text stringlengths 198 433k | conversation_id int64 0 109k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
s=input()
p=s[0]
c=""
q="abcdefghijklmnopqrstuvwxyz"
flag=0
for i in range(1,len(s)):
if s[i] in p:
k=p.find(s[i])
if(len(p)==1):
if(s[i]!=p[0]):
p+=s[i]
elif k==0 :
if p[1]!=s[i-1]:
flag=1
break
elif k==(len(p)-1):
if p[k-1]!=s[i-1]:
flag=1
break
elif p[k-1]!=s[i-1] and p[k+1]!=s[i-1]:
flag=1
break
else:
k=p.find(s[i-1])
if k==0:
c=p
p=s[i]
p+=c
elif k==(len(p)-1):
p+=s[i]
else:
flag=1
break
if flag==1:
print("NO")
else :
print("YES")
for i in q:
if i not in p:
p+=i
print(p)
```
| 15,200 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
t = int(input())
for i in range(t):
s = input()
f = True
st = set()
answer = ''
pos = 0
n = 0
for l in s:
if l not in st:
st.add(l)
n += 1
if pos == n - 1:
answer = answer + l
pos += 1
elif pos == 1:
answer = l + answer
else:
f = False
break
else:
if pos < n:
if answer[pos] == l:
pos += 1
continue
if pos > 1:
if answer[pos - 2] == l:
pos -= 1
continue
f = False
break
if f:
print('YES')
print(answer, end='')
for l in 'abcdefghijklmnopqrstuvwxyz':
if l not in st:
print(l, end='')
print()
else:
print('NO')
```
| 15,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
s = input()
finish = ""
x = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
y = [0]*52
u = 26
d = len(s)
ok = True
s1 = list(s)
for i in range(d):
if (s1[i] in x):
x.remove(s1[i])
if y[u-1]==s1[i]:
u -= 1
elif y[u+1]==s1[i]:
u += 1
else:
if y[u-1]==0:
u -= 1
if (s1[i] in y):
finish += "NO"
ok = False
break
else:
y[u]=s1[i]
elif y[u+1]==0:
u += 1
if (s1[i] in y):
finish += "NO"
ok = False
break
else:
y[u]=s1[i]
else:
finish += "NO"
ok = False
break
if ok:
finish += "YES\n"
for i in range(52):
if y[i]!=0:
finish += y[i]
for i in range(len(x)):
finish += x[i]
print(finish)
```
| 15,202 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
import sys
from collections import deque
alf = set(list('qwertyuiopasdfghjklzxcvbnm'))
input = lambda: sys.stdin.readline().strip()
# print(help(deque))
for i in range(int(input())):
s = input()
ans = deque()
used = set([s[0]])
ans.append(s[0])
f = False
ind = 0
for i in range(len(s)-1):
if s[i+1]in used:
if ind==0:
if ans[1]==s[i+1]:
ind = 1
else:
f = True
break
elif ind == len(ans)-1:
if ans[ind-1]==s[i+1]:
ind -= 1
else:
f = True
break
else:
if ans[ind-1]==s[i+1]:
ind -= 1
elif ans[ind+1]==s[i+1]:
ind += 1
else:
f = True
break
else:
used.add(s[i+1])
if ind==0:
ans.appendleft(s[i+1])
elif ind == len(ans)-1:
ans.append(s[i+1])
ind+=1
else:
f = True
break
if f:
print('NO')
else:
print("YES")
print(''.join(ans)+''.join(list(alf-set(ans))))
```
| 15,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
from collections import Counter
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
import math
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
t, = gil()
for _ in range(t):
s = g()
n = len(s)
graph = {}
flag = True
for i in range(n):
if s[i] not in graph:
graph[s[i]] = set()
if i :
graph[s[i]].add(s[i-1])
if i+1 < n:
graph[s[i]].add(s[i+1])
if len(graph[s[i]]) > 2:
flag = False
if n == 1:
print("YES")
print("abcdefghijklmnopqrstuvwxyz")
elif flag :
for ch in graph:
if len(graph[ch]) == 1:
break
if len(graph[ch]) == 1:
print("YES")
ans = [ch]; l = 1
while len(ans) == l:
for ch in graph[ans[-1]]:
if ch not in ans:
ans.append(ch)
break
l += 1
for ch in "abcdefghijklmnopqrstuvwxyz":
if ch not in ans :
ans.append(ch)
print("".join(ans))
else:
print("NO")
else:
print("NO")
```
| 15,204 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
import heapq
raw_input = stdin.readline
pr = stdout.write
mod=998244353
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return tuple(map(int,stdin.read().split()))
range = xrange # not for python 3.0+
# main code
for t in range(ni()):
s=raw_input().strip()
d=defaultdict(set)
n=len(s)
for i in range(1,n):
if s[i]==s[i-1]:
continue
d[s[i-1]].add(s[i])
d[s[i]].add(s[i-1])
f=0
c=0
st=''
for i in d:
if len(d[i])>2:
f=1
break
elif len(d[i])==2:
c+=1
else:
st=i
if f or c>max(0,len(d)-2):
pr('NO\n')
continue
q=[st]
vis=Counter()
vis[st]=1
ans=''
while q:
x=q.pop(0)
ans+=x
for i in d[x]:
if not vis[i]:
vis[i]=1
q.append(i)
for i in range(97,97+26):
if not d[chr(i)]:
ans+=chr(i)
pr('YES\n')
pr(ans+'\n')
```
| 15,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Submitted Solution:
```
tc = int(input())
while tc:
tc -= 1
s = input().strip()
# print(s)
adj = [ [] for i in range(27)]
for ind in range(1, len(s)):
# print(ord(s[ind - 1])-ord('a'))
adj[ord(s[ind]) - ord('a')].append(ord(s[ind-1]) - ord('a'))
adj[ord(s[ind - 1]) - ord('a')].append(ord(s[ind]) - ord('a'))
flag = 0
# print(adj)
for i in range(26):
adj[i] = list(set(adj[i]))
if len(adj[i]) > 2:
print("NO")
flag = 1
break
if flag:
continue
res = []
fl = [0] * 27
for i in range( 26):
if not fl[i] and len(adj[i]) < 2:
st = i
while not fl[st]:
fl[st] = 1
res.append(st)
for j in range(len(adj[st])):
if not fl[adj[st][j]]:
st = adj[st][j]
break
# print(res)
if len(res) != 26:
print("NO")
else:
print("YES")
ans = ""
for i in range(26):
ch = res[i] + ord('a')
ans += chr(ch)
print(ans)
```
Yes
| 15,206 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Submitted Solution:
```
import string
def cal(s):
for a, b in zip(s[:-1], s[1:]):
if a == b:
return 'NO'
ans = s[0]
i = 0
found = {s[0]}
for a, b in zip(s[:-1], s[1:]):
if b in found:
if i + 1 < len(ans) and ans[i + 1] == b:
i += 1
elif i - 1 >= 0 and ans[i - 1] == b:
i -= 1
else:
return 'NO'
else:
if i == len(ans) - 1:
ans += b
i += 1
elif i == 0:
ans = b + ans
else:
return 'NO'
found.add(b)
return 'YES\n' + ans + str(''.join(set(string.ascii_lowercase) - found))
T = int(input())
for _ in range(T):
s = input()
print(cal(s))
```
Yes
| 15,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Submitted Solution:
```
from math import *
t=int(input())
dd=[chr(97+i) for i in range(26)]
while(t):
t-=1
s=input()
if(len(s)==1):
print("YES")
for i in dd:
print(i,end="")
print()
continue
re=[s[0],s[1]]
p=0
sr=set(s[:2])
flag=0
pr=1
for i in s[2:]:
c=0
try:
if(re[pr-1]==i and pr-1!=-1):
pr=pr-1
continue
else:
c+=1
except:
c+=1
try:
if(re[pr+1]==i):
pr=pr+1
continue
else:
c+=1
except:
c+=1
if(c==2):
if(i in sr or (pr>0 and pr<len(re)-1)):
flag=3
break
if(pr==0):
re.insert(0,i)
pr=0
else:
re.insert(pr+1,i)
pr+=1
sr.add(i)
else:
continue
if(flag==3):
print("NO")
else:
print("YES")
for i in re:
print(i,end="")
for i in dd:
if(i not in sr):
print(i,end="")
print()
```
Yes
| 15,208 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Submitted Solution:
```
def dfs(e,v,res,done):
res.append(v)
done.add(v)
for u in e[v]:
if u not in done:
dfs(e,u,res,done)
def go():
s = input()
if len(s)==1:
return True, 'bacdefghijklmnopqrstuvwxyz'
e = {ss:set() for ss in s}
d = {}
for i in range(len(s) - 1):
a, b = s[i], s[i + 1]
if a not in e[b]:
e[a].add(b)
e[b].add(a)
d[a] = d.get(a, 0) + 1
d[b] = d.get(b, 0) + 1
one = False
start = None
two = False
for a, dd in d.items():
if dd > 2:
return False, ''
if dd == 2:
two = True
if dd == 1:
one = True
start = a
if two and not one:
return False, ''
res=[]
done = set()
dfs(e,start,res, done)
alph = set('bacdefghijklmnopqrstuvwxyz')
ans = ''.join(res) + ''.join(alph-set(res))
return True, ans
t = int(input())
for _ in range(t):
res, ans = go()
if res:
print('YES')
print(ans)
else:
print('NO')
```
Yes
| 15,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Submitted Solution:
```
import sys
import math
input=sys.stdin.readline
def bo(i):
return ord(i)-ord('a')
t=int(input())
while t>0:
t-=1
s=list(input().rstrip())
a=[[] for i in range(27)]
vis=[[0 for i in range(27)] for j in range(27)]
n=len(s)
for i in range(n-1):
#print(bo(s[i]),bo(s[i+1]))
if vis[bo(s[i])][bo(s[i+1])]==0:
vis[bo(s[i])][bo(s[i+1])]=1
vis[bo(s[i+1])][bo(s[i])]=1
a[bo(s[i])].append(bo(s[i+1]))
a[bo(s[i+1])].append(bo(s[i]))
flag=0
for i in range(n):
if len(a[bo(s[i])])>2:
flag=1
break
if flag==1:
print("NO")
continue
#print(a)
q="abcdefghijklmnopqrstuvwxyz"
v=[0 for i in range(27)]
fi=0
z=-9
uu=""
while fi<5:
#print(z,q[z])
if z==-9:
for i in range(27):
if len(a[i])==1:
#print(q[i],end="")
uu+=q[i]
z=a[i][0]
v[i]=1
break
else:
if len(a[z])==1:
#print(q[z],end="")
uu+=q[z]
v[z]=1
break
elif v[a[z][0]]==1 and v[a[z][1]]==1:
#print(q[z],end="")
uu+=q[z]
v[z]=1
break
#print(q[z],end="")
v[z]=1
for j in a[z]:
if v[j]==0:
#print(q[j],end="")
uu+=q[z]
z=j
v[j]=1
break
if z==-9:
fi=-9
break
for i in q:
#print(i,v[bo(i)])
if v[bo(i)]==0:
uu+=i
if fi==-9:
print("NO")
continue
print("YES")
print(uu)
```
No
| 15,210 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Submitted Solution:
```
t = int(input())
for i in range(t):
s = input()
f = True
st = set()
answer = ''
pos = 0
n = 0
for l in s:
if l not in st:
st.add(l)
n += 1
if pos == n - 1:
answer = answer + l
pos += 1
elif pos == 1:
answer = l + answer
else:
if pos < n:
if answer[pos] == l:
pos += 1
continue
if pos > 1:
if answer[pos - 2] == l:
pos -= 1
continue
f = False
break
if f:
print('YES')
print(answer, end='')
for l in 'abcdefghijklmnopqrstuvwxyz':
if l not in st:
print(l, end='')
print()
else:
print('NO')
```
No
| 15,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import defaultdict
t = int(input())
for _ in range(t):
s = input().strip()
d = defaultdict(set)
b = None
for c in s:
if b is None:
b = c
continue
d[c].add(b)
d[b].add(c)
b = c
fns = []
zns = []
r = False
for c in "abcdefghijklmnopqrstuvwxyz":
if len(d[c]) >3:
r = True
break
if len(d[c]) == 1:
fns.append(c)
if not len(d[c]):
zns.append(c)
if r:
print("NO")
continue
if len(fns) != 2:
print("NO")
continue
stack = [fns.pop()]
vs = set()
while stack:
v = stack.pop()
vs.add(v)
zns.append(v)
for u in d[v]:
if u in vs:
continue
stack.append(u)
print("YES")
print("".join(zns))
```
No
| 15,212 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.
Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).
Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.
Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.
Output
For each test case, do the following:
* if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
* otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them.
