message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ujan has been lazy lately, but now has decided to bring his yard to good shape. First, he decided to paint the path from his house to the gate.
The path consists of n consecutive tiles, numbered from 1 to n. Ujan will paint each tile in some color. He will consider the path aesthetic if for any two different tiles with numbers i and j, such that |j - i| is a divisor of n greater than 1, they have the same color. Formally, the colors of two tiles with numbers i and j should be the same if |i-j| > 1 and n mod |i-j| = 0 (where x mod y is the remainder when dividing x by y).
Ujan wants to brighten up space. What is the maximum number of different colors that Ujan can use, so that the path is aesthetic?
Input
The first line of input contains a single integer n (1 ≤ n ≤ 10^{12}), the length of the path.
Output
Output a single integer, the maximum possible number of colors that the path can be painted in.
Examples
Input
4
Output
2
Input
5
Output
5
Note
In the first sample, two colors is the maximum number. Tiles 1 and 3 should have the same color since 4 mod |3-1| = 0. Also, tiles 2 and 4 should have the same color since 4 mod |4-2| = 0.
In the second sample, all five colors can be used.
<image>
Submitted Solution:
```
import math
def findDivisors(n):
i=1
a=[]
while i<=math.sqrt(n):
if(n%i == 0):
if(n/i==i):
if(i!=1 and i!=n):
a.append(i)
else:
if(i!=1 and i!=n):
a.append(i); a.append(n/i)
i=i+1
return a
n = int(input())
a=[]
a=findDivisors(n)
if(len(a)==0):
m=n
else:
m=int(max(a))
print(m)
``` | instruction | 0 | 34,397 | 7 | 68,794 |
No | output | 1 | 34,397 | 7 | 68,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2. | instruction | 0 | 34,492 | 7 | 68,984 |
Tags: constructive algorithms, data structures, dp, greedy, implementation
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n = int(input())
arr = list(map(int,input().split()))
a = [arr[0]]
for i in range(1,n):
if a[-1] != arr[i]:
a.append(arr[i])
n1,n = n,len(a)
if n == 1:
print(1)
return
ne = [n+1]*n
dct = [-1 for _ in range(n1+1)]
for ind,i in enumerate(a):
if dct[i] == -1:
dct[i] = ind
else:
ne[dct[i]] = ind
dct[i] = ind
x,y,ans = 0,1,2
for i in range(2,n):
if a[x] == a[i]:
x = i
elif a[y] == a[i]:
y = i
elif ne[x] < ne[y]:
y = i
ans += 1
else:
x = i
ans += 1
print(ans)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 34,492 | 7 | 68,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2. | instruction | 0 | 34,493 | 7 | 68,986 |
Tags: constructive algorithms, data structures, dp, greedy, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
# ------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l):
print(''.join(map(str, l)))
# sys.setrecursionlimit(100000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# import math
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
n = N()
a = RLL()
a1, a2, ans, s = 0, 0, 0, set()
for v in a:
if v == a1: continue
if v == a2:
s = {a1}
elif v in s:
a1, a2, s = v, a1, {v}
else:
s.add(v)
a1 = v
ans += 1
print(ans)
``` | output | 1 | 34,493 | 7 | 68,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2. | instruction | 0 | 34,494 | 7 | 68,988 |
Tags: constructive algorithms, data structures, dp, greedy, implementation
Correct Solution:
```
from sys import stdin, stdout
from collections import deque
def painting_the_array_2(n, a_a):
dic = {}
ua_a = [a_a[0]]
dic[a_a[0]] = deque()
dic[a_a[0]].append(0)
for i in range(1, n):
if a_a[i] == a_a[i-1]:
continue
ua_a.append(a_a[i])
if a_a[i] not in dic:
dic[a_a[i]] = deque()
dic[a_a[i]].append(i)
r = 0
one = -1
two = -1
# print(dic)
for a in ua_a:
if one == a:
one = a
elif two == a:
two = a
elif one == -1:
one = a
r += 1
elif two == -1:
two = a
r += 1
else:
if len(dic[one]) == 0:
one = a
elif len(dic[two]) == 0:
two = a
elif dic[one][0] > dic[two][0]:
one = a
else:
two = a
r += 1
dic[a].popleft()
return r
n = int(stdin.readline())
a_a = list(map(int, stdin.readline().split()))
r = painting_the_array_2(n, a_a)
stdout.write(str(r))
``` | output | 1 | 34,494 | 7 | 68,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2. | instruction | 0 | 34,495 | 7 | 68,990 |
Tags: constructive algorithms, data structures, dp, greedy, implementation
Correct Solution:
```
import sys
from collections import defaultdict
def main():
n = int(sys.stdin.readline())
a = sys.stdin.readline().split()
data = []
prev = {}
result = 0
current = [{'val': -1, 'prev': -1}, {'val': -1, 'prev': -1}]
for key, val in enumerate(a):
data.append({
'val': val,
'prev': prev[val] if val in prev else -1
})
prev[val] = key
for el in reversed(data):
if el['val'] == current[0]['val']:
current[0] = el
continue
if el['val'] == current[1]['val']:
current[1] = el
continue
result += 1
if current[0]['prev'] < current[1]['prev']:
current[0] = el
else:
current[1] = el
print(result)
return
if __name__ == '__main__':
main()
``` | output | 1 | 34,495 | 7 | 68,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2. | instruction | 0 | 34,496 | 7 | 68,992 |
Tags: constructive algorithms, data structures, dp, greedy, implementation
Correct Solution:
```
import sys
Z=sys.stdin.readline
n=int(Z())
a=[*map(int,Z().split())]
b=[]
p=0
for i in a:
if i!=p:
b.append(i)
p=i
c=0
e=[0]*n
r=[0,0]
lb=len(b)
for i in range(lb):
if e[-b[i]]==c+1:
c+=1
r=[b[i],b[i-1]]
e[-r[0]]=c+1
e[-r[1]]=c+1
else: e[-b[i]]=c+1
print(lb-c)
``` | output | 1 | 34,496 | 7 | 68,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2. | instruction | 0 | 34,497 | 7 | 68,994 |
Tags: constructive algorithms, data structures, dp, greedy, implementation
Correct Solution:
```
#!/usr/bin/env python3
import collections
def solveTestCase():
n = int(input())
A = [int(i) for i in input().split()]
s1 = [0]
s2 = [0]
d = dict()
v = n
B = []
for i in A[::-1]:
nxt = n+1
if i in d:
nxt = d[i]
B.append((i, nxt))
d[i] = v
v -= 1
B = B[::-1]
