message stringlengths 2 16.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 575 109k | cluster float64 16 16 | __index_level_0__ int64 1.15k 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke | instruction | 0 | 27,250 | 16 | 54,500 |
"Correct Solution:
```
N, = map(int, input().split())
G = [set() for _ in range(N+1)]
for i in range(N-1):
A, B = map(int, input().split())
G[A].add(B)
G[B].add(A)
path = []
stack = [1]
vs = set()
while stack:
v = stack.pop()
if v > 0:
vs.add(v)
path.append(v)
for u in G[v]:
if u in vs:
continue
if u == N:
stack = []
path.append(u)
break
stack += [-v, u]
else:
path.pop()
x = path[-(-len(path)//2)]
stack = [1]
vs = set()
while stack:
v = stack.pop()
vs.add(v)
for u in G[v]:
if u in vs:
continue
if u == x:
continue
stack.append(u)
x = len(vs)-1
y = N - 2 - x
if x>y:
print("Fennec")
else:
print("Snuke")
``` | output | 1 | 27,250 | 16 | 54,501 |
Provide a correct Python 3 solution for this coding contest problem.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke | instruction | 0 | 27,251 | 16 | 54,502 |
"Correct Solution:
```
from collections import Counter
N = int(input())
graph =[[] for _ in range(N + 1)]
visited = [None] * (N + 1)
for _ in range(N - 1):
a, b = map(int, input().split())
graph[a].append(b)
graph[b].append(a)
v = 1
adv_v = N
update = True
labels = [0, 1]
vs = [v, adv_v]
prev_vs = [-1, -1]
visited[1] = 0
visited[N] = 1
while update:
update = False
next_cand = []
for label, v, prev_v in zip(labels, vs, prev_vs):
for next_v in graph[v]:
if next_v == prev_v:
continue
if (visited[next_v] is None):
update = True
next_cand.append((next_v, label, v))
vs = []
labels = []
prev_vs = []
for next_v, label, prev_v in next_cand:
if visited[next_v] is None:
visited[next_v] = label
vs.append(next_v)
prev_vs.append(prev_v)
labels.append(label)
c = Counter(visited)
if c[0] > c[1]:
print("Fennec")
else:
print("Snuke")
``` | output | 1 | 27,251 | 16 | 54,503 |
Provide a correct Python 3 solution for this coding contest problem.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke | instruction | 0 | 27,252 | 16 | 54,504 |
"Correct Solution:
```
from collections import deque
n = int(input())
board = [0]*n
edge = [[] for i in range(n)]
for i in range(n-1):
a,b = map(int, input().split())
edge[a-1].append(b-1)
edge[b-1].append(a-1)
q = deque()
q.append((0,1))
q.append((-1,-1))
while q:
from_, color = q.popleft()
if board[from_] != 0:
continue
board[from_] = color
for i in edge[from_]:
q.append((i,color))
if sum(board) > 0:
print("Fennec")
else:
print("Snuke")
``` | output | 1 | 27,252 | 16 | 54,505 |
Provide a correct Python 3 solution for this coding contest problem.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke | instruction | 0 | 27,253 | 16 | 54,506 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
import sys
read = sys.stdin.read
readline = sys.stdin.readline
n,*ab = map(int,read().split())
m = iter(ab)
g = [[] for _ in range(n)]
for a,b in zip(m,m):
g[a-1].append(b-1)
g[b-1].append(a-1)
def dist(g,start):
n = len(g)
dist = [-1]*n
q = [start]
dist[start] = 0
while q:
v = q.pop()
for c in g[v]:
if dist[c] == -1:
dist[c] = dist[v]+1
q.append(c)
return dist
df = dist(g,0)
ds = dist(g,n-1)
cnt = 0
for i,j in zip(df,ds):
if i <= j:
cnt += 1
else:
cnt -= 1
#print(cnt)
if cnt >= 1:
print("Fennec")
else:
print("Snuke")
``` | output | 1 | 27,253 | 16 | 54,507 |
Provide a correct Python 3 solution for this coding contest problem.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke | instruction | 0 | 27,254 | 16 | 54,508 |
"Correct Solution:
```
N=int(input())
edge=[list(map(int,input().split())) for i in range(N-1)]
inf=10**20
c=[[] for i in range(N)]
for i,j in edge:
c[i-1].append(j-1)
c[j-1].append(i-1)
d=[0]+[inf]*(N-1)
v=[0]*N
import sys
sys.setrecursionlimit(10**6)
def dfs(p,v,d):
for n in c[p]:
if v[n]==0:
d[n]=min(d[n],d[p]+1)
v[n]=1
dfs(n,v,d)
dfs(0,v,d)
d2=[inf]*(N-1)+[0]
v=[0]*N
dfs(N-1,v,d2)
c=0
for i,j in zip(d,d2):
c=c+1 if i<=j else c-1
print(['Snuke','Fennec'][c>0])
``` | output | 1 | 27,254 | 16 | 54,509 |
Provide a correct Python 3 solution for this coding contest problem.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke | instruction | 0 | 27,255 | 16 | 54,510 |
"Correct Solution:
```
#!/usr/bin/env python3
import sys
from collections import deque
inf = float('inf')
def solve(N: int, a: "List[int]", b: "List[int]"):
adj = [set() for _ in range(N)]
for u, v in zip(a, b):
adj[u].add(v)
adj[v].add(u)
dist_F = [inf for _ in range(N)]
dist_F[0] = 0
dist_S = [inf for _ in range(N)]
dist_S[N-1] = 0
Q = deque([(0, 0)], N)
while len(Q) > 0:
du, u = Q.pop()
for v in adj[u]:
if dist_F[v] < inf:
continue
dist_F[v] = du + 1
Q.append((du+1, v))
Q = deque([(0, N-1)], N)
while len(Q) > 0:
du, u = Q.pop()
for v in adj[u]:
if dist_S[v] < inf:
continue
dist_S[v] = du + 1
Q.append((du+1, v))
c_S = 0
c_F = 0
for df, ds in zip(dist_F, dist_S):
if df <= ds:
c_F += 1
else:
c_S += 1
if c_F <= c_S:
print('Snuke')
else:
print('Fennec')
return
# Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools (tips: You use the default template now. You can remove this line by using your custom template)
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
N = int(next(tokens)) # type: int
a = [int()] * (N-1) # type: "List[int]"
b = [int()] * (N-1) # type: "List[int]"
for i in range(N-1):
a[i] = int(next(tokens)) - 1
b[i] = int(next(tokens)) - 1
solve(N, a, b)
if __name__ == '__main__':
main()
``` | output | 1 | 27,255 | 16 | 54,511 |
Provide a correct Python 3 solution for this coding contest problem.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke | instruction | 0 | 27,256 | 16 | 54,512 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(2000000)
n = int(input())
p = []
for i in range(n-1):
p.append(list(map(int, input().split()))+ [1])
l = [[] for i in range(n+1)]
for pi in p:
l[pi[0]].append([pi[0],pi[1]])
l[pi[1]].append([pi[1],pi[0]])
kyo = [float("inf")]*(n+1)
kyo[1] = 0
memo = [0] * (n+1)
memo[1] = 1
ruto = []
def dfs(start,ruto):
ruto.extend(l[start])
memo[start] = 1
while ruto:
a = ruto.pop()
if memo[a[1]] == 1:
continue
kyo[a[1]] = kyo[a[0]] + 1
dfs(a[1],ruto)
return
dfs(1,ruto)
nkyo = [float("inf")]*(n+1)
nkyo[n] = 0
memo = [0] * (n+1)
memo[n] = 1
ruto = []
def dfs(start,ruto):
ruto.extend(l[start])
memo[start] = 1
while ruto:
a = ruto.pop()
if memo[a[1]] == 1:
continue
nkyo[a[1]] = nkyo[a[0]] + 1
dfs(a[1],ruto)
return
dfs(n,ruto)
iti = 0
nko = 0
for i in range(1,n+1):
if kyo[i] <= nkyo[i]:
iti += 1
else:
nko += 1
if iti <= nko:
print("Snuke")
else:
print("Fennec")
``` | output | 1 | 27,256 | 16 | 54,513 |
Provide a correct Python 3 solution for this coding contest problem.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke | instruction | 0 | 27,257 | 16 | 54,514 |
"Correct Solution:
```
N = int(input())
ab = [[int(_) for _ in input().split()] for n in range(N - 1)]
from collections import defaultdict
ns = defaultdict(list)
for a, b in ab:
ns[a].append(b)
ns[b].append(a)
from heapq import *
queue = []
heappush(queue, (0, 1))
heappush(queue, (1, N))