Example
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza
Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO
Submitted Solution:
```
from sys import stdin
from bisect import bisect_left as bl
from bisect import bisect_right as br
def input():
return stdin.readline()[:-1]
def intput():
return int(input())
def sinput():
return input().split()
def intsput():
return map(int, sinput())
class RangedList:
def __init__(self, start, stop, val=0):
self.shift = 0 - start
self.start = start
self.stop = stop
self.list = [val] * (stop - start)
def __setitem__(self, key, value):
self.list[key + self.shift] = value
def __getitem__(self, key):
return self.list[key + self.shift]
def __repr__(self):
return str(self.list)
def __iter__(self):
return iter(self.list)
# Code
alphas = 'abcdefghijklmnopqrstuvwxyz'
t = intput()
for _ in range(t):
s = input()
if len(s) == 1:
print(alphas)
continue
nextto = {}
for alph in alphas:
nextto[alph] = set()
for i in range(len(s)):
for j in (i - 1, i + 1):
if j in range(len(s)):
nextto[s[i]].add(s[j])
if any(len(x) >= 3 for x in nextto.values()):
print("NO")
continue
if not any(len(x) == 1 for x in nextto.values()):
print("NO")
continue
for x in nextto:
if len(nextto[x]) == 1:
start = x
ans = [start]
covered = {}
for x in alphas:
covered[x] = False
covered[start] = True
gotcha = False
while not all(covered.values()):
if len(nextto[ans[-1]]) == 0 or len(nextto[ans[-1]]) == 1 and covered[list(nextto[ans[-1]])[0]]:
for k in covered:
if not covered[k] and len(nextto[k]) < 2:
ans.append(k)
break
else:
print("NO")
gotcha = True
break
else:
for d in nextto[ans[-1]]:
if not covered[d]:
ans.append(d)
covered[ans[-1]] = True
'''
print(ans)
print(covered)
input()
'''
if not gotcha:
print("YES")
print(''.join(ans))
```
No
| 15,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Tags: brute force, dfs and similar, graphs, math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
print = sys.stdout.write
MAX = 2*(10**5) + 1
def divisors(a, ans):
div = []
t1 = int(a**.5)
for i in range(1, min(t1+1,ans)):
if a % i == 0:
div.append(i)
div.append(int(a/i))
div.sort()
return div
for ti in range(int(input())):
n = int(input())
p = [int(x)-1 for x in input().split()]
c = [int(x) for x in input().split()]
ans = float('inf')
visited = [0]*n
for i in range(n):
m = 0
cycle = []
cn = i
while(not visited[cn]):
visited[cn] = 1
cycle.append(c[cn])
cn = p[cn]
m += 1
for fi in divisors(m, ans):
for fj in range(fi):
valid = True
for ci in range(fj, m, fi):
valid &= (cycle[ci] == cycle[fj])
if(valid):
ans = min(ans, fi)
print(str(ans))
print('\n')
```
| 15,214 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Tags: brute force, dfs and similar, graphs, math, number theory
Correct Solution:
```
import sys
def bestval(pp, cc):
# print("BESTVAL:")
# print(pp)
# print(cc)
k = len(pp)
k_2 = k//2+1
for f in range(1, k_2):
if k % f == 0:
for offs in range(f):
good = True
num = cc[offs]
# print(f"{f}, {offs}, {num}: ")
upp = (k//f)//2+1
for j in range(1, upp):
v1 = f*j
v2 = k - v1 + offs
v1 += offs
# print(pp[v1], pp[v2])
if cc[v1] != num or cc[v2] != num:
good = False
break
if good:
return f
return k
for q in range(int(sys.stdin.readline())):
n = int(sys.stdin.readline())
p = [int(j)-1 for j in sys.stdin.readline().split()]
c = [int(j)-1 for j in sys.stdin.readline().split()]
fnd = [0]*n
ans = n+1
for i in range(n):
if not fnd[i]:
ppp = [i]
ccc = [c[i]]
fnd[i] = 1
j = p[i]
while j != i:
fnd[j] = 1
ppp.append(j)
ccc.append(c[j])
j = p[j]
# bb =
# print(bb)
ans = min(ans, bestval(ppp, ccc))
sys.stdout.write(str(ans) + '\n')
```
| 15,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Tags: brute force, dfs and similar, graphs, math, number theory
Correct Solution:
```
def uoc(x):
i=1
arr=[]
while i*i<=x:
if x%i==0:
arr.append(i)
if i*i!=x:
arr.append(x//i)
i+=1
return sorted(arr)
def is_ok(arr, n, u):
#assert u is divide by n
for i in range(u):
cur=i
flg=True
for j in range(n//u):
if arr[cur] != arr[(cur+u)%n]:
flg=False
break
cur+=u
if flg==True:
return True
return False
def min_period(arr):
n = len(arr)
u_arr = uoc(n)
for u in u_arr:
if is_ok(arr, n, u):
return u
for _ in range(int(input())):
n = int(input())
p = [-1] + list(map(int, input().split()))
c = [-1] + list(map(int, input().split()))
used=[0]*(n+1)
ans=float('inf')
for x in range(1, n+1):
if used[x]==0:
arr=[c[x]]
used[x]=1
while used[p[x]]==0:
x=p[x]
used[x]=1
arr.append(c[x])
ans=min(ans, min_period(arr))
print(ans)
```
| 15,216 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Tags: brute force, dfs and similar, graphs, math, number theory
Correct Solution:
```
from itertools import combinations
from collections import Counter
MOD = 998244353
def divs(cnt):
res = set()
firstdiv = -1
lastdiv = cnt
for i in range(2, int(cnt ** 0.5 + 1)):
if i > lastdiv:
break
if cnt % i == 0:
res.add(i)
res.add(cnt // i)
if firstdiv == -1:
firstdiv = i
lastdiv = cnt // i
return sorted(res)
def go():
n = int(input())
p = list(map(lambda x: int(x) - 1, input().split()))
c = list(map(int, input().split()))
done = [False] * n
perms = {}
colhist = {}
colmax={}
for i in range(n):
if not done[i]:
colhist_i = Counter()
done[i] = True
colhist_i[c[i]] += 1
j = p[i]
cnt = 1
while j != i:
done[j] = True
colhist_i[c[j]] += 1
j = p[j]
cnt += 1
colhist[i] = colhist_i
colmax[i]= max(colhist_i.values())
perms[i] = cnt
if len(colhist_i) == 1:
return 1
# print(perms)
# print(colhist)
revperms = {}
lim = n
for pp, cnt in perms.items():
revperms.setdefault(cnt, []).append(pp)
lim = min(lim, cnt)
for cnt in sorted(revperms.keys()):
divisors = divs(cnt)
for divisor in divisors:
if divisor>=lim:
break
for perm in revperms[cnt]:
if colmax[perm]<cnt//divisor:
continue
cols=[-1]*divisor
i=perm
for ind in range(divisor):
cols[ind]=c[i]
i=p[i]
for ind in range(divisor,cnt):
if c[i]!=cols[ind%divisor]:
cols[ind%divisor]=-1
i=p[i]
if any(col!=-1 for col in cols):
lim=divisor
break
return lim
t = int(input())
# t=1
# n,k=map(int,input().split())
ans = []
for _ in range(t):
ans.append(str(go()))
# go()
print('\n'.join(ans))
#
# print (divs(24))
# print (divs(2))
# print (divs(21))
```
| 15,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Tags: brute force, dfs and similar, graphs, math, number theory
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from math import sqrt
t=int(input())
for tests in range(t):
n=int(input())
P=list(map(int,input().split()))
C=list(map(int,input().split()))
for i in range(n):
P[i]-=1
USE=[0]*n
ANS=n
for i in range(n):
if USE[i]==1:
continue
cy=i
CYCLE=[]
while USE[cy]!=1:
USE[cy]=1
cy=P[cy]
CYCLE.append(cy)
#print(CYCLE)
LEN=len(CYCLE)
x=LEN
xr=int(sqrt(x))+1
LIST=[]
for k in range(1,xr+1):
if x%k==0:
LIST.append(k)
LIST.append(x//k)
for l in LIST:
if ANS<=LEN//l:
continue
COLORS=[-1]*(LEN//l)
ind=0
for k in range(LEN):
if COLORS[ind]==-1:
COLORS[ind]=C[cy]
elif COLORS[ind]==C[cy]:
True
else:
COLORS[ind]=0
cy=P[cy]
ind+=1
if ind==LEN//l:
ind=0
for c in COLORS:
if c!=0:
ANS=LEN//l
break
#print(CYCLE,l,COLORS)
print(ANS)
```
| 15,218 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Tags: brute force, dfs and similar, graphs, math, number theory
Correct Solution:
```
from sys import stdin
input = stdin.readline
q = int(input())
for rwerew in range(q):
n = int(input())
p = list(map(int,input().split()))
c = list(map(int,input().split()))
for i in range(n):
p[i] -= 1
przyn = [0] * n
grupa = []
i = 0
while i < n:
if przyn[i] == 1:
i += 1
else:
nowa_grupa = [i]
j = p[i]
przyn[i] = 1
while j != i:
przyn[j] = 1
nowa_grupa.append(j)
j = p[j]
grupa.append(nowa_grupa)
grupacol = []
for i in grupa:
cyk = []
for j in i:
cyk.append(c[j])
grupacol.append(cyk)
#print(grupacol)
mini = 234283742834
for cykl in grupacol:
dziel = []
d = 1
while d**2 <= len(cykl):
if len(cykl)%d == 0:
dziel.append(d)
d += 1
dodat = []
for d in dziel:
dodat.append(len(cykl)/d)
dziel_ost = list(map(int,dziel + dodat))
#print(dziel_ost, len(cykl))
for dzielnik in dziel_ost:
for i in range(dzielnik):
indeks = i
secik = set()
chuj = True
while indeks < len(cykl):
secik.add(cykl[indeks])
indeks += dzielnik
if len(secik) > 1:
chuj = False
break
if chuj:
mini = min(mini, dzielnik)
print(mini)
```
| 15,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Tags: brute force, dfs and similar, graphs, math, number theory
Correct Solution:
```
import sys
readline = sys.stdin.readline
readlines = sys.stdin.readlines
int1 = lambda x: int(x) - 1
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int1, readline().split()))
prn = lambda x: print(*x, sep='\n')
def solve():
n = ni()
p = nl()
c = nl()
ans = n
use = [0]*n
for i in range(n):
if use[i]:
continue
v = i
cc = [c[v]]
use[v] = 1
while not use[p[v]]:
v = p[v]
cc.append(c[v])
use[v] = 1
m = len(cc)
ans = min(ans, m)
flag = 0
for j in range(1, min(m//2 + 1, ans)):
if m % j:
continue
ok = [True]*m
for k in range(j):
if len(set(cc[k::j])) == 1:
ans = min(ans, j)
flag = 1
break
if flag:
break
print(ans)
return
# solve()
T = ni()
for _ in range(T):
solve()
```
| 15,220 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Tags: brute force, dfs and similar, graphs, math, number theory
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
print = sys.stdout.write
def divisors(a, ans):
div = []
t1 = int(a**.5)
for i in range(1, min(t1+1,ans)):
if a % i == 0:
div.append(i)
div.append(int(a/i))
div.sort()
return div
for _ in range(int(input())):
n = int(input())
p = [int(i)-1 for i in input().split()]
c = list(map(int, input().split()))
alive = [True]*n
cycle = []
ans = n
for i in p:
if alive[i]:
cycle.append([])
while True:
if alive[i]:
alive[i] = False
cycle[-1].append(i)
i = p[i]
else:
break
for i in cycle:
t2 = len(i)
div = divisors(t2, ans)
for j in div:
if j >= ans:
break
for q in range(j):
t4 = c[i[q]]
for r in range(q+j,t2,j):
if t4 != c[i[r]]:
break
else:
ans = min(ans, j)
print(str(ans) + '\n')
```
| 15,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Submitted Solution:
```
import sys;input=sys.stdin.readline
def divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
divisors.sort()
return divisors
def check(p):
m = len(p)
for k in divisors(m):
for i in range(k):
ty = None
for j in range(m//k):
# print(j)
# print(j, p[i+j*k])
y = Y[p[i+j*k]-1]
if ty == None:
ty = y
elif y != ty:
break
else:
return k
t, = map(int, input().split())
for _ in range(t):
n = int(input())
X = list(map(int, input().split()))
Y = list(map(int, input().split()))
vs = set()
ps = []
for i in range(1, n+1):
if i in vs:
continue
vs.add(i)
stack = [i]
path = [i]
while stack:
v = stack.pop()
u = X[v-1]
if u in vs:
continue
path.append(u)
vs.add(u)
stack.append(u)
ps.append(path)
m = n
for p in ps:
m = min(m, check(p))
print(m)
```
Yes
| 15,222 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Submitted Solution:
```
import sys
input = sys.stdin.readline
input_list = lambda: list(map(int, input().split()))
def factorize(n):
a = []
i = 1
while i*i<=n:
if n % i == 0:
a.append(i)
if i*i != n:
a.append(n//i)
i += 1
return a
def main():
t = int(input())
while t>0:
t -= 1
n = int(input())
p = input_list()
c = input_list()
ans = len(p)
used = [False]*(len(p) + 1)
length = len(p)
for i in range(length):
if used[i]:continue
used[i] = True
j = p[i] - 1
temp = [j]
while j!=i:
j = p[j] - 1
used[j] = True
temp.append(j)
t_len = len(temp)
factors = factorize(t_len)
ans = min(ans, t_len)
for k in factors:
for l in range(k):
index = (l + k)%t_len
while(index != l and c[temp[(index - k + t_len)%t_len]] == c[temp[index]]):
index = (index + k)%t_len
if (index == l):
ans = min(ans, k)
print(ans)
#stop = timeit.default_timer()
#print(stop - start)
def exp(a, n):
if n==0: return 1
result = a
residue = 1
while n>1:
if n%2==1: residue = residue * result % MOD
result = result * result % MOD
return result * residue % MOD
# call to the main function
main()
```
Yes
| 15,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Submitted Solution:
```
t = int(input())
def find_loops():
seen = [0] * n
loops = []
for i in range(n):
if seen[i]:
continue
j = i
l = []
while 1:
if seen[j]:
break
l.append(j)
seen[j] = 1
j = p[j] - 1
loops.append(l)
return loops
def get_factors(number):
factors = []
i = 1
while i ** 2 <= number:
if number % i == 0:
factors.append(i)
if i ** 2 != number:
factors.append(int(number / i))
i = i + 1
return factors
def process_loop(loop):
length = len(loop)
# print("in process loop, loop is {}, length is {}, factors are {}".format(loop, length, get_factors(length)))
min_factor = length
for i in get_factors(length):
# print("checking factors, factor is: {}".format(i))
is_ok = [1] * i
for j in range(length):
# print("color of the {}th element of loop is {}".format(j, c[loop[j]]))
if c[loop[j]] != c[loop[j % i]]:
is_ok[j % i] = 0
for a in is_ok:
if a == 1:
min_factor = min(min_factor, i)
return min_factor
for x in range(t):
n = int(input())
p = list(map(int, input().strip().split()))
c = list(map(int, input().strip().split()))
loops = find_loops()
min_k = n
for loop in loops:
min_k = min(min_k, process_loop(loop))
if min_k == 1:
break
print(min_k)
```
Yes
| 15,224 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
def factor(a):
res=[]
for d in range(1,a+1):
if d**2>a:break
if a%d==0:
res.append(d)
if d**2!=a:res.append(a//d)
return res
def solve(ans):
for f in ff:
if f>=ans:return ans
for sj in range(f):
pc = colors[sj]
for j in range(sj + f, cn, f):
c = colors[j]
if c != pc: break
pc = c
else:return f
q=II()
for _ in range(q):
n=II()
pp=LI1()
cc=LI()
fin=[False]*n
ans=10**9
for i in range(n):
if fin[i]:continue
colors=[]
while not fin[i]:
fin[i]=True
colors.append(cc[i])
i=pp[i]
cn=len(colors)
ff=factor(cn)
ff.sort()
ans=solve(ans)
print(ans)
main()
```
Yes
| 15,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
q = int(input())
for rwerew in range(q):
n = int(input())
p = list(map(int,input().split()))
c = list(map(int,input().split()))
for i in range(n):
p[i] -= 1
przyn = [0] * n
grupa = []
i = 0
while i < n:
if przyn[i] == 1:
i += 1
else:
nowa_grupa = [i]
j = p[i]
przyn[i] = 1
while j != i:
przyn[j] = 1
nowa_grupa.append(j)
j = p[j]
grupa.append(nowa_grupa)
grupacol = []
for i in grupa:
cyk = []
for j in i:
cyk.append(c[j])
grupacol.append(cyk)
#print(grupacol)
mini = 234283742834
for cykl in grupacol:
dziel = []
d = 1
while d**2 <= len(cykl):
if len(cykl)%d == 0:
dziel.append(d)
d += 1
dodat = []
for d in dziel:
dodat.append(len(cykl)/d)
dziel_ost = list(map(int,dziel + dodat))
#print(dziel_ost, len(cykl))
for dzielnik in dziel_ost:
indeks = dzielnik - 1
secik = set()
chuj = True
while indeks < len(cykl):
secik.add(cykl[indeks])
indeks += dzielnik
if len(secik) > 1:
chuj = False
break
if chuj:
mini = min(mini, dzielnik)
print(mini)
```
No
| 15,226 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Submitted Solution:
```
# n = int(input())
# a = list(map(int, input().split()))
# n, k = map(int, input().split())
MOD = 998244353
T = 1
T = int(input())
for t in range(1, T + 1):
# print('Case #' + str(t) + ': ', end = '')
n = int(input())
p = list(map(int, input().split()))
colour = list(map(int, input().split()))
child = {}
parent = {}
for i in p:
child[i] = p[i - 1]
parent[child[i]] = i
visited = set([])
min_dist = MOD
for i in p:
cycle = []
curr = i
while(curr not in visited):
visited.add(curr)
cycle.append(curr)
curr = child[curr]
cycle = 2 * cycle
prev = {}
for i in range(len(cycle)):
if colour[parent[cycle[i]] - 1] in prev:
min_dist = min(i - prev[colour[parent[cycle[i]] - 1]], min_dist)
prev[colour[parent[cycle[i]] - 1]] = i
print(min_dist)
```
No
| 15,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Submitted Solution:
```
answers = []
for _ in range(int(input())):
n = int(input())
P = [0] + list(map(int, input().split()))
C = [0] + list(map(int, input().split()))
ccl = []
used = [0] * (n+1)
for v in range(1, n + 1):
if used[v] == 0:
used[v] = 1
x = P[v]
kk = [C[v]]
while used[x] == 0:
kk.append(C[x])
used[x] = 1
x = P[x]
ccl.append(kk)
ans = n
for i in range(1, n + 1):
if i == P[i]:
ans = 1
for p in ccl:
g = len(p)
for h in range(1, int(g**0.5)+1):
if g % h == 0:
hh = g // h
for Q in [h, hh]:
can = 0
for st in range(Q):
color = p[st]
cac = 1
for z in range(1, g // Q):
if p[st + h * z] != color:
cac = 0
if cac == 1:
can = 1
if can:
ans = min(ans, Q)
print(ans)
```
No
| 15,228 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.
Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).
We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k times}.
Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).
It can be proved that the answer always exists.
Input
The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.
Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.
The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.
It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.
Example
Input
3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4
Output
1
5
2
Note
In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.
In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.
In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
Submitted Solution:
```
from sys import stdin
Pr = [1]
Cm = set()
for i in range(2, 450):
if i in Cm:
continue
Pr.append(i)
for j in range(i, 450, i):
Cm.add(j)
seen = [False for _ in range(200000)]
def check(S,f):
for seq in S:
ln = len(seq)
if ln%f:
continue
for s in range(f):
C = seq[s]
for i in range(s,ln,f):
if seq[i]!= C:
break
else:
return True
T = int(stdin.readline().strip())
ans = []
for _ in range(T):
n = int(stdin.readline().strip())
p = list(map(int, stdin.readline().split()))
c = list(map(int, stdin.readline().split()))
pd,m = 1,n
S = []
for i in range(n):
seen[i] = False
for i in range(n):
if seen[i]:
continue
u,seq = i,[]
while not seen[u]:
seen[u] = True
seq.append(c[u])
u = p[u]-1
S.append(seq)
pd,m = pd*len(seq),min(m,len(seq))
for pr in Pr:
if pr >= m:
ans.append(str(m))
break
if pd%pr==0 and check(S,pr):
ans.append(str(pr))
break
print('\n'.join(ans))
```
No
| 15,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Tags: bitmasks, greedy, math
Correct Solution:
```
n = int(input())
arr = list(map(int,input().split()))
bits = [0 for i in range(21)]
ans = [0 for i in range(n)]
for i in range(n):
for j in range(21):
if ((arr[i]>>j) & 1):
bits[j]+=1
for i in range(21):
for j in range(bits[i]):
x = pow(2,i)
ans[j] = ans[j]|x
f = 0
for x in ans:
f+=(x*x)
print(f)
```
| 15,230 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Tags: bitmasks, greedy, math
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
bit = list(0 for i in range(21))
for i in arr:
k = i
s = list()
while k != 0:
s.append(k % 2)
k //= 2
for j in range(0, len(s)):
bit[j] += s[j]
ans = 0
for i in range(n):
k = 0
for j in range(len(bit)):
if bit[j] > 0:
k += 2**j
bit[j] -= 1
ans += k ** 2
print(ans)
```
| 15,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Tags: bitmasks, greedy, math
Correct Solution:
```
# f = open('test.py')
# def input():
# return f.readline().replace('\n','')
# import heapq
# import bisect
# from collections import deque
from collections import defaultdict
def read_list():
return list(map(int,input().strip().split(' ')))
def print_list(l):
print(' '.join(map(str,l)))
N = int(input())
a = read_list()
dic = defaultdict(int)
for n in a:
i = 0
while n:
if n&1:
dic[i]+=1
i+=1
n>>=1
res = 0
while dic:
tmp = 0
mi = min(dic.values())
dd = []
for i in dic:
tmp+=(1<<i)
dic[i]-=mi
if dic[i]==0:
dd.append(i)
for aa in dd:
del dic[aa]
res+=(tmp**2)*mi
print(res)
```
| 15,232 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Tags: bitmasks, greedy, math
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
@author: Saurav Sihag
"""
rr = lambda: input().strip()
# rri = lambda: int(rr())
rri = lambda: int(stdin.readline())
rrm = lambda: [int(x) for x in rr().split()]
# stdout.write(str()+'\n')
from sys import stdin, stdout
def sol():
n=rri()
v=rrm()
x=[0 for i in range(25)]
for i in v:
z=i
j=0
while z>0:
if (z&1)==1:
x[j]+=1
z=z>>1
j+=1
res=0
for i in range(n):
a=0
for j in range(25):
if x[j]>i:
z=1<<j
a+=z
res+=a*a
print(res)
return
sol()
# T = rri()
# for t in range(1, T + 1):
# ans = sol()
# print("Case #{}: {}".format(t, ans))
```
| 15,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Tags: bitmasks, greedy, math
Correct Solution:
```
n=int(input())
arr=[int(i) for i in input().split()]
binary=[]
num=[0 for i in range(21)]
for i in range(n):
b=bin(arr[i]).replace("0b","")
if len(b)<21:
for k in range(21-len(b)):
b='0'+b
binary.append(b)
for i in range(21):
for j in range(n):
if binary[j][i]=='1':
num[i]+=1
m=max(num)
f=0
ans=0
while(f<=m):
f+=1
a=''
for i in range(21):
if num[i]>0:
num[i]-=1
a=a+'1'
else:
a=a+'0'
ans=ans+(int(a,2))**2
print(ans)
```
| 15,234 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Tags: bitmasks, greedy, math
Correct Solution:
```
n = int(input())
A = list(map(int,input().split()))
bits = [0 for i in range(21)]
for i in range(n):
x = A[i]
for j in range(20):
if(x&(1<<j)):
bits[j]+=1
ans = 0
while(True):
flag = 0
num=0
for i in range(20):
if(bits[i]!=0):
flag=1
bits[i]-=1
num+=2**i
ans+=num*num
if(flag ==0):
break
print(ans)
```
| 15,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Tags: bitmasks, greedy, math
Correct Solution:
```
BITS = 32
n = int(input())
a = list(map(int, input().split()))
bits = [0 for i in range(BITS)]
for x in a:
for i in range(BITS):
if x % 2 == 1:
bits[i] += 1
x //= 2
M = max(bits)
ret = 0
while M > 0:
b = 0
for i in range(BITS):
if bits[i] > 0:
b += 2 ** i
bits[i] -= 1
M -= 1
ret += b * b
print(ret)
```
| 15,236 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Tags: bitmasks, greedy, math
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
n = int(input())
a = list(map(int, input().split()))
cnt = [0] * 20
for i, num in enumerate(a):
for j in range(20):
if num & (1 << j):
cnt[j] += 1
ans = 0
for i in range(n):
tmp = 0
for j in range(20):
if cnt[j] > 0:
tmp += (1 << j)
cnt[j] -= 1
ans += tmp ** 2
print(ans)
```
| 15,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Submitted Solution:
```
n=int(input())
arr=list(map(int,input().split()))
binary=[0]*20
for val in arr:
temp=0
while(val):
if(val & 1):
binary[temp]+=1
val>>=1
temp+=1
ans=0
for i in range(n):
temp=0
for place in range(20):
if(binary[place]):
temp+=(1<<place)
binary[place]-=1
ans+=(temp**2)
print(ans)
```
Yes
| 15,238 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
deg_count = []
for i in range(n):
deg = 0
while a[i] > 0:
# print("a[i]", a[i])
if deg > len(deg_count) - 1:
deg_count.append(0)
if a[i] & 1:
deg_count[deg] += 1
a[i] >>= 1
deg += 1
# print("deg", deg)
# print(*deg_count, sep=" ")
ans = 0
for i in range(n):
num = 0
for deg in range(len(deg_count)):
if deg_count[deg] > 0:
num += 1 << deg
deg_count[deg] -= 1
ans += num*num
print(ans)
```
Yes
| 15,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Submitted Solution:
```
def prog():
n = int(input())
nums = list(map(int,input().split()))
amounts = [0]*20
for num in nums:
num = bin(num)[:1:-1]
for i in range(len(num)):
amounts[i] += "1" == num[i]
totals = []
while True:
left = False
for amount in amounts:
if amount:
left = True
break
if not left:
break
curr = 0
for i in range(20):
if amounts[i]:
amounts[i] -= 1
curr += 1 << i
totals.append(curr)
ans = 0
for total in totals:
ans += total**2
print(ans)
prog()
```
Yes
| 15,240 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Submitted Solution:
```
import sys, math,os
from io import BytesIO, IOBase
# data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from bisect import bisect_left as bl, bisect_right as br, insort
# from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
# from itertools import permutations,combinations
# from decimal import Decimal
def data(): return sys.stdin.buffer.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var): sys.stdout.write(' '.join(map(str, var)) + '\n')
def out(var): sys.stdout.write(str(var) + '\n')
#sys.setrecursionlimit(100000 + 1)
#INF = float('inf')
mod = 998244353
n=int(data())
a=mdata()
d=[[0]*20 for i in range(n+1)]
a.sort()
for i in range(n-1,-1,-1):
d[i]=d[i+1][:]
b=bin(a[i])[2:][::-1]
for j in range(len(b)):
if b[j]=='0':
d[i][j]+=1
ans=0
d1=[0]*20
for i in range(n):
b = list(bin(a[i])[2:][::-1])
for j in range(len(b)):
if b[j]=='1' and d[i][j]+d1[j]>0:
b[j]='0'
d1[j]-=1
else:
if d[i][j]+d1[j]<=0:
b[j]='1'
ans+=int(''.join(b[::-1]),2)**2
out(ans)
```
Yes
| 15,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Submitted Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
from itertools import permutations
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
from math import *
import copy
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
def li():
return [int(x) for x in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def bo(i):
return ord(i)-ord('a')
from copy import *
from bisect import *
def lol(p):
h=bin(p)[2:]
h="0"*(60-len(h))+h
print(h)
n=fi()
a=li()
ans=0
c=[0 for i in range(31)]
for i in range(n):
for j in range(30):
if a[i]&(1<<j):
c[j]+=1
for i in range(n):
x=0
for j in range(30):
if c[j]>0:
c[j]-=1
x|=(1<<j)
ans+=x**2
print(ans)
```
No
| 15,242 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Submitted Solution:
```
import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
n,=I()
l=sorted(I())
an=0
for i in range(n-1):
x=l[i]
l[i]=l[i]&l[i+1]
an+=l[i]*l[i]
l[i+1]|=x
an+=l[n-1]*l[n-1]
print(an)
```
No
| 15,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Submitted Solution:
```
def solve(n, l):
l = sorted(l)[::-1]
ans = 0
i = 0
while i < n:
p_org = l[i-1]*l[i-1] + l[i]*l[i]
g = l[i-1]&l[i]
h = l[i-1]|l[i]
p_new = g*g + h*h
if p_new > p_org:
l[i-1] = h
l[i] = g
l = sorted(l)[::-1]
i += 1
for i in range(n):
ans += l[i]*l[i]
return ans
n = int(input())
l = list(map(int, input().split()))
print(solve(n, l))
```
No
| 15,244 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gottfried learned about binary number representation. He then came up with this task and presented it to you.