next1 = n+2
next2 = n+2
for (v, nxt) in B:
if v == s1[-1]:
next1 = nxt
elif v == s2[-1]:
next2 = nxt
elif s1[-1] == s2[-1]:
s1.append(v)
next1 = nxt
elif next1 < next2:
s2.append(v)
next2 = nxt
else:
s1.append(v)
next1 = nxt
print(len(s1)+len(s2)-2)
if __name__ == "__main__":
solveTestCase()
``` | output | 1 | 34,497 | 7 | 68,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2. | instruction | 0 | 34,498 | 7 | 68,996 |
Tags: constructive algorithms, data structures, dp, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
import bisect
n=int(input())
A=list(map(int,input().split()))
B=[-1]
for a in A:
if a==B[-1]:
continue
else:
B.append(a)
B=B[1:]
LEN=len(B)
X=[[] for i in range(n+1)]
for i in range(LEN):
X[B[i]].append(i)
DP=[1<<30]*LEN
DP[0]=1
for i in range(LEN):
DP[i]=min(DP[i-1]+1,DP[i])
k1=B[i]
x=bisect.bisect(X[k1],i)
#print(i,x,X[k1][x],X[k1][x-1])
if x==len(X[k1]):
True
else:
if i-1>=0 and B[i-1]==B[i+1]:
DP[X[k1][x]]=min(DP[X[k1][x]],DP[i]+X[k1][x]-X[k1][x-1]-2)
else:
DP[X[k1][x]]=min(DP[X[k1][x]],DP[i]+X[k1][x]-X[k1][x-1]-1)
if i-1>=0:
k2=B[i-1]
x=bisect.bisect(X[k2],i-1)
if x==len(X[k2]):
True
else:
DP[X[k2][x]]=min(DP[X[k2][x]],DP[i]+X[k2][x]-X[k2][x-1]-2)
print(DP[-1])
``` | output | 1 | 34,498 | 7 | 68,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2. | instruction | 0 | 34,499 | 7 | 68,998 |
Tags: constructive algorithms, data structures, dp, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
A = list(map(int,input().split()))
if len(set(A))==1:
print(1)
sys.exit()
B = [A[0]]
for i in range(1, n):
if A[i] != A[i-1]:
B.append(A[i])
# print(B)
D = [B[0], B[1]]
E = [0] * (n+10)
E[B[0]] = 1
E[B[1]] = 1
top = [B[0], B[1]]
ans0 = 2
tf = 1
# print(B,top)
for i in range(2, len(B)):
if B[i] in top and tf==1:
continue
else:
tf = 0
if E[B[i]] == 1:
top = [B[i]]
for j in range(len(D)):
E[D[j]] = 0
for j in range(len(D)-1,-1,-1):
if D[j] != B[i]:
top.append(D[j])
break
E[top[0]] = 1
E[top[1]] = 1
D = top[:]
tf = 1
else:
E[B[i]] = 1
D.append(B[i])
ans0 += 1
print(ans0)
``` | output | 1 | 34,499 | 7 | 68,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from itertools import accumulate
from functools import reduce,cmp_to_key,lru_cache
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
# input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
n = val()
l = li()
cnt1 = cnt2 = 0
last1 = -float('inf')
last2 = -1
def compress(l):
l1 = []
for i in l:
if len(l1) and l1[-1] == i:continue
l1.append(i)
return l1
l = compress(l)
if len(l) < 2:
print(1)
exit()
index = defaultdict(deque)
for i in range(len(l)):
index[l[i]].append(i)
inset = set()
myset = set()
ans = 2
myset.add(l[0])
myset.add(l[1])
index[l[0]].popleft()
index[l[1]].popleft()
for i in range(2, len(l)):
index[l[i]].popleft()
if l[i] in myset:
continue
ans += 1
first, second = myset.pop(), myset.pop()
myset.add(l[i])
if index[first] and index[second]:
if index[first][0] < index[second][0]:
myset.add(first)
else:
myset.add(second)
elif index[first]:
myset.add(first)
else:
myset.add(second)
print(ans)
``` | instruction | 0 | 34,500 | 7 | 69,000 |
Yes | output | 1 | 34,500 | 7 | 69,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2.
Submitted Solution:
```
import sys
readline = sys.stdin.readline
N = int(readline())
A = list(map(int, readline().split()))
INF = 10**9+7
res = []
for a in A:
if not res or res[-1] != a:
res.append(a)
cnt = 0
res.append(-1)
dp = [INF]*(N+3)
dp[-1] = 0
geta = 0
mini = 0
for i in range(len(res)-1):
cp = res[i-1]
cn = res[i]
dcp = min(mini + geta + 1, dp[cn] + geta)
geta += 1
dp[cp] = dcp - geta
mini = min(mini, dp[cp])
print(min(dp) + geta)
``` | instruction | 0 | 34,501 | 7 | 69,002 |
Yes | output | 1 | 34,501 | 7 | 69,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2.
Submitted Solution:
```
def main():
n=int(input())
a=readIntArr()
indexes=[[] for _ in range(n+1)]
for i,x in enumerate(a):
indexes[x].append(i)
for x in range(1,n+1):
indexes[x].append(inf)
def getNextIndex(currIndex,x):
temp=indexes[x]
n=len(temp)
i=-1
b=n
while b>0:
while i+b<n and temp[i+b]<=currIndex:
i+=b
b//=2
i+=1
# print(currIndex)
# print(temp)
# print(i,n)
return temp[i]
a0=-1
a1=-1
ans=0
for i,x in enumerate(a):
if x!=a0 and x!=a1 and a0!=a1:
#append to group where next appearance of element is larger
if getNextIndex(i, a0)>getNextIndex(i, a1):
a0=x
else:
a1=x
ans+=1
elif a0==a1: #put in either
if a0!=x:
ans+=1
a0=x
else: #x is either a0 or a1
pass
print(ans)
return
#import sys
#input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
import sys
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
inf=float('inf')
MOD=10**9+7
main()
``` | instruction | 0 | 34,502 | 7 | 69,004 |
Yes | output | 1 | 34,502 | 7 | 69,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2.
Submitted Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
def isprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
n = int(inp())
count = [[] for i in range(n+1)]
arr = lmp()
for i in range(n):
count[arr[i]].append(i)
for i in range(n+1):
count[i].reverse()
a = [arr[0]]
b = [0]
count[arr[0]].pop()
for i in range(1, n):
if arr[i]==a[-1] or arr[i]==b[-1]:
pass
else:
xa = inf if len(count[a[-1]])==0 else count[a[-1]][-1]
xb = inf if len(count[b[-1]])==0 else count[b[-1]][-1]
if xa==inf:
a.append(arr[i])
elif xb==inf:
b.append(arr[i])
else:
if xa<xb: b.append(arr[i])
else: a.append(arr[i])
count[arr[i]].pop()
print(len(a)+len(b)-1)
``` | instruction | 0 | 34,503 | 7 | 69,006 |
Yes | output | 1 | 34,503 | 7 | 69,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2.
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
#start_time = time.time()
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def getMat(n):
return [getInts() for _ in range(n)]
def isInt(s):
return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
Min:
1,2,3,4,2,3
1,2
3,4,3
How can I compress as much as poss?