colors = defaultdict(lambda:None)
black_count, white_count = 0, 0
while queue:
i, index = heappop(queue)
color = i % 2
if colors[index] is None:
colors[index] = color
if color:
white_count += 1
else:
black_count += 1
for t in ns[index]:
if ns[t]:
heappush(queue, (i + 2, t))
if black_count > white_count:
print("Fennec")
else:
print("Snuke")
``` | output | 1 | 27,257 | 16 | 54,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke
Submitted Solution:
```
from collections import deque
n=int(input())
graph=[[] for i in range(n)]
for i in range(n-1):
a,b=map(int,input().split())
a-=1;b-=1
graph[a].append(b)
graph[b].append(a)
distfrom1=[-1]*n
distfromn=[-1]*n
distfrom1[0]=0
distfromn[n-1]=0
d=deque([0])
while d:
v=d.popleft()
for i in graph[v]:
if distfrom1[i]!=-1:continue
distfrom1[i]=distfrom1[v]+1
d.append(i)
d.append(n-1)
while d:
v=d.popleft()
for i in graph[v]:
if distfromn[i]!=-1:continue
distfromn[i]=distfromn[v]+1
d.append(i)
F=0
for i in range(n):
if distfrom1[i]<=distfromn[i]:F+=1
if F>n-F:print("Fennec")
else:print("Snuke")
``` | instruction | 0 | 27,258 | 16 | 54,516 |
Yes | output | 1 | 27,258 | 16 | 54,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke
Submitted Solution:
```
from collections import deque
def dfs(graph, start, goal):
s_stack = deque()
g_stack = deque()
visited = [0 for i in range(N+1)]
s_stack.append([start,0])
g_stack.append([goal, 0])
visited[start] = 1
visited[goal] = 1
s_cnt = 1
g_cnt = 1
round = 0
while s_stack or g_stack:
while s_stack:
a,cost = s_stack.popleft()
if cost > round:
s_stack.appendleft([a,cost])
break
for b in graph[a].keys():
if not visited[b]:
s_stack.append([b,cost+1])
visited[b] += 1
s_cnt += 1
while g_stack:
a, cost = g_stack.popleft()
if cost > round:
g_stack.appendleft([a,cost])
break
for b in graph[a].keys():
if not visited[b]:
g_stack.append([b,cost+1])
visited[b] += 1
g_cnt += 1
round += 1
# print(s_cnt, g_cnt)
return s_cnt > g_cnt
N = int(input())
graph = {}
for i in range(N-1):
a,b = map(int, input().split())
if a not in graph:
graph[a] = {}
if b not in graph:
graph[b] = {}
graph[a][b] = graph[b][a] = 1
print('Fennec' if dfs(graph,1,N) else 'Snuke')
``` | instruction | 0 | 27,259 | 16 | 54,518 |
Yes | output | 1 | 27,259 | 16 | 54,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke
Submitted Solution:
```
import heapq
import sys
input = sys.stdin.readline
def dijlstra_heap(N,s,edge):
d = [float('inf')] * N
used = [True] * N
d[s] = 0
edgelist = []
for a, b in edge[s]:
heapq.heappush(edgelist, a*(10**6)+b)
while len(edgelist):
minedge = heapq.heappop(edgelist)
if not used[minedge%(10**6)]:
continue
v = minedge % (10 ** 6)
d[v] = minedge // (10 ** 6)
used[v] = False
for e in edge[v]:
if used[e[1]]:
heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
return d
def main():
N = int(input())
edge = [[] for i in range(N)]
for i in range(N-1):
a, b = map(int, input().split())
a -= 1
b -= 1
edge[a].append([1,b])
edge[b].append([1,a])
d0 = dijlstra_heap(N,0,edge)
dn = dijlstra_heap(N,N-1,edge)
f_cnt = 0
s_cnt = 0
for i in range(N):
if d0[i] <= dn[i]:
f_cnt += 1
else:
s_cnt += 1
if f_cnt > s_cnt:
print('Fennec')
else:
print('Snuke')
if __name__ == "__main__":
main()
``` | instruction | 0 | 27,260 | 16 | 54,520 |
Yes | output | 1 | 27,260 | 16 | 54,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke
Submitted Solution:
```
N=int(input())
import sys
sys.setrecursionlimit(10**5)
ab=[list(map(int, input().split())) for _ in range (N-1)]
con=[[] for _ in range(N+1)]
#print(ab)
for a,b in ab :
con[a].append(b)
con[b].append(a)
#black Fennec
pc=[0]*(N+1)
q=[(1,0)]
vis=[False]*(N+1)
vis[1]=True
Fcnt=N
while q:
n,c=q.pop()
pc[n]=c
for ch in con[n]:
if vis[ch]==True:
continue
vis[ch]=True
q.append((ch,c+1))
q=[(N,0)]
vis=[False]*(N+1)
vis[N]=True
while q:
n,c=q.pop()
if pc[n]>c:
Fcnt-=1
for ch in con[n]:
if vis[ch]==True:
continue
vis[ch]=True
q.append((ch,c+1))
if Fcnt>N-Fcnt:
print("Fennec")
else:
print("Snuke")
``` | instruction | 0 | 27,261 | 16 | 54,522 |
Yes | output | 1 | 27,261 | 16 | 54,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke
Submitted Solution:
```
from collections import deque
def dfs(graph, start, goal):
s_stack = deque()
g_stack = deque()
visited = [False for i in range(N+1)]
s_stack.append(start)
g_stack.append(goal)
visited[start] = True
visited[goal] = True
s_cnt = 1
g_cnt = 1
while s_stack or g_stack:
if s_stack:
a = s_stack.popleft()
for b in graph[a].keys():
if not visited[b]:
s_stack.append(b)
visited[b] = True
s_cnt += 1
if g_stack:
a = g_stack.popleft()
for b in graph[a].keys():
if not visited[b]:
g_stack.append(b)
visited[b] = True
g_cnt += 1
return s_cnt > g_cnt
N = int(input())
graph = {}
for i in range(N-1):
a,b = map(int, input().split())
if a not in graph:
graph[a] = {}
if b not in graph:
graph[b] = {}
graph[a][b] = graph[b][a] = 1
print('Fennec' if dfs(graph,1,N) else 'Snuke')
``` | instruction | 0 | 27,262 | 16 | 54,524 |
No | output | 1 | 27,262 | 16 | 54,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke
Submitted Solution:
```
from collections import deque
def bfs(start, goal):
queue = deque([start])
visited = []
tekazu = [0 for _ in range(n)] # nは頂点数
while queue:
label = queue.pop()
if label == goal:
visited.append(label)
return tekazu[goal]
if label not in visited:
visited.append(label)
for v in d[label]:
if v not in visited:
tekazu[v] = tekazu[label] + 1
queue.appendleft(v)
return False
n = int(input())
d = {i: [] for i in range(n)}
for _ in range(n - 1):
a, b = map(int, input().split())
d[a - 1].append(b - 1)
d[b - 1].append(a - 1)
Fennec = 0
Snuke = 0
for i in range(n):
if bfs(0, i) <= bfs(n - 1, i):
Fennec += 1
else:
Snuke += 1
if Fennec > Snuke:
print("Fennec")
else:
print("Snuke")
``` | instruction | 0 | 27,263 | 16 | 54,526 |
No | output | 1 | 27,263 | 16 | 54,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke
Submitted Solution:
```
import sys
p = []
def main():
sys.setrecursionlimit(1000000)
N = int(input())
e = [ [] for i in range(N)]
for i in range(N-1):
a,b = map(int,input().split())
a-=1
b-=1
e[a].append(b)
e[b].append(a)
col = [ -1 for i in range(N)]
col[0] = 0
def dfs(n,c,path):
global p
path = path + [n]
if n == N - 1:
p = path
for v in e[n]:
if col[v] == -1:
col[v] = c + 1
dfs(v,c+1,path)
if p:
return
dfs(0,0,[])
if len(p) % 2 ==0:
fn = p[len(p)//2-1]
sn = p[len(p)//2]
else:
fn = p[len(p)//2]
sn = p[len(p)//2+1]
colfn = [ -1 for i in range(N)]
colfn[0] = 0
def dfs2(n,c):
for v in e[n]:
if v == sn:
continue
elif colfn[v] == -1:
colfn[v] = c + 1
dfs2(v,c+1)
dfs2(0,0)
colsn = [ -1 for i in range(N)]
colsn[N-1] = 0
def dfs3(n,c):
for v in e[n]:
if v == fn:
continue
elif colsn[v] == -1:
colsn[v] = c + 1
dfs3(v,c+1)
dfs3(N-1,0)
fennec = -1
snuke = -1
for i in range(N):
if colsn[i] != -1:
snuke += 1
if colfn[i] != -1:
fennec += 1
if fennec > snuke:
print("Fennec")
else:
print("Snuke")
if __name__ == '__main__':
main()
``` | instruction | 0 | 27,264 | 16 | 54,528 |
No | output | 1 | 27,264 | 16 | 54,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fennec and Snuke are playing a board game.
