text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide a correct Python 3 solution for this coding contest problem.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
"Correct Solution:
```
import sys
import itertools
# import numpy as np
import time
import math
import heapq
from collections import defaultdict
sys.setrecursionlimit(10 ** 7)
INF = 10 ** 18
MOD = 10 ** 9 + 7
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
# map(int, input().split())
N = int(input())
adj = [[] for _ in range(N)]
for i in range(N - 1):
x, y = map(int, input().split())
x -= 1
y -= 1
adj[x].append(y)
adj[y].append(x)
def dfs(v, p):
nim = 0
for u in adj[v]:
if u == p:
continue
nim ^= dfs(u, v)
if p == -1:
return nim
nim += 1
return nim
res = dfs(0, -1)
if res:
print("Alice")
else:
print("Bob")
```
| 84,900 |
Provide a correct Python 3 solution for this coding contest problem.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
"Correct Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
N = int(input())
graph = [[] for _ in range(N)]
for _ in range(N-1):
a, b = map(int, input().split())
graph[a-1].append(b-1)
graph[b-1].append(a-1)
checked = [False]*N
def dfs(p):
checked[p] = True
g = 0
for np in graph[p]:
if not checked[np]:
g ^= dfs(np)+1
return g
if dfs(0) == 0:
print("Bob")
else:
print("Alice")
```
| 84,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
Submitted Solution:
```
class Tree():
def __init__(self, n, edge, indexed=1):
self.n = n
self.tree = [[] for _ in range(n)]
for e in edge:
self.tree[e[0] - indexed].append(e[1] - indexed)
self.tree[e[1] - indexed].append(e[0] - indexed)
def setroot(self, root):
self.root = root
self.parent = [None for _ in range(self.n)]
self.parent[root] = -1
self.order = []
self.order.append(root)
stack = [root]
while stack:
node = stack.pop()
for adj in self.tree[node]:
if self.parent[adj] is None:
self.parent[adj] = node
self.order.append(adj)
stack.append(adj)
N = int(input())
edge = [tuple(map(int, input().split())) for _ in range(N - 1)]
tree = Tree(N, edge)
tree.setroot(0)
grundy = [0 for _ in range(N)]
for node in tree.order[::-1]:
for adj in tree.tree[node]:
if tree.parent[node] == adj:
continue
grundy[node] ^= grundy[adj] + 1
print('Alice' if grundy[0] != 0 else 'Bob')
```
Yes
| 84,902 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
N=int(input())
g = [[] for i in range(N)]
for i in range(N-1):
a,b=map(int,input().split())
a -= 1
b -= 1
g[a].append(b)
g[b].append(a)
nim = [-1 for i in range(N)]
def dfs(cur,par):
val = 0
for dst in g[cur]:
if dst == par:
continue
val ^= dfs(dst,cur) + 1
nim[cur] = val
return val
dfs(0 , -1)
#print(nim)
if nim[0] == 0:
print('Bob')
else:
print('Alice')
```
Yes
| 84,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
Submitted Solution:
```
from collections import deque
import heapq
N = int(input())
gr = [None] * N
child = [0] * N
dic = {}
q = []
for i in range(N-1):
x,y = map(int,input().split())
x -= 1
y -= 1
if x not in dic:
dic[x] = []
if y not in dic:
dic[y] = []
dic[x].append(y)
dic[y].append(x)
child[x] += 1
child[y] += 1
dep = [float("inf")] * N
dep[0] = 0
dq = deque([0])
while len(dq) > 0:
n = dq.popleft()
for i in dic[n]:
if dep[i] > dep[n] + 1:
dep[i] = dep[n] + 1
for i in range(N):
if child[i] == 1:
child[i] = 0
gr[i] = 0
heapq.heappush(q,[-1 * dep[i] , i])
while len(q) > 0:
ngr = 0
no = heapq.heappop(q)
now = no[1]
for i in dic[now]:
if gr[i] != None:
ngr ^= (gr[i] + 1)
else:
child[i] -= 1
if child[i] <= 1 and gr[i] == None:
heapq.heappush(q,[-1 * dep[i] , i])
gr[now] = ngr
#print (gr)
if gr[0] == 0:
print ("Bob")
else:
print ("Alice")
```
Yes
| 84,904 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
ab = [list(map(int,input().split())) for i in range(n-1)]
graph = [[] for i in range(n+1)]
deg = [0]*(n+1)
for a,b in ab:
graph[a].append(b)
graph[b].append(a)
deg[a] += 1
deg[b] += 1
deg[1] += 1
dp = [[] for i in range(n+1)]
mark = [0]*(n+1)
stack = []
for i in range(1,n+1):
if deg[i] == 1:
stack.append(i)
mark[i] = 1
while stack:
x = stack.pop()
if dp[x]:
t = 0
for z in dp[x]:
t ^= z
mark[x] = t+1
if x == 1:
break
for y in graph[x]:
if deg[y] > 1:
dp[y].append(mark[x])
deg[y] -= 1
if deg[y] == 1:
stack.append(y)
if mark[1] == 1:
print("Bob")
else:
print("Alice")
```
Yes
| 84,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
n = I()
e = collections.defaultdict(set)
for _ in range(n-1):
x,y = LI()
e[x].add(y)
e[y].add(x)
def search(s):
d = collections.defaultdict(lambda: inf)
d[s] = 0
q = []
heapq.heappush(q, (0, s))
v = collections.defaultdict(bool)
while len(q):
k, u = heapq.heappop(q)
if v[u]:
continue
v[u] = True
for uv in e[u]:
ud = 1
if v[uv]:
continue
vd = k + ud
if d[uv] > vd:
d[uv] = vd
heapq.heappush(q, (vd, uv))
return d
d = search(1)
dd = collections.defaultdict(int)
for i in range(2,n+1):
dd[d[i]] += 1
if dd[1] == 1:
return 'Alice'
k = 0
for i,c in dd.items():
if c % 2 == 1:
return 'Alice'
if k % 2 == 1:
return 'Alice'
return 'Bob'
print(main())
```
No
| 84,906 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
N=int(input())
g = [[] for i in range(N)]
for i in range(N-1):
a,b=map(int,input().split())
a -= 1
b -= 1
g[a].append(b)
g[b].append(a)
nim = [-1 for i in range(N)]
def dfs(cur,par):
val = 0
for dst in g[cur]:
if dst == par:
continue
val ^= dfs(dst,cur) + 1
nim[cur] = val
return val
dfs(0 , -1)
#print(nim)
if nim[0] == 0:
print('Bob')
else:
print('Alice')
```
No
| 84,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
Submitted Solution:
```
N=int(input())
g = [[] for i in range(N)]
for i in range(N-1):
a,b=map(int,input().split())
a -= 1
b -= 1
g[a].append(b)
g[b].append(a)
nim = [-1 for i in range(N)]
def dfs(cur,par):
val = 0
for dst in g[cur]:
if dst == par:
continue
val ^= dfs(dst,cur) + 1
nim[cur] = val
return val
dfs(0 , -1)
#print(nim)
if nim[0] == 0:
print('Bob')
else:
print('Alice')
```
No
| 84,908 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1, 2, ..., N. The edges of the tree are denoted by (x_i, y_i).
On this tree, Alice and Bob play a game against each other. Starting from Alice, they alternately perform the following operation:
* Select an existing edge and remove it from the tree, disconnecting it into two separate connected components. Then, remove the component that does not contain Vertex 1.
A player loses the game when he/she is unable to perform the operation. Determine the winner of the game assuming that both players play optimally.
Constraints
* 2 \leq N \leq 100000
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
Output
Print `Alice` if Alice wins; print `Bob` if Bob wins.
Examples
Input
5
1 2
2 3
2 4
4 5
Output
Alice
Input
5
1 2
2 3
1 4
4 5
Output
Bob
Input
6
1 2
2 4
5 1
6 3
3 2
Output
Alice
Input
7
1 2
3 7
4 6
2 3
2 4
1 5
Output
Bob
Submitted Solution:
```
import sys
input=sys.stdin.readline
sys.setrecursionlimit(1000000)
N=int(input())
edge=[[] for i in range(N)]
for i in range(N-1):
x,y=map(int,input().split())
edge[x-1].append(y-1)
edge[y-1].append(x-1)
def dfs(v,pv):
if v==0:
if len(edge[v])%2==1:
return True
else:
ans=0
for nv in edge[v]:
ans+=int(dfs(nv,v))
return ans%2==1
else:
if len(edge[v])%2==0:
return True
else:
ans=0
for nv in edge[v]:
if nv!=pv:
ans+=int(dfs(nv,v))
else:
continue
return ans%2==1
answer=dfs(0,-1)
if answer:
print("Alice")
else:
print("Bob")
```
No
| 84,909 |
Provide a correct Python 3 solution for this coding contest problem.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
"Correct Solution:
```
import sys
from bisect import bisect
class BinaryIndexedTree:
def __init__(self, n, MOD):
self.size = n + 1
self.tree = [0] * (n + 2)
self.MOD = MOD
def sum(self, i):
i += 1
s = 0
while i > 0:
s = (s + self.tree[i]) % self.MOD
i -= i & -i
return s
def add(self, i, x):
i += 1
while i <= self.size:
self.tree[i] = (self.tree[i] + x) % self.MOD
i += i & -i
def debug_print(self, limit=None):
if limit is None:
limit = self.size
acc = [0] + [self.sum(i) for i in range(limit)]
aaa = [a1 - a0 for a0, a1 in zip(acc, acc[1:])]
print(aaa)
def solve(n, a, b, sss):
if a > b:
a, b = b, a
for s0, s2 in zip(sss, sss[2:]):
if s2 - s0 < a:
return 0
MOD = 10 ** 9 + 7
bit = BinaryIndexedTree(n, MOD)
bit.add(0, 1)
bit.add(1, 1)
sss.insert(0, -10 ** 18 - 1)
pos = 0
for i in range(1, n):
s0 = sss[i]
s1 = sss[i + 1]
lim = bisect(sss, s1 - b) - 1
if lim >= pos:
bit.add(i + 1, bit.sum(lim))
if s1 - s0 < a:
bit.add(i - 1, -bit.sum(i - 1))
pos = i - 1
return bit.sum(n)
n, a, b, *sss = map(int, sys.stdin.read().split())
print(solve(n, a, b, sss))
```
| 84,910 |
Provide a correct Python 3 solution for this coding contest problem.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
"Correct Solution:
```
import sys
input = sys.stdin.readline
from bisect import bisect_left, bisect_right
INF = 10**18 + 100
N,A,B = map(int,input().split())
S = [-INF] + [int(x) for x in sys.stdin.read().split()]
MOD = 10**9 + 7
dpX = [0] * (N+1) # 最後にYを選んだとして、直前に選んだXがどこにあるか
dpY = [0] * (N+1) # 最後にXを選んだとして、直前に選んだYがどこにあるか
dpX[0] = 1
dpY[0] = 1
dpX_cum = [1] * (N+1) + [0] # -1
dpY_cum = [1] * (N+1) + [0] # -1
dpX_left = 0
dpY_left = 0
for n,x in enumerate(S[2:],2):
iA = bisect_right(S,x-A)
iB = bisect_right(S,x-B)
# ....XY
xy = dpY_cum[iB-1] - dpY_cum[dpY_left-1] if iB >= dpY_left else 0
# ....YX
yx = dpX_cum[iA-1] - dpX_cum[dpX_left-1] if iA >= dpX_left else 0
# ....XX が不可能なら捨てる。明示的に捨てるのではなく、生きている番号だけ持つ
if iA != n:
dpY_left = n-1
if iB != n:
dpX_left = n-1
dpX[n-1] = xy
dpX_cum[n-1] = (dpX_cum[n-2] + xy) % MOD
dpX_cum[n] = dpX_cum[n-1]
dpY[n-1] = yx
dpY_cum[n-1] = (dpY_cum[n-2] + yx) % MOD
dpY_cum[n] = dpY_cum[n-1]
answer = dpX_cum[N-1] - dpX_cum[dpX_left-1]
answer += dpY_cum[N-1] - dpY_cum[dpY_left-1]
answer %= MOD
print(answer)
```
| 84,911 |
Provide a correct Python 3 solution for this coding contest problem.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
"Correct Solution:
```
mod=10**9+7
import bisect,sys
input=sys.stdin.readline
N,A,B=map(int,input().split())
S=[int(input()) for i in range(N)]
datax=[i for i in range(N+1)]
datay=[i for i in range(N+1)]
for i in range(2,N+1):
if S[i-1]-S[i-2]>=A:
datax[i]=datax[i-1]
if S[i-1]-S[i-2]>=B:
datay[i]=datay[i-1]
#print(datax)
#print(datay)
dpx=[0]*(N+1)
imosx=[0]*(N+1)
dpy=[0]*(N+1)
imosy=[0]*(N+1)
dpx[0]=1
dpy[0]=1
imosx[0]=1
imosy[0]=1
for i in range(1,N):
id=bisect.bisect_right(S,S[i]-B)
R=min(id+1,i)
L=datax[i]
if R>=L:
dpx[i]=(imosy[R-1]-imosy[L-2]*(L>=2))%mod
imosx[i]=(imosx[i-1]+dpx[i])%mod
else:
imosx[i]=imosx[i-1]
id=bisect.bisect_right(S,S[i]-A)
R=min(id+1,i)
L=datay[i]
if R>=L:
dpy[i]=(imosx[R-1]-imosx[L-2]*(L>=2))%mod
imosy[i]=(imosy[i-1]+dpy[i])%mod
else:
imosy[i]=imosy[i-1]
i=N
R=i
L=datax[i]
if R>=L:
dpx[i]=(imosy[R-1]-imosy[L-2]*(L>=2))%mod
imosx[i]=(imosx[i-1]+dpx[i])%mod
else:
imosx[i]=imosx[i-1]
R=i
L=datay[i]
if R>=L:
dpy[i]=(imosx[R-1]-imosx[L-2]*(L>=2))%mod
imosy[i]=(imosy[i-1]+dpy[i])%mod
else:
imosy[i]=imosy[i-1]
#print(dpx)
#print(dpy)
print((dpx[-1]+dpy[-1])%mod)
```
| 84,912 |
Provide a correct Python 3 solution for this coding contest problem.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
"Correct Solution:
```
"""
https://atcoder.jp/contests/agc009/tasks/agc009_c
A <= B としてよい
dpだろうなぁ
dpA[i][X] = 1つ前を置いたのがBで、Aに置かれた最大がindexXの時の置き方
dpB[i][X] = 同様
if S[i]-S[i-1] >= B:
dpA[i][X] = dpA[i-1][X]
if X == i-1:
dpA[i][X] = ///
推移は、BITですればおｋ
もし差がB以下ならば→直前にBを置いていた場所に重ねおきはできない
→直前にAに置いていて、なおかつ最後にBに置いたのとの差がB以下の場合だけBにおける
→dpA[i][i-1]以外は0になる
#直前とB以上の差があるとき
if X != i-1:
dpBに置く[i][最後にAに置いたのがX] = dpB[i-1][X]
else:
dpB[i][i-1] = ΣdpA[i-1][y] (y<=S-B)
#差がないとき
if X == i-1:
dpB[i][i-1] = ΣdpA[i-1][y] (y<=S-B)
else:
dpB[i][X] = 0
"""
import sys
from sys import stdin
from collections import deque
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,A,B = map(int,stdin.readline().split())
mod = 10**9+7
BITA = [0] * (N+3)
BITB = [0] * (N+3)
bitadd(1,1,BITA)
#bitadd(0,1,BITB)
aq = deque([])
bq = deque([])
Slis = [float("-inf")]
for loop in range(N):
S = int(stdin.readline())
aq.append( (S,loop+2) )
bq.append( (S,loop+2) )
while aq[0][0] <= S-A:
aq.popleft()
while bq[0][0] <= S-B:
bq.popleft()
#dpAへの推移(Bに置く)
#Bに置けるのは、1つ前との差がB以上の場合全部おk
#そうでない場合、前にAにおいていて、かつ差がB以上の場合
"""
#全てokの場合
if S - Slis[-1] >= B:
Aans = bitsum(bq[0][1]-1,BITB)
Aans %= mod
else: #そうでない場合→直前にAに置いていた場合のみ可能→bitをリセットする必要あり
Aans = bitsum(bq[0][1]-1,BITB)
Aans %= mod
if S - Slis[-1] >= A:
Bans = bitsum(aq[0][1]-1,BITA)
Bans %= mod
else:
Bans = bitsum(aq[0][1]-1,BITA)
Bans %= mod
"""
Aans = bitsum(bq[0][1]-1,BITB)
Bans = bitsum(aq[0][1]-1,BITA)
if Aans < 0:
Aans = 0
if Bans < 0:
Bans = 0
Aans %= mod
Bans %= mod
#print (Aans,Bans)
#更新
if S - Slis[-1] >= B:
bitadd(loop+1,Aans,BITA)
else:
nowsum = bitsum(N+2,BITA)
bitadd(1,-1*nowsum,BITA)
bitadd(loop+1,Aans,BITA)
if S - Slis[-1] >= A:
bitadd(loop+1,Bans,BITB)
else:
nowsum = bitsum(N+2,BITB)
bitadd(1,-1*nowsum,BITB)
bitadd(loop+1,Bans,BITB)
Slis.append(S)
if len(Slis) >= 3 and Slis[-1] - Slis[-3] < min(A,B):
print (0)
sys.exit()
print ((bitsum(N+2,BITA) + bitsum(N+2,BITB))% mod)
```
| 84,913 |
Provide a correct Python 3 solution for this coding contest problem.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
"Correct Solution:
```
import sys
readline = sys.stdin.readline
class Lazysegtree:
#RUQ
def __init__(self, A, intv, initialize = True, segf = min):
#区間は 1-indexed で管理
self.N = len(A)
self.N0 = 2**(self.N-1).bit_length()
self.intv = intv
self.segf = segf
self.lazy = [None]*(2*self.N0)
if initialize:
self.data = [intv]*self.N0 + A + [intv]*(self.N0 - self.N)
for i in range(self.N0-1, -1, -1):
self.data[i] = self.segf(self.data[2*i], self.data[2*i+1])
else:
self.data = [intv]*(2*self.N0)
def _ascend(self, k):
k = k >> 1
c = k.bit_length()
for j in range(c):
idx = k >> j
self.data[idx] = self.segf(self.data[2*idx], self.data[2*idx+1])
def _descend(self, k):
k = k >> 1
idx = 1
c = k.bit_length()
for j in range(1, c+1):
idx = k >> (c - j)
if self.lazy[idx] is None:
continue
self.data[2*idx] = self.data[2*idx+1] = self.lazy[2*idx] \
= self.lazy[2*idx+1] = self.lazy[idx]
self.lazy[idx] = None
def query(self, l, r):
L = l+self.N0
R = r+self.N0
self._descend(L//(L & -L))
self._descend(R//(R & -R)-1)
s = self.intv
while L < R:
if R & 1:
R -= 1
s = self.segf(s, self.data[R])
if L & 1:
s = self.segf(s, self.data[L])
L += 1
L >>= 1
R >>= 1
return s
def update(self, l, r, x):
L = l+self.N0
R = r+self.N0
Li = L//(L & -L)
Ri = R//(R & -R)
self._descend(Li)
self._descend(Ri-1)
while L < R :
if R & 1:
R -= 1
self.data[R] = x
self.lazy[R] = x
if L & 1:
self.data[L] = x
self.lazy[L] = x
L += 1
L >>= 1
R >>= 1
self._ascend(Li)
self._ascend(Ri-1)
inf = 10**19
N, A, B = map(int, readline().split())
S = [-inf] + [int(readline()) for _ in range(N)]
MOD = 10**9+7
dpa = Lazysegtree([1] + [0]*N, 0, initialize = True, segf = lambda x, y: (x+y)%MOD)
dpb = Lazysegtree([1] + [0]*N, 0, initialize = True, segf = lambda x, y: (x+y)%MOD)
for i in range(1, N+1):
oka = 0
ng = i
while abs(oka-ng) > 1:
med = (oka+ng)//2
if S[i] - S[med] >= A:
oka = med
else:
ng = med
okb = 0
ng = i
while abs(okb-ng) > 1:
med = (okb+ng)//2
if S[i] - S[med] >= B:
okb = med
else:
ng = med
tb = dpa.query(0, okb+1)
dpa.update(i-1, i, dpb.query(0, oka+1))
dpb.update(i-1, i, tb)
if S[i] - S[i-1] < A:
dpa.update(0, i-1, 0)
if S[i] - S[i-1] < B:
dpb.update(0, i-1, 0)
print((dpa.query(0, N+1) + dpb.query(0, N+1)) % MOD)
```
| 84,914 |
Provide a correct Python 3 solution for this coding contest problem.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
"Correct Solution:
```
from bisect import bisect_left
N,A,B=map(int,input().split())
inf,mod=float("inf"),10**9+7
S=[int(input()) for i in range(N)]
St_A=[0]*N
St_B=[0]*N
J_A,J_B=[0],[0]
for i in range(N):
St_A[i]=bisect_left(S,S[i]+A)-1
St_B[i]=bisect_left(S,S[i]+B)-1
J_A.append(J_A[-1]+int(St_A[i]!=i))
J_B.append(J_B[-1]+int(St_B[i]!=i))
#print(St_A,St_B)
#print(J_A,J_B)
dp_A=[0]*N
dp_B=[0]*N
dp_A[-1],dp_B[-1]=1,1
for i in range(N-1)[::-1]:
if St_A[i]==i:
dp_A[i]=(dp_A[i+1]+dp_B[i+1])%mod
else:
if J_B[St_A[i]]-J_B[i+1]==0:
dp_A[i]=dp_B[St_A[i]]
else:
dp_A[i]=0
if St_B[i]==i:
dp_B[i]=(dp_A[i+1]+dp_B[i+1])%mod
else:
if J_A[St_B[i]]-J_A[i+1]==0:
dp_B[i]=dp_A[St_B[i]]
else:
dp_B[i]=0
#print(dp_A)
#print(dp_B)
print((dp_A[0]+dp_B[0])%mod)
```
| 84,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
Submitted Solution:
```
import sys
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
MOD = 10**9 + 7
def main(N, A, B, S):
if A > B:
A, B = B, A
if np.max(S[2:] - S[:-2]) < A:
return 0
INF = 1 << 62
S = np.append(S, INF)
# A 未満でたどって、Xがどこまで置けるか
X = np.zeros_like(S)
for i in range(1, N + 1):
if S[i] - S[i - 1] >= A:
X[i] = X[i - 1]
else:
X[i] = i
# 直前の Y としてありうる最も右
Y = np.searchsorted(S, S - B, side='right') - 1
dp = np.zeros_like(S)
dp_cum = np.zeros_like(S)
dp[0] = 1
dp_cum[0] = 1
for n in range(1, N + 1):
if X[n - 1] == 0:
# 初出の Y
dp[n] = 1
# 直前の Y の範囲
l, r = max(0, X[n - 1] - 1), min(n - 1, Y[n])
if l <= r:
dp[n] += dp_cum[r] - dp_cum[l] + dp[l]
dp[n] %= MOD
dp_cum[n] = (dp_cum[n - 1] + dp[n]) % MOD
return dp[-1]
signature = '(i8,i8,i8,i8[:],)'
if sys.argv[-1] == 'ONLINE_JUDGE':
from numba.pycc import CC
cc = CC('my_module')
cc.export('main', signature)(main)
cc.compile()
from my_module import main
N, A, B = map(int, readline().split())
S = np.array(read().split(), np.int64)
print(main(N, A, B, S))
```
No
| 84,916 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
Submitted Solution:
```
import sys
input = sys.stdin.readline
from bisect import bisect_left, bisect_right
INF = 10**18 + 100
N,A,B = map(int,input().split())
S = [-INF] + [int(x) for x in sys.stdin.read().split()]
MOD = 10**9 + 7
dpX = [0] * (N+1) # 最後にYを選んだとして、直前に選んだXがどこにあるか
dpY = [0] * (N+1) # 最後にXを選んだとして、直前に選んだYがどこにあるか
dpX[0] = 1
dpY[0] = 1
dpX_cum = [1] * (N+1) + [0]
dpY_cum = [1] * (N+1) + [0]
dpX_left = 0
dpY_left = 0
for n,x in enumerate(S[2:],2):
iA = bisect_right(S,x-A)
iB = bisect_right(S,x-B)
# ....XY
xy = dpY_cum[iB-1] - dpY_cum[dpY_left-1]
# ....YX
yx = dpX_cum[iA-1] - dpX_cum[dpX_left-1]
# ....XX が不可能なら捨てる。明示的に捨てるのではなく、生きている番号だけ持つ
if iA != n:
dpY_left = n-1
if iB != n:
dpX_left = n-1
dpX[n-1] = xy
dpX_cum[n-1] = (dpX_cum[n-2] + xy) % MOD
dpX_cum[n] = dpX_cum[n-1]
dpY[n-1] = yx
dpY_cum[n-1] = (dpY_cum[n-2] + yx) % MOD
dpY_cum[n] = dpY_cum[n-1]
answer = dpX_cum[N-1] - dpX_cum[dpX_left-1]
answer += dpY_cum[N-1] - dpY_cum[dpY_left-1]
answer %= MOD
print(answer)
```
No
| 84,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
Submitted Solution:
```
import sys
import bisect
input = sys.stdin.readline
class Bit:
def __init__(self,n):
self.size = n
self.tree = [0]*(n+1)
def sum(self,i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self,i,x):
while i <= self.size:
self.tree[i] += x
i += i & -i
n,x,y = map(int,input().split())
a = [int(input()) for i in range(n)]
mod = 10**9+7
bsx = [0]*n
bsy = [0]*n
rowx = [0]*n
rowy = [0]*n
for i in range(n):
bsx[i] = bisect.bisect_right(a,a[i]-x)
bsy[i] = bisect.bisect_right(a,a[i]-y)
j = n-i-1
if j == 0:
continue
if a[j-1]+x <= a[j]:
rowx[j-1] += rowx[j]+1
else:
rowx[j-1] = 0
if a[j-1]+y <= a[j]:
rowy[j-1] += rowy[j]+1
else:
rowy[j-1] = 0
sm1 = [0 for i in range(n)]
sm2 = [0 for i in range(n)]
#sm1[i]: dp[i+1][i]
#sm2[i]: dp[i][i+1]
bit1 = Bit(n+2)
bit2 = Bit(n+2)
for i in range(n):
if i == 0:
sm1[i] = 1
sm2[i] = 1
else:
sm1[i] = bit2.sum(min(i,bsx[i]+1))%mod
sm2[i] = bit1.sum(min(i,bsy[i]+1))%mod
bit1.add(i+1,sm1[i])
bit1.add(i+rowx[i]+2,-sm1[i])
bit2.add(i+1,sm2[i])
bit2.add(i+rowy[i]+2,-sm2[i])
print(sm1[-1]+sm2[-1])
```
No
| 84,918 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:
* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.