If we compress indices i and j, everything in between must be left in tact
"""
def solve():
N = getInt()
A = getInts()
B = []
for i in range(N):
if i == 0 or A[i] != B[-1]:
B.append(A[i])
next_ind = [-1]*(N+1)
last_seen = [-1]*(N+1)
for i in range(len(B)-1,-1,-1):
next_ind[B[i]] = last_seen[B[i]]
last_seen[B[i]] = i
new1, new2 = -1, -1
ans = 0
for i in range(len(B)):
if new1 == -1:
assert new2 == -1
new1 = B[i]
ans += 1
elif B[i] == new1:
continue
elif new2 == -1:
new2 = -1
ans += 1
elif B[i] == new2:
continue
else:
x = next_ind[new1[-1]]
y = next_ind[new2[-1]]
if x == -1:
new1 = B[i]
ans += 1
continue
if y == -1:
new2 = B[i]
ans += 1
continue
if x < y:
new2 = B[i]
ans += 1
continue
if y < x:
new1 = B[i]
ans += 1
continue
return ans
#for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)
``` | instruction | 0 | 34,504 | 7 | 69,008 |
No | output | 1 | 34,504 | 7 | 69,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
inf = 10**16
md = 10**9+7
# md = 998244353
from bisect import bisect
n = II()
aa0 = LI()
aa = []
for a in aa0:
if not aa or a != aa[-1]:
aa.append(a)
last = [-1]*(n+1)
lr = []
for i, a in enumerate(aa):
if last[a] != -1:
lr.append((last[a], i))
last[a] = i
lr.sort(key=lambda x: x[1])
# print(lr)
rr = [-1]
vv = [0]
n = len(aa)
for i, (l, r) in enumerate(lr):
j = bisect(rr, l)-1
nv = vv[j]+1
if rr[j] != l and rr[j] != l-1 and l-1 >= 0: nv += aa[l-1] == aa[l+1]
if r+1 < n: nv += aa[r-1] == aa[r+1]
if nv > vv[-1]:
if rr[-1] == r:
vv[-1] = nv
else:
rr.append(r)
vv.append(nv)
# print(rr)
# print(vv)
print(n-vv[-1])
``` | instruction | 0 | 34,505 | 7 | 69,010 |
No | output | 1 | 34,505 | 7 | 69,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
from collections import Counter, defaultdict
import math
mod = 1000000007
for _ in range(1):
n = int(input())
# a,b,c=map(int, input().split())
# d=defaultdict(list)
arr = list(map(int, input().split()))
if n < 4:
if n == 1:
print(1)
elif n == 2:
if arr[0] == arr[1]:
print(1)
else:
print(2)
elif n == 3:
if arr == arr[0] * 3:
print(1)
elif len(set(arr)) == 2:
print(2)
else:
print(3)
sys.exit()
ls=[]
p=-1
for i in arr:
if i!=p:
p=i
ls.append(i)
arr=ls
p1 = -1
p2 = -1
ans = 0
for i in range(len(arr)):
if arr[i]==p1 and arr[i]==p2:
continue
elif arr[i]==p1 or arr[i]==p2:
continue
else:
if (i+1<len(arr) and arr[i+1]==p1) or p2==-1:
p2=arr[i]
ans+=1
else:
p1=arr[i]
ans+=1
print(ans)
``` | instruction | 0 | 34,506 | 7 | 69,012 |
No | output | 1 | 34,506 | 7 | 69,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the two versions is that this version asks the minimal possible answer.
Homer likes arrays a lot. Today he is painting an array a_1, a_2, ..., a_n with two kinds of colors, white and black. A painting assignment for a_1, a_2, ..., a_n is described by an array b_1, b_2, ..., b_n that b_i indicates the color of a_i (0 for white and 1 for black).
According to a painting assignment b_1, b_2, ..., b_n, the array a is split into two new arrays a^{(0)} and a^{(1)}, where a^{(0)} is the sub-sequence of all white elements in a and a^{(1)} is the sub-sequence of all black elements in a. For example, if a = [1,2,3,4,5,6] and b = [0,1,0,1,0,0], then a^{(0)} = [1,3,5,6] and a^{(1)} = [2,4].
The number of segments in an array c_1, c_2, ..., c_k, denoted seg(c), is the number of elements if we merge all adjacent elements with the same value in c. For example, the number of segments in [1,1,2,2,3,3,3,2] is 4, because the array will become [1,2,3,2] after merging adjacent elements with the same value. Especially, the number of segments in an empty array is 0.
Homer wants to find a painting assignment b, according to which the number of segments in both a^{(0)} and a^{(1)}, i.e. seg(a^{(0)})+seg(a^{(1)}), is as small as possible. Find this number.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
Output
Output a single integer, indicating the minimal possible total number of segments.
Examples
Input
6
1 2 3 1 2 2
Output
4
Input
7
1 2 1 2 1 2 1
Output
2
Note
In the first example, we can choose a^{(0)} = [1,1,2,2], a^{(1)} = [2,3] and seg(a^{(0)}) = seg(a^{(1)}) = 2. So the answer is 2+2 = 4.
In the second example, we can choose a^{(0)} = [1,1,1,1], a^{(1)} = [2,2,2] and seg(a^{(0)}) = seg(a^{(1)}) = 1. So the answer is 1+1 = 2.
Submitted Solution:
```
from sys import stdin, stdout
from collections import defaultdict
import math
def main():
n = int(stdin.readline())
arr = list(map(int, stdin.readline().split()))
p = [arr[0]]
for i in range(1, n):
if arr[i] == p[-1]:
continue
p.append(arr[i])
next_pos = defaultdict(list)
for i in range(len(p)-1,-1,-1):
next_pos[p[i]].append(i)
a = 0
b = 0
mn = 0
for x in p:
if x == a or x == b:
next_pos[x].pop()
continue
if len(next_pos[a]) == 0:
a = x
next_pos[a].pop()
mn += 1
continue
if len(next_pos[b]) == 0:
b = x
next_pos[b].pop()
mn += 1
continue
if next_pos[a][-1] <= next_pos[b][-1]:
mn += 1
b = x
continue
else:
mn += 1
a = x
continue
stdout.write(str(mn)+"\n")
main()
``` | instruction | 0 | 34,507 | 7 | 69,014 |
No | output | 1 | 34,507 | 7 | 69,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red). | instruction | 0 | 34,681 | 7 | 69,362 |
Tags: dfs and similar
Correct Solution:
```
import sys
sys.setrecursionlimit(10000)
def iin(x, y):
return 0 <= x < n and 0 <= y < m
def dfs(x, y, fax=-1, fay=-1):
V[x][y] = True
dx = [1, -1, 0, 0]
dy = [0, 0, 1, -1]
for i in range(4):
xx = x + dx[i]
yy = y + dy[i]
if iin(xx, yy) and M[x][y] == M[xx][yy]:
if V[xx][yy] and not (xx == fax and yy == fay):
print("Yes")
sys.exit(0)
if not V[xx][yy]:
dfs(xx, yy, x, y)
n, m = map(int, input().split())
M = [input() for _ in range(n)]
V = [[False] * m for _ in range(n)]
for i in range(n):
for j in range(m):
if not V[i][j]:
dfs(i, j)
print('No')
``` | output | 1 | 34,681 | 7 | 69,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red). | instruction | 0 | 34,682 | 7 | 69,364 |
Tags: dfs and similar
Correct Solution:
```
# B
n, m = map(int, input().split())
board = [input() for _ in range(n)]
visited = [[0]*m for _ in range(n)]
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
# print(board)
def previous(cur, pre):
return cur[0] == -pre[0] and cur[1] == -pre[1]
def search(target, pre, row, line):
print(row, line)
# print(row, line)
if not 0 <= row < n or not 0 <= line < m or board[row][line] != target:
return
if visited[row][line]:
return 1
visited[row][line] = 1
for d in directions:
if not previous(d, pre) and search(target, d, row+d[0], line+d[1]):
return 1
def start(row, line):
if visited[row][line]:
return
visited[row][line] = 1
target = board[row][line]
stack = []
for i in directions:
stack.append((i, row+i[0], line+i[1]))
while stack:
pre, r, l = stack.pop()
if not 0 <= r < n or not 0 <= l < m or board[r][l] != target:
continue
if visited[r][l]:
return 1
visited[r][l] = 1
for d in directions:
if not previous(pre, d):
stack.append((d, r+d[0], l+d[1]))
def main():
for r in range(n):
for li in range(m):
if start(r, li):
# print(r, l)
print('Yes')
return
print('No')
return
main()
``` | output | 1 | 34,682 | 7 | 69,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red). | instruction | 0 | 34,683 | 7 | 69,366 |
Tags: dfs and similar
Correct Solution:
```
from sys import setrecursionlimit
setrecursionlimit(10**6)
class Graph:
'''A graph has a set of vertices and a set of edges, with each
edge being an ordered pair of vertices. '''
def __init__ (self):
self._alist = {}
def add_vertex (self, vertex):
''' Adds 'vertex' to the graph
Preconditions: None
Postconditions: self.is_vertex(vertex) -> True
'''
if vertex not in self._alist:
self._alist[vertex] = set()
def add_edge (self, source, destination):
''' Adds the edge (source, destination)
Preconditions: None
Postconditions:
self.is_vertex(source) -> True,
self.is_vertex(destination),
self.is_edge(source, destination) -> True
'''
self.add_vertex(source)
self.add_vertex(destination)
self._alist[source].add(destination)
def is_edge (self, source, destination):
'''Checks whether (source, destination) is an edge
'''
return (self.is_vertex(source)
and destination in self._alist[source])
def is_vertex (self, vertex):
'''Checks whether vertex is in the graph.
