message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a trampoline park with n trampolines in a line. The i-th of which has strength S_i.
Pekora can jump on trampolines in multiple passes. She starts the pass by jumping on any trampoline of her choice.
If at the moment Pekora jumps on trampoline i, the trampoline will launch her to position i + S_i, and S_i will become equal to max(S_i-1,1). In other words, S_i will decrease by 1, except of the case S_i=1, when S_i will remain equal to 1.
If there is no trampoline in position i + S_i, then this pass is over. Otherwise, Pekora will continue the pass by jumping from the trampoline at position i + S_i by the same rule as above.
Pekora can't stop jumping during the pass until she lands at the position larger than n (in which there is no trampoline). Poor Pekora!
Pekora is a naughty rabbit and wants to ruin the trampoline park by reducing all S_i to 1. What is the minimum number of passes she needs to reduce all S_i to 1?
Input
The first line contains a single integer t (1 β€ t β€ 500) β the number of test cases.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the number of trampolines.
The second line of each test case contains n integers S_1, S_2, ..., S_n (1 β€ S_i β€ 10^9), where S_i is the strength of the i-th trampoline.
It's guaranteed that the sum of n over all test cases doesn't exceed 5000.
Output
For each test case, output a single integer β the minimum number of passes Pekora needs to do to reduce all S_i to 1.
Example
Input
3
7
1 4 2 2 2 2 2
2
2 3
5
1 1 1 1 1
Output
4
3
0
Note
For the first test case, here is an optimal series of passes Pekora can take. (The bolded numbers are the positions that Pekora jumps into during these passes.)
* [1,4,2,2,2,2,2]
* [1,4,1,2,1,2,1]
* [1,3,1,2,1,1,1]
* [1,2,1,2,1,1,1]
For the second test case, the optimal series of passes is show below.
* [2,3]
* [1,3]
* [1,2]
For the third test case, all S_i are already equal to 1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def getvalue(n,seg_el):
i=n+seg_el
ANS=0
ANS=SEG[i]
i>>=1
while i!=0:
ANS+=SEG[i]
i>>=1
return ANS
def updates(l,r,x):
L=l+seg_el
R=r+seg_el
while L<R:
if L & 1:
SEG[L]+=x
L+=1
if R & 1:
R-=1
SEG[R]+=x
L>>=1
R>>=1
t=int(input())
for tests in range(t):
n=int(input())
S=list(map(int,input().split()))
seg_el=1<<((n+1).bit_length())
SEG=[0]*(2*seg_el)
for i in range(n):
SEG[i+seg_el]=S[i]
ANS=0
for i in range(n):
#print([getvalue(k,seg_el) for k in range(n)])
x=getvalue(i,seg_el)
if x>=2:
ANS+=x-1
if i+2<n:
updates(i+2,min(i+2+S[i]-1,n),-1)
print(ANS)
``` | instruction | 0 | 37,943 | 8 | 75,886 |
No | output | 1 | 37,943 | 8 | 75,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a trampoline park with n trampolines in a line. The i-th of which has strength S_i.
Pekora can jump on trampolines in multiple passes. She starts the pass by jumping on any trampoline of her choice.
If at the moment Pekora jumps on trampoline i, the trampoline will launch her to position i + S_i, and S_i will become equal to max(S_i-1,1). In other words, S_i will decrease by 1, except of the case S_i=1, when S_i will remain equal to 1.
If there is no trampoline in position i + S_i, then this pass is over. Otherwise, Pekora will continue the pass by jumping from the trampoline at position i + S_i by the same rule as above.
Pekora can't stop jumping during the pass until she lands at the position larger than n (in which there is no trampoline). Poor Pekora!
Pekora is a naughty rabbit and wants to ruin the trampoline park by reducing all S_i to 1. What is the minimum number of passes she needs to reduce all S_i to 1?
Input
The first line contains a single integer t (1 β€ t β€ 500) β the number of test cases.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the number of trampolines.
The second line of each test case contains n integers S_1, S_2, ..., S_n (1 β€ S_i β€ 10^9), where S_i is the strength of the i-th trampoline.
It's guaranteed that the sum of n over all test cases doesn't exceed 5000.
Output
For each test case, output a single integer β the minimum number of passes Pekora needs to do to reduce all S_i to 1.
Example
Input
3
7
1 4 2 2 2 2 2
2
2 3
5
1 1 1 1 1
Output
4
3
0
Note
For the first test case, here is an optimal series of passes Pekora can take. (The bolded numbers are the positions that Pekora jumps into during these passes.)
* [1,4,2,2,2,2,2]
* [1,4,1,2,1,2,1]
* [1,3,1,2,1,1,1]
* [1,2,1,2,1,1,1]
For the second test case, the optimal series of passes is show below.
* [2,3]
* [1,3]
* [1,2]
For the third test case, all S_i are already equal to 1.