You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).
After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).
Output
Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.
Examples
Input
1
123
Output
15129
Input
3
1 3 5
Output
51
Input
2
349525 699050
Output
1099509530625
Note
In the first sample no operation can be made, thus the answer is 123^2.
In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.
If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
s = 0
l = []
if n == 1:
print(a[0]*a[0])
else:
for i in range(1,len(a),2):
l.append(a[i] & a[i - 1])
l.append(a[i] | a[i-1])
for i in range(len(l)):
l[i] = l[i]*l[i]
if n%2 != 0:
print(sum(l)+1)
else:
print(sum(l))
```
No
| 15,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments [l_1, r_1], [l_2, r_2], ..., [l_n, r_n]. Each segment has one of two colors: the i-th segment's color is t_i.
Let's call a pair of segments i and j bad if the following two conditions are met:
* t_i ≠ t_j;
* the segments [l_i, r_i] and [l_j, r_j] intersect, embed or touch, i. e. there exists an integer x such that x ∈ [l_i, r_i] and x ∈ [l_j, r_j].
Calculate the maximum number of segments that can be selected from the given ones, so that there is no bad pair among the selected ones.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments.
The next n lines contains three integers l_i, r_i, t_i (1 ≤ l_i ≤ r_i ≤ 10^9; t_i ∈ \{1, 2\}) — description of the i-th segment.
Output
Print the maximum number of segments that can be selected, so that there is no bad pair among the selected segments.
Examples
Input
3
1 3 1
4 6 2
2 5 1
Output
2
Input
5
5 8 1
1 3 2
3 4 2
6 6 1
2 10 2
Output
4
Input
7
19 20 1
13 15 2
6 11 2
4 10 1
14 17 1
13 13 2
5 9 1
Output
5
Tags: data structures, dp, graph matchings, sortings
Correct Solution:
```
class LazySegmentTree():
def __init__(self,n,init,merge_func=max,ide_ele=-10**18):
self.n=(n-1).bit_length()
self.merge_func=merge_func
self.ide_ele=ide_ele
self.data=[0 for i in range(1<<(self.n+1))]
self.lazy=[0 for i in range(1<<(self.n+1))]
for i in range(n):
self.data[2**self.n+i]=init[i]
for i in range(2**self.n-1,0,-1):
self.data[i]=self.merge_func(self.data[2*i],self.data[2*i+1])
def propagate_above(self,i):
m=i.bit_length()-1
for bit in range(m,0,-1):
v=i>>bit
add=self.lazy[v]
self.lazy[v]=0
self.data[2*v]+=add
self.data[2*v+1]+=add
self.lazy[2*v]+=add
self.lazy[2*v+1]+=add
def remerge_above(self,i):
while i:
i>>=1
self.data[i]=self.merge_func(self.data[2*i],self.data[2*i+1])+self.lazy[i]
def update(self,l,r,x):
l+=1<<self.n
r+=1<<self.n
l0=l//(l&-l)
r0=r//(r&-r)-1
while l<r:
self.data[l]+=x*(l&1)
self.lazy[l]+=x*(l&1)
l+=(l&1)
self.data[r-1]+=x*(r&1)
self.lazy[r-1]+=x*(r&1)
l>>=1
r>>=1
self.remerge_above(l0)
self.remerge_above(r0)
def query(self,l,r):
l+=1<<self.n
r+=1<<self.n
l0=l//(l&-l)
r0=r//(r&-r)-1
self.propagate_above(l0)
self.propagate_above(r0)
res=self.ide_ele
while l<r:
if l&1:
res=self.merge_func(res,self.data[l])
l+=1
if r&1:
res=self.merge_func(res,self.data[r-1])
l>>=1
r>>=1
return res
import sys
input=sys.stdin.buffer.readline
n=int(input())
a=[]
b=[]
val=set()
for i in range(n):
l,r,t=map(int,input().split())
if t==1:
a.append((l,r))
else:
b.append((l,r))
val.add(l)
val.add(r)
val=sorted(list(val))
comp={i:e for e,i in enumerate(val)}
N=len(val)
querya=[[] for i in range(N)]
queryb=[[] for i in range(N)]
for i in range(len(a)):
l,r=a[i]
a[i]=(comp[l],comp[r])
querya[comp[r]].append(comp[l])
for i in range(len(b)):
l,r=b[i]
b[i]=(comp[l],comp[r])
queryb[comp[r]].append(comp[l])
init=[0]*N
LSTA=LazySegmentTree(N,init)
LSTB=LazySegmentTree(N,init)
dp=[0]*N
for i in range(N):
for l in querya[i]:
LSTA.update(0,l+1,1)
for l in queryb[i]:
LSTB.update(0,l+1,1)
temp1=LSTA.query(0,i+1)
temp2=LSTB.query(0,i+1)
dp[i]=max(temp1,temp2)
if i!=N-1:
LSTA.update(i+1,i+2,dp[i])
LSTB.update(i+1,i+2,dp[i])
print(dp[N-1])
```
| 15,246 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments [l_1, r_1], [l_2, r_2], ..., [l_n, r_n]. Each segment has one of two colors: the i-th segment's color is t_i.
Let's call a pair of segments i and j bad if the following two conditions are met:
* t_i ≠ t_j;
* the segments [l_i, r_i] and [l_j, r_j] intersect, embed or touch, i. e. there exists an integer x such that x ∈ [l_i, r_i] and x ∈ [l_j, r_j].
Calculate the maximum number of segments that can be selected from the given ones, so that there is no bad pair among the selected ones.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments.
The next n lines contains three integers l_i, r_i, t_i (1 ≤ l_i ≤ r_i ≤ 10^9; t_i ∈ \{1, 2\}) — description of the i-th segment.
Output
Print the maximum number of segments that can be selected, so that there is no bad pair among the selected segments.
Examples
Input
3
1 3 1
4 6 2
2 5 1
Output
2
Input
5
5 8 1
1 3 2
3 4 2
6 6 1
2 10 2
Output
4
Input
7
19 20 1
13 15 2
6 11 2
4 10 1
14 17 1
13 13 2
5 9 1
Output
5
Tags: data structures, dp, graph matchings, sortings
Correct Solution:
```
from operator import add
class LazySegmentTree():
def __init__(self,n,init,merge=max,merge_unit=-10**18,operate=add,operate_unit=0):
self.merge=merge
self.merge_unit=merge_unit
self.operate=operate
self.operate_unit=operate_unit
self.n=(n-1).bit_length()
self.data=[0 for i in range(1<<(self.n+1))]
self.lazy=[0 for i in range(1<<(self.n+1))]
for i in range(n):
self.data[2**self.n+i]=init[i]
for i in range(2**self.n-1,0,-1):
self.data[i]=self.merge(self.data[2*i],self.data[2*i+1])
def propagate_above(self,i):
m=i.bit_length()-1
for bit in range(m,0,-1):
v=i>>bit
add=self.lazy[v]
self.lazy[v]=0
self.data[2*v]=self.operate(self.data[2*v],add)
self.data[2*v+1]=self.operate(self.data[2*v+1],add)
self.lazy[2*v]=self.operate(self.lazy[2*v],add)
self.lazy[2*v+1]=self.operate(self.lazy[2*v+1],add)
def remerge_above(self,i):
while i:
i>>=1
self.data[i]=self.operate(self.merge(self.data[2*i],self.data[2*i+1]),self.lazy[i])
def update(self,l,r,x):
l+=1<<self.n
r+=1<<self.n
l0=l//(l&-l)
r0=r//(r&-r)-1
while l<r:
if l&1:
self.data[l]=self.operate(self.data[l],x)
self.lazy[l]=self.operate(self.lazy[l],x)
l+=1
if r&1:
self.data[r-1]=self.operate(self.data[r-1],x)
self.lazy[r-1]=self.operate(self.lazy[r-1],x)
l>>=1
r>>=1
self.remerge_above(l0)
self.remerge_above(r0)
def query(self,l,r):
l+=1<<self.n
r+=1<<self.n
l0=l//(l&-l)
r0=r//(r&-r)-1
self.propagate_above(l0)
self.propagate_above(r0)
res=self.merge_unit
while l<r:
if l&1:
res=self.merge(res,self.data[l])
l+=1
if r&1:
res=self.merge(res,self.data[r-1])
l>>=1
r>>=1
return res
import sys
input=sys.stdin.buffer.readline
n=int(input())
a=[]
b=[]
val=set()
for i in range(n):
l,r,t=map(int,input().split())
if t==1:
a.append((l,r))
else:
b.append((l,r))
val.add(l)
val.add(r)
val=sorted(list(val))
comp={i:e for e,i in enumerate(val)}
N=len(val)
querya=[[] for i in range(N)]
queryb=[[] for i in range(N)]
for i in range(len(a)):
l,r=a[i]
a[i]=(comp[l],comp[r])
querya[comp[r]].append(comp[l])
for i in range(len(b)):
l,r=b[i]
b[i]=(comp[l],comp[r])
queryb[comp[r]].append(comp[l])
init=[0]*N
LSTA=LazySegmentTree(N,init)
LSTB=LazySegmentTree(N,init)
dp=[0]*N
for i in range(N):
for l in querya[i]:
LSTA.update(0,l+1,1)
for l in queryb[i]:
LSTB.update(0,l+1,1)
temp1=LSTA.query(0,i+1)
temp2=LSTB.query(0,i+1)
dp[i]=max(temp1,temp2)
if i!=N-1:
LSTA.update(i+1,i+2,dp[i])
LSTB.update(i+1,i+2,dp[i])
print(dp[N-1])
```
| 15,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments [l_1, r_1], [l_2, r_2], ..., [l_n, r_n]. Each segment has one of two colors: the i-th segment's color is t_i.
Let's call a pair of segments i and j bad if the following two conditions are met:
* t_i ≠ t_j;
* the segments [l_i, r_i] and [l_j, r_j] intersect, embed or touch, i. e. there exists an integer x such that x ∈ [l_i, r_i] and x ∈ [l_j, r_j].
Calculate the maximum number of segments that can be selected from the given ones, so that there is no bad pair among the selected ones.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments.
The next n lines contains three integers l_i, r_i, t_i (1 ≤ l_i ≤ r_i ≤ 10^9; t_i ∈ \{1, 2\}) — description of the i-th segment.
Output
Print the maximum number of segments that can be selected, so that there is no bad pair among the selected segments.
Examples
Input
3
1 3 1
4 6 2
2 5 1
Output
2
Input
5
5 8 1
1 3 2
3 4 2
6 6 1
2 10 2
Output
4
Input
7
19 20 1
13 15 2
6 11 2
4 10 1
14 17 1
13 13 2
5 9 1
Output
5
Submitted Solution:
```
from sys import stdin
def gcd(x, y):
for i in range(2, int(x**0.5)+2):
if x % i == 0 and y % i == 0:
return gcd(x//i, y//i) * i
return 1
def choose2(n):
n = int(n)
if n <= 1:
return 0
return int(n * (n-1) / 2)
t = int(stdin.readline().strip())
for _ in range(t):
m,d,w = list(map(int, stdin.readline().strip().split(' ')))
# count 1 <= x < y <= min(m,d) s.t. w | (d-1)(y-x)
n = w / gcd(w, d-1)
m = min(m,d)
# count 1 <= x < y <= m s.t. n | (y-x)
ans = choose2((m//n)+1) * (m%n) + choose2(m//n) * (n-m%n)
if ans != int(ans):
raise "Shouldn't happen"
print(int(ans))
```
No
| 15,248 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments [l_1, r_1], [l_2, r_2], ..., [l_n, r_n]. Each segment has one of two colors: the i-th segment's color is t_i.
Let's call a pair of segments i and j bad if the following two conditions are met:
* t_i ≠ t_j;
* the segments [l_i, r_i] and [l_j, r_j] intersect, embed or touch, i. e. there exists an integer x such that x ∈ [l_i, r_i] and x ∈ [l_j, r_j].
Calculate the maximum number of segments that can be selected from the given ones, so that there is no bad pair among the selected ones.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments.
The next n lines contains three integers l_i, r_i, t_i (1 ≤ l_i ≤ r_i ≤ 10^9; t_i ∈ \{1, 2\}) — description of the i-th segment.
Output
Print the maximum number of segments that can be selected, so that there is no bad pair among the selected segments.
Examples
Input
3
1 3 1
4 6 2
2 5 1
Output
2
Input
5
5 8 1
1 3 2
3 4 2
6 6 1
2 10 2
Output
4
Input
7
19 20 1
13 15 2
6 11 2
4 10 1
14 17 1
13 13 2
5 9 1
Output
5
Submitted Solution:
```
n = int(input())
strips = []
for i in range(n):
strips.append(list(map(int, input().split())))
strips.sort(key = lambda a: a[1])
new_strips = []
for index, strip in enumerate(strips):
if index == 0:
new_strips.append(strip)
continue
prev = strips[index - 1]
if not (prev[1] > strip[0] and prev[2] != strip[2]):
new_strips.append(strip)
print(len(new_strips))
```
No
| 15,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments [l_1, r_1], [l_2, r_2], ..., [l_n, r_n]. Each segment has one of two colors: the i-th segment's color is t_i.