'''
return vertex in self._alist
def neighbours (self, vertex):
'''Returns the set of neighbours of vertex. DO NOT MUTATE
THIS SET.
Precondition: self.is_vertex(vertex) -> True
'''
return self._alist[vertex]
def vertices (self):
'''Returns a set-like container of the vertices of this
graph.'''
return self._alist.keys()
def depth_first_search(g, v):
'''Return a dictionary whose keys
are all nodes reachable from v.
The dictionary is the search tree
obtained through a depth-first search.
Assumption: g.is_vertex(v) -> True
Running time: O(m), m = # edges
'''
reached = {}
def do_dfs(curr, prev):
# if current node has already been reached, it must be via
# cycle
if curr in reached:
print('Yes')
quit()
reached[curr] = prev
for succ in g.neighbours(curr):
# only continue search with neighbours that aren't the
# immediately previous node
if (succ != prev):
do_dfs(succ, curr)
do_dfs(v, v)
return reached
class UndirectedGraph (Graph):
'''An undirected graph has edges that are unordered pairs of
vertices; in other words, an edge from A to B is the same as one
from B to A.'''
def add_edge (self, a, b):
'''We implement this as a directed graph where every edge has its
opposite also added to the graph'''
super().add_edge (a, b)
super().add_edge (b, a)
def make_graph():
'''Returns an undirected graph with nodes and edges wherever
there is a matching colour pair of neighbours
'''
line_in = input().split()
#read first line, num rows and cols
num_rows = int(line_in[0])
num_cols = int(line_in[1])
#make a list with format [row_idx][col_idx]
board = list()
for x in range(num_rows):
board.append(input())
g = UndirectedGraph()
row_idx = 0
while(row_idx + 1< num_rows):
# turn an item in board into a node if it has an edge.
col_idx = 0
while(col_idx+1 < num_cols):
#check if the node to the right is the same, if so add edge
if (board[row_idx][col_idx] == board[row_idx][col_idx+1]):
g.add_edge(str(row_idx)+','+str(col_idx),
str(row_idx)+','+str(col_idx+1))
#check if the node below is the same, if so add edge
if (board[row_idx][col_idx] == board[row_idx+1][col_idx]):
g.add_edge(str(row_idx)+','+str(col_idx),
str(row_idx+1)+','+str(col_idx))
col_idx += 1
row_idx += 1
# add edges for bottom row
col_idx = 0
while(col_idx + 1 < num_cols):
if(board[row_idx][col_idx] == board[row_idx][col_idx+1]):
g.add_edge(str(row_idx)+','+str(col_idx),
str(row_idx)+','+str(col_idx+1))
col_idx += 1
# add edges for last column
row_idx = 0
while(row_idx +1 < num_rows):
if(board[row_idx][col_idx] == board[row_idx+1][col_idx]):
g.add_edge(str(row_idx)+','+str(col_idx),
str(row_idx+1)+','+str(col_idx))
row_idx += 1
return g
board_g = make_graph()
vertices = board_g.vertices()
reached = set()
for vertex in vertices:
# if a vertex is already in reached and the program hasn't ended,
# it must not be part of a cycle
if vertex not in reached:
#update 'reached' with new list from dfs
reached.update(board_g.depth_first_search(vertex))
# if the program gets through every vertex without finding a cycle
# and quitting, there must be no cycles.
print('No')
``` | output | 1 | 34,683 | 7 | 69,367 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red). | instruction | 0 | 34,684 | 7 | 69,368 |
Tags: dfs and similar
Correct Solution:
```
from collections import defaultdict
[r, c] = [int(item) for item in input().split(' ')]
graph = [input() for i in range(r)]
visited = defaultdict(lambda: 0)
ID = 1
def legal(coor):
(x, y) = coor
return 0 <= x < r and 0 <= y < c
def equal(c1, c2):
return graph[c1[0]][c1[1]] == graph[c2[0]][c2[1]]
def dfs(i, j):
stack = [(i, j)]
p_stack = [(-3, -3)]
while stack:
coor = (x, y) = stack.pop()
p = p_stack.pop()
visited[coor] = True
moves = [(x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)]
for item in moves:
if legal(item) and equal(coor, item) and p != item:
stack.append(item)
p_stack.append(coor)
if visited[item]:
print("Yes")
exit()
for i in range(r):
for j in range(c):
pp = (i, j)
if not visited[pp]:
dfs(i, j)
print("No")
``` | output | 1 | 34,684 | 7 | 69,369 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red). | instruction | 0 | 34,685 | 7 | 69,370 |
Tags: dfs and similar
Correct Solution:
```
import sys
sys.setrecursionlimit(10000)
directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]
def is_valid(x, y):
return (x < rows and y < len(points[x]) and x >= 0 and y >= 0)
def dfs(x, y, fx, fy):
if (visited[x][y]):
print("Yes")
exit()
visited[x][y] = True
for i in range(4):
next_x = x + directions[i][0]
next_y = y + directions[i][1]
if (next_x == fx and next_y == fy):
continue
if (is_valid(x, y) and is_valid(next_x, next_y)):
currPoint = points[x][y]
nextPoint = points[next_x][next_y]
if (currPoint == nextPoint):
dfs(next_x, next_y, x, y)
rows, cols = map(int, input().split())
visited = [[False for x in range(cols)] for x in range(rows)]
points = [['' for x in range(cols)] for x in range(rows)]
for i in range(rows):
points[i] = list(input())
for i in range(rows):
for j in range(cols):
if (not visited[i][j]):
dfs(i, j, i, j)
print("No")
``` | output | 1 | 34,685 | 7 | 69,371 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red). | instruction | 0 | 34,686 | 7 | 69,372 |
Tags: dfs and similar
Correct Solution:
```
rows, cols = [int(i) for i in input().split(' ')]
s = list()
for _ in range(rows):
s.append(list(input()))
a = [[False for _ in range(cols)] for _ in range(rows)]
for r in range(rows):
for c in range(cols):
if not a[r][c]:
p = [[r, c]]
d = [-1, 0]
a[r][c] = True
string = s[r][c]
while p:
row, col = p[-1]
direction = d[-1]
if direction == 4:
del p[-1]
del d[-1]
d[-1] += 1
elif direction == 3 - d[-2]:
d[-1] += 1
else:
if direction == 0:
row -= 1
elif direction == 1:
col -= 1
elif direction == 2:
col += 1
elif direction == 3:
row += 1
if 0 <= row < rows and 0 <= col < cols and string == s[row][col]:
if a[row][col]:
print('Yes')
exit()
else:
a[row][col] = True
p.append([row, col])
d.append(0)
else:
d[-1] += 1
print('No')
``` | output | 1 | 34,686 | 7 | 69,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red). | instruction | 0 | 34,687 | 7 | 69,374 |
Tags: dfs and similar
Correct Solution:
```
import sys
sys.setrecursionlimit(1000000)
def dfs(x,y,px = -1,py = -1):
trace[x][y] = True
dx = [-1,1,0,0]
dy = [0,0,-1,1]
#print(trace)
for i in range(4):
nx = x + dx[i]
ny = y + dy[i]
if (0<=nx<n and 0<=ny<m) and (nx != px or ny != py) and dot[x][y] == dot[nx][ny]:
if trace[nx][ny]:
print('Yes')
exit(0)
else:
dfs(nx,ny,x,y)
n,m = [int(i) for i in input().split()]
dot= []
trace = [[False]*m for _ in range(n)]
for _ in range(n):
dot.append(input())
for i in range(n):
for j in range(m):
if not trace[i][j]:
dfs(i,j)
print('No')
``` | output | 1 | 34,687 | 7 | 69,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red). | instruction | 0 | 34,688 | 7 | 69,376 |
Tags: dfs and similar
Correct Solution:
```
from sys import setrecursionlimit
setrecursionlimit(10**6)
import threading
threading.stack_size(1000000)
class Graph:
'''A graph has a set of vertices and a set of edges, with each
edge being an ordered pair of vertices. '''
def __init__ (self):
self._alist = {}
def add_vertex (self, vertex):
''' Adds 'vertex' to the graph
Preconditions: None
Postconditions: self.is_vertex(vertex) -> True
'''
if vertex not in self._alist:
self._alist[vertex] = set()
def add_edge (self, source, destination):
''' Adds the edge (source, destination)
Preconditions: None
Postconditions:
self.is_vertex(source) -> True,
self.is_vertex(destination),
self.is_edge(source, destination) -> True
'''
self.add_vertex(source)
self.add_vertex(destination)
self._alist[source].add(destination)
def is_edge (self, source, destination):
'''Checks whether (source, destination) is an edge
'''
return (self.is_vertex(source)
and destination in self._alist[source])
def is_vertex (self, vertex):
'''Checks whether vertex is in the graph.