On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.
Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn:
* Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.
* Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.
A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
Output
If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
Examples
Input
7
3 6
1 2
3 1
7 4
5 7
1 4
Output
Fennec
Input
4
1 4
4 2
2 3
Output
Snuke
Submitted Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
tree = [[] for i in range(N+1)]
for i in range(N-1):
a, b = map(int, input().split())
tree[a].append(b)
tree[b].append(a)
dist = [100001] * (N+1)
q = [1]
dist[1] = 0
while q:
now = q.pop()
for next in tree[now]:
if dist[next] == 100001:
dist[next] = dist[now] + 1
q.append(next)
fennec = -dist[N]//2
snuke = -(dist[N]+1)//2
visit = [False] * (N+1)
q = [1]
visit[1] = True
while q:
now = q.pop()
for next in tree[now]:
if not visit[next] and next != N:
visit[next] = True
q.append(next)
fennec += sum(visit)
visit = [False] * (N+1)
q = [N]
visit[N] = True
while q:
now = q.pop()
for next in tree[now]:
if not visit[next] and next != 1:
visit[next] = True
q.append(next)
snuke += sum(visit)
if fennec > snuke:
print("Fennec")
else:
print("Snuke")
``` | instruction | 0 | 27,265 | 16 | 54,530 |
No | output | 1 | 27,265 | 16 | 54,531 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7 | instruction | 0 | 27,266 | 16 | 54,532 |
"Correct Solution:
```
input()
k=len(set(map(int,input().split())))
print(k if k%2 else k-1)
``` | output | 1 | 27,266 | 16 | 54,533 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7 | instruction | 0 | 27,267 | 16 | 54,534 |
"Correct Solution:
```
N=input()
A=list(map(int,input().split()))
if len(set(A))%2==0:
ans=len(set(A))-1
else:
ans=len(set(A))
print(ans)
``` | output | 1 | 27,267 | 16 | 54,535 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7 | instruction | 0 | 27,268 | 16 | 54,536 |
"Correct Solution:
```
input();a=len(set(input().split()));print(a-1+a%2)
``` | output | 1 | 27,268 | 16 | 54,537 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7 | instruction | 0 | 27,269 | 16 | 54,538 |
"Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
a = set(a)
if len(a) % 2 == 0:
print(len(a) - 1)
else:
print(len(a))
``` | output | 1 | 27,269 | 16 | 54,539 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7 | instruction | 0 | 27,270 | 16 | 54,540 |
"Correct Solution:
```
N = int(input())
A = [int(i) for i in input().split()]
l = len(set(A))
print(l - (N - l) % 2)
``` | output | 1 | 27,270 | 16 | 54,541 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7 | instruction | 0 | 27,271 | 16 | 54,542 |
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
s = set(A)
if len(s)%2 == 0:
ans = len(s)-1
else:
ans = len(s)
print(ans)
``` | output | 1 | 27,271 | 16 | 54,543 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7 | instruction | 0 | 27,272 | 16 | 54,544 |
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
aset = set(a)
if (n-len(aset)) % 2 == 0:
print(len(aset))
else:
print(len(aset)-1)
``` | output | 1 | 27,272 | 16 | 54,545 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7 | instruction | 0 | 27,273 | 16 | 54,546 |
"Correct Solution:
```
N = int(input())
nums = list(map(int, input().split()))
unit = set(nums)
if len(unit)%2 == 0:
print(len(unit)-1)
else:
print(len(unit))
``` | output | 1 | 27,273 | 16 | 54,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
x=len(list(set(a)))
if x%2==1:
print(x)
else:
print(x-1)
``` | instruction | 0 | 27,274 | 16 | 54,548 |
Yes | output | 1 | 27,274 | 16 | 54,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7
Submitted Solution:
```
N = int(input())
S = len(set(map(int, input().split())))
print(S - (N - S) % 2)
``` | instruction | 0 | 27,275 | 16 | 54,550 |
Yes | output | 1 | 27,275 | 16 | 54,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7
Submitted Solution:
```
N=int(input())
A=list(map(int,input().split()))
m=len(set(A))
if m%2==0:
print(m-1)
else:
print(m)
``` | instruction | 0 | 27,276 | 16 | 54,552 |
Yes | output | 1 | 27,276 | 16 | 54,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7
Submitted Solution:
```
n = int(input())
a = sorted(list(map(int, input().split())))
ans = 0
for i in range(1, len(a)):
if a[i] == a[i - 1]:
ans += 1
print(n - (ans + 1) // 2 * 2)
``` | instruction | 0 | 27,277 | 16 | 54,554 |
Yes | output | 1 | 27,277 | 16 | 54,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7
Submitted Solution:
```
N = int(input())
A = list(map(int,input().split()))
d = {}
X =[]
for i in range(N):
x = A[i]
if x not in d:
d[x] = 1
X.append(x)
else:
d[x] += 1
ans = 0
for x in X:
a = d[x]
if a%2 == 0:
ans += a//2 - 1
else:
ans += a//2
X.remove(x)
n = len(X)
ans += (n+1)//2
print(ans)
``` | instruction | 0 | 27,278 | 16 | 54,556 |
No | output | 1 | 27,278 | 16 | 54,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7
Submitted Solution:
```
from collections import Counter
import bisect
N = int(input())
A = list(map(int,input().split()))
dups = []
not_dups = []
ans = 0
for key, num in Counter(A).items():
if num == 2: dups.append(key)
else: not_dups.append(key)
dups.sort()
not_dups.sort()
i_min = bisect.bisect(dups, not_dups[0])
i_max = bisect.bisect(dups, not_dups[-1])
print(i_max-i_min+len(not_dups))
``` | instruction | 0 | 27,279 | 16 | 54,558 |
No | output | 1 | 27,279 | 16 | 54,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a.sort()
eq={}
for i in range(n-1):
if a[i]==a[i-1]:
if a[i] in eq:
eq[a[i]]+=1
else:
eq[a[i]]=1
sousa=0
for i in eq.values():
sousa+=i
eat=(sousa+1)//2
print(n-eat*2)
``` | instruction | 0 | 27,280 | 16 | 54,560 |
No | output | 1 | 27,280 | 16 | 54,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
* 3 ≦ N ≦ 10^{5}
* N is odd.
* 1 ≦ A_i ≦ 10^{5}
* A_i is an integer.
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_{N}
Output
Print the answer.
Examples
Input
5
1 2 1 3 7
Output
3
Input
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Output
7
Submitted Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = SR()
return l
mod = 1000000007
#A
#B
#C
"""
x = I()
ans = 2*(x//11)+1
x %= 11
if x == 0:
ans -= 1
if x > 6:
ans += 1
print(ans)
"""
#D
n = I()
a = LI()
d = defaultdict(int)
for i in a:
d[i] += 1
l = list(d.keys())
i = 0
while i < len(l):
if d[l[i]] < 2:
l.pop(i)
i -= 2
i += 1
if len(l) == 0:break
l.sort()
k = 0
for i in l:
k ^= ~(d[i]%2)
d[i] = 1
ans = sum(list(d.values()))
ans -= 1 if k%2 else 0
print(ans)
#E
#F
#G
#H
#I
#J
#K
#L
#M
#N
#O
#P
#Q
#R
#S
#T
``` | instruction | 0 | 27,281 | 16 | 54,562 |
No | output | 1 | 27,281 | 16 | 54,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bridge that connects the left and right banks of a river. There are 2 N doors placed at different positions on this bridge, painted in some colors. The colors of the doors are represented by integers from 1 through N. For each k (1 \leq k \leq N), there are exactly two doors painted in Color k.
Snuke decides to cross the bridge from the left bank to the right bank. He will keep on walking to the right, but the following event will happen while doing so:
* At the moment Snuke touches a door painted in Color k (1 \leq k \leq N), he teleports to the right side of the other door painted in Color k.
It can be shown that he will eventually get to the right bank.
For each i (1 \leq i \leq 2 N - 1), the section between the i-th and (i + 1)-th doors from the left will be referred to as Section i. After crossing the bridge, Snuke recorded whether or not he walked through Section i, for each i (1 \leq i \leq 2 N - 1). This record is given to you as a string s of length 2 N - 1. For each i (1 \leq i \leq 2 N - 1), if Snuke walked through Section i, the i-th character in s is `1`; otherwise, the i-th character is `0`.
<image>
Figure: A possible arrangement of doors for Sample Input 3
Determine if there exists an arrangement of doors that is consistent with the record. If it exists, construct one such arrangement.
Constraints
* 1 \leq N \leq 10^5
* |s| = 2 N - 1
* s consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
N
s
Output
If there is no arrangement of doors that is consistent with the record, print `No`. If there exists such an arrangement, print `Yes` in the first line, then print one such arrangement in the second line, in the following format:
c_1 c_2 ... c_{2 N}
Here, for each i (1 \leq i \leq 2 N), c_i is the color of the i-th door from the left.