Let's call a pair of segments i and j bad if the following two conditions are met:
* t_i ≠ t_j;
* the segments [l_i, r_i] and [l_j, r_j] intersect, embed or touch, i. e. there exists an integer x such that x ∈ [l_i, r_i] and x ∈ [l_j, r_j].
Calculate the maximum number of segments that can be selected from the given ones, so that there is no bad pair among the selected ones.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments.
The next n lines contains three integers l_i, r_i, t_i (1 ≤ l_i ≤ r_i ≤ 10^9; t_i ∈ \{1, 2\}) — description of the i-th segment.
Output
Print the maximum number of segments that can be selected, so that there is no bad pair among the selected segments.
Examples
Input
3
1 3 1
4 6 2
2 5 1
Output
2
Input
5
5 8 1
1 3 2
3 4 2
6 6 1
2 10 2
Output
4
Input
7
19 20 1
13 15 2
6 11 2
4 10 1
14 17 1
13 13 2
5 9 1
Output
5
Submitted Solution:
```
from sys import stdin, stdout
import bisect
def main():
n = int(stdin.readline())
one_l = []
one_r = []
two_l = []
two_r = []
for _ in range(n):
a,b,z = list(map(int, stdin.readline().split()))
if z == 1:
one_l.append(a)
one_r.append(b)
else:
two_l.append(a)
two_r.append(b)
indexes = list(range(len(two_r)))
indexes.sort(key=two_r.__getitem__)
two_l = list(map(two_l.__getitem__, indexes))
two_r = list(map(two_r.__getitem__, indexes))
indexes = list(range(len(one_r)))
indexes.sort(key=one_r.__getitem__)
one_l = list(map(one_l.__getitem__, indexes))
one_r = list(map(one_r.__getitem__, indexes))
o,t = 0,0
mx = -1
val_dp = dict()
val_dp[0] = 0
val_bi = [0]
while o < len(one_l) or t < len(two_l):
if o == len(one_l):
while t < len(two_l):
left = two_l[t]
right = two_r[t]
pos = bisect.bisect_left(val_bi, left) - 1
pos_bi = val_bi[pos]
mx_val = val_dp[pos_bi]
addi = t - bisect.bisect_left(two_r, left) + 1
val_bi.append(right)
val_dp[right] = max(mx, mx_val+addi)
mx = max(mx, val_dp[right])
t+=1
continue
if t == len(two_l):
while o < len(one_l):
left = one_l[o]
right = one_r[o]
pos = bisect.bisect_left(val_bi, left) - 1
pos_bi = val_bi[pos]
mx_val = val_dp[pos_bi]
addi = o - bisect.bisect_left(one_r, left) + 1
val_bi.append(right)
val_dp[right] = max(mx, mx_val+addi)
mx = max(mx, val_dp[right])
o+=1
continue
is_p = False
while o < len(one_l) and one_r[o] <= two_r[t]:
is_p = True
left = one_l[o]
right = one_r[o]
pos = bisect.bisect_left(val_bi, left) - 1
pos_bi = val_bi[pos]
mx_val = val_dp[pos_bi]
addi = o - bisect.bisect_left(one_r, left) + 1
val_bi.append(right)
val_dp[right] = max(mx, mx_val+addi)
mx = max(mx, val_dp[right])
o+=1
if is_p:
continue
while t < len(two_l) and two_r[t] <= one_r[o]:
left = two_l[t]
right = two_r[t]
pos = bisect.bisect_left(val_bi, left) - 1
pos_bi = val_bi[pos]
mx_val = val_dp[pos_bi]
addi = t - bisect.bisect_left(two_r, left) + 1
val_bi.append(right)
val_dp[right] = max(mx, mx_val+addi)
mx = max(mx, val_dp[right])
t+=1
stdout.write(str(mx))
main()
```
No
| 15,250 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments [l_1, r_1], [l_2, r_2], ..., [l_n, r_n]. Each segment has one of two colors: the i-th segment's color is t_i.
Let's call a pair of segments i and j bad if the following two conditions are met:
* t_i ≠ t_j;
* the segments [l_i, r_i] and [l_j, r_j] intersect, embed or touch, i. e. there exists an integer x such that x ∈ [l_i, r_i] and x ∈ [l_j, r_j].
Calculate the maximum number of segments that can be selected from the given ones, so that there is no bad pair among the selected ones.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments.
The next n lines contains three integers l_i, r_i, t_i (1 ≤ l_i ≤ r_i ≤ 10^9; t_i ∈ \{1, 2\}) — description of the i-th segment.
Output
Print the maximum number of segments that can be selected, so that there is no bad pair among the selected segments.
Examples
Input
3
1 3 1
4 6 2
2 5 1
Output
2
Input
5
5 8 1
1 3 2
3 4 2
6 6 1
2 10 2
Output
4
Input
7
19 20 1
13 15 2
6 11 2
4 10 1
14 17 1
13 13 2
5 9 1
Output
5
Submitted Solution:
```
n = int(input())
strips = []
for i in range(n):
strips.append(list(map(int, input().split())))
strips.sort(key = lambda a: a[1])
new_strips = []
for index, strip in enumerate(strips):
if index == 0:
new_strips.append(strip)
continue
prev = new_strips[-1]
if not (prev[1] > strip[0] and prev[2] != strip[2]):
new_strips.append(strip)
print(len(new_strips))
```
No
| 15,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import sys
import math
n = int(sys.stdin.readline())
r = [int(x) for x in sys.stdin.readline().strip().split()]
r = sorted(r)
a = 0
b = n // 3
d = {key:0 for key in r}
for i in range(n):
d[r[i]] += 1
while a != b:
k = math.ceil((a + b) / 2)
rp = 0
for i in d:
rp += min(k, d[i])
if rp < 3 * k:b = k - 1
else:a = k
#print(rp, d, k)
k = a
sys.stdout.write(str(k) + "\n")
rp = []
for i in d:
rp += [i] * (min(k, d[i]))
rp = sorted(rp)
i = 0
for p in range(k):
sys.stdout.write("{} {} {}\n".format(rp[i + 2 * k], rp[i + k], rp[i]))
i += 1
```
| 15,252 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Tags: binary search, data structures, greedy
Correct Solution:
```
from heapq import *
from collections import defaultdict
n = int(input())
r = map(int, input().split())
assoc = defaultdict(int)
for ballsize in r:
assoc[ballsize] += 1
available = [(-1 * val, key) for key, val in assoc.items()]
heapify(available)
ret = []
while len(available) > 2:
a, b, c = heappop(available), heappop(available), heappop(available)
ret.append(sorted([a[1], b[1], c[1]], reverse=True))
# decrease absolute quantities by adding to negative and put them back if not 0
for i, j in (a, b, c):
x, xi = i + 1, j
if x:
heappush(available, (x, xi))
print(len(ret))
for a, b, c in ret:
print(a, b, c)
```
| 15,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Tags: binary search, data structures, greedy
Correct Solution:
```
from io import BytesIO
import os
from collections import Counter
import heapq
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
def main():
count = int(input())
balls = list(map(int, input().split()))
counter = Counter(balls)
pq = []
for value, count in counter.items(): heapq.heappush(pq, (-count, value))
ans = []
while len(pq) > 2:
first = heapq.heappop(pq)
second= heapq.heappop(pq)
third = heapq.heappop(pq)
ans.append(sorted([first[1], second[1], third[1]], reverse=True))
if first[0] < -1: heapq.heappush(pq, (first[0] + 1, first[1]))
if second[0] < -1: heapq.heappush(pq, (second[0] + 1, second[1]))
if third[0] < -1: heapq.heappush(pq, (third[0] + 1, third[1]))
print(len(ans))
for snowman in ans:
print(*snowman)
if __name__ == "__main__":
main()
```
| 15,254 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Tags: binary search, data structures, greedy
Correct Solution:
```
from collections import Counter as c
from heapq import *
input()
a = dict(c([int(x) for x in input().split()]))
d = [(-1 * v, k) for k, v in a.items()]
heapify(d)
ans = []
while len(d) > 2:
a, b, c = heappop(d), heappop(d), heappop(d)
ans.append(sorted([a[1], b[1], c[1]], reverse=True))
for x, y in (a, b, c):
if x+1:
heappush(d, (x+1, y))
print(len(ans))
for x in ans:
print(*x)
```
| 15,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import heapq
from collections import defaultdict
n=int(input())
b=list(map(int,input().split()))
cnt=defaultdict(lambda:0)
for j in b:
cnt[j]+=1
li=[]
for j in list(set(b)):
li.append([-cnt[j],j])
heapq.heapify(li)
l=len(li)
ans=[]
while(l>=3):
ball1=heapq.heappop(li)
ball2 = heapq.heappop(li)
ball3 = heapq.heappop(li)
ans.append([ball1[1],ball2[1],ball3[1]])
cnt1=ball1[0]+1
if cnt1!=0:
heapq.heappush(li,[cnt1,ball1[1]])
else:
l+=-1
cnt2 = ball2[0] + 1
if cnt2 != 0:
heapq.heappush(li, [cnt2, ball2[1]])
else:
l+=-1
cnt3 = ball3[0] + 1
if cnt3 != 0:
heapq.heappush(li, [cnt3, ball3[1]])
else:
l+=-1
print(len(ans))
for j in ans:
j.sort()
print(*j[::-1])
```
| 15,256 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import heapq
n=int(input())
a={}
for r in map(int,input().split()):
a[r]=a.get(r,0)-1
b=[]
for r in a:
b.append((a[r],r))
heapq.heapify(b)
ans,k=[],0
while len(b)>2:
k+=1
c=[heapq.heappop(b),heapq.heappop(b),heapq.heappop(b)]
ans.append([c[0][1],c[1][1],c[2][1]])
for z in c:
if z[0]<-1: heapq.heappush(b,(z[0]+1,z[1]))
print(k)
for z in ans:
z.sort(reverse=True)
print(*z)
```
| 15,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Tags: binary search, data structures, greedy
Correct Solution:
```
from collections import defaultdict
n, t = int(input()), input().split()
p = defaultdict(int)
for i in t:
p[i] += 1
if len(p) < 3: print(0)
else:
q = list(p.items())
a, b = q[0][1], q[1][1]
x, y = 0, 1
if a < b: x, y, a, b = y, x, b, a
for i in range(2, len(q)):
if q[i][1] > b:
if q[i][1] > a: x, y, a, b = i, x, q[i][1], a
else: y, b = i, q[i][1]
k = n // 3
if a > k:
c, s = n - a - b, n - a
if c > b: q[x], q[y], k = (q[x][0], (s >> 1)), (q[y][0], b - (s & 1)), (s >> 1)
else: q[x], q[y], k = (q[x][0], c), (q[y][0], c), c
q = [(int(x), y) for x, y in q]
q.sort(reverse = True)
p, n = [], 2 * k
for i in q:
p += [str(i[0]) + ' '] * i[1]
print(k)
print('\n'.join(p[i] + p[k + i] + p[n + i] for i in range(k)))
```
| 15,258 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
import itertools
import bisect
import heapq
def main():
pass
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")
def binary(n):
return (bin(n).replace("0b", ""))
def decimal(s):
return (int(s, 2))
def pow2(n):
p = 0
while (n > 1):
n //= 2
p += 1
return (p)
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
l.append(i)
n = n / i
if n > 2:
l.append(int(n))
return (l)
def isPrime(n):
if (n == 1):
return (False)
else:
root = int(n ** 0.5)
root += 1
for i in range(2, root):
if (n % i == 0):
return (False)
return (True)
def maxPrimeFactors(n):
maxPrime = -1
while n % 2 == 0:
maxPrime = 2
n >>= 1
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
maxPrime = i
n = n / i
if n > 2:
maxPrime = n
return int(maxPrime)
def countcon(s, i):
c = 0
ch = s[i]
for i in range(i, len(s)):
if (s[i] == ch):
c += 1
else:
break
return (c)
def lis(arr):
n = len(arr)
lis = [1] * n
for i in range(1, n):
for j in range(0, i):
if arr[i] > arr[j] and lis[i] < lis[j] + 1:
lis[i] = lis[j] + 1
maximum = 0
for i in range(n):
maximum = max(maximum, lis[i])
return maximum
def isSubSequence(str1, str2):
m = len(str1)
n = len(str2)
j = 0
i = 0
while j < m and i < n:
if str1[j] == str2[i]:
j = j + 1
i = i + 1
return j == m
def maxfac(n):
root = int(n ** 0.5)
for i in range(2, root + 1):
if (n % i == 0):
return (n // i)
return (n)
def p2(n):
c=0
while(n%2==0):
n//=2
c+=1
return c
def seive(n):
primes=[True]*(n+1)
primes[1]=primes[0]=False
for i in range(2,n+1):
if(primes[i]):
for j in range(i+i,n+1,i):
primes[j]=False
p=[]
for i in range(0,n+1):
if(primes[i]):
p.append(i)
return(p)
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def denofactinverse(n,m):
fac=1
for i in range(1,n+1):
fac=(fac*i)%m
return (pow(fac,m-2,m))
def numofact(n,m):
fac = 1
for i in range(1, n + 1):
fac = (fac * i) % m
return(fac)
def sumnum(n):
s=0
for i in str(n):
s+=int(i)
return s
n=int(input())
l=list(map(int,input().split()))
d={}
for i in l:
if(i not in d):
d.update({i:1})
else:
d[i]+=1
l=[]
for i in d:
l.append([-d[i],i])
l.sort()
#print(l)
ans=[]
while(len(l)>2):