How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.
Constraints
* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)
Input
The input is given from Standard Input in the following format:
N A B
S_1
:
S_N
Output
Print the number of the different divisions under the conditions, modulo 10^9 + 7.
Examples
Input
5 3 7
1
3
6
9
12
Output
5
Input
7 5 3
0
2
4
7
8
11
15
Output
4
Input
8 2 9
3
4
5
13
15
22
26
32
Output
13
Input
3 3 4
5
6
7
Output
0
Submitted Solution:
```
"""
https://atcoder.jp/contests/agc009/tasks/agc009_c
A <= B としてよい
dpだろうなぁ
dpA[i][X] = 1つ前を置いたのがBで、Aに置かれた最大がindexXの時の置き方
dpB[i][X] = 同様
if S[i]-S[i-1] >= B:
dpA[i][X] = dpA[i-1][X]
if X == i-1:
dpA[i][X] = ///
推移は、BITですればおｋ
もし差がB以下ならば→直前にBを置いていた場所に重ねおきはできない
→直前にAに置いていて、なおかつ最後にBに置いたのとの差がB以下の場合だけBにおける
→dpA[i][i-1]以外は0になる
"""
import sys
from sys import stdin
from collections import deque
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,A,B = map(int,stdin.readline().split())
mod = 10**9+7
BITA = [0] * (N+20)
BITB = [0] * (N+20)
bitadd(1,1,BITA)
#bitadd(0,1,BITB)
aq = deque([])
bq = deque([])
Slis = [float("-inf")]
for loop in range(N):
S = int(stdin.readline())
aq.append( (S,loop+2) )
bq.append( (S,loop+2) )
while aq[0][0] <= S-A:
aq.popleft()
while bq[0][0] <= S-B:
bq.popleft()
#dpAへの推移(Bに置く)
#Bに置けるのは、1つ前との差がB以上の場合全部おk
#そうでない場合、前にAにおいていて、かつ差がB以上の場合
#全てokの場合
if S - Slis[-1] >= B:
Aans = bitsum(N+2,BITB)
Aans %= mod
else: #そうでない場合→直前にAに置いていた場合のみ可能→bitをリセットする必要あり
Aans = bitsum(bq[0][1]-1,BITB)
Aans %= mod
if S - Slis[-1] >= A:
Bans = bitsum(N+2,BITA)
Bans %= mod
else:
Bans = bitsum(aq[0][1]-1,BITA)
Bans %= mod
if Aans < 0:
Aans = 0
if Bans < 0:
Bans = 0
#print (Aans,Bans)
#更新
if S - Slis[-1] >= B:
bitadd(loop+1,Aans,BITA)
else:
nowsum = bitsum(N+2,BITA)
bitadd(1,-1*nowsum,BITA)
bitadd(loop+1,Aans,BITA)
if S - Slis[-1] >= A:
bitadd(loop+1,Bans,BITB)
else:
nowsum = bitsum(N+2,BITB)
bitadd(1,-1*nowsum,BITB)
bitadd(loop+1,Bans,BITB)
Slis.append(S)
print ((bitsum(N+2,BITA) + bitsum(N+2,BITB))% mod)
```
No
| 84,919 |
Provide a correct Python 3 solution for this coding contest problem.
Sigma and Sugim are playing a game.
The game is played on a graph with N vertices numbered 1 through N. The graph has N-1 red edges and N-1 blue edges, and the N-1 edges in each color forms a tree. The red edges are represented by pairs of integers (a_i, b_i), and the blue edges are represented by pairs of integers (c_i, d_i).
Each player has his own piece. Initially, Sigma's piece is at vertex X, and Sugim's piece is at vertex Y.
The game is played in turns, where turns are numbered starting from turn 1. Sigma takes turns 1, 3, 5, ..., and Sugim takes turns 2, 4, 6, ....
In each turn, the current player either moves his piece, or does nothing. Here, Sigma can only move his piece to a vertex that is directly connected to the current vertex by a red edge. Similarly, Sugim can only move his piece to a vertex that is directly connected to the current vertex by a blue edge.
When the two pieces come to the same vertex, the game ends immediately. If the game ends just after the operation in turn i, let i be the total number of turns in the game.
Sigma's objective is to make the total number of turns as large as possible, while Sugim's objective is to make it as small as possible.
Determine whether the game will end in a finite number of turns, assuming both players plays optimally to achieve their respective objectives. If the answer is positive, find the number of turns in the game.
Constraints
* 2 ≦ N ≦ 200,000
* 1 ≦ X, Y ≦ N
* X \neq Y
* 1 ≦ a_i, b_i, c_i, d_i ≦ N
* The N-1 edges in each color (red and blue) forms a tree.
Input
The input is given from Standard Input in the following format:
N X Y
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}
c_1 d_1
c_2 d_2
:
c_{N-1} d_{N-1}
Output
If the game will end in a finite number of turns, print the number of turns. Otherwise, print `-1`.
Examples
Input
4 1 2
1 2
1 3
1 4
2 1
2 3
1 4
Output
4
Input
3 3 1
1 2
2 3
1 2
2 3
Output
4
Input
4 1 2
1 2
3 4
2 4
1 2
3 4
1 3
Output
2
Input
4 2 1
1 2
3 4
2 4
1 2
3 4
1 3
Output
-1
Input
5 1 2
1 2
1 3
1 4
4 5
2 1
1 3
1 5
5 4
Output
6
"Correct Solution:
```
import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7 # 998244353
input=lambda:sys.stdin.readline().rstrip()
def resolve():
n,x,y=map(int,input().split())
x-=1; y-=1
E1=[[] for _ in range(n)]
E2=[[] for _ in range(n)]
for _ in range(n-1):
u,v=map(int,input().split())
u-=1; v-=1
E1[u].append(v)
E1[v].append(u)
for _ in range(n-1):
u,v=map(int,input().split())
u-=1; v-=1
E2[u].append(v)
E2[v].append(u)
# E2 において、y-rooted tree とみて、depth,par を計算
par=[None]*n # x の必勝頂点を判定するのに必要
par[y]=y
depth2=[None]*n
depth2[y]=0
Q=[y]
while(Q):
v=Q.pop()
for nv in E2[v]:
if(depth2[nv] is not None): continue
depth2[nv]=depth2[v]+1
par[nv]=v
Q.append(nv)
# E1の辺で、E2での距離が 3 以上のものは必勝
win=[0]*n
for v in range(n):
for nv in E1[v]:
if(par[v]==nv or par[nv]==v or par[v]==par[nv] or par[par[v]]==nv or par[par[nv]]==v): continue
win[nv]=win[v]=1
# E1 において、x-rooted tree とみて探索
# depth1 < depth2 -> 以降も探索できる
# depth1 = depth2 -> そこで捕まる
ans=depth2[x]
depth1=[None]*n
depth1[x]=0
Q=[x]
while(Q):
v=Q.pop()
if(win[v]): # 探索できる状態 & 必勝頂点にいれば勝ち
print(-1)
return
for nv in E1[v]:
if(depth1[nv] is not None): continue
depth1[nv]=depth1[v]+1
ans=max(ans,depth2[nv])
if(depth1[nv]<depth2[nv]): Q.append(nv)
print(2*ans)
resolve()
```
| 84,920 |
Provide a correct Python 3 solution for this coding contest problem.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
"Correct Solution:
```
def check_range(xy, n):
for i in range(2):
if xy[i] == -1:
xy[i] = n - 1
elif xy[i] == n:
xy[i] = 0
return xy
while True:
n = int(input())
if n == 0:
break
ms = [[0 for i in range(n)] for j in range(n)]
xy = [n//2, n//2 - 1]
i = 1
while i <= n*n:
if ms[xy[1]][xy[0]] != 0:
xy[0] -= 1
xy[1] -= 1
xy = check_range(xy, n)
continue
ms[xy[1]][xy[0]] = i
xy[0] += 1
xy[1] -= 1
xy = check_range(xy, n)
i += 1
for i in range(n):
s = ""
for j in range(n):
s += "{0:>4}".format(str(ms[n - i - 1][j]))
print(s)
```
| 84,921 |
Provide a correct Python 3 solution for this coding contest problem.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
"Correct Solution:
```
from sys import stdin
def sqmatrix(n):
return [[0] * n for i in range(n)]
def magicsq(m):
n = len(m)
j = int(n/2)
i = j + 1
for num in range(1, n**2 + 1):
m[i][j] = num
i += 1
j += 1
i = i%n
j = j % n
if m[i][j] != 0:
i += 1
j -= 1
if j < 0:
j = n -1
i = i % n
return m
for line in stdin:
n = int(line)
if n == 0:
break
sm = sqmatrix(n)
for row in magicsq(sm): print (''.join(map(lambda x: str(x).rjust(4, ' '),row)))
```
| 84,922 |
Provide a correct Python 3 solution for this coding contest problem.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
"Correct Solution:
```
def square(n):
mp = [[0 for _ in range(n)] for _ in range(n)]
x, y = n // 2 , n // 2 + 1
mp[y][x] = 1
for i in range(2, n ** 2 + 1):
x = (x + 1) % n
y = (y + 1) % n
while mp[y][x]:
y = (y + 1) % n
x = (x - 1) % n
mp[y][x] = i
for line in mp:
s = ""
for i in line:
si = str(i)
s += si.rjust(4)
print(s)
while True:
n = int(input())
if not n:
break
square(n)
```
| 84,923 |
Provide a correct Python 3 solution for this coding contest problem.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
"Correct Solution:
```
# AOJ 0078 Magic Square
# Python3 2018.6.17 bal4u
MAX = 15
tbl = [0]*MAX
while True:
n = int(input())
if n == 0: break
tbl = [[0 for r in range(n)] for c in range(n)]
r, c = n//2+1, n//2
tbl[r][c] = i = 1
while i < n*n:
r += 1
c += 1
while True:
if c >= n: c = 0;
elif c < 0: c = n - 1;
if r >= n: r = 0;
if tbl[r][c] == 0:
i += 1
tbl[r][c] = i
break
r += 1
c -= 1
for r in range(n):
for c in range(n):
print(format(tbl[r][c], '4d'), end='')
print()
```
| 84,924 |
Provide a correct Python 3 solution for this coding contest problem.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N = int(readline())
if N == 0:
return False
x = N//2; y = x+1
P = [[-1]*N for i in range(N)]
c = 0
N2 = N*N
while c < N2:
if P[y][x] == -1:
c += 1
P[y][x] = c
x = (x + 1) % N; y = (y + 1) % N
else:
x = (x - 1) % N; y = (y + 1) % N
for line in P:
write("".join(map("{:4d}".format, line)))
write("\n")
return True
while solve():
...
```
| 84,925 |
Provide a correct Python 3 solution for this coding contest problem.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
"Correct Solution:
```
import sys
for n in map(int, sys.stdin):
if not n:
break
a = [[-1]*(n+2)]+[[-1]+[0]*n+[-1] for _ in [0]*n]+[[-1]*(n+2)]
x, y = (n+1)//2, (n+1)//2+1
for i in range(1, n**2+1):
while a[y][x]:
if a[y][x] == -1:
if x in {0, n+1}:
x = (1, n)[x==0]
if y in {0, n+1}:
y = (1, n)[y==0]
else:
x, y = x-1, y+1
a[y][x] = i
x, y = x+1, y+1
for l in a[1:-1]:
print("".join(map(lambda x: "{:>4d}".format(x), l[1:-1])))
```
| 84,926 |
Provide a correct Python 3 solution for this coding contest problem.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
"Correct Solution:
```
def print_board(A,n):
for i in range(n):
s = ""
for j in range(n):
st = str(A[i][j])
if len(st) == 1:
s += " "
elif len(st) == 2:
s += " "
elif len(st) == 3:
s += " "
s += str(st)
print(s)
def check_leftdown(A,h,w,n):
if h + 1 > n - 1:
#下に抜けるならば,左に-1してその列に値が入っていない箇所を探す
w -= 1
for x in range(n):
if A[x][w] == 0:
h = x
break
else:
#左に抜けるならば
if w - 1 < 0:
w = n
h += 1
else:
h += 1
w -= 1
return h,w
def check_rightdown(A,h,w,n):
if h + 1 > n - 1:
#下にも右にも抜ける場合
if w + 1 > n - 1:
None
else:
#下に抜けるならば,右に＋1してその列に値が入っていない箇所を探す
w += 1
for x in range(n):
if A[x][w] == 0:
h = x
break
else:
if w + 1 > n - 1:
#右に抜けるならば
w = 0
h += 1
else:
#通常パターン
h += 1
w += 1
if A[h][w] != 0:
#値が既に埋まっている
#左下をチェック
h,w = check_leftdown(A,h,w,n)
return h,w
if __name__ == '__main__':
while True:
try:
n = int(input())
if n == 0:
break
A = [[0 for i in range(n)] for j in range(n)]
cnt = n * n
for x in range(cnt):
if x == 0:
mid = n // 2
h = mid + 1
w = mid
A[h][w] = x + 1
else:
h,w = check_rightdown(A,h,w,n)
A[h][w] = x+1
print_board(A,n)
except EOFError:
break
```
| 84,927 |
Provide a correct Python 3 solution for this coding contest problem.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
"Correct Solution:
```
while 1:
n = int(input())
if n == 0:
break
l = [[ 0 for i in range(n)] for j in range(n)]
i = n//2 + 1
j = n//2
l[i][j] = ' 1'
c = 2
d = '{: >4}'.format(c)
for x in range(n**2 - 1):
i = (i + 1) % n
j = (j + 1) % n
while l[i][j] != 0:
i = (i + 1) % n
j = (j - 1) % n
l[i][j] = d
c += 1
d = '{: >4}'.format(c)
for s in range(n):
print(*l[s],sep="")
```
| 84,928 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
Submitted Solution:
```
while True:
n = int(input())
if n==0:
break
else:
l = [[0 for i in range(n)] for j in range(n)]
x = n//2
y = n//2+1
c = 0
while c < n**2:
c += 1
while l[y][x] != 0:
y += 1
if y >= n:
y = 0
x -= 1
if x <= -1:
x = n-1
l[y][x] = c
x += 1
if x >= n:
x = 0
y += 1
if y >= n:
y = 0
for i in l:
a = ''
for j in i:
if 0 <= j < 10**1:
a = a + ' ' + str(j)
elif 10**1<=j<10**2:
a = a + ' ' + str(j)
elif 10**2<=j<10**3:
a = a + ' ' + str(j)
print(a)
```
Yes
| 84,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
Submitted Solution:
```
while True :
n = int(input())
if n == 0 :
break
magic = list([0] * n for i in range(n))
x = (n+1)//2 - 1
y = (n+1)//2 - 2
for i in range(1, n**2+1) :
x += 1
y += 1
while True :
if x >= n :
x = 0
if y >= n :
y = 0
if magic[x][y] != 0 :
x += 1
y -= 1
if y == -1 :
y = n - 1
if x < n and y < n and magic[x][y] == 0 :
break
magic[x][y] = i
for i in range(n) :
for j in range(n) :
print('{:4}'.format(magic[i][j]), end="")
print()
```
Yes
| 84,930 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
Submitted Solution:
```
while 1:
n=int(input())
if n==0:break
ans=[[0 for i in range(n)] for j in range(n)]
y=n//2
x=n//2-1
for i in range(1,n**2+1):
y=[y+1,y+1-n][y+1==n]
x=[x+1,x+1-n][x+1==n]
if ans[y][x]==0:ans[y][x]=str(i).rjust(4)
else:
y=[y+1,y+1-n][y+1==n]
x=[x-1,x-1+n][x==0]
ans[y][x]=str(i).rjust(4)
for i in ans:print(''.join(map(str,i)))
```
Yes
| 84,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
Submitted Solution:
```
for e in iter(input,'0'):
n=int(e);m,f=[[0]*n for _ in[0]*n],'{:4}'*n
s=lambda x:-~x%n
j=n//2;i=j+1
m[i][j]=1
for k in range(2,n*n+1):
i,j,t=s(i),s(j),j
if m[i][j]:i,j=s(i),t
m[i][j]=k
for r in m:print(f.format(*r))
```
Yes
| 84,932 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
import os
import math
for s in sys.stdin:
n = int(s)
if n == 0:
break
M = [[None for i in range(n)] for j in range(n)]
# 1
x = n // 2
y = n // 2 + 1
M[y][x] = 1
x = x + 1
y = y + 1
counter = 2
while counter < n * n + 1:
if x >= n:
x = 0
elif x < 0:
x = n - 1
if y >= n:
y = 0
elif y < 0:
y = n - 1
if M[y][x] is not None:
x -= 1
y += 1
else:
M[y][x] = counter
counter += 1
x += 1
y += 1
for row in M:
print(*row)
```
No
| 84,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0078
"""
import sys
class MagicSquare():
def __init__(self, size):
self.size = size
self.data = [['@']*self.size for _ in range(self.size)]
self.fill_data()
def fill_data(self):
num = 1
x = self.size // 2
y = x + 1
self.data[y][x] = num
while num < self.size**2:
x += 1
if x >= self.size:
x = 0
y += 1
if y >= self.size:
y = 0
if self.data[y][x] == '@':
num += 1
self.data[y][x] = num
else:
while self.data[y][x] != '@':
x -= 1
if x < 0:
x = self.size - 1
y += 1
if y >= self.size:
y = 0
num += 1
self.data[y][x] = num
def print_data(self):
digits = len(str(self.size**2))
for row in self.data:
print(*row)
def main(args):
while True:
size = int(input())
if size < 3 or size%2 == 0:
break
ms = MagicSquare(size)
ms.print_data()
if __name__ == '__main__':
main(sys.argv[1:])
```
No
| 84,934 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
Submitted Solution:
```
import sys
f = sys.stdin
def set1(sq):
x = len(sq) // 2
y = len(sq) // 2 + 1
square[y][x] = 1
return x, y, 1
def mv_r(sq, x, y):
x = (x + 1) % len(sq)
y = (y + 1) % len(sq)
return x, y
def mv_l(sq, x, y):
x -= 1
if x < 0:
x += len(sq)
y = (y + 1) % len(sq)
return x, y
def setnext(sq, x, y, i):
x, y = mv_r(sq, x,y)
i += 1
while sq[y][x]:
x, y = mv_l(sq, x, y)
sq[y][x] = i
return x, y, i
while True:
n = int(f.readline())
if n == 0:
break
square = [[None for j in range(n)] for i in range(n)]
arg = set1(square)
for i in range(n * n - 1):
arg = setnext(square,*arg)
for line in square:
print(*line)
```
No
| 84,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The numbers 1 to n x n are contained in the n x n square squares one by one, and the sum of the squares in any vertical column and the sum of the squares in any horizontal column are diagonal squares. Those with the same sum of eyes are called magic squares.