'''
return vertex in self._alist
def neighbours (self, vertex):
'''Returns the set of neighbours of vertex. DO NOT MUTATE
THIS SET.
Precondition: self.is_vertex(vertex) -> True
'''
return self._alist[vertex]
def vertices (self):
'''Returns a set-like container of the vertices of this
graph.'''
return self._alist.keys()
def depth_first_search(g, v):
'''Return a dictionary whose keys
are all nodes reachable from v.
The dictionary is the search tree
obtained through a depth-first search.
Assumption: g.is_vertex(v) -> True
Running time: O(m), m = # edges
'''
reached = {}
def do_dfs(curr, prev):
# if current node has already been reached, it must be via
# cycle
if curr in reached:
print('Yes')
quit()
reached[curr] = prev
for succ in g.neighbours(curr):
# only continue search with neighbours that aren't the
# immediately previous node
if (succ != prev):
do_dfs(succ, curr)
do_dfs(v, v)
return reached
class UndirectedGraph (Graph):
'''An undirected graph has edges that are unordered pairs of
vertices; in other words, an edge from A to B is the same as one
from B to A.'''
def add_edge (self, a, b):
'''We implement this as a directed graph where every edge has its
opposite also added to the graph'''
super().add_edge (a, b)
super().add_edge (b, a)
def make_graph():
'''Returns an undirected graph with nodes and edges wherever
there is a matching colour pair of neighbours
'''
line_in = input().split()
#read first line, num rows and cols
num_rows = int(line_in[0])
num_cols = int(line_in[1])
#make a list with format [row_idx][col_idx]
board = list()
for x in range(num_rows):
board.append(input())
g = UndirectedGraph()
row_idx = 0
while(row_idx + 1< num_rows):
# turn an item in board into a node if it has an edge.
col_idx = 0
while(col_idx+1 < num_cols):
#check if the node to the right is the same, if so add edge
if (board[row_idx][col_idx] == board[row_idx][col_idx+1]):
g.add_edge(str(row_idx)+','+str(col_idx),
str(row_idx)+','+str(col_idx+1))
#check if the node below is the same, if so add edge
if (board[row_idx][col_idx] == board[row_idx+1][col_idx]):
g.add_edge(str(row_idx)+','+str(col_idx),
str(row_idx+1)+','+str(col_idx))
col_idx += 1
row_idx += 1
# add edges for bottom row
col_idx = 0
while(col_idx + 1 < num_cols):
if(board[row_idx][col_idx] == board[row_idx][col_idx+1]):
g.add_edge(str(row_idx)+','+str(col_idx),
str(row_idx)+','+str(col_idx+1))
col_idx += 1
# add edges for last column
row_idx = 0
while(row_idx +1 < num_rows):
if(board[row_idx][col_idx] == board[row_idx+1][col_idx]):
g.add_edge(str(row_idx)+','+str(col_idx),
str(row_idx+1)+','+str(col_idx))
row_idx += 1
return g
board_g = make_graph()
vertices = board_g.vertices()
reached = set()
for vertex in vertices:
# if a vertex is already in reached and the program hasn't ended,
# it must not be part of a cycle
if vertex not in reached:
#update 'reached' with new list from dfs
reached.update(board_g.depth_first_search(vertex))
# if the program gets through every vertex without finding a cycle
# and quitting, there must be no cycles.
print('No')
``` | output | 1 | 34,688 | 7 | 69,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
Submitted Solution:
```
'''
Zijian He
1429876
'''
# Import
from sys import setrecursionlimit
setrecursionlimit(10**6)
# Function from class
class Graph:
def __init__ (self):
self._alist = {}
def add_vertex (self, vertex):
if vertex not in self._alist:
self._alist[vertex] = set()
def add_edge (self, source, destination):
self.add_vertex(source)
self.add_vertex(destination)
self._alist[source].add(destination)
def is_edge (self, source, destination):
return (self.is_vertex(source)
and destination in self._alist[source])
def is_vertex (self, vertex):
return vertex in self._alist
def neighbours (self, vertex):
return self._alist[vertex]
def vertices (self):
return self._alist.keys()
class UndirectedGraph (Graph):
def add_edge (self, a, b):
super().add_edge (a, b)
super().add_edge (b, a)
def depth_first_search(g, v):
reached = {}
def do_dfs(curr, prev):
if curr in reached:
return
reached[curr] = prev
for succ in g.neighbours(curr):
do_dfs(succ, curr)
do_dfs(v, v)
return reached
# Variables for checking the cycle
cycle_Detected = False
vertex_Marked = {}
def dfs_v2(g, curr, prev):
global cycle_Detected
global vertex_Marked
if cycle_Detected:
return
vertex_Marked[curr] = True
for succ in g.neighbours(curr):
if vertex_Marked[succ] and succ != prev:
cycle_Detected = True
return
if not vertex_Marked[succ]:
dfs_v2(g, succ, curr)
def cycle(g):
global cycle_Detected
global vertex_Marked
# Initialize all the node as unmarked
for i in g.vertices():
vertex_Marked[i] = False
# Check throughout the node
for j in g.vertices():
if not vertex_Marked[j]:
dfs_v2(g, j, j)
if cycle_Detected:
break
return
# ----------------------------------------------------------------------
# Process the input
row, col = map(int, input().split(" "))
rows = []
# Set node as a dictionary
node = {}
for i in range(row):
rows.append(input())
for i in range(row):
for j in range(col):
node[i * col + j] = rows[i][j]
# Set up the graph
twoDotsGraph = UndirectedGraph()
# Connecting the node with same color
for i in range(row * col):
twoDotsGraph.add_vertex(i)
try:
if node[i]==node[i + 1] and (i % col) != (col - 1):
twoDotsGraph.add_edge(i, i + 1)
if node[i]==node[i + col]:
twoDotsGraph.add_edge(i, i + col)
except:
continue
# Main function
cycle(twoDotsGraph)
if cycle_Detected:
print("Yes")
else:
print("No")
``` | instruction | 0 | 34,689 | 7 | 69,378 |
Yes | output | 1 | 34,689 | 7 | 69,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
Submitted Solution:
```
from sys import setrecursionlimit
setrecursionlimit(10**6)
class Graph:
'''A graph has a set of vertices and a set of edges, with each
edge being an ordered pair of vertices. '''
def __init__ (self):
self._alist = {}
def add_vertex (self, vertex):
''' Adds 'vertex' to the graph
Preconditions: None
Postconditions: self.is_vertex(vertex) -> True
'''
if vertex not in self._alist:
self._alist[vertex] = set()
def add_edge (self, source, destination):
''' Adds the edge (source, destination)
Preconditions: None
Postconditions:
self.is_vertex(source) -> True,
self.is_vertex(destination),
self.is_edge(source, destination) -> True
'''
self.add_vertex(source)
self.add_vertex(destination)
self._alist[source].add(destination)
def is_edge (self, source, destination):
'''Checks whether (source, destination) is an edge
'''
return (self.is_vertex(source)
and destination in self._alist[source])
def is_vertex (self, vertex):
'''Checks whether vertex is in the graph.