t1,t2,t3=heapq.heappop(l),heapq.heappop(l),heapq.heappop(l)
ans.append([t1[1],t2[1],t3[1]])
#print(t1,t2,t3)
if(t1[0]<-1):
t1[0]+=1
heapq.heappush(l,t1)
if (t2[0]<-1):
t2[0]+=1
heapq.heappush(l,t2)
if (t3[0]<-1):
t3[0]+=1
heapq.heappush(l,t3)
print(len(ans))
for i in ans:
i.sort(reverse=True)
print(*i)
```
| 15,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Submitted Solution:
```
from collections import defaultdict
from heapq import *
n = int(input())
bumbas = list(map(int, input().split()))
bumbu_skaits = defaultdict(int)
for b in bumbas:
bumbu_skaits[b] += 1
pieejamaas = []; sniegaviri = []
for a, v in bumbu_skaits.items():
pieejamaas.append( (v * -1, a) )
heapify(pieejamaas)
while len(pieejamaas) > 2:
a, b, c = heappop(pieejamaas), heappop(pieejamaas), heappop(pieejamaas)
sniegaviri.append( sorted([a[1], b[1], c[1]], reverse = True))
for i, j in ( a, b, c):
jaunaissk = i + 1
if jaunaissk != 0:
heappush(pieejamaas, (jaunaissk, j))
print(len(sniegaviri))
for a, b, c in sniegaviri:
print(a, b, c)
```
Yes
| 15,260 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Submitted Solution:
```
import heapq
n = int(input())
arr = list(map(int, input().split()))
mp, res, pq = {}, [], []
for val in arr:
mp.setdefault(val, 0)
mp[val] += 1
for k in mp:
heapq.heappush(pq, [-mp[k], k])
while len(pq) >= 3:
val = [0] * 3
val[0] = heapq.heappop(pq)
val[1] = heapq.heappop(pq)
val[2] = heapq.heappop(pq)
res.append([val[0][1], val[1][1], val[2][1]])
for i in range(3):
if val[i][0] != -1:
val[i][0] += 1
heapq.heappush(pq, val[i])
print(len(res))
for i in res:
print(*sorted(i, reverse=True))
```
Yes
| 15,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Submitted Solution:
```
from collections import defaultdict
n, t = int(input()), input().split()
p = defaultdict(int)
for i in t:
p[i] += 1
if len(p) < 3: print(0)
else:
q = list(p.items())
a, b = q[0][1], q[1][1]
x, y = 0, 1
if a < b: x, y, a, b = y, x, b, a
for i in range(2, len(q)):
if q[i][1] > b:
if q[i][1] > a: x, y, a, b = i, x, q[i][1], a
else: y, b = i, q[i][1]
k = n // 3
if a > k:
c, s = n - a - b, n - a
if c > b: q[x], q[y], k = (q[x][0], (s >> 1)), (q[y][0], b - (s & 1)), (s >> 1)
#(q[y][0], b - (s & 1)), (s >> 1)
else: q[x], q[y], k = (q[x][0], c), (q[y][0], c), c
q = [(int(x), y) for x, y in q]
q.sort(reverse = True)
p, n = [], 2 * k
for i in q:
p += [str(i[0]) + ' '] * i[1]
print(k)
print('\n'.join(p[i] + p[k + i] + p[n + i] for i in range(k)))
#mas facil estaba examen de Cappo
```
Yes
| 15,262 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Submitted Solution:
```
from sys import stdin
import heapq
n = int(stdin.readline())
balls = [int(x) for x in stdin.readline().split()]
balls.sort()
counts = {}
for x in balls:
if x in counts:
counts[x] += 1
else:
counts[x] = 1
q = []
for x in counts:
heapq.heappush(q, [-counts[x], x])
snowmen = []
while len(q) >= 3:
a = heapq.heappop(q)
b = heapq.heappop(q)
c = heapq.heappop(q)
a[0] += 1
b[0] += 1
c[0] += 1
s = [str(x) for x in sorted([a[1], b[1], c[1]], reverse=True)]
snowmen.append(s)
if a[0] < 0:
heapq.heappush(q, a)
if b[0] < 0:
heapq.heappush(q, b)
if c[0] < 0:
heapq.heappush(q, c)
print(len(snowmen))
for x in snowmen:
print(' '.join(x))
```
Yes
| 15,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Submitted Solution:
```
from collections import Counter as c
from bisect import insort
n = int(input())
a = dict(c([int(x) for x in input().split()]))
b = []
for x, y in a.items():
b.append((y, x))
b = sorted(b)
ans = []
temp = []
while b:
temp.append(b.pop())
if len(temp)==3:
x, y = temp[2]
ans += [sorted([temp[0][1], temp[1][1], temp[2][1]], reverse = True)]*x
for i in range(2):
if temp[i][0] > x:
insort(b, (temp[i][0]-x, temp[i][1]))
temp = []
print(len(ans))
for x in ans:
print(*x)
```
No
| 15,264 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Submitted Solution:
```
import heapq
def get_next(li, n):
spill = []
m = heapq.heappop(li)
while m <= n:
spill.append(m)
m = heapq.heappop(li)
for s in spill:
heapq.heappush(li, s)
return m
def get_triad(li):
heapq.heapify(li)
triple = []
n = heapq.heappop(li)
triple.append(n)
triple.append(get_next(li, triple[-1]))
triple.append(get_next(li, triple[-1]))
#print(triple)
def has_triad(li):
s = set(li)
if len(s) > 2:
return True
return False
def solve(li):
triples = 0
while has_triad(li):
get_triad(li)
triples += 1
return triples
n = int(input())
li = [int(i) for i in input().split(' ')]
print(solve(li))
```
No
| 15,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Submitted Solution:
```
import copy
n = int(input())
ara = list(map(int, input().split()))
mp = {}
rec = [[] for _ in range(n)]
ans = []
for val in ara:
mp.setdefault(val, 0)
mp[val] += 1
while 1:
ara = []
tmp = []
for k in mp:
val = mp.get(k, 0)
mp[k] -=1
if mp[k] == 0:
tmp.append(k)
ara.append(k)
if len(ara) == 3:
break
for key in tmp:
del mp[key]
if len(ara) != 3:
break
else:
ara.sort(reverse=True)
ans.append(copy.copy(ara))
print(len(ans))
for i in ans:
for ii in i:
print(ii, end=' ')
print()
```
No
| 15,266 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1, r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1, r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
Submitted Solution:
```
n = int(input())
radis = list(map(int,input().split()))
balls = {}
for i in radis:
if i in balls:
balls[i] += 1
else:
balls[i] = 1
sorted_balls = sorted(balls.items(),key=lambda ball:ball[1])
rds = []
count = []
for r,c in sorted_balls:
rds.append(r)
count.append(c)
# print(rds)
# print(count)
result = 0
snowman = []
while len(rds) >= 3:
result += 1
snowman.append((rds[-1],rds[-2],rds[-3]))
count[-3] -= 1
if count[-3] == 0:
count.pop(-3)
rds.pop(-3)
count[-2] -= 1
if count[-2] == 0:
count.pop(-2)
rds.pop(-2)
count[-1] -= 1
if count[-1] == 0:
count.pop(-1)
rds.pop(-1)
print(result)
for i,j, k in snowman:
ans = sorted([i,j,k])
print(str(ans[2])+" "+str(ans[1])+" "+str(ans[0]))
```
No
| 15,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a directed acyclic graph (a directed graph that does not contain cycles) of n vertices and m arcs. The i-th arc leads from the vertex x_i to the vertex y_i and has the weight w_i.
Your task is to select an integer a_v for each vertex v, and then write a number b_i on each arcs i such that b_i = a_{x_i} - a_{y_i}. You must select the numbers so that:
* all b_i are positive;
* the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible.
It can be shown that for any directed acyclic graph with non-negative w_i, such a way to choose numbers exists.
Input
The first line contains two integers n and m (2 ≤ n ≤ 18; 0 ≤ m ≤ (n(n - 1))/(2)).
Then m lines follow, the i-th of them contains three integers x_i, y_i and w_i (1 ≤ x_i, y_i ≤ n, 1 ≤ w_i ≤ 10^5, x_i ≠ y_i) — the description of the i-th arc.
It is guaranteed that the lines describe m arcs of a directed acyclic graph without multiple arcs between the same pair of vertices.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_v ≤ 10^9), which must be written on the vertices so that all b_i are positive, and the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible. If there are several answers, print any of them. It can be shown that the answer always exists, and at least one of the optimal answers satisfies the constraints 0 ≤ a_v ≤ 10^9.
Examples
Input
3 2
2 1 4
1 3 2
Output
1 2 0
Input
5 4
1 2 1
2 3 1
1 3 6
4 5 8
Output
43 42 41 1337 1336
Input
5 5
1 2 1
2 3 1
3 4 1
1 5 1
5 4 10
Output
4 3 2 1 2
Tags: bitmasks, dfs and similar, dp, flows, graphs, math
Correct Solution:
```
from heapq import heappush, heappop
class MinCostFlow:
INF = 10**18
def __init__(self, N):
self.N = N
self.G = [[] for i in range(N)]
def add_edge(self, fr, to, cap, cost):
forward = [to, cap, cost, None]
backward = forward[3] = [fr, 0, -cost, forward]
self.G[fr].append(forward)
self.G[to].append(backward)
def flow(self, s, t, f):
N = self.N; G = self.G
INF = MinCostFlow.INF
res = 0
H = [0]*N
prv_v = [0]*N
prv_e = [None]*N
d0 = [INF]*N
dist = [INF]*N
while f:
dist[:] = d0
dist[s] = 0
que = [(0, s)]
while que:
c, v = heappop(que)
if dist[v] < c:
continue
r0 = dist[v] + H[v]
for e in G[v]:
w, cap, cost, _ = e
if cap > 0 and r0 + cost - H[w] < dist[w]:
dist[w] = r = r0 + cost - H[w]
prv_v[w] = v; prv_e[w] = e
heappush(que, (r, w))
if dist[t] == INF:
return None
for i in range(N):
H[i] += dist[i]
d = f; v = t
while v != s:
d = min(d, prv_e[v][1])
v = prv_v[v]
f -= d
res += d * H[t]
v = t
while v != s:
e = prv_e[v]
e[1] -= d
e[3][1] += d
v = prv_v[v]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
n,m = map(int,input().split())
G = MinCostFlow(n+2)
coef = [0 for i in range(n)]
edge = []
for _ in range(m):
x,y,b = map(int,input().split())
G.add_edge(y,x,10**18,-1)
coef[x-1] += b
coef[y-1] -= b
edge.append((x,y))
s = 0
for i in range(n):
if coef[i]<0:
G.add_edge(0,i+1,-coef[i],0)
s -= coef[i]
elif coef[i]>0:
G.add_edge(i+1,n+1,coef[i],0)
#G.add_edge(0,n+1,10**18,0)
f = G.flow(0,n+1,s)
#print(-f)
Edge = [[] for i in range(n)]
use = [False]*m
uf = UnionFindVerSize(n)
for i in range(m):
u,v = edge[i]
for e in G.G[u]:
to = e[0]
if to==v and e[1]:
Edge[v-1].append((u-1,1))
Edge[u-1].append((v-1,-1))
use[i] = True
uf.unite(u-1,v-1)
edge = [(edge[i][0],edge[i][1]) for i in range(m) if not use[i]]
for u,v in edge:
if not uf.is_same_group(u-1,v-1):
Edge[v-1].append((u-1,1))
Edge[u-1].append((v-1,-1))
uf.unite(u-1,v-1)
used_1 = [False]*n
used_2 = [False]*n
lazy = [0 for i in range(n)]
a = [0 for i in range(n)]
def dfs(v,pv):
lazy[v] = min(lazy[v],a[v])
for nv,c in Edge[v]:
if not used_1[nv]:
used_1[nv] = True
a[nv] = a[v] + c
dfs(nv,v)
lazy[v] = min(lazy[v],lazy[nv])
def add(v,pv,ff):
a[v] += ff
for nv,c in Edge[v]:
if not used_2[nv]:
used_2[nv] = True
add(nv,v,ff)
for i in range(n):
if not used_1[i]:
used_1[i] = True
dfs(i,-1)
used_2[i] = True
add(i,-1,-lazy[i]+1)
#print(used_1)
#print(lazy)
print(*a)
```
| 15,268 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a directed acyclic graph (a directed graph that does not contain cycles) of n vertices and m arcs. The i-th arc leads from the vertex x_i to the vertex y_i and has the weight w_i.
Your task is to select an integer a_v for each vertex v, and then write a number b_i on each arcs i such that b_i = a_{x_i} - a_{y_i}. You must select the numbers so that:
* all b_i are positive;
* the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible.
It can be shown that for any directed acyclic graph with non-negative w_i, such a way to choose numbers exists.
Input
The first line contains two integers n and m (2 ≤ n ≤ 18; 0 ≤ m ≤ (n(n - 1))/(2)).
Then m lines follow, the i-th of them contains three integers x_i, y_i and w_i (1 ≤ x_i, y_i ≤ n, 1 ≤ w_i ≤ 10^5, x_i ≠ y_i) — the description of the i-th arc.
It is guaranteed that the lines describe m arcs of a directed acyclic graph without multiple arcs between the same pair of vertices.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_v ≤ 10^9), which must be written on the vertices so that all b_i are positive, and the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible. If there are several answers, print any of them. It can be shown that the answer always exists, and at least one of the optimal answers satisfies the constraints 0 ≤ a_v ≤ 10^9.