There are the following methods to create a magic square with an odd number of squares on each side.
1. Put 1 in the square just below the center square.
2. Put the following numbers in the square at the bottom right.
However, if the square you are trying to insert a number is out of the square, or if the number is already filled, search for the square to insert the number according to the following method.
* If it extends to the right, it will be on the left side of the same row, if it extends to the left, it will be on the right side of the same row, and if it extends below, it will be on the top of the same column. Put in.
* If the square you are trying to enter is filled, put it in the square diagonally below the left of the filled square.
3. Repeat 2 until all the squares are filled.
Follow this method to create a program that takes the number n of squares on one side as input and outputs magic squares of that size. However, n is an odd number between 3 and 15. Output each number in the square as a right-justified 4-digit number.
input
Multiple inputs are given. Each input gives n (a positive integer) on one line. The input ends with 0. The number of inputs does not exceed 10.
output
Output n x n magic squares for each input.
Example
Input
3
5
0
Output
4 9 2
3 5 7
8 1 6
11 24 7 20 3
4 12 25 8 16
17 5 13 21 9
10 18 1 14 22
23 6 19 2 15
Submitted Solution:
```
from sys import stdin
def sqmatrix(n):
return [[0] * n for i in range(n)]
def magicsq(m):
n = len(m)
j = int(n/2)
i = j + 1
for num in range(1, n**2 + 1):
m[i][j] = num
i += 1
j += 1
i = i%n
j = j % n
if m[i][j] != 0:
i += 1
j -= 1
if j < 0:
j = n -1
i = i % n
return m
for line in stdin:
n = int(line)
if n == 0:
break
sm = sqmatrix(n)
for row in magicsq(sm): print (' '.join(map(str,row)))
```
No
| 84,936 |
Provide a correct Python 3 solution for this coding contest problem.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
"Correct Solution:
```
from sys import stdin
# from line_profiler import LineProfiler
def main():
head = [[8,1,2,4],[1,2,4,8],[2,4,8,1],[4,8,1,2]]
while(True):
W,H = map(int, stdin.readline().split())
m = [[0]*W for _ in range(H)]
ps = []
if not (W or H): break
for h in range(H):
s = stdin.readline()
for w in range(W):
if s[w] in [".", "X", "E", "N", "W", "S"]:
m[h][max(0,w-1)] |= 1
m[min(H-1,h+1)][w] |= 2
m[h][min(W-1,w+1)] |= 4
m[max(0,h-1)][w] |= 8
if s[w] in ["E", "N", "W", "S"]:
ps.append([h,w,["E", "N", "W", "S"].index(s[w])])
if s[w] == "X":
m[h][w] |= 16
if W <= 2 or H <= 2:
print(1)
continue
ttt = 0
while(True):
ttt += 1
if ttt > 180: print("NA"); break
dest = []
positions = [a[:2] for a in ps]
for pi,p in enumerate(ps):
for i in range(4):
if head[p[2]][i] & m[p[0]][p[1]]:
d = head[p[2]][i]
if d == 1 and not [p[0],p[1]+1] in positions: p[2] = 0; dest.append([pi,2,p[0],p[1]+1]); break
elif d == 2 and not [p[0]-1,p[1]] in positions: p[2] = 1; dest.append([pi,3,p[0]-1,p[1]]); break
elif d == 4 and not [p[0],p[1]-1] in positions: p[2] = 2; dest.append([pi,0,p[0],p[1]-1]); break
elif d == 8 and not [p[0]+1,p[1]] in positions: p[2] = 3; dest.append([pi,1,p[0]+1,p[1]]); break
else:
dest.append([pi,(p[2]+2)&3,p[0],p[1]])
dest = sorted(dest,key=lambda x: (x[2:],x[1]))
dellist = []
for pi,d,dy,dx in dest:
if not [dy,dx] in positions: positions.append([dy,dx]); ps[pi][:2] = [dy,dx]
if m[ps[pi][0]][ps[pi][1]] & 16: dellist.append(pi)
for idx in sorted(dellist)[::-1]: del ps[idx]
if not len(ps): print(ttt); break
if __name__ == "__main__":
# prf = LineProfiler()
# prf.add_function(main)
# prf.runcall(main)
# prf.print_stats()
main()
```
| 84,937 |
Provide a correct Python 3 solution for this coding contest problem.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
"Correct Solution:
```
from sys import stdin
from itertools import chain
head = [[8,1,2,4],[1,2,4,8],[2,4,8,1],[4,8,1,2]]
while(True):
W,H = map(int, stdin.readline().split())
m = [[0]*W for _ in range(H)]
ps = []
if not (W or H): break
flag = True
for h in range(H):
s = stdin.readline()
if "X" in s: flag = False
for w in range(W):
if s[w] in [".", "X", "E", "N", "W", "S"]:
m[h][max(0,w-1)] |= 1
m[min(H-1,h+1)][w] |= 2
m[h][min(W-1,w+1)] |= 4
m[max(0,h-1)][w] |= 8
if s[w] in ["E", "N", "W", "S"]:
ps.append([h,w,["E", "N", "W", "S"].index(s[w])])
if s[w] == "X":
m[h][w] |= 16
# if flag:
# print("NA")
# continue
if W <= 2 or H <= 2:
print(1)
continue
# print(ps)
# for n in m:
# print(n)
ttt = 0
while(True):
ttt += 1
if ttt > 180: print("NA"); break
# import time
# time.sleep(1)
mt = [[0]*W for _ in range(H)]
for p in ps:
mt[p[0]][p[1]] |= 1
dest = []
# for mmm in mt:
# print(mmm)
# print()
# print([a[:2] for a in ps])
for pi,p in enumerate(ps):
for i in range(4):
if head[p[2]][i] & m[p[0]][p[1]]:
d = head[p[2]][i]
if d == 1 and not [p[0],p[1]+1] in [a[:2] for a in ps]: p[2] = 0; dest.append([pi,2,p[0],p[1]+1]); break
elif d == 2 and not [p[0]-1,p[1]] in [a[:2] for a in ps]: p[2] = 1; dest.append([pi,3,p[0]-1,p[1]]); break
elif d == 4 and not [p[0],p[1]-1] in [a[:2] for a in ps]: p[2] = 2; dest.append([pi,0,p[0],p[1]-1]); break
elif d == 8 and not [p[0]+1,p[1]] in [a[:2] for a in ps]: p[2] = 3; dest.append([pi,1,p[0]+1,p[1]]); break
else:
dest.append([pi,(p[2]+2)&3,p[0],p[1]])
dest = sorted(dest,key=lambda x: (x[2:],x[1]))
# print(dest)
# for mmm in mt:
# print(mmm)
# print("ps = ",ps,dest)
dellist = []
for pi,d,dy,dx in dest:
# print(W,H,dy,dx,mt)
if not mt[dy][dx]:
# print("move",W,H,ps[pi][:2],dy,dx)
mt[dy][dx] |= 1
ps[pi][:2] = [dy,dx]
if m[ps[pi][0]][ps[pi][1]] & 16:
# print("OUT")
# print(ttt)
# time.sleep(3)
dellist.append(pi)
# print(dellist)
for idx in sorted(dellist)[::-1]:
del ps[idx]
if not len(ps):
print(ttt)
break
# ms = [stdin.readline() for _ in range(H)]
# print(m)
# from pprint import pprint
# print(ps)
# for n in m:
# print(n)
# print()
```
| 84,938 |
Provide a correct Python 3 solution for this coding contest problem.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
"Correct Solution:
```
while True:
dic = {"#":-3, ".":-1, "X":-2, "E":0, "N":1, "W":2, "S":3}
def fix(c):
return dic[c]
w, h = map(int, input().split())
if w == 0:
break
mp = [list(map(fix, list(input()))) for _ in range(h)]
persons = []
for y in range(h):
for x in range(w):
if mp[y][x] in (0, 1, 2, 3):
persons.append([x, y, mp[y][x]])
vx = (1, 0, -1, 0)
vy = (0, -1, 0, 1)
time = 0
while time < 180:
if not persons:
print(time)
break
reserve = {}
for index, state in enumerate(persons):
x, y, d = state
for i in range(d - 1, d + 3):
dx, dy = vx[i % 4], vy[i % 4]
if mp[y + dy][x + dx] in (-1, -2):
persons[index][2] = i % 4
if (x + dx, y + dy) in reserve:
if reserve[(x + dx, y + dy)][1] > (i + 2) % 4:
reserve[(x + dx, y + dy)] = (index, (i + 2) % 4, mp[y + dy][x + dx])
else:
reserve[(x + dx, y + dy)] = (index, (i + 2) % 4, mp[y + dy][x + dx])
break
reserve = sorted(reserve.items(), key=lambda a:-a[1][0])
for k, v in reserve:
nx, ny = k
index, nd, panel = v
px, py, _ = persons[index]
mp[py][px] = -1
if panel == -1:
persons[index] = [nx, ny, (nd + 2) % 4]
mp[ny][nx] = (nd + 2) % 4
else:
persons.pop(index)
time += 1
else:
print("NA")
```
| 84,939 |
Provide a correct Python 3 solution for this coding contest problem.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
"Correct Solution:
```
dd = ((1, 0), (0, -1), (-1, 0), (0, 1))
ENWS = "ENWS"
ds = [-1, 0, 1, 2]
ss = [2, 3, 0, 1]
while 1:
W, H = map(int, input().split())
if W == H == 0:
break
S = [input() for i in range(H)]
ps = []; U = [[-1]*W for i in range(H)]
for i in range(H):
for j, c in enumerate(S[i]):
if c in ENWS:
ps.append([i, j, ENWS.index(c)])
U[i][j] = 1
elif c == '#':
U[i][j] = W*H
rest = len(ps)
t = 0
while rest > 0 and t <= 180:
for p in ps:
y, x, c = p
if c == -1:
continue
for d in ds:
dx, dy = dd[(c+d) % 4]
nx = x + dx; ny = y + dy
if U[ny][nx] == -1:
p[2] = (c+d) % 4
break
ps.sort(key = lambda x: ss[x[2]])
qs = []
for p in ps:
y, x, c = p
if c == -1:
continue
dx, dy = dd[c]
nx = x + dx; ny = y + dy
if U[ny][nx] == -1:
U[ny][nx] = 1
if S[ny][nx] == 'X':
rest -= 1
p[2] = -1
qs.append((ny, nx))
else:
p[0] = ny; p[1] = nx
qs.append((y, x))
for y, x in qs:
U[y][x] = -1
t += 1
if t <= 180:
print(t)
else:
print("NA")
```
| 84,940 |
Provide a correct Python 3 solution for this coding contest problem.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
"Correct Solution:
```
from collections import defaultdict
from typing import Dict
class Players(object):
def __init__(self):
self.players: Dict[int, "Player"] = {}
self.reservation = defaultdict(list)
self.square = None
self.goal = []
def set_goal(self, x, y):
self.goal.append((x, y))
def add(self, player: "Player"):
self.players[len(self.players)] = player
def search_all(self):
for i, player in self.players.items():
x, y, d = player.search(self.square)
self.reservation[(x, y)].append(((d-2)%4, i))
for k, v in self.reservation.items():
if len(v) == 1:
continue
self.reservation[k] = [min(v)]
def move_all(self):
for (x, y), l in self.reservation.items():
_id = l[0][1]
_x, _y = self.players[_id].pos
self.players[_id].move(x, y)
self.square[_y][_x], self.square[y][x] = ".", "o"
self.reservation.clear()
for i, player in self.players.copy().items():
x, y = player.pos
if (x, y) in self.goal:
del self.players[i]
self.square[y][x] = "."
@property
def is_finished(self):
return len(self.players) == 0
class Player(object):
dir_dict = {"E": 0, "N": 1, "W": 2, "S": 3}
dir_diff = ((1, 0), (0, -1), (-1, 0), (0, 1))
def __init__(self, x, y, direction):
self.x = x
self.y = y
self.dir = self.dir_dict[direction]
def search(self, square):
for i in range(self.dir-1, self.dir+3):
dx, dy = self.dir_diff[i % 4]
if square[self.y+dy][self.x+dx] in (".", "X"):
self.dir = i % 4
return self.x+dx, self.y+dy, self.dir
else:
return self.x, self.y, self.dir
def move(self, x, y):
self.x, self.y = x, y
@property
def pos(self):
return self.x, self.y
while True:
W, H = map(int, input().split())
if W == 0:
break
square = []
players = Players()
for y in range(H):
square.append(list(input()))
for x, c in enumerate(square[-1]):
if c in "ENWS":
players.add(Player(x, y, c))
elif c == "X":
players.set_goal(x, y)
players.square = square
for i in range(1, 121):
players.search_all()
players.move_all()
if players.is_finished:
print(i)
break
else:
print("NA")
```
| 84,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
Submitted Solution:
```
from sys import stdin
head = [[8,1,2,4],[1,2,4,8],[2,4,8,1],[4,8,1,2]]
while(True):
W,H = map(int, stdin.readline().split())
m = [[0]*W for _ in range(H)]
ps = []
if not (W or H): break
for h in range(H):
s = stdin.readline()
for w in range(W):
if s[w] in [".", "X", "E", "N", "W", "S"]:
m[h][max(0,w-1)] |= 1
m[min(H-1,h+1)][w] |= 2
m[h][min(W-1,w+1)] |= 4
m[max(0,h-1)][w] |= 8
if s[w] in ["E", "N", "W", "S"]:
ps.append([h,w,["E", "N", "W", "S"].index(s[w])])
if s[w] == "X":
m[h][w] |= 16
if W <= 2 or H <= 2:
print(1)
continue
# print(ps)
# for n in m:
# print(n)
ttt = 0
while(True):
ttt += 1
if ttt > 180: print("NA"); break
# import time
# time.sleep(1)
mt = [[0]*W for _ in range(H)]
for p in ps:
mt[p[0]][p[1]] |= 1
dest = []
# for mmm in mt:
# print(mmm)
# print()
# print([a[:2] for a in ps])
for pi,p in enumerate(ps):
for i in range(4):
if head[p[2]][i] & m[p[0]][p[1]]:
d = head[p[2]][i]
if d == 1 and not [p[0],p[1]+1] in [a[:2] for a in ps]: p[2] = 0; dest.append([pi,2,p[0],p[1]+1]); break
elif d == 2 and not [p[0]-1,p[1]] in [a[:2] for a in ps]: p[2] = 1; dest.append([pi,3,p[0]-1,p[1]]); break
elif d == 4 and not [p[0],p[1]-1] in [a[:2] for a in ps]: p[2] = 2; dest.append([pi,0,p[0],p[1]-1]); break
elif d == 8 and not [p[0]+1,p[1]] in [a[:2] for a in ps]: p[2] = 3; dest.append([pi,1,p[0]+1,p[1]]); break
else:
dest.append([pi,(p[2]+2)&3,p[0],p[1]])
# print(dest)
dest = sorted(dest,key=lambda x: (x[2:],x[1]))
# for mmm in mt:
# print(mmm)
# print("ps = ",ps,dest)
dellist = []
for pi,d,dy,dx in dest:
# print(W,H,dy,dx,mt)
if not mt[dy][dx]:
mt[dy][dx] |= 1
ps[pi][:2] = [dy,dx]
if m[dy][dx] & 16:
# print("OUT")
# print(ttt)
# time.sleep(3)
dellist.append(pi)
for idx in sorted(dellist)[::-1]:
del ps[idx]
if not len(ps):
print(ttt)
break
# ms = [stdin.readline() for _ in range(H)]
# print(m)
# from pprint import pprint
# print(ps)
# for n in m:
# print(n)
# print()
```
No
| 84,942 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
Submitted Solution:
```
from sys import stdin
readline = stdin.readline
from itertools import product
N, W, S, E = (0, -1), (-1, 0), (0, 1), (1, 0)
BLANK = ('.', 'X')
DIRECTION = {
'N': N,
'W': W,
'S': S,
'E': E,
}
check_priority = {
N: (E, N, W, S),
W: (N, W, S, E),
S: (W, S, E, N),
E: (S, E, N, W),
}
step_priority = (E, N, W, S)
def search_people(sq):
people = []
for y, x in product(range(h), range(w)):
if sq[y][x] in DIRECTION:
sq[y][x] = person = Person(x, y, DIRECTION[sq[y][x]], sq)
people.append(person)
return people
class Person:
def __init__(self, x, y, direction, sq):
self.x, self.y = x, y
self.direction = direction
self.sq = sq
self.evacuating = True
def check_direction(self):
for direction in check_priority[self.direction]:
dx, dy = direction
if self.sq[self.y + dy][self.x + dx] in BLANK:
self.direction = direction
break
def check_stepable(self):
dx, dy = self.direction
x, y = self.x + dx, self.y + dy
if self.sq[y][x] == 'X':
self.stepable = True
return
for dx, dy in step_priority:
tmp = self.sq[y + dy][x + dx]
if not isinstance(tmp, Person):
continue
if tmp == self:
self.stepable = True
break
dx, dy = tmp.direction
if tmp.x + dx == x and tmp.y + dy == y:
self.stepable = False
break
def __repr__(self):
return "'P'"
def step(self):
if self.stepable:
dx, dy = self.direction
self.sq[self.y][self.x] = '.'
self.x += dx
self.y += dy
if self.sq[self.y][self.x] == '.':
self.sq[self.y][self.x] = self
else:
self.evacuating = False
while True:
w, h = map(int, readline().split())
if w == 0:
break
squares = [list(readline().strip()) for _ in range(h)]
people = search_people(squares)
for i in range(180):
for pi in people:
pi.check_direction()
for pi in people:
pi.check_stepable()
for pi in people:
pi.step()
people = [pi for pi in people if pi.evacuating]
if len(people) == 0:
print(i + 1)
break
else:
print('NA')
```
No
| 84,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
Submitted Solution:
```
from sys import stdin
readline = stdin.readline
from itertools import product
N, W, S, E = (0, -1), (-1, 0), (0, 1), (1, 0)
BLANK = ('.', 'X')
DIRECTION = {
'N': N,
'W': W,
'S': S,
'E': E,
}
check_priority = {
N: (E, N, W, S),
W: (N, W, S, E),
S: (W, S, E, N),
E: (S, E, N, W),
}
step_priority = (E, N, W, S)
def search_people(sq):
people = []
for y, x in product(range(h), range(w)):
if sq[y][x] in DIRECTION:
sq[y][x] = person = Person(x, y, DIRECTION[sq[y][x]], sq)
people.append(person)
return people
class Person:
def __init__(self, x, y, direction, sq):
self.x, self.y = x, y
self.direction = direction
self.sq = sq
self.evacuating = True
def check_direction(self):
for direction in check_priority[self.direction]:
dx, dy = direction
if self.sq[self.y + dy][self.x + dx] in BLANK:
self.direction = direction
break
def check_stepable(self):
dx, dy = self.direction
x, y = self.x + dx, self.y + dy
if self.sq[y][x] == 'X':
return True
for dx, dy in step_priority:
tmp = self.sq[y + dy][x + dx]
if not isinstance(tmp, Person):
continue
if tmp == self:
self.stepable = True
break
dx, dy = tmp.direction
if tmp.x + dx == x and tmp.y + dy == y:
self.stepable = False
break
def __repr__(self):
return "'P'"
def step(self):
if self.stepable:
dx, dy = self.direction
self.sq[self.y][self.x] = '.'
self.x += dx
self.y += dy
if self.sq[self.y][self.x] == '.':
self.sq[self.y][self.x] = self
else:
self.evacuating = False
while True:
w, h = map(int, readline().split())
if w == 0:
break
squares = [list(readline().strip()) for _ in range(h)]
people = search_people(squares)
for i in range(180):
for pi in people:
pi.check_direction()
for pi in people:
pi.check_stepable()
for pi in people:
pi.step()
people = [pi for pi in people if pi.evacuating]
if len(people) == 0:
print(i + 1)
break
else:
print('NA')
```
No
| 84,944 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The huge maze The Squares has been newly completed in the famous theme park. Evacuation drills must be conducted under the guidance of the fire department, but the time required for the drills cannot be predicted due to the huge maze. Therefore, you decided to develop an evacuation drill simulator based on the following specifications.
As shown in Fig. 1, the giant maze is represented by W × H squares of horizontal W and vertical H. Each square is either a passage (white square), a wall (brown square), or an emergency exit (green square). The circles in the figure represent people, and the lowercase letters (E, W, S, N) in them represent the direction in which the person is facing (north, south, east, and west). The figure is drawn with the upward direction facing north.
<image>
Figure 1
People in the giant maze initially stand facing either north, south, east, or west. Each person attempts to move in 1-second increments at the same time, following the steps below.
1. Look at the right, front, left, and back squares in the direction you are currently facing, and turn to the first vacant aisle or emergency exit you find. If there is no such square, the direction will not change.
2. If the square in front of you is open and not in front of another person, move it. If there are multiple people with the same square in front of you, the selected one will move in the order of the people in that square, east, north, west, and south.
Those who arrive at the emergency exit after moving will evacuate safely and disappear from the maze.
Create a program that inputs the given huge maze and the location information of people and outputs the time when all people finish evacuating. If it takes more than 180 seconds to escape, output NA. Maze and person location information is given by the characters in rows H and columns W. The meaning of each character is as follows.
: Wall
.: Floor
X: Emergency exit
E: People facing east
N: People facing north
W: People facing west
S: People facing south
The boundary between the maze and the outside is either the wall # or the emergency exit X. In addition, there is always one or more people in the huge maze.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
W H
str1
str2
::
strH
The first line gives the horizontal size W of the maze and the vertical size H (1 ≤ W, H ≤ 30). The following H line is given the string stri (length W) that represents the i-th line of the maze.
