *= 4
if v % 4 == 1:
ans ^= r * 2
else:
ans ^= r
print('Alice' if ans else 'Bob')
``` | output | 1 | 95,116 | 0 | 190,233 |
Provide a correct Python 3 solution for this coding contest problem.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob | instruction | 0 | 95,117 | 0 | 190,234 |
"Correct Solution:
```
from collections import defaultdict
def solve(l, ss):
xor = 0
for d in range(min(ss), l + 1):
sl = ss[d]
sl.sort()
while sl:
s = sl.pop()
ps = s[:-1]
ss[d + 1].append(ps)
if s[-1] == '1' and sl and sl[-1][:-1] == ps:
sl.pop()
else:
xor ^= d & -d
del ss[d]
return xor
n, l = map(int, input().split())
ss = defaultdict(list)
for s in (input() for _ in range(n)):
ss[l - len(s) + 1].append(s)
print('Alice' if solve(l, ss) else 'Bob')
``` | output | 1 | 95,117 | 0 | 190,235 |
Provide a correct Python 3 solution for this coding contest problem.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob | instruction | 0 | 95,118 | 0 | 190,236 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(100000)
def dfs(cur, dep=0):
if cur == -1:
x = l - dep
return x & -x
return dfs(trie[cur][0], dep + 1) ^ dfs(trie[cur][1], dep + 1)
n, l = map(int, input().split())
trie = [[-1, -1] for _ in range(100001)]
idx = 1
for s in (input() for _ in range(n)):
cur = 0
for c in map(int, s):
if trie[cur][c] == -1:
trie[cur][c] = idx
idx += 1
cur = trie[cur][c]
xor = dfs(trie[0][0]) ^ dfs(trie[0][1])
print('Alice' if xor else 'Bob')
``` | output | 1 | 95,118 | 0 | 190,237 |
Provide a correct Python 3 solution for this coding contest problem.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob | instruction | 0 | 95,119 | 0 | 190,238 |
"Correct Solution:
```
import sys
#from collections import defaultdict
sys.setrecursionlimit(10**6)
def input():
return sys.stdin.readline()[:-1]
n, l = map(int, input().split())
ss = [list(input())[::-1] for _ in range(n)]
def addTree(tree, sentence):
if not sentence:
return None
if sentence[-1] not in tree:
tree[sentence[-1]] = {}
p = sentence.pop()
tree[p] = addTree(tree[p], sentence)
return tree
def createTree(sentences):
tree = {}
for sentence in sentences:
tree = addTree(tree, sentence)
return tree
tree = createTree(ss)
#print(tree)
grundy = 0
def dfs(cur, level):
global grundy
if cur is None:
return
elif len(cur) == 1:
grundy ^= (l-level) & (-l+level)
for k in cur.keys():
dfs(cur[k], level+1)
return
dfs(tree, 0)
if grundy == 0:
print("Bob")
else:
print("Alice")
``` | output | 1 | 95,119 | 0 | 190,239 |
Provide a correct Python 3 solution for this coding contest problem.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob | instruction | 0 | 95,120 | 0 | 190,240 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(1000000)
def getGrundyNumber(x):
ans = 1
while x % (ans * 2) == 0:
ans *= 2
return ans
def dfs(iT, Hgt):
num = 0
for c in Trie[iT]:
if c != -1:
dfs(c, Hgt - 1)
num += 1
if num == 1:
Hgts[Hgt - 1] = Hgts.get(Hgt - 1, 0) + 1
N, L = map(int, input().split())
Ss = [input() for i in range(N)]
# トライ木を作成する
Trie = [[-1, -1]]
for S in Ss:
iT = 0
for c in map(int, S):
if Trie[iT][c] == -1:
Trie += [[-1, -1]]
Trie[iT][c] = len(Trie) - 1
iT = Trie[iT][c]
# 子が１つの頂点を探す
Hgts = {}
dfs(0, L + 1)
# Grundy数のXORを求める
ans = 0
for Hgt, num in Hgts.items():
if num % 2:
ans ^= getGrundyNumber(Hgt)
if ans:
print('Alice')
else:
print('Bob')
``` | output | 1 | 95,120 | 0 | 190,241 |
Provide a correct Python 3 solution for this coding contest problem.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob | instruction | 0 | 95,121 | 0 | 190,242 |
"Correct Solution:
```
def main():
import sys
input = sys.stdin.readline
class TreiNode:
def __init__(self, char_num, depth):
self.end = False
self.child = [None] * char_num
self.depth = depth
def __setitem__(self, i, x):
self.child[i] = x
def __getitem__(self, i):
return self.child[i]
class Trei:
def __init__(self, char_num):
self.root = TreiNode(char_num, 0)
self.char_num = char_num
def add(self, S):
v = self.root
for s in S:
if v[s] is None:
v[s] = TreiNode(self.char_num, v.depth + 1)
v = v[s]
v.end = True
def exist(self, S):
v = self.root
for s in S:
if v[s] is None:
return False
v = v[s]
if v.end:
return True
else:
return False
N, L = map(int, input().split())
T = Trei(2)
for _ in range(N):
S = input().rstrip('\n')
S = [int(s) for s in S]
T.add(S)
g = 0
st = [T.root]
while st:
v = st.pop()
for i in range(2):
if v[i] is None:
d = L - v.depth
g ^= d & -d
else:
st.append(v[i])
if g:
print('Alice')
else:
print('Bob')
if __name__ == '__main__':
main()
``` | output | 1 | 95,121 | 0 | 190,243 |
Provide a correct Python 3 solution for this coding contest problem.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob | instruction | 0 | 95,122 | 0 | 190,244 |
"Correct Solution:
```
import sys
from collections import deque
sys.setrecursionlimit(10000)
INF = float('inf')
N, L = list(map(int, input().split()))
S = set()
for _ in range(N):
S.add(input())
# 追加できるのが 01... だけで
# L が 4 とすると
# 01: 次はない
# 010: 011 がある
# 011: 010 がある
# 0100: 011, 0101 がある
# 0101: 011, 0100
# 0110: 010, 0111
# 0111: 010, 0110
# 候補の長さと L だけによって次の候補の長さと数が決まる
def grundy(size):
"""
:param int size: 候補の長さ
:return:
"""
# gs = {0}
# g = 0
# for sz in range(size + 1, L + 1):
# g ^= grundy(sz)
# gs.add(g)
#
# i = 0
# while i in gs:
# i += 1
# return i
i = 1
while (L - size + 1) % i == 0:
i *= 2
return i // 2
# S の各要素からトライ木つくる
trie = {}
for s in S:
t = trie
for c in s:
if c not in t:
t[c] = {}
t = t[c]
# 候補文字列の長さのリスト
ok_size_list = []
children = deque()
children.append((trie, 0))
while len(children) > 0:
node, size = children.popleft()
# 0 か 1 か片方しか行けなかったら、行けない方は S の prefix にならない
# 両方行けない場合はどう追加しても S のいずれかが prefix になってしまうからダメ
if len(node) == 1:
ok_size_list.append(size + 1)
for c, child in node.items():
children.append((child, size + 1))
g = 0
for size in ok_size_list:
g ^= grundy(size)
if g != 0:
print('Alice')
else:
print('Bob')
``` | output | 1 | 95,122 | 0 | 190,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob
Submitted Solution:
```
import sys
readline = sys.stdin.readline
class Node:
def __init__(self, sigma, depth):
self.end = False
self.child = [None] * sigma
self.depth = depth
def __setitem__(self, i, x):
self.child[i] = x
def __getitem__(self, i):
return self.child[i]
class Trie():
def __init__(self, sigma):
self.sigma = sigma
self.root = Node(sigma, 0)
def add(self, S):
vn = self.root
for cs in S:
if vn[cs] is None:
vn[cs] = Node(self.sigma, vn.depth + 1)
vn = vn[cs]
vn.end = True
ans = 0
N, L = map(int, readline().split())
Tr = Trie(2)
for _ in range(N):
S = list(map(int, readline().strip()))
Tr.add(S)
cnt = 0
stack = [Tr.root]
while stack:
vn = stack.pop()
for i in range(2):
if vn[i] is None:
r = L - vn.depth
cnt ^= -r&r
else:
stack.append(vn[i])
print('Alice' if cnt else 'Bob')
``` | instruction | 0 | 95,123 | 0 | 190,246 |
Yes | output | 1 | 95,123 | 0 | 190,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob
Submitted Solution:
```
import sys
sys.setrecursionlimit(2 * 10 ** 5)
class TrieNode:
def __init__(self, char):
self.char = char
self.nextnode = dict()
self.is_indict = False
class Trie:
def __init__(self, charset):
self.charset = charset
self.root = TrieNode('')
def add(self, a_str):
node = self.root
for i, char in enumerate(a_str):
if char not in node.nextnode:
node.nextnode[char] = TrieNode(char)
node = node.nextnode[char]
if i == len(a_str) - 1:
node.is_indict = True
def dfs(self, node, dep):
ret, cnt = 0, 0
if node.is_indict:
return 0
for s in '01':
if s not in node.nextnode:
cnt += 1
else:
ret ^= self.dfs(node.nextnode[s], dep + 1)
height = L - dep
if cnt % 2:
power2 = 0
while height > 0 and height % 2 == 0:
power2 += 1
height //= 2
ret ^= 2 ** power2
return ret
def debug_output(self, node, now):
print(node.char, list(node.nextnode.items()), node.is_indict, now)
if node.is_indict:
print(now)
for n in node.nextnode.values():
self.debug_output(n, now + n.char)
N, L = map(int, input().split())
T = Trie('01')
for _ in range(N):
T.add(input())
# T.debug_output(T.root, '')
print("Alice" if T.dfs(T.root, 0) else "Bob")
``` | instruction | 0 | 95,124 | 0 | 190,248 |
Yes | output | 1 | 95,124 | 0 | 190,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob
Submitted Solution:
```
from functools import cmp_to_key
def f(s, t):
m = min(len(s), len(t))
if s[:m] > t[:m]:
return 1
elif s[:m] < t[:m]:
return -1
else:
if len(s) > len(t):
return -1
else:
return 1
def ms(s, t):
i = 0
for c1, c2 in zip(s, t):
if c1 != c2:
return i
i += 1
return i
def xxor(x):
if x & 3 == 0:
return x
if x & 3 == 1:
return 1
if x & 3 == 2:
return x + 1
if x & 3 == 3:
return 0
def gray(x):
return x ^ (x // 2)
n, l = map(int, input().split())
s = [input() for _ in range(n)]
s.sort(key=cmp_to_key(f))
g = gray(l) ^ gray(l - len(s[0]))
for i in range(n - 1):
b = ms(s[i], s[i + 1]) + 1
g ^= gray(l - b + 1) ^ gray(l - b)
t = len(s[i + 1]) - b
g ^= gray(l - b) ^ gray(l - b - t)
if g == 0:
print('Bob')
else:
print('Alice')
``` | instruction | 0 | 95,125 | 0 | 190,250 |
Yes | output | 1 | 95,125 | 0 | 190,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob
Submitted Solution:
```
import sys
input=sys.stdin.readline
sys.setrecursionlimit(10**9)
from collections import deque
class Node:
def __init__(self,depth):
self.depth=depth
self.left=None
self.right=None
def insert(node,s):
n=node
for i in range(len(s)):
t=s[i]
if t=='0':
if n.left is None:
n.left=Node(i+1)
n=n.left
else:
if n.right is None:
n.right=Node(i+1)
n=n.right
class Trie:
def __init__(self):
self.root=Node(0)
def insert(self,s:str):
insert(self.root,s)
n,l=map(int,input().split())
S=[input().strip() for _ in range(n)]
trie=Trie()
for s in S:
trie.insert(s)
Data=[]
q=deque([trie.root])
def dfs(node):
if node.right is None and node.left is None:
return
if node.right is None or node.left is None:
Data.append(l-node.depth)
if node.right:
q.append(node.right)
if node.left:
q.append(node.left)
while q:
dfs(q.popleft())
xor=0