Examples
Input
3 2
2 1 4
1 3 2
Output
1 2 0
Input
5 4
1 2 1
2 3 1
1 3 6
4 5 8
Output
43 42 41 1337 1336
Input
5 5
1 2 1
2 3 1
3 4 1
1 5 1
5 4 10
Output
4 3 2 1 2
Submitted Solution:
```
n, m = map(int, input().split())
adj = [[] for i in range(n)]
radj = [[] for i in range(n)]
out = [0] * n
inc = [0] * n
diff = [0] * n
for _ in range(m):
u, v, w = map(int, input().split())
u-=1;v-=1
adj[u].append((v,w))
out[u] += v
diff[u] += w
diff[v] -= w
inc[v] += 1
radj[v].append((u,w))
found = [0] * n
topo = []
stack = [i for i in range(n) if inc[i] == 0]
while stack:
nex = stack.pop()
topo.append(nex)
for v, _ in adj[nex]:
found[v] += 1
if inc[v] == found[v]:
stack.append(v)
best = [-1] * n
bestV = 10 ** 9
out = [n+1] * n
for v in topo:
smol = n + 4
for u, _ in radj[v]:
smol = min(out[u], smol)
out[v] = smol - 1
import itertools
things = [[0] if inc[topo[i]] == 0 else (0,1) for i in range(n)]
for tup in itertools.product(*things):
copy = out[::]
for v in topo[::-1]:
tol = 0
for u, _ in adj[v]:
tol = max(copy[u], tol)
copy[v] = tol + 1
curr = 0
for i in range(n):
curr += copy[i] * diff[i]
if curr < bestV:
bestV = curr
best = copy
print(' '.join(map(str,best)))
```
No
| 15,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a directed acyclic graph (a directed graph that does not contain cycles) of n vertices and m arcs. The i-th arc leads from the vertex x_i to the vertex y_i and has the weight w_i.
Your task is to select an integer a_v for each vertex v, and then write a number b_i on each arcs i such that b_i = a_{x_i} - a_{y_i}. You must select the numbers so that:
* all b_i are positive;
* the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible.
It can be shown that for any directed acyclic graph with non-negative w_i, such a way to choose numbers exists.
Input
The first line contains two integers n and m (2 ≤ n ≤ 18; 0 ≤ m ≤ (n(n - 1))/(2)).
Then m lines follow, the i-th of them contains three integers x_i, y_i and w_i (1 ≤ x_i, y_i ≤ n, 1 ≤ w_i ≤ 10^5, x_i ≠ y_i) — the description of the i-th arc.
It is guaranteed that the lines describe m arcs of a directed acyclic graph without multiple arcs between the same pair of vertices.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_v ≤ 10^9), which must be written on the vertices so that all b_i are positive, and the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible. If there are several answers, print any of them. It can be shown that the answer always exists, and at least one of the optimal answers satisfies the constraints 0 ≤ a_v ≤ 10^9.
Examples
Input
3 2
2 1 4
1 3 2
Output
1 2 0
Input
5 4
1 2 1
2 3 1
1 3 6
4 5 8
Output
43 42 41 1337 1336
Input
5 5
1 2 1
2 3 1
3 4 1
1 5 1
5 4 10
Output
4 3 2 1 2
Submitted Solution:
```
a,b=map(int,input().split());c=0
while a<=b:a*=3;b*=2;c+=1
print(c)
```
No
| 15,270 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a directed acyclic graph (a directed graph that does not contain cycles) of n vertices and m arcs. The i-th arc leads from the vertex x_i to the vertex y_i and has the weight w_i.
Your task is to select an integer a_v for each vertex v, and then write a number b_i on each arcs i such that b_i = a_{x_i} - a_{y_i}. You must select the numbers so that:
* all b_i are positive;
* the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible.
It can be shown that for any directed acyclic graph with non-negative w_i, such a way to choose numbers exists.
Input
The first line contains two integers n and m (2 ≤ n ≤ 18; 0 ≤ m ≤ (n(n - 1))/(2)).
Then m lines follow, the i-th of them contains three integers x_i, y_i and w_i (1 ≤ x_i, y_i ≤ n, 1 ≤ w_i ≤ 10^5, x_i ≠ y_i) — the description of the i-th arc.
It is guaranteed that the lines describe m arcs of a directed acyclic graph without multiple arcs between the same pair of vertices.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_v ≤ 10^9), which must be written on the vertices so that all b_i are positive, and the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible. If there are several answers, print any of them. It can be shown that the answer always exists, and at least one of the optimal answers satisfies the constraints 0 ≤ a_v ≤ 10^9.
Examples
Input
3 2
2 1 4
1 3 2
Output
1 2 0
Input
5 4
1 2 1
2 3 1
1 3 6
4 5 8
Output
43 42 41 1337 1336
Input
5 5
1 2 1
2 3 1
3 4 1
1 5 1
5 4 10
Output
4 3 2 1 2
Submitted Solution:
```
n, m = map(int, input().split())
D = [0]*(n+1)
def cD(n):
if type(D[n]) == int:
return D[n]
else:
result = D[n]()
D[n] = result
return result
def cD_(n):
return cD(n)
def wrap(k):
if type(k) == int:
return lambda : k
else:
return k
c = 0
for _ in range(m):
x, y, w = map(int, input().split())
p = D[x]
D[x] = (lambda k, q: (lambda : max(cD_(k) + 1, wrap(q)())))(y, p)
for _ in range(n+1):
cD(_)
print(' '.join(str(x) for x in D[1:]))
```
No
| 15,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a directed acyclic graph (a directed graph that does not contain cycles) of n vertices and m arcs. The i-th arc leads from the vertex x_i to the vertex y_i and has the weight w_i.
Your task is to select an integer a_v for each vertex v, and then write a number b_i on each arcs i such that b_i = a_{x_i} - a_{y_i}. You must select the numbers so that:
* all b_i are positive;
* the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible.
It can be shown that for any directed acyclic graph with non-negative w_i, such a way to choose numbers exists.
Input
The first line contains two integers n and m (2 ≤ n ≤ 18; 0 ≤ m ≤ (n(n - 1))/(2)).
Then m lines follow, the i-th of them contains three integers x_i, y_i and w_i (1 ≤ x_i, y_i ≤ n, 1 ≤ w_i ≤ 10^5, x_i ≠ y_i) — the description of the i-th arc.
It is guaranteed that the lines describe m arcs of a directed acyclic graph without multiple arcs between the same pair of vertices.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_v ≤ 10^9), which must be written on the vertices so that all b_i are positive, and the value of the expression ∑ _{i = 1}^{m} w_i b_i is the lowest possible. If there are several answers, print any of them. It can be shown that the answer always exists, and at least one of the optimal answers satisfies the constraints 0 ≤ a_v ≤ 10^9.
Examples
Input
3 2
2 1 4
1 3 2
Output
1 2 0
Input
5 4
1 2 1
2 3 1
1 3 6
4 5 8
Output
43 42 41 1337 1336
Input
5 5
1 2 1
2 3 1
3 4 1
1 5 1
5 4 10
Output
4 3 2 1 2
Submitted Solution:
```
n, m = map(int, input().split())
adj = [[] for i in range(n)]
radj = [[] for i in range(n)]
out = [0] * n
inc = [0] * n
diff = [0] * n
for _ in range(m):
u, v, w = map(int, input().split())
u-=1;v-=1
adj[u].append((v,w))
out[u] += v
diff[u] += w
diff[v] -= w
inc[v] += 1
radj[v].append((u,w))
found = [0] * n
topo = []
stack = [i for i in range(n) if inc[i] == 0]
while stack:
nex = stack.pop()
topo.append(nex)
for v, _ in adj[nex]:
found[v] += 1
if inc[v] == found[v]:
stack.append(v)
best = [-1] * n
bestV = 10 ** 9
out = [n+1] * n
for v in topo:
tol = n
for u, _ in radj[v]:
tol = min(out[u], tol)
out[v] = tol - 1
import itertools
things = [(0,1)]*n
for tup in itertools.product(*things):
#tup = list(map(int, '111101010011001110'))
copy = out[::]
for v in topo[::-1]:
if tup[v] and len(adj[v]) > 0:
smol = 0
for u, _ in adj[v]:
smol = max(copy[u], smol)
copy[v] = smol + 1
curr = 0
for i in range(n):
curr += copy[i] * diff[i]
#print(copy, curr)
if curr < bestV:
bestV = curr
best = copy
out = [n+1] * n
for v in topo[::-1]:
tol = 0
for u, _ in adj[v]:
tol = max(out[u], tol)
out[v] = tol + 1
import itertools
things = [(0,1)]*n
for tup in itertools.product(*things):
copy = out[::]
for v in topo:
if tup[v] and len(radj[v]) > 0:
smol = n + 5
for u, _ in radj[v]:
smol = min(copy[u], smol)
copy[v] = smol - 1
curr = 0
for i in range(n):
curr += copy[i] * diff[i]
#print(copy, curr)
if curr < bestV:
bestV = curr
best = copy
print(' '.join(map(str,best)))
```
No
| 15,272 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Tags: greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = [*map(int,input().split())]
d = dict()
if(len(set(a))==1):
print(0)
continue
elif(len(set(a))==n):
print(1)
continue
t = []
for i in range(n):
if(i==0):
t.append(a[i])
else:
if(t[-1]==a[i]):
continue
t.append(a[i])
d = dict()
for i in t:
try:
d[i]+=1
except KeyError:
d[i] = 1
vis =dict()
cnt = 0
mn = int(1e9)
for i in t:
try:
if(vis[i]):
continue
except KeyError:
ans = 0
vis[i] = 1
if(cnt==0):
if(t[-1]==i):
ans = max(0,d[i]-1)
else:
ans = d[i]
else:
if (t[-1] == i):
ans = max(0, d[i])
else:
ans = d[i]+1
cnt+=1
mn = min(mn,ans)
print(mn)
```
| 15,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Tags: greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
mp={}
if len(set(arr))==1:
print(0)
else:
for i in range(n):
if arr[i] in mp:
mp[arr[i]]=mp[arr[i]]+[i]
else:
mp[arr[i]]=[i]
ans = 10**9
for i in mp:
curr = 0
if mp[i][0]!=0:
curr+=1
if mp[i][-1]!=n-1:
curr+=1
for j in range(1,len(mp[i])):
if mp[i][j]!=mp[i][j-1]+1:
curr+=1
ans=min(ans,curr)
print (ans)
```
| 15,274 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
def solve():
n = int(input())
arr = list(map(int, input().split()))
d = dict()
for i in range(n):
el = arr[i]
if not (el in d.keys()):
d[el] = []
d[el].append(i)
ans = 2 ** 31
for i in d.keys():
regions = 0
a = d[i]
for i in range(len(a)):
if(i == 0):
if(a[i] > 0):
regions += 1
if(i == len(a) - 1):
if(a[i] < n - 1):
regions += 1
if(i != len(a) - 1):
if(a[i + 1] - a[i] > 1):
regions += 1
ans = min(ans, regions)
print(ans)
for i in range(t):
solve()
```
| 15,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Tags: greedy, implementation
Correct Solution:
```
T = int(input())
for t in range(T):
n = int(input())
l = list(map(int, input().split()))
new_l = [l[0]]
prev = l[0]
counts = {l[0]: 1}
for i in range(1, n):
cur = l[i]
if cur != prev:
new_l.append(cur)
r = counts.get(cur, 0)
counts[cur] = r + 1
prev = cur
# print(new_l)
# print("counts", counts)
counts[new_l[0]] = counts.get(new_l[0]) - 1
counts[new_l[-1]] = counts.get(new_l[-1]) - 1
# print("counts 2", counts)
print(min(counts.values()) + 1)
```
| 15,276 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Tags: greedy, implementation
Correct Solution:
```
from collections import Counter
def unique(lis):
prev = -1
res = []
for i in lis:
if i != prev:
res.append(i)
prev = i
return res
for _ in range(int(input())):
n = int(input())
lis = list(map(int, input().split()))
lis = unique(lis)
dic = Counter(lis)
mi = float('inf')
for i in dic:
ans = dic[i] + 1
if lis[0] == i:
ans -= 1
if lis[-1] == i:
ans -= 1
mi = min(mi, ans)
print(mi)
```
| 15,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Tags: greedy, implementation
Correct Solution:
```
import sys
import math
import collections
t=int(input())
for w in range(t):
n=int(input())
l=[int(i) for i in input().split()]
d={}
for i in range(n):
if(l[i] in d):
d[l[i]]+=[i]
else:
d[l[i]]=[-1,i]
for i in d:
d[i].append(n)
m=20000000
for i in d:
c=0
for j in range(1,len(d[i])):
if(d[i][j]-d[i][j-1]>1):
c+=1
m=min(m,c)
print(m)
```
| 15,278 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Tags: greedy, implementation
Correct Solution:
```
#import sys, math
#input = sys.stdin.readline
import os
import sys
from io import BytesIO, IOBase
import heapq as h
import bisect
from types import GeneratorType
BUFSIZE = 8192
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index+1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
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")
import collections as col
import math, string
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())
MOD = 10**9+7
mod=10**9+7
t=int(input())
#t=1
p=10**9+7
def ncr_util():
inv[0]=inv[1]=1
fact[0]=fact[1]=1
for i in range(2,300001):
inv[i]=(inv[i%p]*(p-p//i))%p
for i in range(1,300001):
inv[i]=(inv[i-1]*inv[i])%p
fact[i]=(fact[i-1]*i)%p
def z_array(s1):
n = len(s1)
z=[0]*(n)
l, r, k = 0, 0, 0
for i in range(1,n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and s1[r - l] == s1[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and s1[r - l] == s1[r]:
r += 1
z[i] = r - l
r -= 1
return z
'''
MAXN1=100001
spf=[0]*MAXN1
def sieve():
spf[1]=1
for i in range(2,MAXN1):
spf[i]=i
for i in range(4,MAXN1,2):
spf[i]=2
for i in range(3,math.ceil(math.sqrt(MAXN1))):
if spf[i]==i:
for j in range(i*i,MAXN1,i):
if spf[j]==j:
spf[j]=i
def factor(x):
d1={}
x1=x
while x!=1:
d1[spf[x]]=d1.get(spf[x],0)+1
x//=spf[x]
return d1
'''
def solve():
d={}
for i in l:
d[i]=[]
for i in range(n):
d[l[i]].append(i)
mini=n+1
for i in d:
cnt=0
if d[i][0]!=0:
cnt+=1
if d[i][-1]!=n-1:
cnt+=1
for j in range(len(d[i])-1):
if d[i][j+1]-d[i][j]>1:
cnt+=1
mini=min(mini,cnt)
return mini
for _ in range(t):
#x=int(input())
#d={}
#x,y=map(int,input().split())
n=int(input())
#s=input()
#s2=input()
l=list(map(int,input().split()))
#l.sort()
#l.sort(revrese=True)
#l2=list(map(int,input().split()))
#l=str(n)
#l.sort(reverse=True)
#l2.sort(reverse=True)
#l1.sort(reverse=True)
#(solve())
print(solve())
```
| 15,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Tags: greedy, implementation
Correct Solution:
```
from collections import Counter
def main(xs):
xs2 = [xs[0]]
# remove consequtive
for i in range(1, len(xs)):
if xs2[-1] != xs[i]:
xs2.append(xs[i])
counter = Counter(xs2)
for k, v in counter.items():
counter[k] += 1
counter[xs2[0]] -= 1
counter[xs2[-1]] -= 1
first = True
for k, v in counter.items():
if first:
m = v
first = False
else:
if v < m:
m = v
if m < 0:
return 0
else:
return m
if __name__ == "__main__":
t = int(input())
for _ in range(t):
input()
as_ = [int(x) for x in input().split()]
res = main(as_)
print(res)
def test1():
assert True == False
def test2():
assert True == False
def test_min():
assert True == False
def test_max():
assert True == False
```
| 15,280 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
dic={}
x = list(map(int,input().split()))
#x=(list(x[0])+x).append(x[-1])
for j in range(1,n+1):
dic.setdefault(x[j-1],[0,2,0])
dic[x[j-1]][0]+=1
dic[x[j - 1]][2] += 1
for j in range(1,n):
if(x[j]==x[j-1]):
dic[x[j]][2]-=1
#print(dic)
z=0
for key, val in dic.items():
dic[key][2] += 1
if(n==val[0]):
print(0)
z=1
break
if(x[0]==key and val[2]>1):
dic[key][2]-=1
if (x[-1] == key and val[2] > 1):
dic[key][2] -= 1
min=n
if(z==0):
for key, val in dic.items():
if(val[2]<min):
min=val[2]
#print(key,val)
print(min)
```
Yes
| 15,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Submitted Solution:
```
# fuck les dict qui prennent plus de temps que des array wtf
import sys
input=sys.stdin.readline
def main():
n = int(input())
for _ in range(n):
size = int(input())
arr = [*map(int,input().split())]
counter = 9**9
optim = [[] for i in range(size)]
for i, val in enumerate(arr):
optim[val-1].append(i)
for val in optim:
if len(val)==0:
continue
count = 0
if val[0]!=0:count=1
for val1, val2 in zip(val[:-1],val[1:]):
if val2-val1>1:count+=1
if val[-1]!=size-1:count+=1
counter=min(counter,count)
print(counter)
'''
for val in set(arr):
#arr2 = []
count = 0
#dist = 0
b=0
for num in arr:
if num!=val and b == 0:count+=1;b = 1
elif num==val:
b = 0
#if dist>0:arr2 += [j]
#dist=-1
#dist += 1
counter = min(counter,count)
print(counter)
'''
main()
```
Yes
| 15,282 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Submitted Solution:
```
"""
Author: Enivar
Date:
"""
from sys import exit, stderr
from collections import defaultdict
def debug(*args):
for i in args:
stderr.write(str(i)+' ')
stderr.write('\n')
for _ in range(int(input())):
n = int(input())
a = [int(x) for x in input().split()]
st = set(a)
if len(st)==1:
print(0)
continue
d = defaultdict(list)
narr = []
for i in range(n):
try:
prev = d[a[i]][-1]
#debug('prien i',i,prev)
if i-prev==1: d[a[i]][-1] = i
else: d[a[i]].append(i)
continue
except:
d[a[i]].append(i)
continue
mn = 10**18
fg = True
for ke in d:
ln = len(d[ke])
#debug(ke,d[ke])
if ln==1:
if d[ke][0]==0 or d[ke][0]==n-1 or a[0] == ke:
fg = False
print(1)
break
else: mn = min(mn, 2)
else:
if d[ke][-1]==n-1: ln-=1
if a[0]==ke: ln-=1
mn = min(mn,ln+1)
# debug(d[ke])
if fg: print(mn)
```
Yes
| 15,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Submitted Solution:
```
"""
Author - Satwik Tiwari .