Examples
Input
2
010
Output
Yes
1 1 2 2
Input
2
001
Output
No
Input
3
10110
Output
Yes
1 3 2 1 2 3
Input
3
10101
Output
No
Input
6
00111011100
Output
Yes
1 6 1 2 3 4 4 2 3 5 6 5
Submitted Solution:
```
import sys
sys.setrecursionlimit(100000)
def dfs(v, t, cnt, goal, color, links, ans):
# print(v, t, ans)
if v == t:
if cnt == goal:
ans[v - 1] = color
return True
else:
return False
memo_u = 0
if v:
memo_u = ans[v - 1]
if not memo_u:
ans[v - 1] = color
memo_v = ans[v]
if not memo_v:
color += 1
ans[v] = color
# print(ans)
for l in links[v]:
if ans[l - 1] and ans[l - 1] != ans[v]:
continue
res = dfs(l, t, cnt + 1, goal, color, links, ans)
if res:
return True
ans[v - 1] = memo_u
ans[v] = memo_v
return False
n = int(input())
s = input()
to_, from_ = [set(), set()], [set(), set()]
sl = [1] + list(map(int, s)) + [1]
cnt = sum(sl) - 1
for i, (pc, c) in enumerate(zip(sl, sl[1:])):
if pc:
from_[c].add(i)
if c:
to_[pc].add(i + 1)
# print(from_, to_)
links = [None] * (2 * n)
for f, t in zip(from_, to_):
for l in f:
links[l] = {k for k in t if k - 1 != l}
# print(links)
ans = [0] * (n * 2)
res = dfs(0, 2 * n, 0, cnt, 0, links, ans)
if res:
print('Yes')
mx = max(ans)
if mx != n:
mx = mx * 2 + 1
for i, v in enumerate(ans):
if v == 0:
ans[i] = mx // 2
mx += 1
print(*ans)
else:
print('No')
``` | instruction | 0 | 28,112 | 16 | 56,224 |
No | output | 1 | 28,112 | 16 | 56,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bridge that connects the left and right banks of a river. There are 2 N doors placed at different positions on this bridge, painted in some colors. The colors of the doors are represented by integers from 1 through N. For each k (1 \leq k \leq N), there are exactly two doors painted in Color k.
Snuke decides to cross the bridge from the left bank to the right bank. He will keep on walking to the right, but the following event will happen while doing so:
* At the moment Snuke touches a door painted in Color k (1 \leq k \leq N), he teleports to the right side of the other door painted in Color k.
It can be shown that he will eventually get to the right bank.
For each i (1 \leq i \leq 2 N - 1), the section between the i-th and (i + 1)-th doors from the left will be referred to as Section i. After crossing the bridge, Snuke recorded whether or not he walked through Section i, for each i (1 \leq i \leq 2 N - 1). This record is given to you as a string s of length 2 N - 1. For each i (1 \leq i \leq 2 N - 1), if Snuke walked through Section i, the i-th character in s is `1`; otherwise, the i-th character is `0`.
<image>
Figure: A possible arrangement of doors for Sample Input 3
Determine if there exists an arrangement of doors that is consistent with the record. If it exists, construct one such arrangement.
Constraints
* 1 \leq N \leq 10^5
* |s| = 2 N - 1
* s consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
N
s
Output
If there is no arrangement of doors that is consistent with the record, print `No`. If there exists such an arrangement, print `Yes` in the first line, then print one such arrangement in the second line, in the following format:
c_1 c_2 ... c_{2 N}
Here, for each i (1 \leq i \leq 2 N), c_i is the color of the i-th door from the left.
Examples
Input
2
010
Output
Yes
1 1 2 2
Input
2
001
Output
No
Input
3
10110
Output
Yes
1 3 2 1 2 3
Input
3
10101
Output
No
Input
6
00111011100
Output
Yes
1 6 1 2 3 4 4 2 3 5 6 5
Submitted Solution:
```
import sys
sys.setrecursionlimit(100000)
def dfs(v, t, cnt, goal, links, ans):
# print(v, t, ans)
if v == t:
if cnt == goal:
ans[v - 1] = cnt
return True
else:
return False
memo_u, memo_v = ans[v - 1], ans[v]
if not memo_u:
ans[v - 1] = cnt
if not memo_v:
ans[v] = cnt + 1
# print(ans)
for l in links[v]:
if ans[l - 1] and ans[l - 1] != ans[v]:
continue
res = dfs(l, t, cnt + 1, goal, links, ans)
if res:
return True
ans[v - 1] = memo_u
ans[v] = memo_v
return False
n = int(input())
s = input()
to_, from_ = [set(), set()], [set(), set()]
sl = [1] + list(map(int, s)) + [1]
cnt = sum(sl) - 1
for i, (pc, c) in enumerate(zip(sl, sl[1:])):
if pc:
from_[c].add(i)
if c:
to_[pc].add(i + 1)
# print(from_, to_)
links = [None] * (2 * n)
for f, t in zip(from_, to_):
for l in f:
links[l] = {k for k in t if k - 1 != l}
# print(links)
ans = [0] * (n * 2)
res = dfs(0, 2 * n, 0, cnt, links, ans)
if res:
print('Yes')
dic = {0: 0}
cur = 1
flg_z = True
for i, v in enumerate(ans):
if v == 0:
if flg_z:
del dic[0]
flg_z = not flg_z
if v in dic:
ans[i] = dic[v]
else:
ans[i] = dic[v] = cur
cur += 1
print(*ans)
else:
print('No')
``` | instruction | 0 | 28,113 | 16 | 56,226 |
No | output | 1 | 28,113 | 16 | 56,227 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No | instruction | 0 | 28,972 | 16 | 57,944 |
"Correct Solution:
```
n = int(input())
a = []
for i in range(3):
aa = list(map(int,input().split()))
a.append(aa)
for i in range(n):
x,y,z = a[0][i],a[1][i],a[2][i]
x,z = x-y,z-y
if i % 2 == 0:
if y % 6 != 2:
print("No")
exit()
else:
if y % 6 != 5:
print("No")
exit()
y = 0
if (x,y,z) != (-1,0,1) and (x,y,z) != (1,0,-1):
print("No")
exit()
e,o = 0,0
for i in range(n):
if i % 2 == 0:
if a[0][i] > a[2][i]: e += 1
else:
if a[0][i] > a[2][i]: o += 1
e %= 2
o %= 2
el = []
ol = []
bite = [0]
bito = [0]
for i in range(n):
if(i % 2 == 0):
el.append(a[1][i] // 6 + 1)
bite.append(0)
else:
ol.append(a[1][i] // 6 + 1)
bito.append(0)
def sum(i,bit):
s = 0
while(i > 0):
s += bit[i]
i = i&(i-1)
return s
def add(i,x,bit):
while(i < len(bit)):
bit[i] += x
i += i & (-i)
signe = 0
signo = 0
for i in range(len(el)):
signe += (i - sum(el[i],bite))
add(el[i],1,bite)
for i in range(len(ol)):
signo += (i - sum(ol[i],bito))
add(ol[i],1,bito)
if (signe % 2 == o) and (signo % 2 == e) :print("Yes")
else :print("No")
``` | output | 1 | 28,972 | 16 | 57,945 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No | instruction | 0 | 28,973 | 16 | 57,946 |
"Correct Solution:
```
def merge_and_count_rec(A, W, l, r):
if l+1 >= r:
return 0
m = (l+r)//2
cnt = merge_and_count_rec(A, W, l, m)
cnt += merge_and_count_rec(A, W, m, r)
i,j,k = l,m,l
while i<m and j<r:
if A[i]<=A[j]:
W[k] = A[i]
i+=1
else:
W[k] = A[j]
j+=1
cnt += m-i
k+=1
if i<m:
W[k:r] = A[i:m]
if j<r:
W[k:r] = A[j:r]
A[l:r] = W[l:r]
return cnt
N = int(input())
A = tuple(list(map(int,input().split())) for _ in range(3))
O = []
ok = True
flip = [0,0]
for i in range(N):
ok &= (A[1][i] % 3) == 2
ok &= sorted((A[0][i], A[1][i], A[2][i]))==[A[1][i]-1, A[1][i], A[1][i]+1]
dist = abs(A[1][i]//3 - i)
ok &= (dist % 2) == 0
order = [A[0][i], A[1][i], A[2][i]] == [A[1][i]-1, A[1][i], A[1][i]+1]
if ((dist//2)%2 == 0) != order:
flip[i%2] += 1
B = A[1][0::2]
hb = merge_and_count_rec(B, A[0], 0, len(B))
C = A[1][1::2]
hc = merge_and_count_rec(C, A[0], 0, len(C))
ok &= (hb % 2) == (flip[1]%2)
ok &= (hc % 2) == (flip[0]%2)
print("Yes" if ok else "No")