def Grundy(n):
ret=1
while n%2==0:
n//=2
ret*=2
return ret
for i in Data:
xor^=Grundy(i)
print('Alice' if xor else 'Bob')
``` | instruction | 0 | 95,126 | 0 | 190,252 |
Yes | output | 1 | 95,126 | 0 | 190,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob
Submitted Solution:
```
from collections import defaultdict
n, l = map(int, input().split())
nums = defaultdict(set)
for s in (input() for _ in range(n)):
nums[l - len(s) + 1].add(int(s, 2) + (1 << len(s)))
xor = 0
for i in range(min(nums), l + 1):
ni = nums[i]
for k in ni:
if k ^ 1 not in ni:
xor ^= i & -i
nums[i + 1].add(k // 2)
del nums[i]
print('Alice' if xor else 'Bob')
``` | instruction | 0 | 95,127 | 0 | 190,254 |
No | output | 1 | 95,127 | 0 | 190,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob
Submitted Solution:
```
import sys
sys.setrecursionlimit(2 * 10 ** 5)
class TrieNode:
def __init__(self, char):
self.char = char
self.nextnode = dict()
self.is_indict = False
class Trie:
def __init__(self, charset):
self.charset = charset
self.root = TrieNode('')
def add(self, a_str):
node = self.root
for i, char in enumerate(a_str):
if char not in node.nextnode:
node.nextnode[char] = TrieNode(char)
node = node.nextnode[char]
if i == len(a_str) - 1:
node.is_indict = True
def dfs(self, node, dep):
ret, cnt = 0, 0
for s in '01':
if s not in node.nextnode:
cnt += 1
else:
ret ^= self.dfs(node.nextnode[s], dep + 1)
dep += 1
if cnt % 2:
power2 = 0
while dep > 0 and dep % 2 == 0:
power2 += 1
dep //= 2
ret ^= 2 ** power2
return ret
def debug_output(self, node, now):
print(node.char, list(node.nextnode.items()), node.is_indict, now)
if node.is_indict:
print(now)
for n in node.nextnode.values():
self.debug_output(n, now + n.char)
N, L = map(int, input().split())
T = Trie('01')
for _ in range(N):
T.add(input())
# T.debug_output(T.root, '')
print("Alice" if T.dfs(T.root, 0) else "Bob")
``` | instruction | 0 | 95,128 | 0 | 190,256 |
No | output | 1 | 95,128 | 0 | 190,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob
Submitted Solution:
```
n,l = map(int,input().split())
s = [input() for _ in range(n)]
s2 = [(len(si),si) for si in s]
s2.sort()
max_d = s2[-1][0]
cnt = [0] * (max_d+1)
done = set()
stack = []
for i in range(max_d,-1,-1):
next = set()
while(stack):
sj = stack.pop()
if(not (sj + '0') in done):
cnt[max_d-len(sj)] += 1
if(not (sj + '1') in done):
cnt[max_d-len(sj)] += 1
if(len(sj)>0):
next.add(sj[:-1])
done.add(sj)
if(s2):
while(s2[-1][0] == i):
_,sj = s2.pop()
if(len(sj)>0):
next.add(sj[:-1])
done.add(sj)
if(len(s2)==0):
break
stack = list(next)
cnt_odd = 0
bef = cnt[0]
for ci in cnt[1:]:
ci %= 2
if(bef==0)&(ci==1):
cnt_odd += 1
bef = ci
if(cnt_odd%2==0):
print('Bob')
else:
print('Alice')
``` | instruction | 0 | 95,129 | 0 | 190,258 |
No | output | 1 | 95,129 | 0 | 190,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.
Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:
* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.
We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:
* Add a new string to S. After addition, S must still be a good string set.
The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5
Input
Input is given from Standard Input in the following format:
N L
s_1
s_2
:
s_N
Output
If Alice will win, print `Alice`; if Bob will win, print `Bob`.