The number of datasets does not exceed 50.
Output
For each input dataset, the time when all people finish evacuating is output on one line.
Examples
Input
10 3
##########
#E.......X
##########
4 4
####
#N.#
#..X
####
5 5
#####
#N..#
###.X
#S..#
#####
6 6
######
#..#X#
#.EE.#
####N#
#....#
######
8 8
##X#####
#....E.#
#####.##
#.#...##
#.W.#..#
#.#.N#.X
#X##.#.#
########
0 0
Output
8
NA
9
16
10
Input
10 3
E.......X
4 4
N.#
..X
5 5
N..#
.X
S..#
6 6
..#X#
.EE.#
N#
....#
8 8
X#####
....E.#
.##
.#...##
.W.#..#
.#.N#.X
X##.#.#
0 0
Output
8
NA
9
16
10
Submitted Solution:
```
from sys import stdin
head = [[8,1,2,4],[1,2,4,8],[2,4,8,1],[4,8,1,2]]
while(True):
W,H = map(int, stdin.readline().split())
m = [[0]*W for _ in range(H)]
ps = []
if not (W or H): break
for h in range(H):
s = stdin.readline()
for w in range(W):
if s[w] in [".", "X", "E", "N", "W", "S"]:
m[h][max(0,w-1)] |= 1
m[min(H-1,h+1)][w] |= 2
m[h][min(W-1,w+1)] |= 4
m[max(0,h-1)][w] |= 8
if s[w] in ["E", "N", "W", "S"]:
ps.append([h,w,["E", "N", "W", "S"].index(s[w])])
if s[w] == "X":
m[h][w] |= 16
# print(ps)
# for n in m:
# print(n)
ttt = 0
while(True):
ttt += 1
if ttt > 180: print("NA"); break
# import time
# time.sleep(1)
mt = [[0]*W for _ in range(H)]
for p in ps:
mt[p[0]][p[1]] |= 1
dest = []
# for mmm in mt:
# print(mmm)
# print()
# print([a[:2] for a in ps])
for pi,p in enumerate(ps):
for i in range(4):
if head[p[2]][i] & m[p[0]][p[1]]:
d = head[p[2]][i]
if d == 1 and not [p[0],p[1]+1] in [a[:2] for a in ps]: p[2] = 0; dest.append([pi,2,p[0],p[1]+1]); break
elif d == 2 and not [p[0]-1,p[1]] in [a[:2] for a in ps]: p[2] = 1; dest.append([pi,3,p[0]-1,p[1]]); break
elif d == 4 and not [p[0],p[1]-1] in [a[:2] for a in ps]: p[2] = 2; dest.append([pi,0,p[0],p[1]-1]); break
elif d == 8 and not [p[0]+1,p[1]] in [a[:2] for a in ps]: p[2] = 3; dest.append([pi,1,p[0]+1,p[1]]); break
else:
dest.append([pi,(p[2]+2)&3,p[0],p[1]])
# print(dest)
dest = sorted(dest,key=lambda x: (x[2:],x[1]))
# for mmm in mt:
# print(mmm)
# print("ps = ",ps,dest)
dellist = []
for pi,d,dy,dx in dest:
if not mt[dy][dx]:
mt[dy][dx] |= 1
ps[pi][:2] = [dy,dx]
if m[dy][dx] & 16:
# print("OUT")
# print(ttt)
# time.sleep(3)
dellist.append(pi)
for idx in sorted(dellist)[::-1]:
del ps[idx]
if not len(ps):
print(ttt)
break
# ms = [stdin.readline() for _ in range(H)]
# print(m)
# from pprint import pprint
# print(ps)
# for n in m:
# print(n)
# print()
```
No
| 84,945 |
Provide a correct Python 3 solution for this coding contest problem.
Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have either 0 or 2 children and leaves have the same depth) with sufficiently large depth. Then, starting from its root, write on it a number, say, 1 for each position of first tree, and draw different number, say, 2 for second tree. The “intersection” of two trees is a tree with nodes numbered both 1 and 2, and the “union” is a tree with nodes numbered either 1 or 2, or both. For example, the intersection of trees in Figures 1 and 2 is a tree in Figure 3, and the union of them is a tree in Figure 4.
<image>
A node of a tree is expressed by a sequence of characters, “(,)“. If a node has a left child, the expression of the child is inserted between ’(’ and ’,’. The expression of a right child is inserted between ’,’ and ’)’. For exam- ple, the expression of trees in Figures 1 and 2 are “((,),(,))“ and “((,(,)),)“, respectively.
Input
Each line of the input contains an operation. An operation starts with a character which specifies the type of operation, either ’i’ or ’u’: ’i’ means intersection, and ’u’ means union. Following the character and a space, two tree expressions are given, separated by a space. It is assumed that 1 <= #nodes in a tree <= 100, and no tree expression contains spaces and syntax errors. Input is terminated by EOF.
Output
For each line of the input, output a tree expression, without any space, for the result of the operation.
Example
Input
i ((,),(,)) ((,(,)),)
u ((,),(,)) ((,(,)),)
Output
((,),)
((,(,)),(,))
"Correct Solution:
```
# AOJ 1001: Binary Tree Intersection And Union
# Python3 2018.7.5 bal4u
def parse(p, i):
global sz
node[i][2] += 1
del p[0]
if p[0] != ',':
if node[i][0] == 0:
node[i][0] = sz
sz += 1
parse(p, node[i][0])
del p[0]
if p[0] != ')':
if node[i][1] == 0:
node[i][1] = sz
sz += 1
parse(p, node[i][1])
del p[0]
def act(i, k):
global ans
if node[i][2] < k: return
ans += '('
if node[i][0] > 0: act(node[i][0], k)
ans += ','
if node[i][1] > 0: act(node[i][1], k)
ans += ')'
while True:
try: op, a, b = input().split()
except: break
sz = 1
node = [[0 for j in range(3)] for i in range(210)]
parse(list(a), 0)
parse(list(b), 0)
ans = ''
act(0, 2 if op== 'i' else 1)
print(ans)
```
| 84,946 |
Provide a correct Python 3 solution for this coding contest problem.
Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have either 0 or 2 children and leaves have the same depth) with sufficiently large depth. Then, starting from its root, write on it a number, say, 1 for each position of first tree, and draw different number, say, 2 for second tree. The “intersection” of two trees is a tree with nodes numbered both 1 and 2, and the “union” is a tree with nodes numbered either 1 or 2, or both. For example, the intersection of trees in Figures 1 and 2 is a tree in Figure 3, and the union of them is a tree in Figure 4.
<image>
A node of a tree is expressed by a sequence of characters, “(,)“. If a node has a left child, the expression of the child is inserted between ’(’ and ’,’. The expression of a right child is inserted between ’,’ and ’)’. For exam- ple, the expression of trees in Figures 1 and 2 are “((,),(,))“ and “((,(,)),)“, respectively.
Input
Each line of the input contains an operation. An operation starts with a character which specifies the type of operation, either ’i’ or ’u’: ’i’ means intersection, and ’u’ means union. Following the character and a space, two tree expressions are given, separated by a space. It is assumed that 1 <= #nodes in a tree <= 100, and no tree expression contains spaces and syntax errors. Input is terminated by EOF.
Output
For each line of the input, output a tree expression, without any space, for the result of the operation.
Example
Input
i ((,),(,)) ((,(,)),)
u ((,),(,)) ((,(,)),)
Output
((,),)
((,(,)),(,))
"Correct Solution:
```
class Node:
def __init__(self, left, right):
self.left = left
self.right = right
self.val = ''
@classmethod
def fromString(cls, s):
pos = 1
if s[pos] == '(':
e = getEnd(s, 1)
left = cls.fromString(s[1:e+1])
pos = e + 2
else:
left = None
pos += 1
if s[pos] == '(':
e = getEnd(s, pos)
right = Node.fromString(s[pos:e+1])
else:
right = None
return cls(left, right)
def __str__(self):
lstr = "" if self.left == None else str(self.left)
rstr = "" if self.right == None else str(self.right)
return "({},{})".format(lstr, rstr)
def getEnd(s, pos):
count = 0
while True:
if s[pos] == '(':
count += 1
elif s[pos] == ')':
count -= 1
if count == 0:
break
pos += 1
return pos
def intersection(root1, root2):
if root1.left == None or root2.left == None:
root1.left = None
else:
intersection(root1.left, root2.left)
if root1.right == None or root2.right == None:
root1.right = None
else:
intersection(root1.right, root2.right)
def union(root1, root2):
if root1.left == None and root2.left == None:
pass
elif root1.left != None and root2.left == None:
pass
elif root1.left == None and root2.left != None:
root1.left = root2.left
else:
union(root1.left, root2.left)
if root1.right == None and root2.right == None:
pass
elif root1.right != None and root2.right == None:
pass
elif root1.right == None and root2.right != None:
root1.right = root2.right
else:
union(root1.right, root2.right)
if __name__ == '__main__':
try:
while True:
line = input().strip().split()
c = line[0]
a = Node.fromString(line[1])
b = Node.fromString(line[2])
if c == 'i':
intersection(a, b)
else:
union(a, b)
print(a)
except EOFError:
pass
```
| 84,947 |
Provide a correct Python 3 solution for this coding contest problem.
Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have either 0 or 2 children and leaves have the same depth) with sufficiently large depth. Then, starting from its root, write on it a number, say, 1 for each position of first tree, and draw different number, say, 2 for second tree. The “intersection” of two trees is a tree with nodes numbered both 1 and 2, and the “union” is a tree with nodes numbered either 1 or 2, or both. For example, the intersection of trees in Figures 1 and 2 is a tree in Figure 3, and the union of them is a tree in Figure 4.
<image>
A node of a tree is expressed by a sequence of characters, “(,)“. If a node has a left child, the expression of the child is inserted between ’(’ and ’,’. The expression of a right child is inserted between ’,’ and ’)’. For exam- ple, the expression of trees in Figures 1 and 2 are “((,),(,))“ and “((,(,)),)“, respectively.
Input
Each line of the input contains an operation. An operation starts with a character which specifies the type of operation, either ’i’ or ’u’: ’i’ means intersection, and ’u’ means union. Following the character and a space, two tree expressions are given, separated by a space. It is assumed that 1 <= #nodes in a tree <= 100, and no tree expression contains spaces and syntax errors. Input is terminated by EOF.
Output
For each line of the input, output a tree expression, without any space, for the result of the operation.
Example
Input
i ((,),(,)) ((,(,)),)
u ((,),(,)) ((,(,)),)
Output
((,),)
((,(,)),(,))
"Correct Solution:
```
import re
class bintree():
def __init__(self, str1):
self.str1 = str1
if self.str1[1] != ",":
c = 0
counter = 0
for s in str1[1:]:
counter += 1
if s == "(":
c += 1
elif s == ")":
c -= 1
if c == 0:
break
self.left = bintree(str1[1:counter+1])
#match1 = re.search("\((\(.*\)),.*\)",str1)
#if match1 != None:
# self.left = bintree(str(match1.group(1)))
else:
self.left = ""
if self.str1[-2] != ",":
str1 = str1[0]+str1[1+len(str(self.left)):]
match2 = re.search("\(,(\(.*\))\)",str1)
if match2 != None:
self.right = bintree(str(str1[2:-1]))
else:
self.right = ""
def __str__(self):
return self.str1
def inter(bin1, bin2):
if bin1.left != "" and bin2.left != "":
strleft = inter(bin1.left, bin2.left)
else:
strleft = ""
if bin1.right != "" and bin2.right != "":
strright = inter(bin1.right, bin2.right)
else:
strright = ""
return "(" + strleft + "," + strright + ")"
def union(bin1, bin2):
if bin1.left != "" or bin2.left != "":
if bin1.left == "":
strleft = str(bin2.left)
elif bin2.left == "":
strleft = str(bin1.left)
else:
strleft = union(bin1.left, bin2.left)
else:
strleft = ""
if bin1.right != "" or bin2.right != "":
if bin1.right == "":
strright = str(bin2.right)
elif bin2.right == "":
strright = str(bin1.right)
else:
strright = union(bin1.right, bin2.right)
else:
strright = ""
#print("(" + strleft + "," + strright + ")")
return "(" + strleft + "," + strright + ")"
while True:
try:
inputs = map(str,input().split())
order = next(inputs)
if order == "i":
t1 = bintree(next(inputs))
t2 = bintree(next(inputs))
#print(t1.left,t1.right)
print(inter(t1, t2))# == "((((((,),),),),),)")
elif order == "u":
t1 = bintree(next(inputs))
t2 = bintree(next(inputs))
#print(t1.left,t1.right)
#print(t1.left,t1.right)
print(union(t1,t2))# == "((((,),(,)),((,),(,))),(((,),(,)),((,),(,))))")
except EOFError as exception:
break
```
| 84,948 |
Provide a correct Python 3 solution for this coding contest problem.
Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have either 0 or 2 children and leaves have the same depth) with sufficiently large depth. Then, starting from its root, write on it a number, say, 1 for each position of first tree, and draw different number, say, 2 for second tree. The “intersection” of two trees is a tree with nodes numbered both 1 and 2, and the “union” is a tree with nodes numbered either 1 or 2, or both. For example, the intersection of trees in Figures 1 and 2 is a tree in Figure 3, and the union of them is a tree in Figure 4.
<image>
A node of a tree is expressed by a sequence of characters, “(,)“. If a node has a left child, the expression of the child is inserted between ’(’ and ’,’. The expression of a right child is inserted between ’,’ and ’)’. For exam- ple, the expression of trees in Figures 1 and 2 are “((,),(,))“ and “((,(,)),)“, respectively.
Input
Each line of the input contains an operation. An operation starts with a character which specifies the type of operation, either ’i’ or ’u’: ’i’ means intersection, and ’u’ means union. Following the character and a space, two tree expressions are given, separated by a space. It is assumed that 1 <= #nodes in a tree <= 100, and no tree expression contains spaces and syntax errors. Input is terminated by EOF.
Output
For each line of the input, output a tree expression, without any space, for the result of the operation.
Example
Input
i ((,),(,)) ((,(,)),)
u ((,),(,)) ((,(,)),)
Output
((,),)
((,(,)),(,))
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import sys
def tree_left(node, it):
c = next(it)
if c == "(":
node[0] = [None, None]
tree_left(node[0], it)
c = next(it)
if c == ",":
tree_right(node, it)
else:
assert False
def tree_right(node, it):
c = next(it)
if c == "(":
node[1] = [None, None]
tree_left(node[1], it)
c = next(it)
if c == ")":
pass
else:
assert False
def tree_parse(s):
root = [None, None]
s = s[1:]
it = iter(s)
tree_left(root, it)
return root
def tree_dump(node):
s = ""
s += "("
if node[0]:
s += tree_dump(node[0])
s += ","
if node[1]:
s += tree_dump(node[1])
s += ")"
return s
def intersection(t1, t2):
if t1 is None or t2 is None:
return None
node = [None, None]
node[0] = intersection(t1[0], t2[0])
node[1] = intersection(t1[1], t2[1])
return node
def union(t1, t2):
if t1 is None and t2 is None:
return None
elif t1 is None:
return t2
elif t2 is None:
return t1
node = [None, None]
node[0] = union(t1[0], t2[0])
node[1] = union(t1[1], t2[1])
return node
def main():
for line in sys.stdin:
op, s1, s2 = line.split()
tree1 = tree_parse(s1)
tree2 = tree_parse(s2)
#print(tree_dump(tree1))
#print(tree_dump(tree2))
if op == "i":
ans = intersection(tree1, tree2)
print(tree_dump(ans))
elif op == "u":
ans = union(tree1, tree2)
print(tree_dump(ans))
if __name__ == "__main__": main()
```
| 84,949 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n,w = LI()
if n == 0:
break
a = [I() for _ in range(n)]
ma = max(a)
l = (ma // w + 1)
b = [0] * l
for c in a:
b[c//w] += 1
h = max(b)
r = 0.01
for i in range(l):
bi = b[i]
d = (l-i-1) / (l-1)
r += d * bi / h
rr.append(r)
return '\n'.join(map(str, rr))
print(main())
```
| 84,950 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
"Correct Solution:
```
while True:
n, w = map(int, input().split())
if n is 0 and w is 0:
break
l = [int(input()) for i in range(n)]
max_num = 0
for i in range(n):
max_num = max(max_num, l[i])
hist_size = int((max_num/w) + 1)
hist = [0.0] * int(hist_size)
for i in range(n):
hist[int(l[i]/w)] += 1
max_height= 0
for i in range(hist_size):
max_height = max(max_height,hist[i])
ans = 0.0
for i in range(hist_size-1):
ans += hist[i]/max_height*(hist_size - 1 - i)/(hist_size-1)
ans += 0.01
print(ans)
```
| 84,951 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
"Correct Solution:
```
# coding: utf-8
while 1:
n,w=map(int,input().split())
if n==0:
break
dic={}
mx=-1
max_section=-1
for i in range(1+100//w):
dic[i]=0
for i in range(n):
x=int(input())
dic[x//w]+=1
mx=max(mx,dic[x//w])
max_section=max(max_section,x//w)
ans=0
for i in range(1+100//w):
ans+=(dic[i]/mx)*(1-i*(1/(max_section)))
print(ans+0.01)
```
| 84,952 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
"Correct Solution:
```
while True:
n, w = list(map(int, input().split()))
if n==0:
break
v = [int(input()) for i in range(n)]
h = [0 for i in range(100)]
ma = 0
mi = 1e9
for i in range(n):
h[v[i]//w] += 1
ma = max(ma, v[i]//w)
mi = min(mi, v[i]//w)
sm = ma-mi
h = [h[i]/max(h) for i in range(100)]
ans = 0.01
for i in range(mi, ma):
ans += (sm-i+mi)/(sm+mi)*h[i]
print(ans)
```
| 84,953 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
"Correct Solution:
```
while True:
N,W = map(int,input().split())
if N == 0: break
src = [int(input()) for i in range(N)]
highest = max(src)
bands = highest//W + 1
hist = [0 for i in range(bands)]
for i in range(N):
hist[src[i] // W] += 1
ans = maxn = 0
for i,n in enumerate(hist):
ans += n * (bands-1-i)
maxn = max(n, maxn)
print(ans / maxn / (bands-1) + 0.01)
```
| 84,954 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
"Correct Solution:
```
while True:
N, W = map(int, input().split())
if not N:
break
s = list(sorted([int(input()) for i in range(N)]))
m = s[-1] // W + 1
ss = [0] * m
for c in s:
ss[c // W] += 1
unit = max(ss)
a = 0
for i, c in enumerate(ss):
a += (1 - i / (m - 1)) * (c / unit)
print(a + 0.01)
```
| 84,955 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
"Correct Solution:
```
while 1:
n,w=map(int,input().split())
if n==w==0:break
a=[0]*100
s=h=int(0)
while n:
b=int(input())
a[b//w]+=1
s=max(s,b//w)
h=max(h,a[b//w])
n-=1
ans=0.01
for i in range(int(s)+1):
ans+=a[i]/h*(s-i)/s
print(ans)
```
| 84,956 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
Submitted Solution:
```
from heapq import *
def calcInk(wdt,val):
maxval=val[-1]
histnum=maxval//wdt
hist=[]
for i in range(histnum+1):
hist.append([])
vl=len(val)
for i in range(vl):
v=heappop(val)
hist[v//wdt].append(v)
highest=0
for i in range(len(hist)):
highest=max(highest,len(hist[i]))
ink=0
for i in range(len(hist)):
ink+=(1-i/histnum)*(len(hist[i])/highest)
return ink+0.01
if __name__=='__main__':
while True:
n,w=map(int,input().split())
if n==w==0:
break
V=[]
for _ in range(n):
heappush(V,int(input()))
```
No
| 84,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
Submitted Solution:
```
from heapq import *
```
No
| 84,958 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
Submitted Solution:
```
from heapq import *
def calcInk(wdt,val):
maxval=val[-1]
histnum=maxval//wdt
hist=[]
for i in range(histnum+1):
hist.append([])
vl=len(val)
for i in range(vl):
v=heappop(val)
hist[v//wdt].append(v)
highest=0
for i in range(len(hist)):
highest=max(highest,len(hist[i]))
ink=0
for i in range(len(hist)):
ink+=(1-i/histnum)*(len(hist[i])/highest)
return ink+0.01
```
No
| 84,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
Submitted Solution:
```
if __name__=='__main__':
while True:
n,w=map(int,input().split())
if n==w==0:
break
V=[]
for _ in range(n):
heappush(V,int(input()))
maxV=V[-1]
histnum=maxV//w
hist=[]
for i in range(histnum+1):
hist.append([])
vl=len(V)
for i in range(vl):
v=heappop(V)
hist[v//w].append(v)
highest=0
for i in range(len(hist)):
highest=max(highest,len(hist[i]))
ink=0
for i in range(len(hist)):
ink+=(1-i/histnum)*(len(hist[i])/highest)
print(ink)
```
No
| 84,960 |
Provide a correct Python 3 solution for this coding contest problem.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
"Correct Solution:
```
#!/usr/bin/env python3
#aoj b
import re
from collections import defaultdict
def check(i):#メモ化再帰
if d[i] > 0:
return d[i]
d[i] = 1#チェック状態
for y in range(p[i][0],p[i][1]+1):
for x in range(p[i][2],p[i][3]+1):
if s[y][x] == ".":#長方形内部に空の部分が存在
d[i] = 3 #SUSPICIOUS
return d[i]
elif s[y][x] != i:#長方形内部に他の物質が存在
if d[s[y][x]] == 1:#その物質がチェック状態
d[i] = 3 #SUSPICIOUS(循環参照 テストケース4)
return d[i]
c = check(s[y][x])#そいつの状態確認の再帰
if c == 3:
d[i] == 3
return d[i]
d[i] = 2#SAFE
return 2
N = int(input())
for _ in range(N):
h,w = map(int,input().split())
s = [list(input()) for _ in range(h)]
p = defaultdict(list)
#p[0]:上端,p[1]:下端,p[2]:左端,p[3]:右端
#y座標の追加は横方向で見る(上端は上から、下端は下から)
for y in range(h):#上端
for x in range(w):
if s[y][x] != '.' and len(p[s[y][x]]) < 1:
p[s[y][x]].append(y)
for y in range(h)[::-1]:#下端
for x in range(w):
if s[y][x] != '.' and len(p[s[y][x]]) < 2:
p[s[y][x]].append(y)
#x座標の追加は縦方向で見る(左端は左から、右端は右から)
for x in range(w):#左端
for y in range(h):
if s[y][x] != '.' and len(p[s[y][x]]) < 3:
p[s[y][x]].append(x)
for x in range(w)[::-1]:#右端
for y in range(h):
if s[y][x] != '.' and len(p[s[y][x]]) < 4:
p[s[y][x]].append(x)
d = defaultdict(lambda : 0)#メモ化再帰用dict
for i in p.keys():
if check(i) == 3:
print('SUSPICIOUS')
break
else:
print('SAFE')
```
| 84,961 |
Provide a correct Python 3 solution for this coding contest problem.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
"Correct Solution:
```
def can_fill(image, k, v):
for y in range(v[1], v[3] + 1):
for x in range(v[0], v[2] + 1):
m = image[y][x]
if m != k and m != '*':
return False
return True
def is_safe(image, p):
keys = list(p.keys())
i = 0
while i < len(keys):
k = keys[i]
v = p[k]
if can_fill(image, k, v):
for y in range(v[1], v[3] + 1):
for x in range(v[0], v[2] + 1):
image[y][x] = '*'
keys.remove(k)
i = 0
else:
i += 1
return len(keys) == 0
n = int(input())
for i in range(n):
h, w = map(int, input().split(' '))
image = [list(input()) for y in range(h)]
p = {}
for y in range(h):
for x in range(w):
m = image[y][x]
if m != '.':
if m not in p:
p[m] = [x, y, x, y]
else:
p[m][0] = min(p[m][0], x)
p[m][1] = min(p[m][1], y)
p[m][2] = max(p[m][2], x)
p[m][3] = max(p[m][3], y)
print('SAFE' if is_safe(image, p) else 'SUSPICIOUS')
```
| 84,962 |
Provide a correct Python 3 solution for this coding contest problem.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
n = I()
ni = 0
while ni < n:
ni += 1
h,w = LI()
s = [[c for c in S()] for _ in range(h)]
d = collections.defaultdict(lambda: [inf,-inf,inf,-inf])
for i in range(h):
for j in range(w):
if s[i][j] == '.':
continue
t = d[s[i][j]]
if t[0] > i:
t[0] = i
if t[1] < i:
t[1] = i
if t[2] > j:
t[2] = j
if t[3] < j:
t[3] = j
f = True
k = set(d.keys())
while f:
f = False
for t in list(k):
hi,ha,wi,wa = d[t]
ff = True
for i in range(hi,ha+1):
for j in range(wi,wa+1):
if s[i][j] != t and s[i][j] != '?':
ff = False
break
if not ff:
break
if ff:
k.remove(t)
f = True
for i in range(hi,ha+1):
for j in range(wi,wa+1):
s[i][j] = '?'