``` | output | 1 | 28,973 | 16 | 57,947 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No | instruction | 0 | 28,974 | 16 | 57,948 |
"Correct Solution:
```
class BIT():
def __init__(self,n):
self.BIT=[0]*(n+1)
self.num=n
def query(self,idx):
res_sum = 0
while idx > 0:
res_sum += self.BIT[idx]
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
while idx <= self.num:
self.BIT[idx] += x
idx += idx&(-idx)
return
N=int(input())
a=[list(map(int,input().split())) for i in range(3)]
odd=[]
even=[]
for i in range(N):
x,y,z=a[0][i],a[1][i],a[2][i]
if x<z:
if y!=x+1 or z!=y+1:
print("No")
exit()
else:
if i%2==0:
if x%6==1:
even.append(x//6+1)
else:
print("No")
exit()
else:
if x%6==4:
odd.append(x//6+1)
else:
print("No")
exit()
else:
if y!=x-1 or z!=y-1:
print("No")
exit()
else:
if i%2==0:
if z%6==1:
even.append(-(z//6)-1)
else:
print("No")
exit()
else:
if z%6==4:
odd.append(-(z//6)-1)
else:
print("No")
exit()
oddsign=0
ODD=[(abs(odd[i]),i) for i in range(len(odd))]
ODD.sort()
for i in range(len(odd)):
val,id=ODD[i]
check=(id-i)%2
if check==1:
oddsign+=(odd[id]>0)
else:
oddsign+=(odd[id]<0)
evensign=0
EVEN=[(abs(even[i]),i) for i in range(len(even))]
EVEN.sort()
for i in range(len(even)):
val,id=EVEN[i]
check=(id-i)%2
if check==1:
evensign+=(even[id]>0)
else:
evensign+=(even[id]<0)
oddbit=BIT(len(odd))
check=0
for i in range(len(odd)):
v=abs(odd[i])
check+=i-oddbit.query(v)
oddbit.update(v,1)
if check%2!=evensign%2:
print("No")
exit()
evenbit=BIT(len(even))
check=0
for i in range(len(even)):
v=abs(even[i])
check+=i-evenbit.query(v)
evenbit.update(v,1)
if check%2!=oddsign%2:
print("No")
exit()
print("Yes")
``` | output | 1 | 28,974 | 16 | 57,949 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No | instruction | 0 | 28,975 | 16 | 57,950 |
"Correct Solution:
```
import sys
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split())) #空白あり
N = I()
a = [LI() for i in range(3)]
A = []
B = []
c = 0
d = 0
for i in range(N):
if i % 2 == 0:
if a[0][i] % 6 == 1 and a[1][i] % 6 == 2 and a[2][i] % 6 == 3:
A.append(a[1][i]//6)
elif a[0][i] % 6 == 3 and a[1][i] % 6 == 2 and a[2][i] % 6 == 1:
c += 1
A.append(a[1][i]//6)
else:
print('No')
exit()
else:
if a[0][i] % 6 == 4 and a[1][i] % 6 == 5 and a[2][i] % 6 == 0:
B.append(a[1][i]//6)
elif a[0][i] % 6 == 0 and a[1][i] % 6 == 5 and a[2][i] % 6 == 4:
d += 1
B.append(a[1][i]//6)
else:
print('No')
exit()
class SegmentTree():
def __init__(self,init_value,seg_func,unit_elt):
n = len(init_value)
self.seg_func = seg_func
self.unit_elt = unit_elt
self.num = 1 << (n-1).bit_length() # n以上の最小2冪
self.tree = [unit_elt]*(2*self.num)
for i in range(n):
self.tree[self.num + i] = init_value[i]
for i in range(self.num-1,0,-1):
self.tree[i] = self.seg_func(self.tree[2*i],self.tree[2*i+1])
def update(self,i,x): # i(0-index)番目の値をxに更新
i += self.num
self.tree[i] = x
while i > 1:
i >>= 1
self.tree[i] = self.seg_func(self.tree[2*i],self.tree[2*i+1])
def quary(self,p,q): # [p,q)に関するseg_funcの結果を返す(p,qは0-index)
p += self.num
q += self.num
res = self.unit_elt
while p < q:
if p & 1:
res = self.seg_func(res,self.tree[p])
p += 1
if q & 1:
res = self.seg_func(res,self.tree[q-1])
p >>= 1
q >>= 1
return res
def seg_func(x,y):
return x + y
seg1 = SegmentTree([0]*(len(A)),seg_func,unit_elt=0)
seg2 = SegmentTree([0]*(len(B)),seg_func,unit_elt=0)
a = 0 # Aの転倒数
for i in range(len(A)):
a += i-seg1.quary(0,A[i])
seg1.update(A[i],1)
b = 0 # Bの転倒数
for i in range(len(B)):
b += i-seg2.quary(0,B[i])
seg2.update(B[i],1)
if (a-d) % 2 == 0 and (b-c) % 2 == 0:
print('Yes')
else:
print('No')
``` | output | 1 | 28,975 | 16 | 57,951 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No | instruction | 0 | 28,976 | 16 | 57,952 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from operator import itemgetter, mul
from functools import reduce
N = int(readline())
A = list(map(int,readline().split()))
B = list(map(int,readline().split()))
C = list(map(int,readline().split()))
sgn = [0] * N
for i,(a,b,c) in enumerate(zip(A,B,C)):
if a == b-1 and c == b+1:
sgn[i] = 1
elif a == b+1 and c == b-1:
sgn[i] = -1
else:
print('No')
exit()
if any(x%6 != 2 for x in B[::2]) or any(x%6 != 5 for x in B[1::2]):
print('No')
exit()
class BinaryIndexedTree():
def __init__(self, seq):
self.size = len(seq)
self.depth = self.size.bit_length()
self.build(seq)
def build(self,seq):
data = seq
size = self.size
for i,x in enumerate(data):
j = i+(i&(-i))
if j < size:
data[j] += data[i]
self.data = data
def __repr__(self):
return self.data.__repr__()
def get_sum(self,i):
data = self.data
s = 0
while i:
s += data[i]
i -= i & -i
return s
def add(self, i, x):
data = self.data
size = self.size
while i < size:
data[i] += x
i += i & -i
def find_kth_element(self,k):
data = self.data; size = self.size
x,sx = 0,0
dx = 1 << (self.depth)
for i in range(self.depth - 1, -1, -1):
dx = (1 << i)
if x + dx >= size:
continue
y = x + dx
sy = sx + data[y]
if sy < k:
x,sx = y,sy
return x + 1
def Inversion(seq):
# seqは、1,2,...,Nの順列
N = len(seq)
bit = BinaryIndexedTree([0] * (N+1))
inv = N*(N-1)//2
for x in seq:
inv -= bit.get_sum(x)
bit.add(x,1)
return inv
OD = sorted(enumerate(B[::2],1), key=itemgetter(1))
ind, val = zip(*OD)
od_inv = Inversion(ind)
EV = sorted(enumerate(B[1::2],1), key=itemgetter(1))
ind, val = zip(*EV)
ev_inv = Inversion(ind)
od_sgn = reduce(mul, sgn[::2])
ev_sgn = reduce(mul, sgn[1::2])
condition = (od_sgn == (-1)**ev_inv) and (ev_sgn == (-1)**od_inv)
answer = 'Yes' if condition else 'No'
print(answer)
``` | output | 1 | 28,976 | 16 | 57,953 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No | instruction | 0 | 28,977 | 16 | 57,954 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = [list(map(int, input().split())) for _ in range(3)]
for j in range(n):
if not ((a[0][j] - 1 == a[1][j] and a[2][j] + 1 == a[1][j]) or (a[0][j] + 1 == a[1][j] and a[2][j] - 1 == a[1][j])):
print("No")
exit()
for j in range(n):
if (j % 2) ^ (((a[0][j] - 1) // 3) % 2) == 1:
print("No")
exit()
# BIT
def add(a, w, bit):
x = a
while x <= 3 * n:
bit[x] += w
x += x & -x
def solve(a, bit):
ret = 0
x = a
while x > 0:
ret += bit[x]
x -= x & -x
return ret
bit_odd = [0] * (3*n+1)
inv_odd = 0
flip_even = 0
for j in range(n):
if j % 2 == 1:
if a[0][j] > a[1][j]:
flip_even += 1
else:
inv_odd += j//2 - solve(a[0][j], bit_odd)
add(a[0][j], 1, bit_odd)
if flip_even % 2 != inv_odd % 2:
print("No")
exit()
bit_even = [0] * (3*n+1)
inv_even = 0
flip_odd = 0
for j in range(n):
if j % 2 == 0:
if a[0][j] > a[1][j]:
flip_odd += 1
else:
inv_even += j//2 - solve(a[0][j], bit_even)
add(a[0][j], 1, bit_even)
if flip_odd % 2 != inv_even % 2:
print("No")
exit()
print("Yes")
``` | output | 1 | 28,977 | 16 | 57,955 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No | instruction | 0 | 28,978 | 16 | 57,956 |
"Correct Solution:
```
"""
https://atcoder.jp/contests/agc006/tasks/agc006_e
まず、ちゃんと並んでない場合はだめ
3x+0
3x+1