Examples
Input
2 2
00
01
Output
Alice
Input
2 2
00
11
Output
Bob
Input
3 3
0
10
110
Output
Alice
Input
2 1
0
1
Output
Bob
Input
1 2
11
Output
Alice
Input
2 3
101
11
Output
Bob
Submitted Solution:
```
def main():
import sys
from collections import defaultdict
input = sys.stdin.readline
mod = 10**9+7
mod2 = 10**9+9
mod3 = 998244353
N, L = map(int, input().split())
dic = defaultdict(int)
dic2 = defaultdict(int)
dic3 = defaultdict(int)
h_list = []
h2_list = []
h3_list = []
pair = {}
pair2 = {}
pair3 = {}
M = 0
for _ in range(N):
s = input().rstrip('\n')
h = 0
h2 = 0
h3 = 0
for i in range(len(s)):
M += 1
h = (h*1007 + int(s[i]) + 1) % mod
pair[h] = (h + 1)%mod if s[i] == '0' else (h-1)%mod
h2 = (h2 * 2009 + int(s[i]) + 1) % mod2
pair2[h2] = (h2 + 1)%mod2 if s[i] == '0' else (h2-1)%mod2
h3 = (h3 * 3001 + int(s[i]) + 1) % mod3
pair3[h3] = (h3 + 1) % mod3 if s[i] == '0' else (h3 - 1) % mod3
dic[h] = i+1
dic2[h2] = i+1
dic[h3] = i+1
h_list.append(h)
h2_list.append(h2)
h3_list.append(h3)
g = 0
seen = defaultdict(int)
seen2 = defaultdict(int)
seen3 = defaultdict(int)
for i in range(M):
s, s2, s3 = h_list[i], h2_list[i], h3_list[i]
if seen[s] and seen2[s2] and seen3[s3]:
continue
t = pair[s]
t2 = pair2[s2]
t3 = pair3[s3]
if not (dic[t] and dic2[t2] and dic3[t3]):
if not seen[s]:
tmp = L - dic[s] + 1
elif not seen[s2]:
tmp = L - dic2[s2] + 1
else:
tmp = L - dic3[s3] + 1
cnt = 0
while tmp % 2 == 0:
tmp //= 2
cnt += 1
g ^= (2**cnt)
#print(g, s, s2, t, t2, dic[t], dic2[t2])
seen[s] = 1
seen2[s2] = 1
seen3[s3] = 1
if g:
print('Alice')
else:
print('Bob')
if __name__ == '__main__':
main()
``` | instruction | 0 | 95,130 | 0 | 190,260 |
No | output | 1 | 95,130 | 0 | 190,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b". | instruction | 0 | 95,391 | 0 | 190,782 |
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
from collections import defaultdict
def solve():
ss = list(input())
k = int(input())
n = len(ss)
if n*(n+1)//2 < k:
print ("No such line.")
return
kiss = defaultdict(int)
for i,s in enumerate( ss ):
kiss[s] += n-i
alpha = 'abcdefghijklmnopqrstuvwxyz'
ans = ""
pos = [ i for i in range(n) ]
while True:
cur = ""
for a in alpha:
if kiss[a] >= k:
cur = a
break
else:
k -= kiss[a]
# print(kiss['c'])
ans += cur
pos = [ i for i in pos if ss[i] == cur ]
#single character
if len(pos) >= k:
# print("brok here", pos , k)
break
k -= len(pos)
# get the next elements
for a in alpha:
kiss[a] = 0
nchar = [ (ss[i+1] , i+1) for i in pos if i+1 < n ]
# print(pos , nchar)
pos = []
for c,i in nchar:
kiss[c] += n-i
pos.append(i)
print(ans)
solve()
``` | output | 1 | 95,391 | 0 | 190,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b". | instruction | 0 | 95,392 | 0 | 190,784 |
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
from heapq import *
l=input()
k=int(input())
n=len(l)
if k>n*(n+1)/2:
print("No such line.")
quit()
ss=[(l[i],i) for i in range(n)]
heapify(ss)
while k:
k-=1
t=heappop(ss)
if k==0:
print(t[0])
else:
if t[1]<n-1:
heappush(ss,(t[0]+l[t[1]+1],t[1]+1))
# Made By Mostafa_Khaled
``` | output | 1 | 95,392 | 0 | 190,785 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.