24th NOV , 2020 - Tuesday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O 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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
for pp in range(t):
solve(pp)
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
def solve(case):
n = int(inp())
a = lis()
pos = {}
for i in range(n):
if(a[i] not in pos):
pos[a[i]] = [i]
else:
pos[a[i]].append(i)
# print(pos)
ans = inf
for i in pos:
temp = 0
# temp+=len(pos[i])-1
if(pos[i][0] !=0):
temp+=1
if(pos[i][len(pos[i])-1] != n-1):
temp+=1
for j in range(1,len(pos[i])):
if(pos[i][j] != pos[i][j-1] + 1):
temp+=1
ans = min(ans,temp)
print(ans)
# testcase(1)
testcase(int(inp()))
```
Yes
| 15,284 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
seq = list(map(int, input().split()))
str_seq = ",".join(list(map(str, seq)))
if i == 679:
print(seq[5])
x = set(seq)
mn = float("inf")
if len(x) == 1:
print(0)
else:
for num in x:
y = [j for j in str_seq.split(str(num)) if j != "" and j != ","]
mn = min(mn, len(y))
print(mn)
```
No
| 15,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
d={}
k={}
for i in range(n):
if arr[i] in d:
if arr[i-1]!=arr[i]:
d[arr[i]].append(i)
k[arr[i]].append(i)
else:
d[arr[i]]=[i]
k[arr[i]]=[i]
m=n
for j in d:
out=len(d[j])+1
for i in k[j]:
if i==0:
out-=1
elif i==n-1:
out-=1
m=min(out,m)
print(m)
```
No
| 15,286 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Submitted Solution:
```
def counts(arr):
stat = []
n = len(arr)
for i in arr:
if i not in stat:
stat.append(i)
minislands = n
if len(stat) == 1:
return 0
elif len(stat) == n:
return 1
for x in stat:
islands = 0
changed = False
for i in range(n):
if arr[i] == x:
if changed:
changed = False
elif not changed:
islands += 1
changed = True
minislands = min(minislands, islands)
if not minislands or minislands == 2:
return minislands
return minislands
def main():
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().strip().split()))
ind = counts(arr)
print(ind)
if __name__ == "__main__":
main()
```
No
| 15,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a, initially consisting of n integers.
You want to transform this sequence so that all elements in it are equal (i. e. it contains several occurrences of the same element).
To achieve this, you choose some integer x that occurs at least once in a, and then perform the following operation any number of times (possibly zero): choose some segment [l, r] of the sequence and remove it. But there is one exception: you are not allowed to choose a segment that contains x. More formally, you choose some contiguous subsequence [a_l, a_{l + 1}, ..., a_r] such that a_i ≠ x if l ≤ i ≤ r, and remove it. After removal, the numbering of elements to the right of the removed segment changes: the element that was the (r+1)-th is now l-th, the element that was (r+2)-th is now (l+1)-th, and so on (i. e. the remaining sequence just collapses).
Note that you can not change x after you chose it.
For example, suppose n = 6, a = [1, 3, 2, 4, 1, 2]. Then one of the ways to transform it in two operations is to choose x = 1, then:
1. choose l = 2, r = 4, so the resulting sequence is a = [1, 1, 2];
2. choose l = 3, r = 3, so the resulting sequence is a = [1, 1].
Note that choosing x is not an operation. Also, note that you can not remove any occurrence of x.
Your task is to find the minimum number of operations required to transform the sequence in a way described above.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of operations required to transform the given sequence in a way described in the problem statement. It can be proven that it is always possible to perform a finite sequence of operations so the sequence is transformed in the required way.
Example
Input
5
3
1 1 1
5
1 2 3 4 5
5
1 2 3 2 1
7
1 2 3 1 2 3 1
11
2 2 1 2 3 2 1 2 3 1 2
Output
0
1
1
2
3
Submitted Solution:
```
from sys import stdin, stdout, stderr, maxsize
# mod = int(1e9 + 7)
# import re # can use multiple splits
tup = lambda: map(int, stdin.buffer.readline().split())
I = lambda: int(stdin.buffer.readline())
lint = lambda: [int(x) for x in stdin.buffer.readline().split()]
S = lambda: stdin.readline().replace('\n', '').strip()
# def grid(r, c): return [lint() for i in range(r)]
# def debug(*args, c=6): print('\033[3{}m'.format(c), *args, '\033[0m', file=stderr)
stpr = lambda x: stdout.write(f'{x}' + '\n')
star = lambda x: print(' '.join(map(str, x)))
# from math import ceil, floor
from collections import defaultdict
# from bisect import bisect_left, bisect_right
#popping from the end is less taxing,since you don't have to shift any elements
#from collections import Counter
for _ in range(I()):
n = I()
ls = lint()
if _ == 679:
print(*ls )
t = set(ls)
m = maxsize
x = ''.join(map(str , ls))
for i in t:
a = x.split(str(i))
s = sum(map(lambda x : 1 if len(x) > 0 else 0, a))
#print(s,a)
m = min(m , s)
print(m)
```
No
| 15,288 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Tags: brute force, greedy
Correct Solution:
```
from collections import Counter
# region fastio
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")
# endregion
def func(arr):
freq = Counter(arr)
ans = max(Counter(arr).values())
return ans
def main():
for _ in range(int(input())):
_ = input()
arr = list(map(int, input().split()))
print(func(arr))
return
if __name__ == '__main__':
main()
```
| 15,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Tags: brute force, greedy
Correct Solution:
```
t = int(input())
while t>0 :
n = int(input())
a = list(map(int,input().strip().split()))[:n]
ans=1
x = 1
for i in range(1,n,1):
if a[i]==a[i-1]:
x+=1
ans = max(x,ans)
else:
x=1
print(ans)
t= t -1
```
| 15,290 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Tags: brute force, greedy
Correct Solution:
```
t=int(input())
for j in range(0,t):
n=int(input())
s=list(map(int,input().split()))
p={}
for i in range(0,n):
if(p.get(s[i],-1)==-1): p[s[i]]=1
else: p[s[i]]+=1
zd=0
for i in p:
zd=max(zd,p[i])
print(zd)
```
| 15,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Tags: brute force, greedy
Correct Solution:
```
ans = []
for _ in range(int(input())):
n = int(input())
u = list(map(int, input().split()))
mx = 1
k = 1
for i in range(1, n):
if u[i] == u[i - 1]:
k += 1
else:
mx = max(mx, k)
k = 1
mx = max(mx, k)
ans.append(mx)
print('\n'.join(map(str, ans)))
```
| 15,292 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Tags: brute force, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
def println(val):
sys.stdout.write(str(val) + '\n')
ans = []
def solve():
global ans
n = int(input())
a = list(map(int, input().split()))
import collections
cnt = collections.Counter(a)
res = 0
for key in cnt:
if cnt[key] > res:
res = cnt[key]
ans += [str(res)]
for _ in range(int(input()) if 1 else 1):
solve()
print('\n'.join(ans))
```
| 15,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Tags: brute force, greedy
Correct Solution:
```
for tt in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
used = [set() for i in range(150)]
ans = 0
for x in a:
for i in range(150):
if x not in used[i]:
ans = max(ans,i+1)
used[i].add(x)
break
print(ans)
```
| 15,294 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Tags: brute force, greedy
Correct Solution:
```
from collections import deque, Counter,defaultdict as dft
from heapq import heappop ,heappush
from math import log,sqrt,factorial,cos,tan,sin,radians,log2,ceil,floor
from bisect import bisect,bisect_left,bisect_right
from decimal import *
import sys,threading
from itertools import permutations, combinations
from copy import deepcopy
input = sys.stdin.readline
ii = lambda: int(input())
si = lambda: input().rstrip()
mp = lambda: map(int, input().split())
ms= lambda: map(str,input().strip().split(" "))
ml = lambda: list(mp())
mf = lambda: map(float, input().split())
def solve():
n=ii()
arr=ml()
mx=1
pre=arr[0]
count=1
for i in range(1,n):
#print(pre,count,arr[i])
if arr[i]==pre:
count+=1
mx=max(mx,count)
else:
pre=arr[i]
count=1
print(mx)
if __name__ == "__main__":
#tc=1
tc = ii()
for i in range(tc):
solve()
```
| 15,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Tags: brute force, greedy
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if True else 1):
n = int(input())
#a, b = map(int, input().split())
#c, d = map(int, input().split())
a=list(map(int, input().split()))
count = [0]*(n+69)
for i in a:count[i]+=1
print(max(count))
```
| 15,296 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
l = list(map(int,input().split()))
l1 = set(l)
a = []
for i in l1:
a.append(l.count(i))
print(max(a))
```
Yes
| 15,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
def main():
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
c = Counter(a)
ans = 0
for val in c.values():
ans = max(ans, val)
print(ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
Yes
| 15,298 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar has n balls, numbered with integers 1, 2, …, n. Numbers a_1, a_2, …, a_n are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that a_i ≤ a_{i+1} for all 1 ≤ i < n.
Nezzar wants to color the balls using the minimum number of colors, such that the following holds.
* For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.
Please help Nezzar determine the minimum number of colors.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1,a_2,…,a_n (1 ≤ a_i ≤ n). It is guaranteed that a_1 ≤ a_2 ≤ … ≤ a_n.
Output
For each test case, output the minimum number of colors Nezzar can use.
Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
Note
Let's match each color with some numbers. Then:
In the first test case, one optimal color assignment is [1,2,3,3,2,1].
In the second test case, one optimal color assignment is [1,2,1,2,1].
Submitted Solution:
```
for i in range(int(input())):
n = int(input())
l = list(map(int,input().split()))
freq={}
res = 0
for i in l:
if i in freq:
freq[i]+=1
else:
freq[i]=1
if len(freq)==1:
print(list(freq.values())[0])
else:
j = max(freq.values())
print(j)
```
Yes
| 15,299 |
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