if not k:
rr.append('SAFE')
else:
rr.append('SUSPICIOUS')
return '\n'.join(map(str,rr))
print(main())
```
| 84,963 |
Provide a correct Python 3 solution for this coding contest problem.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
"Correct Solution:
```
import sys
from collections import defaultdict
def check(i): #メモ化再帰でやります
if d[i] > 0:
return d[i]
d[i] = 1 #チェック状態
"""長方形と仮定し、矛盾があればSUSPICUOUS"""
for y in range(p[i][0],p[i][1]+1):
for x in range(p[i][2],p[i][3]+1):
if s[y][x] == ".": #長方形内部に空の部分が存在
d[i] = 3 #SUSPICUOUS
return d[i]
elif s[y][x] != i: #長方形内部に他の物質が存在
if d[s[y][x]] == 1: #その物質が現在チェック状態
d[i] = 3 #SUSPICUOUS(循環参照であるため。例:テストケース4)
return d[i]
c = check(s[y][x]) #再帰でそいつの状態をみる
if c == 3: #SUSPICUOUS
d[i] = 3 #SUSPICUOUS
return d[i]
d[i] = 2 #SAFE
return 2
n = int(sys.stdin.readline())
for _ in range(n):
h,w = map(int, sys.stdin.readline().split())
s = [sys.stdin.readline() for i in range(h)]
p = defaultdict(list)
"""p[0]:上端,p[1]:下端,p[2]:左端,p[3]:右端"""
for y in range(h):
for x in range(w):
if s[y][x] != "." and len(p[s[y][x]]) < 1:
p[s[y][x]].append(y)
for y in range(h)[::-1]:
for x in range(w):
if s[y][x] != "." and len(p[s[y][x]]) < 2:
p[s[y][x]].append(y)
for x in range(w):
for y in range(h):
if s[y][x] != "." and len(p[s[y][x]]) < 3:
p[s[y][x]].append(x)
for x in range(w)[::-1]:
for y in range(h):
if s[y][x] != "." and len(p[s[y][x]]) < 4:
p[s[y][x]].append(x)
d = defaultdict(lambda : 0) #メモ化再帰用のdict
for i in p.keys():
if check(i) == 3:
print("SUSPICIOUS")
break
else:
print("SAFE")
```
| 84,964 |
Provide a correct Python 3 solution for this coding contest problem.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
"Correct Solution:
```
n = int(input())
for _ in range(n):
h, w = map(int, input().split())
mp = [list(input()) for _ in range(h)]
range_dic = {}
keys = []
for y in range(h):
for x in range(w):
c = mp[y][x]
if c in range_dic:
x1, x2, y1, y2 = range_dic[c]
range_dic[c] = (min(x, x1), max(x, x2), min(y, y1), max(y, y2))
else:
range_dic[c] = (x, x, y, y)
keys.append(c)
while keys:
tmp = keys[:]
for key in keys:
break_flag = False
x1, x2, y1, y2 = range_dic[key]
for x in range(x1, x2 + 1):
for y in range(y1, y2 + 1):
if not mp[y][x] in (key, "#"):
break_flag = True
break
if break_flag:
break
else:
for y in range(y1, y2 + 1):
mp[y][x1:x2 + 1] = ["#"] * (x2 - x1 + 1)
keys.remove(key)
if tmp == keys:
print("SUSPICIOUS")
break
else:
print("SAFE")
```
| 84,965 |
Provide a correct Python 3 solution for this coding contest problem.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
"Correct Solution:
```
n = int(input())
for _ in range(n):
h, w = map(int, input().split())
mp = [list(input()) for _ in range(h)]
range_dic = {}
keys = []
for y in range(h):
for x in range(w):
c = mp[y][x]
if c in range_dic:
x1, x2, y1, y2 = range_dic[c]
range_dic[c] = (min(x, x1), max(x, x2), min(y, y1), max(y, y2))
else:
range_dic[c] = (x, x, y, y)
keys.append(c)
while keys:
tmp = keys[:]
for key in keys:
break_flag = False
x1, x2, y1, y2 = range_dic[key]
for x in range(x1, x2 + 1):
for y in range(y1, y2 + 1):
if not mp[y][x] in (key, "#"):
break_flag = True
break
if break_flag:
break
else:
for x in range(x1, x2 + 1):
for y in range(y1, y2 + 1):
mp[y][x] = "#"
keys.remove(key)
if tmp == keys:
print("SUSPICIOUS")
break
else:
print("SAFE")
```
| 84,966 |
Provide a correct Python 3 solution for this coding contest problem.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
"Correct Solution:
```
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(input())
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())
def dfs(s,cnt):
global ans
if cnt > 10:
ans = False
return
for t in lines[s]:
dfs(t,cnt+1)
N = inp()
for _ in range(N):
H,W = inpl()
MAP = [['.']*(W+2)] + [['.']+list(input())+['.'] for y in range(H)] + [['.']*(W+2)]
als = set([])
almm = [[INF,0,INF,0] for _ in range(30)]
for y in range(1,H+1):
for x in range(1,W+1):
tmp = MAP[y][x]
if tmp != '.':
tmp = ord(tmp) - ord('A')
als.add(tmp)
almm[tmp][0] = min(almm[tmp][0],x)
almm[tmp][1] = max(almm[tmp][1],x)
almm[tmp][2] = min(almm[tmp][2],y)
almm[tmp][3] = max(almm[tmp][3],y)
ans = True
lines = defaultdict(set)
for a in als:
xl,xr,yl,yr = almm[a]
alpha = chr(a+ord('A'))
for x in range(xl,xr+1):
for y in range(yl,yr+1):
if MAP[y][x] == '.':
ans = False
break
elif MAP[y][x] != alpha:
lines[a].add(ord(MAP[y][x])-ord('A'))
for a in als:
dfs(a,0)
if not ans:
print('SUSPICIOUS')
else:
print('SAFE')
```
| 84,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Submitted Solution:
```
from itertools import permutations
from collections import defaultdict
def judge(field, c):
for y in xrange(corner[c][2], corner[c][3] + 1):
for x in xrange(corner[c][0], corner[c][1] + 1):
if field[y][x] != c and field[y][x] != "*":
return False
field[y][x] = "*"
return True
N = int(raw_input())
for loop in xrange(N):
h, w = map(int, raw_input().split())
field = [raw_input() for i in xrange(h)]
chars = set([])
corner = defaultdict(lambda :[49,0,49,0])
for y in xrange(h):
for x in xrange(w):
c = field[y][x]
if c == ".": continue
chars.add(c)
corner[c][0] = min(corner[c][0], x)
corner[c][1] = max(corner[c][1], x)
corner[c][2] = min(corner[c][2], y)
corner[c][3] = max(corner[c][3], y)
for order in permutations(chars):
cp = map(list, field)
for c in order:
if not judge(cp, c):
break
else:
print("SAFE")
break
else:
print("SUSPICIOUS")
```
No
| 84,968 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Submitted Solution:
```
class Baggage:
height = 0
width = 0
contentsdata = [[]]
def __init__(self, H, W):
self.height = H
self.width = W
self.contentsdata = [["." for i in range(0, H)] for j in range(0, W)]
def setContents(self, x, y, cnt):
self.contentsdata[x][y] = cnt
def contentscheck(baggage, checkwords): # ????????? => True???????????¢???=> False
left = 10000
right = 10000
up = 1000
bottom = 1000
for x in range(0, baggage.width): # ???????????????????????¨?????????????????????????????????
if(baggage.contentsdata[x].count(checkwords) > 0):
left = x
break
for x in sorted([int(i) for i in range(0, baggage.width)], reverse=True): # ???????????????????????¨?????????????????????????????????
if(baggage.contentsdata[x].count(checkwords) > 0):
right = x
break
for y in range(0, baggage.height): # ???????????????????????¨?????????????????????????????????
for x in range(0, baggage.width):
if(baggage.contentsdata[x][y] == checkwords):
up = y
break
else:
continue
break
for y in sorted([int(i) for i in range(0, baggage.height)], reverse=True): # ???????????????????????¨?????????????????????????????????
for x in range(0, baggage.width):
if(baggage.contentsdata[x][y] == checkwords):
bottom = y
break
else:
continue
break
# print(checkwords + ": (left, right, up, bottom) = (" + str(left) + ", " + str(right) + ", " + str(up) + ", " + str(bottom) + ")")
for x in range(left, right + 1):
for y in range(up, bottom + 1):
if(baggage.contentsdata[x][y] == "."):
return True
break
else:
continue
break
return False
n = int(input())
usedwordslist = []
for k in range(0, n):
try:
size = [int(i) for i in input().split(" ")] # size = [H, W]
checkingbaggage = Baggage(size[0], size[1])
for i in range(0, size[0]):
data = list(input())
for j in range(0, size[1]):
checkingbaggage.contentsdata[j][i] = data[j]
if(data[j] != "." and usedwordslist.count(data[j]) == 0):
usedwordslist.append(data[j])
usedwordslist.sort()
for i in usedwordslist:
if(contentscheck(checkingbaggage, i)):
print("SUSPICIOUS")
break
else:
print("SAFE")
except:
raise
exit()
'''???????????\????¢??????¨
for i in range(0, size[0]):
for j in range(0, size[1]):
print(checkingbaggage.contentsdata[j][i])
print(usedwordslist)'''
usedwordslist.clear()
```
No
| 84,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Submitted Solution:
```
class Baggage:
height = 0
width = 0
contentsdata = [[]]
def __init__(self, H, W):
self.height = H
self.width = W
self.contentsdata = [["." for i in range(0, H)] for j in range(0, W)]
def setContents(self, x, y, cnt):
self.contentsdata[x][y] = cnt
def contentscheck(baggage, checkwords): # ????????? => 0???????????¢???????????§???=> 1 ?????¨???????????¢???=> 2
left = 0
right = 0
up = 0
bottom = 0
for x in range(0, baggage.width): # ???????????????????????¨?????????????????????????????????
if(baggage.contentsdata[x].count(checkwords) > 0):
left = x
break
for x in sorted([int(i) for i in range(0, baggage.width)], reverse=True): # ???????????????????????¨?????????????????????????????????
if(baggage.contentsdata[x].count(checkwords) > 0):
right = x
break
for y in range(0, baggage.height): # ???????????????????????¨?????????????????????????????????
for x in range(0, baggage.width):
if(baggage.contentsdata[x][y] == checkwords):
up = y
break
else:
continue
break
for y in sorted([int(i) for i in range(0, baggage.height)], reverse=True): # ???????????????????????¨?????????????????????????????????
for x in range(0, baggage.width):
if(baggage.contentsdata[x][y] == checkwords):
bottom = y
break
else:
continue
break
# print(checkwords + ": (left, right, up, bottom) = (" + str(left) + ", " + str(right) + ", " + str(up) + ", " + str(bottom) + ")")
complite = 1
for x in range(left, right + 1):
for y in range(up, bottom + 1):
# print(checkwords + ":" + baggage.contentsdata[x][y])
if(baggage.contentsdata[x][y] != checkwords):
complite = 0
if(baggage.contentsdata[x][y] == "."):
return 0
break
else:
continue
break
if(complite == 0):
return 1
else:
return 2
n = int(input())
usedwordslist = []
for k in range(0, n):
cmp = False
try:
size = [int(i) for i in input().split(" ")] # size = [H, W]
checkingbaggage = Baggage(size[0], size[1])
for i in range(0, size[0]):
data = list(input())
for j in range(0, size[1]):
checkingbaggage.contentsdata[j][i] = data[j]
if(data[j] != "." and usedwordslist.count(data[j]) == 0):
usedwordslist.append(data[j])
usedwordslist.sort()
for i in usedwordslist:
jdg = int(contentscheck(checkingbaggage, i))
if(jdg == 0):
print("SUSPICIOUS")
break
if(jdg == 2):
cmp = True
else:
print("cmp " + str(cmp))
if(cmp):
print("SAFE")
elif(len(usedwordslist) == 0):
print("SAFE")
else:
print("SUSPICIOUS")
except:
raise
exit()
'''???????????\????¢??????¨
for i in range(0, size[0]):
for j in range(0, size[1]):
print(checkingbaggage.contentsdata[j][i])
print(usedwordslist)'''
usedwordslist.clear()
```
No
| 84,970 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the story of 20XX. The number of air passengers increased as a result of the stable energy supply by the renewable power network and the invention of liquefied synthetic fuel. However, the threat of terrorism by aircraft still exists, and the importance of fast and highly reliable automatic baggage inspection systems is increasing. Since it is actually difficult for an inspector to inspect all the baggage, we would like to establish a mechanism for the inspector to inspect only the baggage judged to be suspicious by the automatic inspection.
At the request of the aviation industry, the International Cabin Protection Company investigated recent passenger baggage in order to develop a new automated inspection system. As a result of the investigation, it was found that the baggage of recent passengers has the following tendency.
* Baggage is shaped like a rectangular parallelepiped with only one side short.
* Items that ordinary passengers pack in their baggage and bring into the aircraft include laptop computers, music players, handheld game consoles, and playing cards, all of which are rectangular.
* Individual items are packed so that their rectangular sides are parallel to the sides of the baggage.
* On the other hand, weapons such as those used for terrorism have a shape very different from a rectangle.
Based on the above survey results, we devised the following model for baggage inspection. Each piece of baggage is considered to be a rectangular parallelepiped container that is transparent to X-rays. It contains multiple items that are opaque to X-rays. Here, consider a coordinate system with the three sides of the rectangular parallelepiped as the x-axis, y-axis, and z-axis, irradiate X-rays in the direction parallel to the x-axis, and take an image projected on the y-z plane. The captured image is divided into grids of appropriate size, and the material of the item reflected in each grid area is estimated by image analysis. Since this company has a very high level of analysis technology and can analyze even the detailed differences in materials, it can be considered that the materials of the products are different from each other. When multiple items overlap in the x-axis direction, the material of the item that is in the foreground for each lattice region, that is, the item with the smallest x-coordinate is obtained. We also assume that the x-coordinates of two or more items are never equal.
Your job can be asserted that it contains non-rectangular (possibly a weapon) item when given the results of the image analysis, or that the baggage contains anything other than a rectangular item. It is to create a program that determines whether it is presumed that it is not included.
Input
The first line of input contains a single positive integer, which represents the number of datasets. Each dataset is given in the following format.
> H W
> Analysis result 1
> Analysis result 2
> ...
> Analysis result H
>
H is the vertical size of the image, and W is an integer representing the horizontal size (1 <= h, w <= 50). Each line of the analysis result is composed of W characters, and the i-th character represents the analysis result in the grid region i-th from the left of the line. For the lattice region where the substance is detected, the material is represented by uppercase letters (A to Z). At this time, the same characters are used if they are made of the same material, and different characters are used if they are made of different materials. The lattice region where no substance was detected is represented by a dot (.).
For all datasets, it is guaranteed that there are no more than seven material types.
Output
For each data set, output "SUSPICIOUS" if it contains items other than rectangles, and "SAFE" if not, on one line.
Sample Input
6
1 1
..
3 3
...
.W.
...
10 10
..........
.DDDDCC ..
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
..........
.DDDDDD ...
.DDDDCC ..
.DDDDCC ..
ADDDDCCC ..
AAA .. CCC ..
AAABB BBC ..
AAABBBB ...
..BBBBB ...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE ....
EEEEEEE ...
.EEEEEEE ..
..EEEEEEE.
... EEEEEEE
.... EEEEE.
..... EEE ..
...... E ...
16 50
.................................................................
......... AAAAAAAAAAAAAAAAA ............................
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA .....
.... PPP ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ....
.... PPP .............. AAAAA.AAAAAAAAAAAAAAAA .......
.... PPP ................ A .... AAA.AAAAAAAAAA ........
.... PPP ........... IIIIIAAIIAIII.AAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIAAAAAAAAAAAAAAAAAA ........
..CCCCCCCCCCCCCC ... IIIIIIIIIIIII ... AAAAAAAAAAA ......
.... PPP .................. AAAAAAAAAAA .....
MMMMPPPMMMMMMMMMMMMMMM ............. AAAAAAAAAAA ....
MMMMPPPMMMMMMMMMMMMMMM .............. AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............... AAAAAAAAAAA ...
MMMMMMMMMMMMMMMMMMMMMM ............................
Output for the Sample Input
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Example
Input
6
1 1
.
3 3
...
.W.
...
10 10
..........
.DDDDCCC..
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
..........
.DDDDDD...
.DDDDCCC..
.DDDDCCC..
ADDDDCCC..
AAA..CCC..
AAABBBBC..
AAABBBB...
..BBBBB...
..........
10 10
R..E..C.T.
R.EEE.C.T.
.EEEEE....
EEEEEEE...
.EEEEEEE..
..EEEEEEE.
...EEEEEEE
....EEEEE.
.....EEE..
......E...
16 50
..................................................
.........AAAAAAAAAAAAAAAA.........................
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.....
....PPP...AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA....
....PPP..............AAAAA.AAAAAAAAAAAAAAAA.......
....PPP................A....AAA.AAAAAAAAAA........
....PPP...........IIIIIAAIIAIII.AAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIAAAAAAAAAAAAAAAAAA........
..CCCCCCCCCCCCC...IIIIIIIIIIIII...AAAAAAAAAA......
....PPP............................AAAAAAAAAA.....
MMMMPPPMMMMMMMMMMMMMMM.............AAAAAAAAAAA....
MMMMPPPMMMMMMMMMMMMMMM..............AAAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM...............AAAAAAAAAA...
MMMMMMMMMMMMMMMMMMMMMM............................
Output
SAFE
SAFE
SAFE
SUSPICIOUS
SUSPICIOUS
SUSPICIOUS
Submitted Solution:
```
class Baggage:
height = 0
width = 0
contentsdata = [[]]
def __init__(self, H, W):
self.height = H
self.width = W
self.contentsdata = [["." for i in range(0, H)] for j in range(0, W)]
def setContents(self, x, y, cnt):
self.contentsdata[x][y] = cnt
def contentscheck(baggage, checkwords): # ????????? => 0???????????¢???????????§???=> 1 ?????¨???????????¢???=> 2
left = 0
right = 0
up = 0
bottom = 0
for x in range(0, baggage.width): # ???????????????????????¨?????????????????????????????????
if(baggage.contentsdata[x].count(checkwords) > 0):
left = x
break
for x in sorted([int(i) for i in range(0, baggage.width)], reverse=True): # ???????????????????????¨?????????????????????????????????
if(baggage.contentsdata[x].count(checkwords) > 0):
right = x
break
for y in range(0, baggage.height): # ???????????????????????¨?????????????????????????????????
for x in range(0, baggage.width):
if(baggage.contentsdata[x][y] == checkwords):
up = y
break
else:
continue
break
for y in sorted([int(i) for i in range(0, baggage.height)], reverse=True): # ???????????????????????¨?????????????????????????????????
for x in range(0, baggage.width):
if(baggage.contentsdata[x][y] == checkwords):
bottom = y
break
else:
continue
break
# print(checkwords + ": (left, right, up, bottom) = (" + str(left) + ", " + str(right) + ", " + str(up) + ", " + str(bottom) + ")")
complite = 1
for x in range(left, right + 1):
for y in range(up, bottom + 1):
# print(checkwords + ":" + baggage.contentsdata[x][y])
if(baggage.contentsdata[x][y] != checkwords):
complite = 0
if(baggage.contentsdata[x][y] == "."):
return 0
break
else:
continue
break
if(complite == 0):
return 1
else:
return 2
n = int(input())
usedwordslist = []
for k in range(0, n):
cmp = False
try:
size = [int(i) for i in input().split(" ")] # size = [H, W]
checkingbaggage = Baggage(size[0], size[1])
for i in range(0, size[0]):
data = list(input())
for j in range(0, size[1]):
checkingbaggage.contentsdata[j][i] = data[j]
if(data[j] != "." and usedwordslist.count(data[j]) == 0):
usedwordslist.append(data[j])
usedwordslist.sort()
for i in usedwordslist:
jdg = int(contentscheck(checkingbaggage, i))
if(jdg == 0):
print("SUSPICIOUS")
break
if(jdg == 2):
cmp = True
else:
if(cmp):
print("SAFE")
elif(len(usedwordslist) == 0):
print("SAFE")
else:
print("SUSPICIOUS")
except:
raise
exit()
'''???????????\????¢??????¨
for i in range(0, size[0]):
for j in range(0, size[1]):
print(checkingbaggage.contentsdata[j][i])
print(usedwordslist)'''
usedwordslist.clear()
```
No
| 84,971 |
Provide a correct Python 3 solution for this coding contest problem.