3x+2
またはこれが上下反転されているもの以外が存在したら即NO
これを xと表すことにする。上下反転を-xと表すこととする
すると初期状態はxで表すと
1,2,3,4,5…,N となる。
操作は、連続する3つを選び、両端を入れ替え、3つの項に-1を掛ける
となる。
そのため、偶奇が違う数があるとout
次に転倒数を考える。転倒数は交換するたびに1減る or 1増える
偶奇番目を分けて考える
交換すると、両端の偶奇の正負のパリティは変化しないが、反対側が変化する
よって、偶奇が逆の集合の転倒数を求める
→奇数ならば積は負・偶数なら積は正 が保たれる そうでない場合はNO
それ以外はすべて作れるのか…？証明したい
1 2 3 4
-3-2-1 4
-3-4 1 2
分からん…とりあえず出してみよう
"""
from sys import stdin
import sys
def bitadd(a,w,bit): #aにwを加える(1-origin)
x = a
while x <= (len(bit)-1):
bit[x] += w
x += x & (-1 * x)
def bitsum(a,bit): #ind 1～aまでの和を求める
ret = 0
x = a
while x > 0:
ret += bit[x]
x -= x & (-1 * x)
return ret
N = int(stdin.readline())
a = []
for i in range(3):
tmp = list(map(int,stdin.readline().split()))
a.append(tmp)
lis = []
bi = []
for j in range(N):
a[0][j] -= 1
a[1][j] -= 1
a[2][j] -= 1
if a[0][j]//3 == a[1][j]//3 == a[2][j]//3 and a[0][j]//3%2 == j%2:
if a[0][j] > a[1][j] > a[2][j]:
lis.append(a[0][j]//3)
bi.append(-1)
elif a[0][j] < a[1][j] < a[2][j]:
lis.append(a[0][j]//3)
bi.append(1)
else:
print ("No")
sys.exit()
else:
print ("No")
sys.exit()
evenall = 1
for i in range(0,N,2):
evenall *= bi[i]
oddall = 1
for i in range(1,N,2):
oddall *= bi[i]
a = lis
#print (a,bi)
#even
BIT = [0] * (N+2)
ans = 0
for i in range(0,N,2):
ans += i//2 - bitsum(a[i]+1,BIT)
bitadd(a[i]+1,1,BIT)
#print (ans , oddall)
if (ans % 2 == 0 and oddall == -1) or (ans % 2 == 1 and oddall == 1):
print ("No")
sys.exit()
#odd
BIT = [0] * (N+2)
ans = 0
for i in range(1,N,2):
ans += i//2 - bitsum(a[i]+1,BIT)
bitadd(a[i]+1,1,BIT)
#print (ans , evenall)
if (ans % 2 == 0 and evenall == -1) or (ans % 2 == 1 and evenall == 1):
print ("No")
sys.exit()
print ("Yes")
``` | output | 1 | 28,978 | 16 | 57,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = [list(map(int, input().split())) for _ in range(3)]
for j in range(n):
if not ((a[0][j] - 1 == a[1][j] and a[2][j] + 1 == a[1][j]) or (a[0][j] + 1 == a[1][j] and a[2][j] - 1 == a[1][j])):
print("No")
exit()
for j in range(n):
if (j % 2) * (((a[0][j] - 1) // 3) % 2) == 1:
print("No")
exit()
# BIT
def add(a, w, bit):
x = a
while x <= n:
bit[x] += w
x += x & -x
def solve(a, bit):
ret = 0
x = a
while x > 0:
ret += bit[x]
x -= x & -x
return ret
bit_odd = [0] * (3*n+1)
inv_odd = 0
flip_even = 0
for j in range(n):
if j % 2 == 1:
if a[0][j] > a[1][j]:
flip_even += 1
else:
inv_odd += j - solve(a[0][j], bit_odd)
add(a[0][j], 1, bit_odd)
if flip_even % 2 != inv_odd % 2:
print("No")
exit()
bit_even = [0] * (3*n+1)
inv_even = 0
flip_odd = 0
for j in range(n):
if j % 2 == 0:
if a[0][j] > a[1][j]:
flip_odd += 1
else:
inv_even += j - solve(a[0][j], bit_even)
add(a[0][j], 1, bit_even)
if flip_odd % 2 != inv_even % 2:
print("No")
exit()
print("Yes")
``` | instruction | 0 | 28,979 | 16 | 57,958 |
No | output | 1 | 28,979 | 16 | 57,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with 3 rows and N columns. The cell at the i-th row and j-th column is denoted (i, j). Initially, each cell (i, j) contains the integer i+3j-3.
<image>
A grid with N=5 columns
Snuke can perform the following operation any number of times:
* Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.
<image>
An example sequence of operations (each chosen subrectangle is colored blue)
Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}. Determine whether it is achievable.
Constraints
* 5≤N≤10^5
* 1≤a_{i,j}≤3N
* All a_{i,j} are distinct.
Input
The input is given from Standard Input in the following format:
N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}
Output
If Snuke's objective is achievable, print `Yes`. Otherwise, print `No`.
Examples
Input
5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3
Output
Yes
Input
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Output
No
Input
5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
Output
Yes
Input
6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18
Output
Yes
Input
7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9
Output
No
Submitted Solution:
```
n=int(input())
input()
t=list(range(2,2+(n-1)*3 +1,3))
k=sorted([int(i) for i in input().strip().split(" ")])
input()
if t==k:
print("Yes")
else:
print("No")
``` | instruction | 0 | 28,980 | 16 | 57,960 |
No | output | 1 | 28,980 | 16 | 57,961 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider a grid of squares with 10^9 rows and N columns. Let (i, j) be the square at the i-th column (1 \leq i \leq N) from the left and j-th row (1 \leq j \leq 10^9) from the bottom.
Snuke has cut out some part of the grid so that, for each i = 1, 2, ..., N, the bottom-most h_i squares are remaining in the i-th column from the left. Now, he will paint the remaining squares in red and blue. Find the number of the ways to paint the squares so that the following condition is satisfied:
* Every remaining square is painted either red or blue.
* For all 1 \leq i \leq N-1 and 1 \leq j \leq min(h_i, h_{i+1})-1, there are exactly two squares painted red and two squares painted blue among the following four squares: (i, j), (i, j+1), (i+1, j) and (i+1, j+1).
Since the number of ways can be extremely large, print the count modulo 10^9+7.
Constraints
* 1 \leq N \leq 100
* 1 \leq h_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
h_1 h_2 ... h_N
Output
Print the number of the ways to paint the squares, modulo 10^9+7.
Examples
Input
9
2 3 5 4 1 2 4 2 1
Output
12800
Input
2
2 2
Output
6
Input
5
2 1 2 1 2
Output
256
Input
9
27 18 28 18 28 45 90 45 23
Output
844733013 | instruction | 0 | 29,823 | 16 | 59,646 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 10 ** 9 + 7
n = I()
H = LI() + [1]
dp = [1] * (n + 1)
for k in range(n):
new_dp = [0] * (n + 1)
for i in range(n + 1):
if H[i] > H[k]:
new_dp[i] = dp[k] * 2
elif H[k - 1] <= H[i]:
new_dp[i] = dp[i] * 2 * pow(2, H[k] - H[i], mod)
elif H[k - 1] > H[k]:
new_dp[i] = dp[i] + dp[k]
else:
new_dp[i] = (dp[i] + dp[k - 1]) * pow(2, H[k] - H[k - 1], mod)
new_dp[i] %= mod
dp = new_dp
print(dp[-1])
``` | output | 1 | 29,823 | 16 | 59,647 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider a grid of squares with 10^9 rows and N columns. Let (i, j) be the square at the i-th column (1 \leq i \leq N) from the left and j-th row (1 \leq j \leq 10^9) from the bottom.
Snuke has cut out some part of the grid so that, for each i = 1, 2, ..., N, the bottom-most h_i squares are remaining in the i-th column from the left. Now, he will paint the remaining squares in red and blue. Find the number of the ways to paint the squares so that the following condition is satisfied:
* Every remaining square is painted either red or blue.
* For all 1 \leq i \leq N-1 and 1 \leq j \leq min(h_i, h_{i+1})-1, there are exactly two squares painted red and two squares painted blue among the following four squares: (i, j), (i, j+1), (i+1, j) and (i+1, j+1).
Since the number of ways can be extremely large, print the count modulo 10^9+7.