'''
return vertex in self._alist
def neighbours (self, vertex):
'''Returns the set of neighbours of vertex. DO NOT MUTATE
THIS SET.
Precondition: self.is_vertex(vertex) -> True
'''
return self._alist[vertex]
def vertices (self):
'''Returns a set-like container of the vertices of this
graph.'''
return self._alist.keys()
class UndirectedGraph (Graph):
'''An undirected graph has edges that are unordered pairs of
vertices; in other words, an edge from A to B is the same as one
from B to A.'''
def add_edge (self, a, b):
'''We implement this as a directed graph where every edge has its
opposite also added to the graph'''
super().add_edge (a, b)
super().add_edge (b, a)
def depth_first_search(g, v):
'''Return a dictionary whose keys
are all nodes reachable from v.
The dictionary is the search tree
obtained through a depth-first search.
Assumption: g.is_vertex(v) -> True
Running time: O(m), m = # edges
'''
reached = {}
def do_dfs(curr, prev):
if curr in reached:
return
reached[curr] = prev
for succ in g.neighbours(curr):
do_dfs(succ, curr)
do_dfs(v, v)
return reached
# first line is number of rows and columns
rowscols = input().split()
rows = int(rowscols[0])
cols = int(rowscols[1])
g = UndirectedGraph()
colours = {}
# create all of the vertices
name = 0
for i in range(rows):
r = input()
for e in range(cols):
g.add_vertex(name)
colours[name] = r[e] # save the colour for the vertex name
name += 1
# create edges horizontally
name = 0
for i in range(rows):
for e in range(cols-1):
n_c = name
n_n = name+1
if colours[n_c] == colours[n_n]:
g.add_edge(n_c, n_n)
name += 1
name += 1 # skip one, since rows aren't linked to each other horizontally
# create edges vertically
name = 0
for e in range(cols):
for i in range(rows-1):
n_c = name
n_n = name+cols # the next one is below, so add the column amount
if colours[n_c] == colours[n_n]:
g.add_edge(n_c, n_n)
name += 1
# check for cycle
def is_cycle(g):
for v in g.vertices():
searched = depth_first_search(g, v)
for i in g.neighbours(v):
if searched[i] != v: # if a different path was found on any point
print('Yes')
return
print('No')
is_cycle(g)
``` | instruction | 0 | 34,690 | 7 | 69,380 |
Yes | output | 1 | 34,690 | 7 | 69,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
Submitted Solution:
```
class Graph:
def __init__ (self):
self._alist = {}
def add_vertex (self, vertex):
if vertex not in self._alist:
self._alist[vertex] = set()
def add_edge (self, source, destination):
self.add_vertex(source)
self.add_vertex(destination)
self._alist[source].add(destination)
def is_edge (self, source, destination):
return (self.is_vertex(source)
and destination in self._alist[source])
def is_vertex (self, vertex):
return vertex in self._alist
def neighbours (self, vertex):
return self._alist[vertex]
def vertices (self):
return self._alist.keys()
class UndirectedGraph (Graph):
def add_edge (self, a, b):
super().add_edge (a, b)
super().add_edge (b, a)
#from search import depth_first_search
def depth_first_search(g, v):
reached = {}
def do_dfs(curr, prev):
if curr in reached:
return
reached[curr] = prev
for succ in g.neighbours(curr):
do_dfs(succ, curr)
do_dfs(v, v)
return reached
from sys import setrecursionlimit
setrecursionlimit(10**6)
#------------------------------------------------------------------
def cycleExists(g): # - G is an undirected graph.
vertexMarked = { u : False for u in g.vertices() } # - All nodes are initially unmarked.
cycleDetected = [False] # - Define cycleDetected as a list so we can change
# its value per reference, see:
# http://stackoverflow.com/questions/11222440/python-variable-reference-assignment
for u in g.vertices(): # - Visit all nodes.
if not vertexMarked[u]:
dfs_visit(g, u, cycleDetected, u, vertexMarked) # - u is its own predecessor initially
if cycleDetected[0]:
break
return cycleDetected[0]
#--------
def dfs_visit(g, u, cycleDetected, pred_node, vertexMarked):
if cycleDetected[0]: # - Stop dfs if cycle is found.
return
vertexMarked[u] = True # - Mark node.
for v in g.neighbours(u): # - Check neighbors, where G[u] is the adjacency list of u.
if vertexMarked[v] and v != pred_node: # - If neighbor is marked and not predecessor,
cycleDetected[0] = True # then a cycle exists.
return
if not vertexMarked[v]: # - Call dfs_visit recursively.