You are recording a result of a secret experiment, which consists of a large set of N-dimensional vectors. Since the result may become very large, you are thinking of compressing it. Fortunately you already have a good compression method for vectors with small absolute values, all you have to do is to preprocess the vectors and make them small.
You can record the set of vectors in any order you like. Let's assume you process them in the order v_1, v_2,..., v_M. Each vector v_i is recorded either as is, or as a difference vector. When it is recorded as a difference, you can arbitrarily pick up an already chosen vector v_j (j<i) and a real value r. Then the actual vector value recorded is (v_i - r v_j). The values of r and j do not affect the compression ratio so much, so you don't have to care about them.
Given a set of vectors, your task is to write a program that calculates the minimum sum of the squared length of the recorded vectors.
Input
The input is like the following style.
N M
v_{1,1} v_{1,2} ... v_{1,N}
...
v_{M,1} v_{M,2} ... v_{M,N}
The first line contains two integers N and M (1 \leq N, M \leq 100), where N is the dimension of each vector, and M is the number of the vectors. Each of the following M lines contains N floating point values v_{i,j} (-1.0 \leq v_{i,j} \leq 1.0) which represents the j-th element value of the i-th vector.
Output
Output the minimum sum of the squared length of the recorded vectors. The output should not contain an absolute error greater than 10^{-6}.
Examples
Input
2 3
1.0 1.0
-1.0 0.0
0.5 0.5
Output
1.0
Input
1 1
1.0
Output
1.0
Input
4 3
1.0 1.0 0.0 0.0
-1.0 0.0 -1.0 0.0
0.5 0.5 0.5 0.5
Output
3.0
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def calc(V, es, r):
mins = [(10**18, -1)]*V
for s, t, w in es:
mins[t] = min(mins[t], (w, s))
mins[r] = (-1, -1)
group = [0]*V
comp = [0]*V
cnt = 0
used = [0]*V
for v in range(V):
if not used[v]:
chain = []
cur = v
while cur!=-1 and not used[cur]:
chain.append(cur)
used[cur] = 1
cur = mins[cur][1]
if cur!=-1:
cycle = 0
for e in chain:
group[e] = cnt
if e==cur:
cycle = 1
comp[cnt] = 1
if not cycle:
cnt += 1
if cycle:
cnt += 1
else:
for e in chain:
group[e] = cnt
cnt += 1
if cnt == V:
return sum(map(lambda x:x[0], mins)) + 1
res = sum(mins[v][0] for v in range(V) if v!=r and comp[group[v]])
n_es = []
for s, t, w in es:
gs = group[s]; gt = group[t]
if gs == gt:
continue
if comp[gt]:
n_es.append((gs, gt, w - mins[t][0]))
else:
n_es.append((gs, gt, w))
return res + calc(cnt, n_es, group[r])
def solve():
N, M = map(int, readline().split())
V = []
D = []
for i in range(M):
*Vi, = map(float, readline().split())
d = sum(e**2 for e in Vi)
if d <= 1e-9:
continue
V.append(Vi)
D.append(d)
M = len(V)
E = []
for i in range(M):
Vi = V[i]
for j in range(M):
if i == j:
continue
Vj = V[j]
t = 0
for k in range(N):
t += Vi[k] * Vj[k]
r = t / (D[j])
c = 0
for k in range(N):
c += (Vi[k] - r*Vj[k])**2
E.append((j+1, i+1, c))
E.append((0, i+1, D[i]))
write("%.16f\n" % calc(M+1, E, 0))
solve()
```
| 84,972 |
Provide a correct Python 3 solution for this coding contest problem.
Sunuke-kun's dictionary contains the words s1, ..., sn, which consist of n lowercase letters. This satisfies s1 <... <sn when compared in lexicographical order. Unfortunately, some characters are faint and unreadable. Unreadable characters are represented by?. Find out how many ways to restore the dictionary by replacing? With lowercase letters, even with mod 1,000,000,007.
Constraints
* 1 ≤ n ≤ 50
* 1 ≤ | si | ≤ 20
* The characters that appear in si are lowercase letters or?
Input
n
s1
.. ..
sn
Output
Print the answer on one line.
Examples
Input
2
?sum??mer
c??a??mp
Output
703286064
Input
3
snuje
????e
snule
Output
1
"Correct Solution:
```
# seishin.py
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
MOD = 10**9 + 7
N = int(readline())
M = 20
L = 26
ca = ord('a')
cq = ord('?')
S = [[ca-1]*M for i in range(N)]
for i in range(N):
s = readline().strip()
S[i][:len(s)] = map(ord, s)
memo = [[[[-1]*(L+2) for i in range(M+1)] for j in range(N+1)] for k in range(N+1)]
for i in range(N+1):
for p in range(M+1):
for c in range(L+2):
memo[i][i][p][c] = 1
for i in range(N+1):
for j in range(i+1, N+1):
for p in range(M+1):
memo[i][j][p][L+1] = 0
for c in range(L+2):
memo[i][j][M][c] = (i+1 == j)
def dfs(l, r, p, c):
if memo[l][r][p][c] != -1:
return memo[l][r][p][c]
res = dfs(l, r, p, c+1)
for i in range(l+1, r+1):
if (S[i-1][p] != ca + c - 1) if S[i-1][p] != cq else (c == 0):
break
res += dfs(l, i, p+1, 0) * dfs(i, r, p, c+1) % MOD
memo[l][r][p][c] = res = res % MOD
return res
write("%d\n" % dfs(0, N, 0, 0))
solve()
```
| 84,973 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
4
Durett 7
Gayles 3
Facenda 6
Daughtery 0
1
+ Mccourtney 2
Output
Mccourtney is not working now.
Durett is working hard now.
"Correct Solution:
```
from heapq import heappush, heappop, heapify
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N = int(readline())
P = []; Q = []
E = []; R = {}
L = N + 20000
for i in range(N):
s, a = readline().split(); a = int(a) * L + i
E.append(s)
Q.append((-a, i))
R[s] = (i, a)
heapify(Q)
for i in range(N // 5):
b, i = heappop(Q)
P.append((-b, i))
heapify(P)
pn = len(P); qn = len(Q)
ln = N
cur = N
M = int(readline())
for i in range(M):
p, *g = readline().split()
db = dk = -1
if p == "+":
t, b = g; b = int(b) * L + cur
E.append(t)
R[t] = (cur, b)
ln += 1
if ln >= 5 and -Q[0][0] < b:
da = 1
pn += 1
heappush(P, (b, cur))
if pn > ln // 5:
while 1:
c, k = heappop(P)
if E[k] is not None:
if c == b:
da = 0
else:
db = 0
dk = k
heappush(Q, (-c, k))
break
pn -= 1; qn += 1
else:
da = 0
qn += 1
heappush(Q, (-b, cur))
if pn < ln // 5:
while 1:
c, k = heappop(Q)
if E[k] is not None:
if -b == c:
da = 1
else:
db = 1
dk = k
heappush(P, (-c, k))
break
pn += 1; qn -= 1
if da:
write("%s is working hard now.\n" % t)
else:
write("%s is not working now.\n" % t)
cur += 1
else:
t, = g
j, b = R[t]
E[j] = None
ln -= 1
if P and P[0][0] <= b:
pn -= 1
if pn < ln // 5:
while 1:
c, k = heappop(Q)
if E[k] is not None:
heappush(P, (-c, k))
db = 1; dk = k
break
pn += 1; qn -= 1
else:
qn -= 1
if pn > ln // 5:
while 1:
c, k = heappop(P)
if E[k] is not None:
heappush(Q, (-c, k))
db = 0; dk = k
break
qn += 1; pn -= 1
if db != -1:
if db:
write("%s is working hard now.\n" % E[dk])
else:
write("%s is not working now.\n" % E[dk])
solve()
```
| 84,974 |
Provide a correct Python 3 solution for this coding contest problem.
E-training
Nene is writing a program to look up $ N $ integers $ V_1, V_2, V_3, \ cdots, V_N $ for programming training.
As told by his instructor, Umiko, Nene wrote a program to look up multiples of 2, 3, and 6.
Multiples of 2 were $ A $, multiples of 3 were $ B $, and multiples of 6 were $ C $.
Umiko told me to look up the number of "numbers that are neither multiples of 2 nor multiples of 3".
However, Nene was tired, so she decided to cheat only for the answer.
Based only on the values of $ N, A, B, and C $, you can find the number of "numbers that are neither multiples of 2 nor multiples of 3". Create a program that asks for this.
input
$ N, A, B, C $ are given separated by blanks.
output
Output the number of "numbers that are neither multiples of 2 nor multiples of 3" in the data. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A, B, C $ are integers greater than or equal to $ 0 $ and less than or equal to $ N $
* No inconsistent data is given, such as $ A $ being greater than $ N $
Input example 1
6 3 2 1
Output example 1
2
For example, if your data is $ 2, 3, 4, 5, 6, 7 $, then $ 5 $ and $ 7 $ are "numbers that are neither multiples of 2 nor multiples of 3".
Input example 2
10 9 9 9
Output example 2
1
Example
Input
6 3 2 1
Output
2
"Correct Solution:
```
N, A, B, C = [int(x) for x in input().split()]
ans = N - (A + B) + C
print(ans)
```
| 84,975 |
Provide a correct Python 3 solution for this coding contest problem.
E-training
Nene is writing a program to look up $ N $ integers $ V_1, V_2, V_3, \ cdots, V_N $ for programming training.
As told by his instructor, Umiko, Nene wrote a program to look up multiples of 2, 3, and 6.
Multiples of 2 were $ A $, multiples of 3 were $ B $, and multiples of 6 were $ C $.
Umiko told me to look up the number of "numbers that are neither multiples of 2 nor multiples of 3".
However, Nene was tired, so she decided to cheat only for the answer.
Based only on the values of $ N, A, B, and C $, you can find the number of "numbers that are neither multiples of 2 nor multiples of 3". Create a program that asks for this.
input
$ N, A, B, C $ are given separated by blanks.
output
Output the number of "numbers that are neither multiples of 2 nor multiples of 3" in the data. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A, B, C $ are integers greater than or equal to $ 0 $ and less than or equal to $ N $
* No inconsistent data is given, such as $ A $ being greater than $ N $
Input example 1
6 3 2 1
Output example 1
2
For example, if your data is $ 2, 3, 4, 5, 6, 7 $, then $ 5 $ and $ 7 $ are "numbers that are neither multiples of 2 nor multiples of 3".
Input example 2
10 9 9 9
Output example 2
1
Example
Input
6 3 2 1
Output
2
"Correct Solution:
```
N,A,B,C=map(int,input().split())
print(N-A-B+C)
```
| 84,976 |
Provide a correct Python 3 solution for this coding contest problem.
E-training
Nene is writing a program to look up $ N $ integers $ V_1, V_2, V_3, \ cdots, V_N $ for programming training.
As told by his instructor, Umiko, Nene wrote a program to look up multiples of 2, 3, and 6.
Multiples of 2 were $ A $, multiples of 3 were $ B $, and multiples of 6 were $ C $.
Umiko told me to look up the number of "numbers that are neither multiples of 2 nor multiples of 3".
However, Nene was tired, so she decided to cheat only for the answer.
Based only on the values of $ N, A, B, and C $, you can find the number of "numbers that are neither multiples of 2 nor multiples of 3". Create a program that asks for this.
input
$ N, A, B, C $ are given separated by blanks.
output
Output the number of "numbers that are neither multiples of 2 nor multiples of 3" in the data. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A, B, C $ are integers greater than or equal to $ 0 $ and less than or equal to $ N $
* No inconsistent data is given, such as $ A $ being greater than $ N $
Input example 1
6 3 2 1
Output example 1
2
For example, if your data is $ 2, 3, 4, 5, 6, 7 $, then $ 5 $ and $ 7 $ are "numbers that are neither multiples of 2 nor multiples of 3".
Input example 2
10 9 9 9
Output example 2
1
Example
Input
6 3 2 1
Output
2
"Correct Solution:
```
n,a,b,c = map(int,input().split())
print(n - (a+b-c))
```
| 84,977 |
Provide a correct Python 3 solution for this coding contest problem.
E-training
Nene is writing a program to look up $ N $ integers $ V_1, V_2, V_3, \ cdots, V_N $ for programming training.
As told by his instructor, Umiko, Nene wrote a program to look up multiples of 2, 3, and 6.
Multiples of 2 were $ A $, multiples of 3 were $ B $, and multiples of 6 were $ C $.
Umiko told me to look up the number of "numbers that are neither multiples of 2 nor multiples of 3".
However, Nene was tired, so she decided to cheat only for the answer.
Based only on the values of $ N, A, B, and C $, you can find the number of "numbers that are neither multiples of 2 nor multiples of 3". Create a program that asks for this.
input
$ N, A, B, C $ are given separated by blanks.
output
Output the number of "numbers that are neither multiples of 2 nor multiples of 3" in the data. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A, B, C $ are integers greater than or equal to $ 0 $ and less than or equal to $ N $
* No inconsistent data is given, such as $ A $ being greater than $ N $
Input example 1
6 3 2 1
Output example 1
2
For example, if your data is $ 2, 3, 4, 5, 6, 7 $, then $ 5 $ and $ 7 $ are "numbers that are neither multiples of 2 nor multiples of 3".
Input example 2
10 9 9 9
Output example 2
1
Example
Input
6 3 2 1
Output
2
"Correct Solution:
```
n, a, b, c = map(int, input().split())
print(n-(c+a-c+b-c))
```
| 84,978 |
Provide a correct Python 3 solution for this coding contest problem.
E-training
Nene is writing a program to look up $ N $ integers $ V_1, V_2, V_3, \ cdots, V_N $ for programming training.
As told by his instructor, Umiko, Nene wrote a program to look up multiples of 2, 3, and 6.
Multiples of 2 were $ A $, multiples of 3 were $ B $, and multiples of 6 were $ C $.
Umiko told me to look up the number of "numbers that are neither multiples of 2 nor multiples of 3".
However, Nene was tired, so she decided to cheat only for the answer.
Based only on the values of $ N, A, B, and C $, you can find the number of "numbers that are neither multiples of 2 nor multiples of 3". Create a program that asks for this.
input
$ N, A, B, C $ are given separated by blanks.
output
Output the number of "numbers that are neither multiples of 2 nor multiples of 3" in the data. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A, B, C $ are integers greater than or equal to $ 0 $ and less than or equal to $ N $
* No inconsistent data is given, such as $ A $ being greater than $ N $
Input example 1
6 3 2 1
Output example 1
2
For example, if your data is $ 2, 3, 4, 5, 6, 7 $, then $ 5 $ and $ 7 $ are "numbers that are neither multiples of 2 nor multiples of 3".
Input example 2
10 9 9 9
Output example 2
1
Example
Input
6 3 2 1
Output
2
"Correct Solution:
```
n,a,b,c=map(int,input().split())
print((n+c)-(a+b))
```
| 84,979 |
Provide a correct Python 3 solution for this coding contest problem.
E-training
Nene is writing a program to look up $ N $ integers $ V_1, V_2, V_3, \ cdots, V_N $ for programming training.
As told by his instructor, Umiko, Nene wrote a program to look up multiples of 2, 3, and 6.
Multiples of 2 were $ A $, multiples of 3 were $ B $, and multiples of 6 were $ C $.
Umiko told me to look up the number of "numbers that are neither multiples of 2 nor multiples of 3".
However, Nene was tired, so she decided to cheat only for the answer.
Based only on the values of $ N, A, B, and C $, you can find the number of "numbers that are neither multiples of 2 nor multiples of 3". Create a program that asks for this.
input
$ N, A, B, C $ are given separated by blanks.
output
Output the number of "numbers that are neither multiples of 2 nor multiples of 3" in the data. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A, B, C $ are integers greater than or equal to $ 0 $ and less than or equal to $ N $
* No inconsistent data is given, such as $ A $ being greater than $ N $
Input example 1
6 3 2 1
Output example 1
2
For example, if your data is $ 2, 3, 4, 5, 6, 7 $, then $ 5 $ and $ 7 $ are "numbers that are neither multiples of 2 nor multiples of 3".
Input example 2
10 9 9 9
Output example 2
1
Example
Input
6 3 2 1
Output
2
"Correct Solution:
```
n, a, b, c = map(int, input().split())
print(n-a-b+c)
```
| 84,980 |
Provide a correct Python 3 solution for this coding contest problem.
E-training
Nene is writing a program to look up $ N $ integers $ V_1, V_2, V_3, \ cdots, V_N $ for programming training.
As told by his instructor, Umiko, Nene wrote a program to look up multiples of 2, 3, and 6.
Multiples of 2 were $ A $, multiples of 3 were $ B $, and multiples of 6 were $ C $.
Umiko told me to look up the number of "numbers that are neither multiples of 2 nor multiples of 3".
However, Nene was tired, so she decided to cheat only for the answer.
Based only on the values of $ N, A, B, and C $, you can find the number of "numbers that are neither multiples of 2 nor multiples of 3". Create a program that asks for this.
input
$ N, A, B, C $ are given separated by blanks.
output
Output the number of "numbers that are neither multiples of 2 nor multiples of 3" in the data. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A, B, C $ are integers greater than or equal to $ 0 $ and less than or equal to $ N $
* No inconsistent data is given, such as $ A $ being greater than $ N $
Input example 1
6 3 2 1
Output example 1
2
For example, if your data is $ 2, 3, 4, 5, 6, 7 $, then $ 5 $ and $ 7 $ are "numbers that are neither multiples of 2 nor multiples of 3".
Input example 2
10 9 9 9
Output example 2
1
Example
Input
6 3 2 1
Output
2
"Correct Solution:
```
n,a,b,c=map(int,input().split());print(n-a-b+c)
```
| 84,981 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
"Correct Solution:
```
def cross(a: complex, b: complex) -> float:
return a.real * b.imag - a.imag * b.real
def cross_point(c: complex, d: complex) -> complex:
global lt, lv
vec = d - c
v1 = cross(lv, vec)
v2 = cross(lv, lt - c)
return c + v2 / v1 * vec
if __name__ == "__main__":
n = int(input())
points = [complex(*map(int, input().split())) for _ in range(n)]
point0 = points.pop(0)
points.append(point0)
q = int(input())
while q:
x1, y1, x2, y2 = map(int, input().split())
ls, lt = (x1 + 1j * y1, x2 + 1j * y2)
lv = lt - ls
area = 0.0
prev = point0
prev_flag = 0 <= cross(lv, prev - ls)
cp1, cp2 = None, None
for p in points:
curr_flag = 0 <= cross(lv, p - ls)
if prev_flag and curr_flag:
area += cross(prev, p)
elif prev_flag != curr_flag:
cp = cross_point(prev, p)
if prev_flag:
area += cross(prev, cp)
cp1 = cp
else:
area += cross(cp, p)
cp2 = cp
prev, prev_flag = p, curr_flag
if cp1 is not None and cp2 is not None:
area += cross(cp1, cp2)
print(area / 2)
q -= 1
```
| 84,982 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def cross3(O, A, B):
ox, oy = O; ax, ay = A; bx, by = B
return (ax - ox) * (by - oy) - (bx - ox) * (ay - oy)
def cross_point(p0, p1, q0, q1):
x0, y0 = p0; x1, y1 = p1
x2, y2 = q0; x3, y3 = q1
dx0 = x1 - x0; dy0 = y1 - y0
dx1 = x3 - x2; dy1 = y3 - y2
s = (y0-y2)*dx1 - (x0-x2)*dy1
sm = dx0*dy1 - dy0*dx1
if -EPS < sm < EPS:
return None
return x0 + s*dx0/sm, y0 + s*dy0/sm
EPS = 1e-9
def convex_cut(P, line):
q0, q1 = line
N = len(P)
Q = []
for i in range(N):
p0 = P[i-1]; p1 = P[i]
cv0 = cross3(q0, q1, p0)
cv1 = cross3(q0, q1, p1)
if cv0 * cv1 < EPS:
v = cross_point(q0, q1, p0, p1)
if v is not None:
Q.append(v)
if cv1 > -EPS:
Q.append(p1)
return Q
def polygon_area(P):
s = 0
N = len(P)
for i in range(N):
p0 = P[i-1]; p1 = P[i]
s += p0[0]*p1[1] - p0[1]*p1[0]
return abs(s) / 2
def solve():
N = int(readline())
P = [list(map(int, readline().split())) for i in range(N)]
Q = int(readline())
for i in range(Q):
x0, y0, x1, y1 = map(int, readline().split())
P0 = convex_cut(P, ((x0, y0), (x1, y1)))
write("%.16f\n" % polygon_area(P0))
solve()
```
| 84,983 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
"Correct Solution:
```
import cmath
EPS = 1e-6
#外積
def OuterProduct(one, two):
tmp = one.conjugate() * two
return tmp.imag
#点が直線上にあるか
def IsOnLine(point, begin, end):
return abs(OuterProduct(begin-point, end-point)) <= EPS
#3点が反時計回りか
#一直線上のときの例外処理できていない→F
def CCW(p, q, r):
one, two = q-p, r-q
if OuterProduct(one, two) > EPS:
return True
else:
return False
def Crosspoint(a, b, c, d):
if abs(OuterProduct(b-a, d-c)) <= EPS:
return False
else:
u = OuterProduct(c-a, d-a) / OuterProduct(b-a, d-c)
return (1-u)*a + u*b
#凹多角形の面積(多角形は反時計回りに与えられる)
def Area(dots):
res = 0
for i in range(len(dots)-1):
res += OuterProduct(dots[i], dots[i+1])
res += OuterProduct(dots[-1], dots[0])
return res/2
n = int(input())
dots = []
for _ in range(n):
x, y = map(float, input().split())
dots.append(complex(x, y))
q = int(input())
for _ in range(q):
x, y, z, w = map(int, input().split())
p1, p2 = complex(x, y), complex(z, w)
res = []
for i in range(-1, n-1):
if IsOnLine(dots[i], p1, p2) or CCW(p1, p2, dots[i]):
res.append(dots[i])
if not IsOnLine(dots[i], p1, p2) and not IsOnLine(dots[i+1], p1, p2) and (CCW(p1, p2, dots[i]) != CCW(p1, p2, dots[i+1])):
res.append(Crosspoint(dots[i], dots[i+1], p1, p2))
if not res:
print(0)
else:
print(Area(res))
```
| 84,984 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
"Correct Solution:
```
#!/usr/bin/env python3
# CGL_4_C: Convex Polygon Convex Cut
def cut(ps, p0, p1):
polygon = []
for p2, p3 in zip(ps, ps[1:] + [ps[0]]):
if cross(p0, p1, p0, p2) >= 0:
polygon.append(p2)
if intersect(p0, p1, p2, p3):
p = intersect_point(p0, p1, p2, p3)
if not same(p, p2):
polygon.append(p)
return polygon
def intersect(p0, p1, p2, p3):
if cross(p0, p1, p0, p2) >= 0 and cross(p0, p1, p0, p3) < 0:
return True
elif cross(p0, p1, p0, p2) <= 0 and cross(p0, p1, p0, p3) > 0:
return True
return False
def same(p0, p1):
x0, y0 = p0
x1, y1 = p1
return abs(x0 - x1) < 1e-10 and abs(y0 - y1) < 1e-10
def intersect_point(p0, p1, p2, p3):
d1 = abs(cross(p0, p1, p0, p2))
d2 = abs(cross(p0, p1, p0, p3))
t = d1 / (d1 + d2)
x2, y2 = p2
x3, y3 = p3
return x2 + (x3 - x2) * t, y2 + (y3 - y2) * t
def area(ps):
"""Calculate area of a polygon.