Constraints
* 1 \leq N \leq 100
* 1 \leq h_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
h_1 h_2 ... h_N
Output
Print the number of the ways to paint the squares, modulo 10^9+7.
Examples
Input
9
2 3 5 4 1 2 4 2 1
Output
12800
Input
2
2 2
Output
6
Input
5
2 1 2 1 2
Output
256
Input
9
27 18 28 18 28 45 90 45 23
Output
844733013 | instruction | 0 | 29,824 | 16 | 59,648 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 10 ** 9 + 7
n = I()
H = LI() + [1]
dp = [1] * (n + 1)
for k in range(n):
new_dp = [0] * (n + 1)
for i in range(n + 1):
if H[i] > H[k]:
new_dp[i] = dp[k] * 2
elif H[k - 1] <= H[i]:
new_dp[i] = dp[i] * 2 * pow(2, H[k] - H[i], mod)
elif H[k - 1] > H[k]:
new_dp[i] = dp[i] - dp[k] + dp[k] * 2
else:
new_dp[i] = (dp[i] - dp[k - 1] + dp[k - 1] * 2) * pow(2, H[k] - H[k - 1], mod)
new_dp[i] %= mod
dp = new_dp
print(dp[-1])
``` | output | 1 | 29,824 | 16 | 59,649 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider a grid of squares with 10^9 rows and N columns. Let (i, j) be the square at the i-th column (1 \leq i \leq N) from the left and j-th row (1 \leq j \leq 10^9) from the bottom.
Snuke has cut out some part of the grid so that, for each i = 1, 2, ..., N, the bottom-most h_i squares are remaining in the i-th column from the left. Now, he will paint the remaining squares in red and blue. Find the number of the ways to paint the squares so that the following condition is satisfied:
* Every remaining square is painted either red or blue.
* For all 1 \leq i \leq N-1 and 1 \leq j \leq min(h_i, h_{i+1})-1, there are exactly two squares painted red and two squares painted blue among the following four squares: (i, j), (i, j+1), (i+1, j) and (i+1, j+1).
Since the number of ways can be extremely large, print the count modulo 10^9+7.
Constraints
* 1 \leq N \leq 100
* 1 \leq h_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
h_1 h_2 ... h_N
Output
Print the number of the ways to paint the squares, modulo 10^9+7.
Examples
Input
9
2 3 5 4 1 2 4 2 1
Output
12800
Input
2
2 2
Output
6
Input
5
2 1 2 1 2
Output
256
Input
9
27 18 28 18 28 45 90 45 23
Output
844733013 | instruction | 0 | 29,825 | 16 | 59,650 |
"Correct Solution:
```
import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
MOD=10**9+7
def powmod(b,e):
global MOD
if e==0: return 1
if e&1: return (powmod((b*b)%MOD,e//2)*b)%MOD
return powmod((b*b)%MOD,e//2)
n=int(input())
h=[1]+list(map(int,input().split()))
"""
# naive solution, O(sum h_i)
archive=[]
# do rows
a1=[1]
for i in range(1,n+1):
if h[i]<=h[i-1]:
c=0
while len(a1)>h[i]:
c+=a1.pop(-1)
a1[0]=((a1[0]+c)*2)%MOD
else:
a1[0]=(a1[0]*2)%MOD
mul=powmod(2,h[i]-h[i-1])
for j in range(1,len(a1)):
a1[j]=(a1[j]*mul)%MOD
extras=[a1[0]]
for _ in range(h[i]-h[i-1]-1):
extras.append(extras[-1]*2)
a1+=extras[::-1]
archive.append(a1)
print(sum(a1)%MOD)
"""
# smarter solution? i think it's O(N^2)
root=1 # number of times that this column just alternates
pos=[] # [x,y] means it starts at x, and its size is y blocks
sz=1
for i in range(1,n+1):
if h[i]<=h[i-1]:
c=0
while pos and sz-pos[-1][1]>=h[i]:
a,b=pos.pop(-1)
c+=(a*(powmod(2,b)-1))%MOD
sz-=b
if sz>h[i]:
a,b=pos.pop(-1)
pos.append([(a*powmod(2,sz-h[i]))%MOD,b-(sz-h[i])])
c+=(a*(powmod(2,sz-h[i])-1))%MOD
sz=h[i]
root=((root+c)*2)%MOD
else:
root=(root*2)%MOD
mul=powmod(2,h[i]-h[i-1])
for j in range(len(pos)):
pos[j][0]=(pos[j][0]*mul)%MOD
pos.append([root,h[i]-h[i-1]])
sz=h[i]
# get answer
for a,b in pos:
root+=(a*(powmod(2,b)-1))%MOD
print(root%MOD)
``` | output | 1 | 29,825 | 16 | 59,651 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider a grid of squares with 10^9 rows and N columns. Let (i, j) be the square at the i-th column (1 \leq i \leq N) from the left and j-th row (1 \leq j \leq 10^9) from the bottom.
Snuke has cut out some part of the grid so that, for each i = 1, 2, ..., N, the bottom-most h_i squares are remaining in the i-th column from the left. Now, he will paint the remaining squares in red and blue. Find the number of the ways to paint the squares so that the following condition is satisfied:
* Every remaining square is painted either red or blue.
* For all 1 \leq i \leq N-1 and 1 \leq j \leq min(h_i, h_{i+1})-1, there are exactly two squares painted red and two squares painted blue among the following four squares: (i, j), (i, j+1), (i+1, j) and (i+1, j+1).
Since the number of ways can be extremely large, print the count modulo 10^9+7.
Constraints
* 1 \leq N \leq 100
* 1 \leq h_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
h_1 h_2 ... h_N
Output
Print the number of the ways to paint the squares, modulo 10^9+7.
Examples
Input
9
2 3 5 4 1 2 4 2 1
Output
12800
Input
2
2 2
Output
6
Input
5
2 1 2 1 2
Output
256
Input
9
27 18 28 18 28 45 90 45 23
Output
844733013 | instruction | 0 | 29,826 | 16 | 59,652 |
"Correct Solution:
```
mod=10**9+7
n=int(input())
H=list(map(int,input().split()))
def solve(h):
if not h:
return 1
elif len(h)==1:
return pow(2,h[0],mod)
N=len(h)
a=[h[i] for i in range(N)]
a=list(set(a))
a.sort()
comp={i:e+1 for e,i in enumerate(a)}
data={comp[e]:e for e in comp.keys()}
data[0]=0
data[0]=1
#print(comp)
dp=[[0 for i in range(len(a)+1)] for j in range(N)]
i=0
if h[i+1]>=h[i]:
id=comp[h[i]]
id2=comp[h[i+1]]
for j in range(id,id2+1):
dp[i][j]=2
for j in range(0,id):
dp[i][j]=1
else:
id=comp[h[i+1]]
for j in range(0,id):
dp[i][j]=pow(2,h[i]-h[i+1],mod)
dp[i][id]=2*(pow(2,h[i]-h[i+1],mod)-1)
dp[i][id]%=mod
id2=comp[h[i]]
dp[i][id]+=2
dp[i][id]%=mod
for i in range(1,N-1):
if h[i+1]>=h[i]:
id=comp[h[i]]
id2=comp[h[i+1]]
for j in range(id,id2+1):
dp[i][j]=(2*dp[i-1][id])%mod
for j in range(0,id):
dp[i][j]=dp[i-1][j]
else:
id=comp[h[i+1]]
id2=comp[h[i]]
for j in range(0,id):
dp[i][j]=(pow(2,h[i]-h[i+1],mod)*dp[i-1][j])%mod
for j in range(id,id2):
low=data[j]
up=data[j+1]-1
dp[i][id]+=dp[i-1][j]*pow(2,h[i]-up,mod)*(pow(2,up-low+1,mod)-1)
dp[i][id]%=mod
dp[i][id]+=2*dp[i-1][id2]
dp[i][id]%=mod
ans=0
id=comp[h[-1]]
for i in range(0,id):
low=data[i]
up=data[i+1]-1
ans+=dp[N-2][i]*pow(2,h[-1]-up,mod)*(pow(2,up-low+1,mod)-1)
ans%=mod
ans+=2*dp[N-2][id]
ans%=mod
return ans
ans=pow(2,H.count(1),mod)
check=[i for i in range(n) if H[i]==1]
check=[-1]+check+[n]
for i in range(len(check)-1):
l,r=check[i],check[i+1]
ans*=solve(H[l+1:r])
ans%=mod
print(ans)
``` | output | 1 | 29,826 | 16 | 59,653 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider a grid of squares with 10^9 rows and N columns. Let (i, j) be the square at the i-th column (1 \leq i \leq N) from the left and j-th row (1 \leq j \leq 10^9) from the bottom.