dfs_visit(g, v, cycleDetected, u, vertexMarked)
#---------------------------------------------------------------------
# get dimension of matrix from user input
rowNum, colNum = map(int, input().split(" "))
rows = []
nodeDict = {}
# store rows into a list for later use
for i in range(rowNum):
rows.append(input())
# store elements into a dictionary
# element form: integer:letter(key:value)
for i in range(rowNum):
for j in range(colNum):
nodeDict[i* colNum+j] = rows[i][j]
# create an undirected graph
twoDotsGraph = UndirectedGraph()
# add edges connecting vertices with the same letter value(color) in dictionary
for v in range(rowNum*colNum):
twoDotsGraph.add_vertex(v)
try:
if nodeDict[v]==nodeDict[v+1] and v%colNum!=(colNum-1):
twoDotsGraph.add_edge(v, v+1)
if nodeDict[v]==nodeDict[v+colNum]:
twoDotsGraph.add_edge(v, v+colNum)
except:
continue
if cycleExists(twoDotsGraph):
print("Yes")
else:
print("No")
``` | instruction | 0 | 34,691 | 7 | 69,382 |
Yes | output | 1 | 34,691 | 7 | 69,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
Submitted Solution:
```
import sys
sys.setrecursionlimit(10000)
def dfs(x,y,xprev,yprev):
global vst
if vst[x][y]:
return
vst[x][y] = True
for dx, dy in [(0,1),(0,-1),(1,0),(-1,0)]:
xprime = x + dx
yprime = y + dy
if 0<=xprime<n and 0<=yprime<m:
if color[x][y] == color[xprime][yprime]\
and ((xprime,yprime) != (xprev, yprev)):
if vst[xprime][yprime]:
#print(vst)
#print(x,y,xprime, yprime)
print('Yes')
sys.exit(0)
else:
dfs(xprime, yprime, x, y)
n, m = tuple(map(int, input().split()))
color = [input() for _ in range(n)]
vst = [[False for _ in range(m)]for _ in range(n)]
for i in range(n):
for j in range(m):
dfs(i,j,-1,-1)
print('No')
``` | instruction | 0 | 34,692 | 7 | 69,384 |
Yes | output | 1 | 34,692 | 7 | 69,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
Submitted Solution:
```
# coding: utf-8
rows, cols = input().split(" ")
rows, cols = int(rows), int(cols)
m = []
START = 0
UP = 1
DOWN = 2
LEFT = 3
RIGH = 4
for i in range(rows):
row = [c for c in input()]
m.append(row)
def in_range(x, y):
return x < rows and x >= 0 and y < cols and y >= 0
def up(r, c, m):
return m[r-1][c]
def down(r, c, m):
return m[r+1][c]
def left(r, c, m):
return m[r][c-1]
def right(r, c, m):
return m[r][c+1]
def check(position, direction, m, coords):
row, col = position
current_color = m[row][col]
if position in coords:
return True
coords.append(position)
result = False
if in_range(r+1, c) and m[r-1][c] == current_color and direction != DOWN:
result = result or check((r-1, c), UP, m, coords)
if in_range(r+1, c) and m[r+1][c] == current_color and direction != UP:
result = result or check((r+1, c), DOWN, m, coords)
if in_range(r, c-1) and m[r][c-1] == current_color and direction != RIGHT:
result = result or check((r, c-1), LEFT, m, coords)
if in_range(r, c+1) and m[r][c+1] == current_color and direction != LEFT:
result = result or check((r, c+1), RIGHT, m, coords)
return result
found = False
for r in range(rows):
for c in range(cols):
if check((r,c), START, m, []):
found = True
break
if found:
break
print("YES" if found else "NO")
``` | instruction | 0 | 34,693 | 7 | 69,386 |
No | output | 1 | 34,693 | 7 | 69,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
Submitted Solution:
```
n, m = map(int , input().split())
l=[None for i in range(n)]
vis=[[0 for i in range(m)] for j in range(n)]
vis1=[[0 for i in range(m)] for j in range(n)]
b=1
s=0
r=0
x=0
for i in range(0,n):
l[i]=input()
def f(i,j,k):
b=0
if k>=4 and ((abs(i-s)==1 and abs(j-r)==0 )or (abs(i-s)==0 and abs(j-r)==1)):
vis1[0][0]=1
vis[i][j]=1
return
vis[i][j]=1
if(i<n-1) and (l[i][j]==l[i+1][j] )and (vis[i+1][j]!=1 ):
b=1
f(i+1,j,k+1)
if (i>0) and (l[i][j]==l[i-1][j] ) and (vis[i-1][j]!=1 ):
b=1
f(i-1,j,k+1)
if(j<m-1) and (l[i][j]==l[i][j+1]) and (vis[i][j+1]!=1 ):
b=1
f(i,j+1,k+1)
if(j>0) and (l[i][j]==l[i][j-1]) and (vis[i][j-1]!=1 ):
b=1
f(i,j-1,k+1)
if(b==0):
return
f(0,0,0)
for i in range (0,n):
for j in range (0,m):
k=0
f(i,j,k)
if (vis1[0][0]):
x=1
print("Yes")
break
if (x==0):
print("No")
``` | instruction | 0 | 34,694 | 7 | 69,388 |
No | output | 1 | 34,694 | 7 | 69,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
Submitted Solution:
```
import sys
#sys.stdin = open('in.txt')
sys.setrecursionlimit(10000)
isin = lambda x, y: 0 <= x < n and 0 <= y < m
def dfs(curx, cury, fax = -1, fay = -1):
global found
vis[curx][cury] = True
dx = [1, -1, 0, 0]
dy = [0, 0, 1, -1]
for i in range(4):
newx = curx + dx[i]
newy = cury + dy[i]
if isin(newx, newy):
if G[curx][cury] == G[newx][newy] and not vis[newx][newy]:
dfs(newx, newy, curx, cury)
if G[curx][cury] == G[newx][newy] and vis [newx][newy] and not (newx ==
fax and newy == fay):
found = True
#print("Yes")
#sys.exit(0)
n, m = map(int, input().split())
vis = [[False] * m for _ in range(n)]
G = [input() for _ in range(n)]
found = False
for i in range(n):
for j in range(m):
if not vis[i][j]:
dfs(i, j)
if found:
print('Yes')
sys.exit(0)
print('No')
``` | instruction | 0 | 34,695 | 7 | 69,390 |
No | output | 1 | 34,695 | 7 | 69,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
<image>
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Examples
Input
3 4
AAAA
ABCA
AAAA
Output
Yes
Input
3 4
AAAA
ABCA
AADA
Output
No
Input
4 4
YYYR
BYBY
BBBY
BBBY
Output
Yes
Input
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Output
Yes
Input
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
Output
No
Note
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
Submitted Solution:
```
n, m = map(int, input().split(' '))
s = [list(input()+'.') for i in range(n)]
s += [list('.'*(m + 1))]
def solve():
flag = False
for i in range(n):
for j in range(m):
if s[i][j] != '.':
d = 0
d += (s[i + 1][j] == s[i][j])
d += (s[i - 1][j] == s[i][j])
d += (s[i][j + 1] == s[i][j])
d += (s[i][j - 1] == s[i][j])
if d < 2:
s[i][j] = '.'