>>> area([(0, 0), (2, 2), (-1, 1)])
2.0
>>> area([])
0.0
>>> area([(0, 0)])
0.0
>>> area([(0, 0), (1, 0)])
0.0
"""
area = 0.0
if ps:
for p0, p1 in zip(ps, ps[1:]+[ps[0]]):
area += cross((0, 0), p0, p0, p1)
return area / 2
def cross(p0, p1, p2, p3):
x0, y0 = p0
x1, y1 = p1
x2, y2 = p2
x3, y3 = p3
return (x1-x0)*(y3-y2) - (x3-x2)*(y1-y0)
def run():
n = int(input())
ps = []
for _ in range(n):
x, y = [int(i) for i in input().split()]
ps.append((x, y))
q = int(input())
for _ in range(q):
x1, y1, x2, y2 = [int(i) for i in input().split()]
print("{:.10f}".format(area(cut(ps, (x1, y1), (x2, y2)))))
if __name__ == '__main__':
run()
```
| 84,985 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
"Correct Solution:
```
from sys import stdin
readline = stdin.readline
class vector:
def cross(a, b):
return a.real * b.imag - a.imag * b.real
def dot(a, b):
return a.real * b.real + a.imag * b.imag
def ccw(a, b, c):
b -= a
c -= a
if vector.cross(b, c) > 0:
return 1
if vector.cross(b, c) < 0:
return 2
if vector.dot(b, c) < 0:
return 3
if abs(b) < abs(c):
return 4
return 5
def polygon(p):
if len(p) < 3:
return 0
return 0.5 * sum(vector.cross(p[i - 1], p[i]) for i in range(len(p)))
def intersection(p1, p2, p3, p4):
a1 = p4 - p2
b1 = p2 - p3
b2 = p1 - p2
s1 = vector.cross(a1, b2) / 2
s2 = vector.cross(a1, b1) / 2
if s1 + s2 == 0:
return 0
c1 = p1 + (p3 - p1) * s1 / (s1 + s2)
return c1
def main():
n = int(readline())
p = [map(float, readline().split()) for _ in range(n)]
p = [x + y * 1j for x, y in p]
q = int(readline())
for _ in range(q):
p1x, p1y, p2x, p2y = map(float, readline().split())
p1, p2 = p1x + p1y * 1j, p2x + p2y * 1j
pre_tmp = vector.ccw(p[-1], p1, p2)
left = []
for i in range(len(p)):
tmp = vector.ccw(p[i], p1, p2)
if pre_tmp != tmp and all(i in (1, 2) for i in (pre_tmp, tmp)):
c1 = vector.intersection(p1, p[i - 1], p2, p[i])
left.append(c1)
if tmp != 2:
left.append(p[i])
pre_tmp = tmp
print('{:.6f}'.format(vector.polygon(left)))
main()
```
| 84,986 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
"Correct Solution:
```
def cross(a, b):
return a.real * b.imag - a.imag * b.real
def cross_point(c, d):
l = d - c
v1 = cross(lv, l)
v2 = cross(lv, lt - c)
return c + v2 / v1 * l
n = int(input())
points = [complex(*map(int, input().split())) for _ in range(n)]
point0 = points.pop(0)
points.append(point0)
q = int(input())
while q:
x1, y1, x2, y2 = map(int, input().split())
ls, lt = (x1 + 1j * y1, x2 + 1j * y2)
lv = lt - ls
area = 0
prev = point0
prev_flag = cross(lv, prev - ls) >= 0
cp1, cp2 = None, None
for p in points:
curr_flag = cross(lv, p - ls) >= 0
if prev_flag and curr_flag:
area += cross(prev, p)
elif prev_flag != curr_flag:
cp = cross_point(prev, p)
if prev_flag:
area += cross(prev, cp)
cp1 = cp
else:
area += cross(cp, p)
cp2 = cp
prev, prev_flag = p, curr_flag
if cp1 is not None and cp2 is not None:
area += cross(cp1, cp2)
print(area / 2)
q -= 1
```
| 84,987 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
"Correct Solution:
```
# cross point
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def cross_point(p1, p2, p3, p4):
# p1 and p2 are points on a segment.
# p3 and p4 are points on the other segment.
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = cross(base, hypo1) / abs(base)
d2 = cross(base, hypo2) / abs(base)
cp = p1 + d1 / (d1 - d2) * (p2 - p1)
return cp
# area of a triangle
def _area_of_triangle(c1, c2, c3):
v1 = c2 - c1
v2 = c3 - c1
return abs(v1.real * v2.imag - v1.imag * v2.real) / 2
# convex cut
def convex_cut(points, c1, c2):
points.append(points[0])
ref_vec = c2 - c1
cross_point1 = None
flag = 0
# Detection of one intersection point
for i, segment in enumerate(zip(points, points[1:])):
p1, p2 = segment
cross1 = cross(ref_vec, p1 - c1)
cross2 = cross(ref_vec, p2 - c1)
flag += cross1
if cross1 <= 0 and cross2 > 0:
cross_point1 = cross_point(c1, c2, p1, p2)
points = points[i+1:]
break
elif cross1 > 0 and cross2 <= 0:
cross_point1 = cross_point(c1, c2, p1, p2)
points = points[i::-1] + points[:i:-1]
break
# Processing when there is no intersection point
if cross_point1 == None:
if flag > 0:
cross_point1 = points[0]
points = points[1:]
else:
return 0
# find area
cut_area = 0
for p1, p2 in zip(points, points[1:]):
if cross(ref_vec, p1 - c1) * cross(ref_vec, p2 - c1) <= 0:
cross_point2 = cross_point(c1, c2, p1, p2)
cut_area += _area_of_triangle(cross_point1, cross_point2, p1)
break
else:
cut_area += _area_of_triangle(cross_point1, p1, p2)
return cut_area
# acceptance of input
import sys
file_input = sys.stdin
n = int(file_input.readline())
def string_to_complex(s):
x, y = map(float, s.split())
return x + y * 1j
G = [string_to_complex(file_input.readline()) for i in range(n)]
# output
q = int(file_input.readline())
for line in file_input:
p1x, p1y, p2x, p2y = map(int, line.split())
p1 = p1x + p1y * 1j
p2 = p2x + p2y * 1j
ans = convex_cut(G.copy(), p1, p2)
print("{:f}".format(ans))
```
| 84,988 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
"Correct Solution:
```
from sys import stdin
readline = stdin.readline
class vector:
def cross(a, b):
return a.real * b.imag - a.imag * b.real
def dot(a, b):
return a.real * b.real + a.imag * b.imag
def ccw(a, b, c):
b -= a
c -= a
if vector.cross(b, c) > 0:
return 1
if vector.cross(b, c) < 0:
return 2
if vector.dot(b, c) < 0:
return 3
if abs(b) < abs(c):
return 4
return 5
def polygon(p):
if len(p) < 3:
return 0
return 0.5 * sum(vector.cross(p[i - 1], p[i]) for i in range(len(p)))
def intersection(p1, p2, p3, p4):
a1 = p4 - p2
b1 = p2 - p3
b2 = p1 - p2
s1 = vector.cross(a1, b2) / 2
s2 = vector.cross(a1, b1) / 2
if s1 + s2 == 0:
return 0
c1 = p1 + (p3 - p1) * s1 / (s1 + s2)
return c1
def main():
n = int(readline())
p = [map(float, readline().split()) for _ in range(n)]
p = [x + y * 1j for x, y in p]
q = int(readline())
for _ in range(q):
p1x, p1y, p2x, p2y = map(float, readline().split())
p1, p2 = p1x + p1y * 1j, p2x + p2y * 1j
pre_tmp = vector.ccw(p[-1], p1, p2)
left = []
for i in range(len(p)):
tmp = vector.ccw(p[i], p1, p2)
if pre_tmp != tmp and all(i in (1, 2) for i in (pre_tmp, tmp)):
c1 = vector.intersection(p1, p[i - 1], p2, p[i])
left.append(c1)
if tmp != 2:
left.append(p[i])
pre_tmp = tmp
print('{:.6f}'.format(vector.polygon(left)))
import sys
if sys.hexversion >= 0x030203F0:
main()
else:
print('Error')
```
| 84,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
Submitted Solution:
```
#!/usr/bin/python3
import array
from fractions import Fraction
import functools
import itertools
import math
import os
import sys
def main():
N = read_int()
P = [Vec(*read_ints()) for _ in range(N)]
Q = read_int()
for _ in range(Q):
x1, y1, x2, y2 = read_ints()
print(solve(N, P, Vec(x1, y1), Vec(x2, y2)))
def solve(N, P, A, B):
b = B - A
P = [p - A for p in P]
ccw = 0
cw = 0
for p in P:
c = b.cross(p)
if c >= 0:
ccw += 1
if c <= 0:
cw += 1
if ccw == N:
return float(poly_area(P))
if cw == N:
return 0
cross_points = []
for i in range(N):
j = (i + 1) % N
p = P[i]
q = P[j]
qp = q - p
cross_qp_b = qp.cross(b)
if cross_qp_b == 0:
continue
k = Fraction(b.cross(p), cross_qp_b)
if 0 < k <= 1:
t = Fraction(p.cross(qp), b.cross(qp))
cross_points.append((t, i, k))
cross_points.sort()
_, i1, k1 = cross_points[0]
_, i2, k2 = cross_points[1]
x1 = P[i1] + k1 * (P[(i1 + 1) % N] - P[i1])
x2 = P[i2] + k2 * (P[(i2 + 1) % N] - P[i2])
Q = [x2]
j = (i2 + 1) % N
while j != i1:
Q.append(P[j])
j = (j + 1) % N
Q.append(P[i1])
Q.append(x1)
return float(poly_area(Q))
def poly_area(P):
N = len(P)
a = 0
for i in range(1, N - 1):
a += Fraction((P[i + 1] - P[i]).cross(P[0] - P[i + 1]), 2)
return a
###############################################################################
# AUXILIARY FUNCTIONS
class Vec(object):
def __init__(self, x, y):
self.x = x
self.y = y
super().__init__()
def __add__(self, other):
return Vec(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Vec(self.x - other.x, self.y - other.y)
def __mul__(self, scalar):
return Vec(self.x * scalar, self.y * scalar)
def __rmul__(self, scalar):
return Vec(self.x * scalar, self.y * scalar)
def __truediv__(self, scalar):
return Vec(self.x / scalar, self.y / scalar)
def __iadd__(self, other):
self.x += other.x
self.y += other.y
return self
def __isub__(self, other):
self.x -= other.x
self.y -= other.y
return self
def __imul__(self, scalar):
self.x *= scalar
self.y *= scalar
return self
def __idiv__(self, scalar):
self.x /= scalar
self.y /= scalar
return self
def __neg__(self):
return Vec(-self.x, -self.y)
def __eq__(self, other):
return self.x == other.x and self.y == other.y
def __ne__(self, other):
return not self.__eq__(other)
def __hash__(self):
return hash('Vec', self.x, self.y)
def dot(self, other):
return self.x * other.x + self.y * other.y
def cross(self, other):
return self.x * other.y - self.y * other.x
def abs2(self):
return self.x * self.x + self.y * self.y
def __abs__(self):
return math.sqrt(float(self.abs2()))
def __repr__(self):
return '({}, {})'.format(self.x, self.y)
DEBUG = 'DEBUG' in os.environ
def inp():
return sys.stdin.readline().rstrip()
def read_int():
return int(inp())
def read_ints():
return [int(e) for e in inp().split()]
def dprint(*value, sep=' ', end='\n'):
if DEBUG:
print(*value, sep=sep, end=end)
if __name__ == '__main__':
main()
```
Yes
| 84,990 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
Submitted Solution:
```
import cmath
import math
import os
import sys
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353
PI = cmath.pi
TAU = cmath.pi * 2
EPS = 1e-10
class Point:
"""
2次元空間上の点
"""
# 反時計回り側にある
CCW_COUNTER_CLOCKWISE = 1
# 時計回り側にある
CCW_CLOCKWISE = -1
# 線分の後ろにある
CCW_ONLINE_BACK = 2
# 線分の前にある
CCW_ONLINE_FRONT = -2
# 線分上にある
CCW_ON_SEGMENT = 0
def __init__(self, c: complex):
self.c = c
@property
def x(self):
return self.c.real
@property
def y(self):
return self.c.imag
@staticmethod
def from_rect(x: float, y: float):
return Point(complex(x, y))
@staticmethod
def from_polar(r: float, phi: float):
return Point(cmath.rect(r, phi))
def __add__(self, p):
"""
:param Point p:
"""
return Point(self.c + p.c)
def __iadd__(self, p):
"""
:param Point p:
"""
self.c += p.c
return self
def __sub__(self, p):
"""
:param Point p:
"""
return Point(self.c - p.c)
def __isub__(self, p):
"""
:param Point p:
"""
self.c -= p.c
return self
def __mul__(self, f: float):
return Point(self.c * f)
def __imul__(self, f: float):
self.c *= f
return self
def __truediv__(self, f: float):
return Point(self.c / f)
def __itruediv__(self, f: float):
self.c /= f
return self
def __repr__(self):
return "({}, {})".format(round(self.x, 10), round(self.y, 10))
def __neg__(self):
return Point(-self.c)
def __eq__(self, p):
return abs(self.c - p.c) < EPS
def __abs__(self):
return abs(self.c)
@staticmethod
def ccw(a, b, c):
"""
線分 ab に対する c の位置
線分上にあるか判定するだけなら on_segment とかのが速い
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_C&lang=ja
:param Point a:
:param Point b:
:param Point c:
"""
b = b - a
c = c - a
det = b.det(c)
if det > EPS:
return Point.CCW_COUNTER_CLOCKWISE
if det < -EPS:
return Point.CCW_CLOCKWISE
if b.dot(c) < -EPS:
return Point.CCW_ONLINE_BACK
if c.norm() - b.norm() > EPS:
return Point.CCW_ONLINE_FRONT
return Point.CCW_ON_SEGMENT
def dot(self, p):
"""
内積
:param Point p:
:rtype: float
"""
return self.x * p.x + self.y * p.y
def det(self, p):
"""
外積
:param Point p:
:rtype: float
"""
return self.x * p.y - self.y * p.x
def dist(self, p):
"""
距離
:param Point p:
:rtype: float
"""
return abs(self.c - p.c)
def norm(self):
"""
原点からの距離
:rtype: float
"""
return abs(self.c)
def phase(self):
"""
原点からの角度
:rtype: float
"""
return cmath.phase(self.c)
def angle(self, p, q):
"""
p に向いてる状態から q まで反時計回りに回転するときの角度
-pi <= ret <= pi
:param Point p:
:param Point q:
:rtype: float
"""
return (cmath.phase(q.c - self.c) - cmath.phase(p.c - self.c) + PI) % TAU - PI
def area(self, p, q):
"""
p, q となす三角形の面積
:param Point p:
:param Point q:
:rtype: float
"""
return abs((p - self).det(q - self) / 2)
def projection_point(self, p, q, allow_outer=False):
"""
線分 pq を通る直線上に垂線をおろしたときの足の座標
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_A&lang=ja
:param Point p:
:param Point q:
:param allow_outer: 答えが線分の間になくても OK
:rtype: Point|None
"""
diff_q = q - p
# 答えの p からの距離
r = (self - p).dot(diff_q) / abs(diff_q)
# 線分の角度
phase = diff_q.phase()
ret = Point.from_polar(r, phase) + p
if allow_outer or (p - ret).dot(q - ret) < EPS:
return ret
return None
def reflection_point(self, p, q):
"""
直線 pq を挟んで反対にある点
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_B&lang=ja
:param Point p:
:param Point q:
:rtype: Point
"""
# 距離
r = abs(self - p)
# pq と p-self の角度
angle = p.angle(q, self)
# 直線を挟んで角度を反対にする
angle = (q - p).phase() - angle
return Point.from_polar(r, angle) + p
def on_segment(self, p, q, allow_side=True):
"""
点が線分 pq の上に乗っているか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_C&lang=ja
:param Point p:
:param Point q:
:param allow_side: 端っこでギリギリ触れているのを許容するか
:rtype: bool
"""
if not allow_side and (self == p or self == q):
return False
# 外積がゼロ: 面積がゼロ == 一直線
# 内積がマイナス: p - self - q の順に並んでる
return abs((p - self).det(q - self)) < EPS and (p - self).dot(q - self) < EPS
class Line:
"""
2次元空間上の直線
"""
def __init__(self, a: float, b: float, c: float):
"""
直線 ax + by + c = 0
"""
self.a = a
self.b = b
self.c = c
@staticmethod
def from_gradient(grad: float, intercept: float):
"""
直線 y = ax + b
:param grad: 傾き
:param intercept: 切片
:return:
"""
return Line(grad, -1, intercept)
@staticmethod
def from_segment(p1, p2):
"""
:param Point p1:
:param Point p2:
"""
a = p2.y - p1.y
b = p1.x - p2.x
c = p2.y * (p2.x - p1.x) - p2.x * (p2.y - p1.y)
return Line(a, b, c)
@property
def gradient(self):
"""
傾き
"""
return INF if self.b == 0 else -self.a / self.b
@property
def intercept(self):
"""
切片
"""
return INF if self.b == 0 else -self.c / self.b
def is_parallel_to(self, l):
"""
平行かどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A&lang=ja
:param Line l:
"""
# 法線ベクトル同士の外積がゼロ
return abs(Point.from_rect(self.a, self.b).det(Point.from_rect(l.a, l.b))) < EPS
def is_orthogonal_to(self, l):
"""
直行しているかどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A&lang=ja
:param Line l:
"""
# 法線ベクトル同士の内積がゼロ
return abs(Point.from_rect(self.a, self.b).dot(Point.from_rect(l.a, l.b))) < EPS
def intersection_point(self, l):
"""
交差する点
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_C&lang=ja
:param Line l:
:rtype: Point|None
"""
a1, b1, c1 = self.a, self.b, self.c
a2, b2, c2 = l.a, l.b, l.c
det = a1 * b2 - a2 * b1
if abs(det) < EPS:
# 並行
return None
x = (b1 * c2 - b2 * c1) / det
y = (a2 * c1 - a1 * c2) / det
return Point.from_rect(x, y)
def dist(self, p):
"""
他の点との最短距離
:param Point p:
"""
raise NotImplementedError()
def has_point(self, p):
"""
p が直線上に乗っているかどうか
:param Point p:
"""
return abs(self.a * p.x + self.b * p.y + self.c) < EPS
class Segment:
"""
2次元空間上の線分
"""
def __init__(self, p1, p2):
"""
:param Point p1:
:param Point p2:
"""
self.p1 = p1
self.p2 = p2
def norm(self):
"""
線分の長さ
"""
return abs(self.p1 - self.p2)
def phase(self):
"""
p1 を原点としたときの p2 の角度
"""
return cmath.phase(self.p2 - self.p1)
def is_parallel_to(self, s):
"""
平行かどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A&lang=ja
:param Segment s:
:return:
"""
# 外積がゼロ
return abs((self.p1 - self.p2).det(s.p1 - s.p2)) < EPS
def is_orthogonal_to(self, s):
"""
直行しているかどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A&lang=ja
:param Segment s:
:return:
"""
# 内積がゼロ
return abs((self.p1 - self.p2).dot(s.p1 - s.p2)) < EPS
def intersects_with(self, s, allow_side=True):
"""
交差するかどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_B&lang=ja
:param Segment s:
:param allow_side: 端っこでギリギリ触れているのを許容するか
"""
if self.is_parallel_to(s):
# 並行なら線分の端点がもう片方の線分の上にあるかどうか
return (s.p1.on_segment(self.p1, self.p2, allow_side) or
s.p2.on_segment(self.p1, self.p2, allow_side) or
self.p1.on_segment(s.p1, s.p2, allow_side) or
self.p2.on_segment(s.p1, s.p2, allow_side))
else:
# allow_side ならゼロを許容する
det_lower = EPS if allow_side else -EPS
ok = True
# self の両側に s.p1 と s.p2 があるか
ok &= (self.p2 - self.p1).det(s.p1 - self.p1) * (self.p2 - self.p1).det(s.p2 - self.p1) < det_lower
# s の両側に self.p1 と self.p2 があるか
ok &= (s.p2 - s.p1).det(self.p1 - s.p1) * (s.p2 - s.p1).det(self.p2 - s.p1) < det_lower
return ok
def closest_point(self, p):
"""
線分上の、p に最も近い点
:param Point p:
"""
# p からおろした垂線までの距離
d = (p - self.p1).dot(self.p2 - self.p1) / self.norm()
# p1 より前
if d < EPS:
return self.p1
# p2 より後
if -EPS < d - self.norm():
return self.p2
# 線分上
return Point.from_polar(d, (self.p2 - self.p1).phase()) + self.p1
def dist(self, p):
"""
他の点との最短距離
:param Point p:
"""
return abs(p - self.closest_point(p))
def dist_segment(self, s):
"""
他の線分との最短距離
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_D&lang=ja
:param Segment s:
"""
if self.intersects_with(s):
return 0.0
return min(
self.dist(s.p1),
self.dist(s.p2),
s.dist(self.p1),
s.dist(self.p2),
)
def has_point(self, p, allow_side=True):
"""
p が線分上に乗っているかどうか
:param Point p:
:param allow_side: 端っこでギリギリ触れているのを許容するか
"""
return p.on_segment(self.p1, self.p2, allow_side=allow_side)
class Polygon:
"""
2次元空間上の多角形
"""
def __init__(self, points):
"""
:param list of Point points:
"""
self.points = points
def iter2(self):
"""
隣り合う2点を順に返すイテレータ
:rtype: typing.Iterator[(Point, Point)]
"""
return zip(self.points, self.points[1:] + self.points[:1])
def iter3(self):
"""
隣り合う3点を順に返すイテレータ
:rtype: typing.Iterator[(Point, Point, Point)]
"""
return zip(self.points,
self.points[1:] + self.points[:1],
self.points[2:] + self.points[:2])
def area(self):
"""
面積