Snuke has cut out some part of the grid so that, for each i = 1, 2, ..., N, the bottom-most h_i squares are remaining in the i-th column from the left. Now, he will paint the remaining squares in red and blue. Find the number of the ways to paint the squares so that the following condition is satisfied:
* Every remaining square is painted either red or blue.
* For all 1 \leq i \leq N-1 and 1 \leq j \leq min(h_i, h_{i+1})-1, there are exactly two squares painted red and two squares painted blue among the following four squares: (i, j), (i, j+1), (i+1, j) and (i+1, j+1).
Since the number of ways can be extremely large, print the count modulo 10^9+7.
Constraints
* 1 \leq N \leq 100
* 1 \leq h_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
h_1 h_2 ... h_N
Output
Print the number of the ways to paint the squares, modulo 10^9+7.
Examples
Input
9
2 3 5 4 1 2 4 2 1
Output
12800
Input
2
2 2
Output
6
Input
5
2 1 2 1 2
Output
256
Input
9
27 18 28 18 28 45 90 45 23
Output
844733013 | instruction | 0 | 29,827 | 16 | 59,654 |
"Correct Solution:
```
"""
https://atcoder.jp/contests/agc026/tasks/agc026_d
10101100011
01010011100
→1列目を決めたとする
→2列目の置き方は、左端を決めればすべて決まる
偶数列目を反転してみる？
10 → 11
01 00
10 → 11
10 11
すなわち、偶数列目が反転して表示されるとすると
全て01を入れ替える or 全てそのまま しか許されない
→すべてそのままが可能なのは全て0 or 1のときのみ
dp[i][0][X] = i列目の下からX個まですべて0の時の通り数
dp[i][1][X] = i列目の下からX個まですべて1の時の通り数
→答えは dp[i][0][0]
if h[i-1] >= h[i]:
dp[i][0][X] = dp[i-1][1][X] + dp[i-1][0][h[i]]
dp[i][1][X] = dp[i-1][0][X] + dp[i-1][1][h[i]]
else:
if X <= h[i-1]:
dp[i][0][X] = (dp[i-1][1][X] + dp[i-1][0][h[i-1]]) * pow(2,h[i]-h[i-1],mod)
dp[i][1][X] = (dp[i-1][0][X] + dp[i-1][1][h[i-1]]) * pow(2,h[i]-h[i-1],mod)
else:
dp[i][0][X] = (dp[i-1][1][h[i-1]] + dp[i-1][0][h[i-1]]) * pow(2,h[i]-X,mod)
dp[i][1][X] = (dp[i-1][0][h[i-1]] + dp[i-1][1][h[i-1]]) * pow(2,h[i]-X,mod)
"""
from sys import stdin
N = int(stdin.readline())
h = list(map(int,stdin.readline().split()))
mod = 10**9+7
dp = [ [{},{}] for i in range(N) ]
numlis = set()
numlis.add(1)
for i in h:
numlis.add(i)
for i in numlis:
if i <= h[0]:
dp[0][0][i] = pow(2,h[0]-i,mod)
dp[0][1][i] = pow(2,h[0]-i,mod)
for i in range(1,N):
for X in numlis:
if X > h[i]:
continue
if h[i-1] >= h[i]:
dp[i][0][X] = (dp[i-1][1][X] + dp[i-1][0][h[i]]) % mod
dp[i][1][X] = (dp[i-1][0][X] + dp[i-1][1][h[i]]) % mod
else:
if X <= h[i-1]:
dp[i][0][X] = (dp[i-1][1][X] + dp[i-1][0][h[i-1]]) * pow(2,h[i]-h[i-1],mod) % mod
dp[i][1][X] = (dp[i-1][0][X] + dp[i-1][1][h[i-1]]) * pow(2,h[i]-h[i-1],mod) % mod
else:
dp[i][0][X] = (dp[i-1][1][h[i-1]] + dp[i-1][0][h[i-1]]) * pow(2,h[i]-X,mod) % mod
dp[i][1][X] = (dp[i-1][0][h[i-1]] + dp[i-1][1][h[i-1]]) * pow(2,h[i]-X,mod) % mod
print ((dp[N-1][0][1] + dp[N-1][1][1]) % mod)
``` | output | 1 | 29,827 | 16 | 59,655 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider a grid of squares with 10^9 rows and N columns. Let (i, j) be the square at the i-th column (1 \leq i \leq N) from the left and j-th row (1 \leq j \leq 10^9) from the bottom.
Snuke has cut out some part of the grid so that, for each i = 1, 2, ..., N, the bottom-most h_i squares are remaining in the i-th column from the left. Now, he will paint the remaining squares in red and blue. Find the number of the ways to paint the squares so that the following condition is satisfied:
* Every remaining square is painted either red or blue.
* For all 1 \leq i \leq N-1 and 1 \leq j \leq min(h_i, h_{i+1})-1, there are exactly two squares painted red and two squares painted blue among the following four squares: (i, j), (i, j+1), (i+1, j) and (i+1, j+1).
Since the number of ways can be extremely large, print the count modulo 10^9+7.
Constraints
* 1 \leq N \leq 100
* 1 \leq h_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
h_1 h_2 ... h_N
Output
Print the number of the ways to paint the squares, modulo 10^9+7.
Examples
Input
9
2 3 5 4 1 2 4 2 1
Output
12800
Input
2
2 2
Output
6
Input
5
2 1 2 1 2
Output
256
Input
9
27 18 28 18 28 45 90 45 23
Output
844733013 | instruction | 0 | 29,828 | 16 | 59,656 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 10 ** 9 + 7
n = I()
H = LI() + [1]
dp = [0] * (n + 1)
ret = pow(2, H[0], mod)
for kk in range(n + 1):
if H[0] >= H[kk]:
dp[kk] = pow(2, H[0] - H[kk], mod) * 2
else:
dp[kk] = 2
for k in range(1, n):
new_dp = [0] * (n + 1)
for i in range(n + 1):
if H[i] <= H[k]:
if H[k - 1] <= H[i]:
new_dp[i] = dp[i] * 2 * pow(2, H[k] - H[i], mod)
elif H[k - 1] > H[k]:
new_dp[i] = dp[i] - dp[k] + dp[k] * 2
else:
new_dp[i] = (dp[i] - dp[k - 1] + dp[k - 1] * 2) * pow(2, H[k] - H[k - 1], mod)
else:
new_dp[i] = dp[k] * 2
new_dp[i] %= mod
dp = new_dp
print(dp[-1])
``` | output | 1 | 29,828 | 16 | 59,657 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider a grid of squares with 10^9 rows and N columns. Let (i, j) be the square at the i-th column (1 \leq i \leq N) from the left and j-th row (1 \leq j \leq 10^9) from the bottom.
Snuke has cut out some part of the grid so that, for each i = 1, 2, ..., N, the bottom-most h_i squares are remaining in the i-th column from the left. Now, he will paint the remaining squares in red and blue. Find the number of the ways to paint the squares so that the following condition is satisfied:
* Every remaining square is painted either red or blue.
* For all 1 \leq i \leq N-1 and 1 \leq j \leq min(h_i, h_{i+1})-1, there are exactly two squares painted red and two squares painted blue among the following four squares: (i, j), (i, j+1), (i+1, j) and (i+1, j+1).
Since the number of ways can be extremely large, print the count modulo 10^9+7.
Constraints
* 1 \leq N \leq 100
* 1 \leq h_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
h_1 h_2 ... h_N
Output
Print the number of the ways to paint the squares, modulo 10^9+7.
Examples
Input
9
2 3 5 4 1 2 4 2 1
Output
12800
Input
2
2 2
Output
6
Input
5
2 1 2 1 2
Output
256
Input
9
27 18 28 18 28 45 90 45 23
Output
844733013 | instruction | 0 | 29,829 | 16 | 59,658 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
MOD = 10**9+7
def compress(L):
L2 = list(set(L))
L2.sort()
C = {v : k for k, v in enumerate(L2)}
return L2, C
N = int(readline())
height = [0]
H = list(map(int, readline().split()))
height.extend(H)
orih, C = compress(height)
L = len(orih)
dp = [0]*L
dp[0] = 1
phidx = 0
for i in range(1, N+1):
hidx = C[H[i-1]]
dp2 = [0]*L
if phidx >= hidx:
for h in range(hidx+1):
dp2[h] = (dp[hidx]+dp[h])%MOD
else:
if phidx == 0:
dp2[0] = pow(2, orih[hidx])%MOD
for h in range(1, hidx+1):
dp2[h] = pow(2, orih[hidx]-orih[h]+1, MOD)
else:
for h in range(hidx+1):
dp2[h] = (dp[phidx] + dp[min(phidx, h)])*pow(2, orih[hidx]-orih[max(h, phidx)], MOD)%MOD
dp = dp2[:]
phidx = hidx
#print(dp)
print(dp[0])
``` | output | 1 | 29,829 | 16 | 59,659 |
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