flag = True
return flag
for _ in range(50):
solve()
for i in range(n):
for j in range(m):
if s[i][j] != '.':
print ('Yes')
exit()
print('No')
``` | instruction | 0 | 34,696 | 7 | 69,392 |
No | output | 1 | 34,696 | 7 | 69,393 |
Provide a correct Python 3 solution for this coding contest problem.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | instruction | 0 | 34,947 | 7 | 69,894 |
"Correct Solution:
```
r,l = map(int,input().split())
n,m = map(int,input().split())
print((r-n)*(l-m))
``` | output | 1 | 34,947 | 7 | 69,895 |
Provide a correct Python 3 solution for this coding contest problem.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | instruction | 0 | 34,948 | 7 | 69,896 |
"Correct Solution:
```
h, w = map(int, input().split())
y, x = map(int, input().split())
print((h - y) * (w - x))
``` | output | 1 | 34,948 | 7 | 69,897 |
Provide a correct Python 3 solution for this coding contest problem.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | instruction | 0 | 34,949 | 7 | 69,898 |
"Correct Solution:
```
a=map(int,open(0).read().split());a,b,c,d=a;print((a-c)*(b-d))
``` | output | 1 | 34,949 | 7 | 69,899 |
Provide a correct Python 3 solution for this coding contest problem.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | instruction | 0 | 34,950 | 7 | 69,900 |
"Correct Solution:
```
h, w=map(int, input().split())
H, W=map(int, input().split())
print((H-h)*(W-w))
``` | output | 1 | 34,950 | 7 | 69,901 |
Provide a correct Python 3 solution for this coding contest problem.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | instruction | 0 | 34,951 | 7 | 69,902 |
"Correct Solution:
```
H,W=map(int,input().split())
h,w=map(int,input().split())
print(H*W-((H*w+W*h-h*w)))
``` | output | 1 | 34,951 | 7 | 69,903 |
Provide a correct Python 3 solution for this coding contest problem.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | instruction | 0 | 34,952 | 7 | 69,904 |
"Correct Solution:
```
H,W = map(int,input().split())
h,w = map(int,input().split())
print(H*W+h*w-h*W-w*H)
``` | output | 1 | 34,952 | 7 | 69,905 |
Provide a correct Python 3 solution for this coding contest problem.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | instruction | 0 | 34,953 | 7 | 69,906 |
"Correct Solution:
```
a,b,c,d=map(int,open(0).read().split())
print((a-c)*(b-d))
``` | output | 1 | 34,953 | 7 | 69,907 |
Provide a correct Python 3 solution for this coding contest problem.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0 | instruction | 0 | 34,954 | 7 | 69,908 |
"Correct Solution:
```
n,m=map(int,input().split())
x,y=map(int,input().split())
print(n*m-x*m-y*(n-x))
``` | output | 1 | 34,954 | 7 | 69,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0
Submitted Solution:
```
h0,w0 = map(int,input().split())
h,w= map(int,input().split())
print((h0-h)*(w0-w))
``` | instruction | 0 | 34,955 | 7 | 69,910 |
Yes | output | 1 | 34,955 | 7 | 69,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0
Submitted Solution:
```
H,W=map(int,input().split())
h,w=map(int,input().split())
print(H*W-(h*W+H*w-w*h))
``` | instruction | 0 | 34,956 | 7 | 69,912 |
Yes | output | 1 | 34,956 | 7 | 69,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0
Submitted Solution:
```
H,W=map(int,input().split())
h,w=map(int,input().split())
print((H*W)-(H*w)-(W*h)+(h*w))
``` | instruction | 0 | 34,957 | 7 | 69,914 |
Yes | output | 1 | 34,957 | 7 | 69,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0
Submitted Solution:
```
H,W=map(int,input().split())
h,w=map(int,input().split())
sum=H*W-H*w-W*h+w*h
print(sum)
``` | instruction | 0 | 34,958 | 7 | 69,916 |
Yes | output | 1 | 34,958 | 7 | 69,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0
Submitted Solution:
```
p,q= input().split()
a,b =(int(p), int(q))
P,Q= input().split()
A,B =(int(P), int(Q))
print(int(a*b - A*B))
``` | instruction | 0 | 34,959 | 7 | 69,918 |
No | output | 1 | 34,959 | 7 | 69,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0
Submitted Solution:
```
H,W=map(int, input().split())
h,w=map(int, input().split())
print (H*W - (h*w))
``` | instruction | 0 | 34,960 | 7 | 69,920 |
No | output | 1 | 34,960 | 7 | 69,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0
Submitted Solution:
```
a=list(input().split())
b=list(input(),split())
c=int(a[0])
d=int(a[1])
e=int(b[0])
f=int(b[1])
g=c*d-d*e-c*f+e*f
print(g)
``` | instruction | 0 | 34,961 | 7 | 69,922 |
No | output | 1 | 34,961 | 7 | 69,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are H rows and W columns of white square cells.
You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.
How many white cells will remain?
It can be proved that this count does not depend on what rows and columns are chosen.
Constraints
* All values in input are integers.
* 1 \leq H, W \leq 20
* 1 \leq h \leq H
* 1 \leq w \leq W
Input
Input is given from Standard Input in the following format:
H W
h w
Output
Print the number of white cells that will remain.
Examples
Input
3 2
2 1
Output
1
Input
5 5
2 3
Output
6
Input
2 4
2 4
Output
0
Submitted Solution:
```
HW = list(map(int, input().split))
hw = list(map(int, input().split))
print(HW[0]*HW[1] - hw[0]*HW[1] - HW[0]*hw[1] + hw[0]*hw[1])
``` | instruction | 0 | 34,962 | 7 | 69,924 |
No | output | 1 | 34,962 | 7 | 69,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You might have remembered Theatre square from the [problem 1A](https://codeforces.com/problemset/problem/1/A). Now it's finally getting repaved.
The square still has a rectangular shape of n × m meters. However, the picture is about to get more complicated now. Let a_{i,j} be the j-th square in the i-th row of the pavement.
You are given the picture of the squares:
* if a_{i,j} = "*", then the j-th square in the i-th row should be black;
* if a_{i,j} = ".", then the j-th square in the i-th row should be white.
The black squares are paved already. You have to pave the white squares. There are two options for pavement tiles:
* 1 × 1 tiles — each tile costs x burles and covers exactly 1 square;
* 1 × 2 tiles — each tile costs y burles and covers exactly 2 adjacent squares of the same row. Note that you are not allowed to rotate these tiles or cut them into 1 × 1 tiles.
You should cover all the white squares, no two tiles should overlap and no black squares should be covered by tiles.
What is the smallest total price of the tiles needed to cover all the white squares?
Input
The first line contains a single integer t (1 ≤ t ≤ 500) — the number of testcases. Then the description of t testcases follow.
The first line of each testcase contains four integers n, m, x and y (1 ≤ n ≤ 100; 1 ≤ m ≤ 1000; 1 ≤ x, y ≤ 1000) — the size of the Theatre square, the price of the 1 × 1 tile and the price of the 1 × 2 tile.
Each of the next n lines contains m characters. The j-th character in the i-th line is a_{i,j}. If a_{i,j} = "*", then the j-th square in the i-th row should be black, and if a_{i,j} = ".", then the j-th square in the i-th row should be white.
It's guaranteed that the sum of n × m over all testcases doesn't exceed 10^5.
Output
For each testcase print a single integer — the smallest total price of the tiles needed to cover all the white squares in burles.
Example
Input
4
1 1 10 1
.
1 2 10 1
..
2 1 10 1
.
.
3 3 3 7
..*
*..
.*.
Output
10
1
20
18
Note
In the first testcase you are required to use a single 1 × 1 tile, even though 1 × 2 tile is cheaper. So the total price is 10 burles.
In the second testcase you can either use two 1 × 1 tiles and spend 20 burles or use a single 1 × 2 tile and spend 1 burle. The second option is cheaper, thus the answer is 1.
The third testcase shows that you can't rotate 1 × 2 tiles. You still have to use two 1 × 1 tiles for the total price of 20.
In the fourth testcase the cheapest way is to use 1 × 1 tiles everywhere. The total cost is 6 ⋅ 3 = 18. | instruction | 0 | 35,336 | 7 | 70,672 |
Tags: brute force, dp, greedy, implementation, two pointers
Correct Solution:
```
def cost (count, x, y):
q,r = divmod(count,2)
return q*y + r*x
t = int(input())
for _ in range (t):
n,m,x,y = map (int, input().split())
info = list()
for _ in range (n):
info.append (input())
if 2*x <= y:
count = 0
for s in info:
for char in s:
if char == ".":
count += 1
ans = count*x
else:
L = [[0]*m for i in range (n)]
for i in range (n):
for j in range (m):
if info[i][j] == ".":
if j == 0:
L[i][j] = 1
else:
L[i][j] = L[i][j-1] + 1
ans = 0
for i in range (n):
for j in range (m-1):
if L[i][j]>0 and L[i][j+1]==0:
ans += cost(L[i][j], x, y)
if L[i][m-1] > 0:
ans += cost(L[i][m-1], x, y)
print (ans)
``` | output | 1 | 35,336 | 7 | 70,673 |
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