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_3_A&lang=ja
"""
# 外積の和 / 2
dets = []
for p, q in self.iter2():
dets.append(p.det(q))
return abs(math.fsum(dets)) / 2
def is_convex(self, allow_straight=False, allow_collapsed=False):
"""
凸多角形かどうか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_3_B&lang=ja
:param allow_straight: 3点がまっすぐ並んでるのを許容するかどうか
:param allow_collapsed: 面積がゼロの場合を許容するか
"""
ccw = []
for a, b, c in self.iter3():
ccw.append(Point.ccw(a, b, c))
ccw = set(ccw)
if len(ccw) == 1:
if ccw == {Point.CCW_CLOCKWISE}:
return True
if ccw == {Point.CCW_COUNTER_CLOCKWISE}:
return True
if allow_straight and len(ccw) == 2:
if ccw == {Point.CCW_ONLINE_FRONT, Point.CCW_CLOCKWISE}:
return True
if ccw == {Point.CCW_ONLINE_FRONT, Point.CCW_COUNTER_CLOCKWISE}:
return True
if allow_collapsed and len(ccw) == 3:
return ccw == {Point.CCW_ONLINE_FRONT, Point.CCW_ONLINE_BACK, Point.CCW_ON_SEGMENT}
return False
def has_point_on_edge(self, p):
"""
指定した点が辺上にあるか
:param Point p:
:rtype: bool
"""
for a, b in self.iter2():
if p.on_segment(a, b):
return True
return False
def contains(self, p, allow_on_edge=True):
"""
指定した点を含むか
Winding Number Algorithm
https://www.nttpc.co.jp/technology/number_algorithm.html
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_3_C&lang=ja
:param Point p:
:param bool allow_on_edge: 辺上の点を許容するか
"""
angles = []
for a, b in self.iter2():
if p.on_segment(a, b):
return allow_on_edge
angles.append(p.angle(a, b))
# 一周以上するなら含む
return abs(math.fsum(angles)) > EPS
@staticmethod
def convex_hull(points, allow_straight=False):
"""
凸包。x が最も小さい点のうち y が最も小さい点から反時計回り。
Graham Scan O(N log N)
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_4_A&lang=ja
:param list of Point points:
:param allow_straight: 3点がまっすぐ並んでるのを許容するかどうか
:rtype: list of Point
"""
points = points[:]
points.sort(key=lambda p: (p.x, p.y))
# allow_straight なら 0 を許容する
det_lower = -EPS if allow_straight else EPS
sz = 0
#: :type: list of (Point|None)
ret = [None] * (N * 2)
for p in points:
while sz > 1 and (ret[sz - 1] - ret[sz - 2]).det(p - ret[sz - 1]) < det_lower:
sz -= 1
ret[sz] = p
sz += 1
floor = sz
for p in reversed(points[:-1]):
while sz > floor and (ret[sz - 1] - ret[sz - 2]).det(p - ret[sz - 1]) < det_lower:
sz -= 1
ret[sz] = p
sz += 1
ret = ret[:sz - 1]
if allow_straight and len(ret) > len(points):
# allow_straight かつ全部一直線のときに二重にカウントしちゃう
ret = points
return ret
@staticmethod
def diameter(points):
"""
直径
凸包構築 O(N log N) + カリパー法 O(N)
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_4_B&lang=ja
:param list of Point points:
"""
# 反時計回り
points = Polygon.convex_hull(points, allow_straight=False)
if len(points) == 1:
return 0.0
if len(points) == 2:
return abs(points[0] - points[1])
# x軸方向に最も遠い点対
si = points.index(min(points, key=lambda p: (p.x, p.y)))
sj = points.index(max(points, key=lambda p: (p.x, p.y)))
n = len(points)
ret = 0.0
# 半周回転
i, j = si, sj
while i != sj or j != si:
ret = max(ret, abs(points[i] - points[j]))
ni = (i + 1) % n
nj = (j + 1) % n
# 2つの辺が並行になる方向にずらす
if (points[ni] - points[i]).det(points[nj] - points[j]) > 0:
j = nj
else:
i = ni
return ret
def convex_cut_by_line(self, line_p1, line_p2):
"""
凸多角形を直線 line_p1-line_p2 でカットする。
凸じゃないといけません
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_4_C&lang=ja
:param line_p1:
:param line_p2:
:return: (line_p1-line_p2 の左側の多角形, line_p1-line_p2 の右側の多角形)
:rtype: (Polygon|None, Polygon|None)
"""
n = len(self.points)
line = Line.from_segment(line_p1, line_p2)
# 直線と重なる点
on_line_points = []
for i, p in enumerate(self.points):
if line.has_point(p):
on_line_points.append(i)
# 辺が直線上にある
has_on_line_edge = False
if len(on_line_points) >= 3:
has_on_line_edge = True
elif len(on_line_points) == 2:
# 直線上にある点が隣り合ってる
has_on_line_edge = abs(on_line_points[0] - on_line_points[1]) in [1, n - 1]
# 辺が直線上にある場合、どっちか片方に全部ある
if has_on_line_edge:
for p in self.points:
ccw = Point.ccw(line_p1, line_p2, p)
if ccw == Point.CCW_COUNTER_CLOCKWISE:
return Polygon(self.points[:]), None
if ccw == Point.CCW_CLOCKWISE:
return None, Polygon(self.points[:])
ret_lefts = []
ret_rights = []
d = line_p2 - line_p1
for p, q in self.iter2():
det_p = d.det(p - line_p1)
det_q = d.det(q - line_p1)
if det_p > -EPS:
ret_lefts.append(p)
if det_p < EPS:
ret_rights.append(p)
# 外積の符号が違う == 直線の反対側にある場合は交点を追加
if det_p * det_q < -EPS:
intersection = line.intersection_point(Line.from_segment(p, q))
ret_lefts.append(intersection)
ret_rights.append(intersection)
# 点のみの場合を除いて返す
l = Polygon(ret_lefts) if len(ret_lefts) > 1 else None
r = Polygon(ret_rights) if len(ret_rights) > 1 else None
return l, r
N = int(sys.stdin.buffer.readline())
XY = [list(map(float, sys.stdin.buffer.readline().split())) for _ in range(N)]
points = []
for x, y in XY:
points.append(Point(complex(x, y)))
polygon = Polygon(points)
Q = int(sys.stdin.buffer.readline())
POINTS = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(Q)]
for x1, y1, x2, y2 in POINTS:
l, r = polygon.convex_cut_by_line(Point.from_rect(x1, y1), Point.from_rect(x2, y2))
if l:
print(l.area())
else:
print(0)
```
Yes
| 84,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
Submitted Solution:
```
from sys import stdin
readline = stdin.readline
from enum import Enum
direction = Enum('direction', 'CCW CW CAB ABC ACB')
class vector:
def cross(a, b):
return a.real * b.imag - a.imag * b.real
def dot(a, b):
return a.real * b.real + a.imag * b.imag
def ccw(a, b, c):
b -= a
c -= a
if vector.cross(b, c) > 0:
return direction.CCW
if vector.cross(b, c) < 0:
return direction.CW
if vector.dot(b, c) < 0:
return direction.CAB
if abs(b) < abs(c):
return direction.ABC
return direction.ACB
def polygon(p):
if len(p) < 3:
return 0
return 0.5 * sum(vector.cross(p[i - 1], p[i]) for i in range(len(p)))
def intersection(p1, p2, p3, p4):
a1 = p4 - p2
b1 = p2 - p3
b2 = p1 - p2
s1 = vector.cross(a1, b2) / 2
s2 = vector.cross(a1, b1) / 2
if s1 + s2 == 0:
return 0
c1 = p1 + (p3 - p1) * s1 / (s1 + s2)
return c1
def main():
n = int(readline())
p = [map(float, readline().split()) for _ in range(n)]
p = [x + y * 1j for x, y in p]
q = int(readline())
for _ in range(q):
p1x, p1y, p2x, p2y = map(float, readline().split())
p1, p2 = p1x + p1y * 1j, p2x + p2y * 1j
pre_tmp = vector.ccw(p[-1], p1, p2)
left = []
for i in range(len(p)):
tmp = vector.ccw(p[i], p1, p2)
if pre_tmp != tmp and all(i in (direction.CW, direction.CCW) for i in (pre_tmp, tmp)):
c1 = vector.intersection(p1, p[i - 1], p2, p[i])
left.append(c1)
if tmp != direction.CW:
left.append(p[i])
pre_tmp = tmp
print('{:.6f}'.format(vector.polygon(left)))
main()
```
No
| 84,992 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
Submitted Solution:
```
# cross point
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def cross_point(p1, p2, p3, p4):
# p1 and p2 are points on a segment.
# p3 and p4 are points on the other segment.
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = cross(base, hypo1) / abs(base)
d2 = cross(base, hypo2) / abs(base)
cp = p1 + d1 / (d1 - d2) * (p2 - p1)
return cp
# area of a triangle
def _area_of_triangle(c1, c2, c3):
v1 = c2 - c1
v2 = c3 - c1
return abs(v1.real * v2.imag - v1.imag * v2.real) / 2
# convex cut
def convex_cut(points, c1, c2):
points.append(points[0])
ref_vec = c2 - c1
for i, segment in enumerate(zip(points, points[1:])):
p1, p2 = segment
cross1 = cross(ref_vec, p1 - c1)
cross2 = cross(ref_vec, p2 - c1)
if cross1 <= 0 and cross2 > 0:
cross_point1 = cross_point(c1, c2, p1, p2)
points = points[i+1:]
break
elif cross1 > 0 and cross2 <= 0:
cross_point1 = cross_point(c1, c2, p1, p2)
points = points[i::-1] + points[:i:-1]
break
cut_area = 0
for p1, p2 in zip(points, points[1:]):
if cross(ref_vec, p1 - c1) * cross(ref_vec, p2 - c1) <= 0:
cross_point2 = cross_point(c1, c2, p1, p2)
cut_area += _area_of_triangle(cross_point1, cross_point2, p1)
break
else:
cut_area += _area_of_triangle(cross_point1, p1, p2)
return cut_area
# acceptance of input
import sys
file_input = sys.stdin
n = int(file_input.readline())
def string_to_complex(s):
x, y = map(float, s.split())
return x + y * 1j
G = [string_to_complex(file_input.readline()) for i in range(n)]
# output
q = int(file_input.readline())
for line in file_input:
p1x, p1y, p2x, p2y = map(int, line.split())
p1 = p1x + p1y * 1j
p2 = p2x + p2y * 1j
ans = convex_cut(G.copy(), p1, p2)
print("{:f}".format(ans))
```
No
| 84,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
Submitted Solution:
```
from sys import stdin
readline = stdin.readline
from enum import Enum
direction = Enum('direction', 'CCW CW CAB ABC ACB')
class vector:
def cross(a, b):
return a.real * b.imag - a.imag * b.real
def dot(a, b):
return a.real * b.real + a.imag * b.imag
def ccw(a, b, c):
b -= a
c -= a
if vector.cross(b, c) > 0:
return direction.CCW
if vector.cross(b, c) < 0:
return direction.CW
if vector.dot(b, c) < 0:
return direction.CAB
if vector.abs(b) < abs(c):
return direction.ABC
return direction.ACB
def polygon(p):
if len(p) < 3:
return 0
return 0.5 * sum(vector.cross(p[i - 1], p[i]) for i in range(len(p)))
def intersection(p1, p2, p3, p4):
a1 = p4 - p2
b1 = p2 - p3
b2 = p1 - p2
s1 = vector.cross(a1, b2) / 2
s2 = vector.cross(a1, b1) / 2
c1 = p1 + (p3 - p1) * s1 / (s1 + s2)
return c1
def main():
n = int(readline())
p = [map(float, readline().split()) for _ in range(n)]
p = [x + y * 1j for x, y in p]
q = int(readline())
for _ in range(q):
p1x, p1y, p2x, p2y = map(float, readline().split())
p1, p2 = p1x + p1y * 1j, p2x + p2y * 1j
pre_tmp = vector.ccw(p[-1], p1, p2)
left = []
for i in range(len(p)):
tmp = vector.ccw(p[i], p1, p2)
if pre_tmp != tmp and all(i in (direction.CW, direction.CCW) for i in (pre_tmp, tmp)):
c1 = vector.intersection(p1, p[i - 1], p2, p[i])
left.append(c1)
if tmp != direction.CW:
left.append(p[i])
pre_tmp = tmp
print('{:.6f}'.format(vector.polygon(left)))
main()
```
No
| 84,994 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
As shown in the figure above, cut a convex polygon g by a line p1p2 and print the area of the cut polygon which is on the left-hand side of the line.
g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n−1) are sides of the convex polygon. The line segment connecting pn and p1 is also a side of the polygon.
Constraints
* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* -10000 ≤ p1x,p1y,p2x,p2y ≤ 10000
* No point in g will occur more than once.
* p1 ≠ p2
Input
The input is given in the following format:
g (the sequence of the points of the polygon)
q (the number of queries = the number of target lines)
1st query
2nd query
:
qth query
g is given as a sequence of points p1,..., pn in the following format:
n
x1 y1
x2 y2
:
xn yn
The first integer n is the number of points. The coordinate of the i-th point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Note that all interior angles of given convex polygons are less than or equal to 180.
For each query, a line represented by two points p1 and p2 is given. The coordinates of the points are given by four integers p1x, p1y, p2x and p2y.
Output
For each query, print the area of the cut polygon. The output values should be in a decimal fraction with an error less than 0.00001.
Example
Input
4
1 1
4 1
4 3
1 3
2
2 0 2 4
2 4 2 0
Output
2.00000000
4.00000000
Submitted Solution:
```
from sys import stdin
readline = stdin.readline
from enum import Enum
direction = Enum('direction', 'CCW CW CAB ABC ACB')
class vector:
def cross(a, b):
return a.real * b.imag - a.imag * b.real
def dot(a, b):
return a.real * b.real + a.imag * b.imag
def ccw(a, b, c):
b -= a
c -= a
if vector.cross(b, c) > 0:
return direction.CCW
if vector.cross(b, c) < 0:
return direction.CW
if vector.dot(b, c) < 0:
return direction.CAB
if vector.abs(b) < abs(c):
return direction.ABC
return direction.ACB
def polygon(p):
return 0.5 * sum(vector.cross(p[i - 1], p[i]) for i in range(len(p)))
def intersection(p1, p2, p3, p4):
a1 = p4 - p2
b1 = p2 - p3
b2 = p1 - p2
s1 = vector.cross(a1, b2) / 2
s2 = vector.cross(a1, b1) / 2
c1 = p1 + (p3 - p1) * s1 / (s1 + s2)
return c1
def main():
n = int(readline())
p = [map(int, readline().split()) for _ in range(n)]
p = [x + y * 1j for x, y in p]
q = int(readline())
for _ in range(q):
p1x, p1y, p2x, p2y = map(int, readline().split())
p1, p2 = p1x + p1y * 1j, p2x + p2y * 1j
pre_tmp = vector.ccw(p[-1], p1, p2)
left = []
for i in range(len(p)):
tmp = vector.ccw(p[i], p1, p2)
if pre_tmp != tmp and all(i in (direction.CW, direction.CCW) for i in (pre_tmp, tmp)):
c1 = vector.intersection(p1, p[i - 1], p2, p[i])
left.append(c1)
if tmp != direction.CW:
left.append(p[i])
pre_tmp = tmp
print(vector.polygon(left))
main()
```
No
| 84,995 |
Provide a correct Python 3 solution for this coding contest problem.
For given a sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$, print the previous permutation and the next permutation in lexicographic order.
Constraints
* $1 \leq n \leq 9$
* $a_i$ consist of $1, 2, ..., n$
Input
A sequence is given in the following format.
$n$
$a_0 \; a_1 \; ... \; a_{n-1}$
Output
Print the previous permutation, the given sequence and the next permutation in the 1st, 2nd and 3rd lines respectively. Separate adjacency elements by a space character. Note that if there is no permutation, print nothing in the corresponding line.
Examples
Input
3
2 1 3
Output
1 3 2
2 1 3
2 3 1
Input
3
3 2 1
Output
3 1 2
3 2 1
"Correct Solution:
```
from itertools import permutations
n = int(input())
arr = tuple(map(int, input().split()))
p = list(permutations(sorted(arr)))
idx = p.index(arr)
for i in range(max(0, idx - 1), min(len(p), idx + 2), 1):
print(*p[i])
```
| 84,996 |
Provide a correct Python 3 solution for this coding contest problem.
For given a sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$, print the previous permutation and the next permutation in lexicographic order.
Constraints
* $1 \leq n \leq 9$
* $a_i$ consist of $1, 2, ..., n$
Input
A sequence is given in the following format.
$n$
$a_0 \; a_1 \; ... \; a_{n-1}$
Output
Print the previous permutation, the given sequence and the next permutation in the 1st, 2nd and 3rd lines respectively. Separate adjacency elements by a space character. Note that if there is no permutation, print nothing in the corresponding line.
Examples
Input
3
2 1 3
Output
1 3 2
2 1 3
2 3 1
Input
3
3 2 1
Output
3 1 2
3 2 1
"Correct Solution:
```
import heapq
from collections import deque
from enum import Enum
import sys
import math
from _heapq import heappush, heappop
BIG_NUM = 2000000000
MOD = 1000000007
EPS = 0.000000001
N = int(input())
table = [None] * N
check = [False] * (N+1)
MAP = {}
rev_MAP = {}
perm_index = 0
def recursive(index):
global perm_index
if index == N:
tmp = 0
for i in range(N):
tmp = 10*tmp+table[i]
MAP[perm_index] = tmp
rev_MAP[tmp] = perm_index
perm_index += 1
return
for i in range(1,(N+1)):
if check[i]:
continue
check[i] = True
table[index] = i
recursive(index+1)
check[i] = False
def outPut(num):
S = deque()
while True:
S.appendleft(num%10)
num //= 10
if num == 0:
break
print("%d"%(S.popleft()),end="")
while len(S) > 0:
print(" %d"%(S.popleft()),end="")
print()
recursive(0)
tmp = 0
for i in list(map(int,input().split())):
tmp = 10*tmp+i
index = rev_MAP[tmp]
if index > 0:
outPut(MAP[index-1])
outPut(tmp)
if index+1 in MAP:
outPut(MAP[index+1])
```
| 84,997 |
Provide a correct Python 3 solution for this coding contest problem.
For given a sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$, print the previous permutation and the next permutation in lexicographic order.
Constraints
* $1 \leq n \leq 9$
* $a_i$ consist of $1, 2, ..., n$
Input
A sequence is given in the following format.
$n$
$a_0 \; a_1 \; ... \; a_{n-1}$
Output
Print the previous permutation, the given sequence and the next permutation in the 1st, 2nd and 3rd lines respectively. Separate adjacency elements by a space character. Note that if there is no permutation, print nothing in the corresponding line.
Examples
Input
3
2 1 3
Output
1 3 2
2 1 3
2 3 1
Input
3
3 2 1
Output
3 1 2
3 2 1
"Correct Solution:
```
import itertools
n = int(input())
num = list(map(int, input().split()))
per = list(itertools.permutations(num))
per.sort()
index = per.index(tuple(num))
if index != 0:
print(' '.join(str(n) for n in per[index-1]))
print(' '.join(str(n) for n in per[index]))
if index != len(per)-1:
print(' '.join(str(n) for n in per[index+1]))
```
| 84,998 |
Provide a correct Python 3 solution for this coding contest problem.
For given a sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$, print the previous permutation and the next permutation in lexicographic order.
Constraints
* $1 \leq n \leq 9$
* $a_i$ consist of $1, 2, ..., n$
Input
A sequence is given in the following format.
$n$
$a_0 \; a_1 \; ... \; a_{n-1}$
Output
Print the previous permutation, the given sequence and the next permutation in the 1st, 2nd and 3rd lines respectively. Separate adjacency elements by a space character. Note that if there is no permutation, print nothing in the corresponding line.
Examples
Input
3
2 1 3
Output
1 3 2
2 1 3
2 3 1
Input
3
3 2 1
Output
3 1 2
3 2 1
"Correct Solution:
```
import itertools
if __name__ == '__main__':
n = int(input())
now = tuple(map(int,input().split()))
seq = [i+1 for i in range(n)]
seq2 = list(itertools.permutations(seq))
ind = 0
for j,k in enumerate(seq2):
if k == now:
ind = j
if ind == 0:
print(*seq2[ind])
if len(seq2) != 1:
print(*seq2[ind+1])
elif ind+1 == len(seq2):
print(*seq2[ind-1])
print(*seq2[ind])
else:
print(*seq2[ind-1])
print(*seq2[ind])
print(*seq2[ind+1])
```
| 84,999 |
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