Submitted Solution:
```
#######################################################################################################################
# Author: BlackFyre
# Language: PyPy 3.7
#######################################################################################################################
from sys import stdin, stdout, setrecursionlimit
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log, log2
from random import seed, randint
from datetime import datetime
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
from collections import defaultdict as dd
mod = pow(10, 9) + 7
mod2 = 998244353
# setrecursionlimit(3000)
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var) + "\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x % y else 0
def ceil(a, b): return (a + b - 1) // b
def def_value(): return 0
def def_inf(): return inf
for _ in range(iinp()):
n = iinp()
a = lmp()
c = l1d(n+1)
ans=0
for i in range(n):
t = c[i]
if t<a[i]-1:
ans += (a[i]-t-1)
t = (a[i]-1)
c[i] = t-a[i]+1
for j in range(i+2,min(n,i+a[i]+1)):
c[j]+=1
print(ans)
``` | instruction | 0 | 37,944 | 8 | 75,888 |
No | output | 1 | 37,944 | 8 | 75,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
The University where A and B study is a set of rooms connected by corridors. Overall, the University has n rooms connected by n - 1 corridors so that you can get from any room to any other one by moving along the corridors. The rooms are numbered from 1 to n.
Every day Π and B write contests in some rooms of their university, and after each contest they gather together in the same room and discuss problems. A and B want the distance from the rooms where problems are discussed to the rooms where contests are written to be equal. The distance between two rooms is the number of edges on the shortest path between them.
As they write contests in new rooms every day, they asked you to help them find the number of possible rooms to discuss problems for each of the following m days.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of rooms in the University.
The next n - 1 lines describe the corridors. The i-th of these lines (1 β€ i β€ n - 1) contains two integers ai and bi (1 β€ ai, bi β€ n), showing that the i-th corridor connects rooms ai and bi.
The next line contains integer m (1 β€ m β€ 105) β the number of queries.
Next m lines describe the queries. The j-th of these lines (1 β€ j β€ m) contains two integers xj and yj (1 β€ xj, yj β€ n) that means that on the j-th day A will write the contest in the room xj, B will write in the room yj.
Output
In the i-th (1 β€ i β€ m) line print the number of rooms that are equidistant from the rooms where A and B write contest on the i-th day.
Examples
Input
4
1 2
1 3
2 4
1
2 3
Output
1
Input
4
1 2
2 3
2 4
2
1 2
1 3
Output
0
2
Note
in the first sample there is only one room at the same distance from rooms number 2 and 3 β room number 1.
Submitted Solution:
```
class Problem:
def __init__(self, ribs):
self.paths = {}
for r in ribs:
c1, c2 = r
self.paths[c1] = self.paths.get(c1, []) + [c2]
self.paths[c2] = self.paths.get(c2, []) + [c1]
def get_successors(self, cab):
return self.paths.get(cab, [])
def search(problem, start_node, prev_res=None):
closed = set()
fringe = [(start_node, 0)]
res = {}
while fringe:
node = fringe.pop()
if node[0] not in closed:
closed.add(node[0])
for successor in problem.get_successors(node[0]):
fringe.append((successor, node[1] + 1))
if res.get(successor, 99999999) > node[1] + 1 and successor != start_node:
res[successor] = node[1] + 1
if prev_res is not None:
counter = sum([1 if (prev_res.get(k) == res[k]) else 0 for k in res])
print(counter)
return res
def solution(ribs, cab_a, cab_b):
problem = Problem(ribs)
search(problem, cab_b, search(problem, cab_a))
def main():
cabinets_count = int(input().strip())
if 1 == cabinets_count:
print(1)
exit(0)
ribs = [tuple(map(int, input().strip().split())) for _ in range(cabinets_count - 1)]
days_count = int(input().strip())
days = [tuple(map(int, input().strip().split())) for _ in range(days_count)]
for day in days:
solution(ribs, day[0], day[1])
main()
``` | instruction | 0 | 38,083 | 8 | 76,166 |
No | output | 1 | 38,083 | 8 | 76,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked. | instruction | 0 | 38,207 | 8 | 76,414 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
import sys
from collections import deque
def debug(x, table):
for name, val in table.items():
if x is val:
print('DEBUG:{} -> {}'.format(name, val), file=sys.stderr)
return None
def solve():
n, m = map(int, sys.stdin.readline().split())
r = [int(i) for i in sys.stdin.readline().split()]
es = [[] for i in range(n)]
for i in range(m):
line = [int(j) for j in sys.stdin.readline().split()]
for u in line[1:]:
es[u - 1].append(i)
Adj = [[] for i in range(m)]
for u, v in es:
Adj[u].append(v)
Adj[v].append(u)
edges = dict()
for i, e in enumerate(es):
e.sort()
if tuple(e) not in edges:
edges[tuple(e)] = r[i]
elif edges[tuple(e)] != r[i]:
print('NO')
return None
else:
pass
cols = [None] * m
for u in range(m):
if cols[u] is None:
if not bfs(Adj, edges, cols, u):
print('NO')
return None
else:
pass
print('YES')
def bfs(Adj, edges, cols, u):
nxts = deque([u])
cols[u] = 0
while nxts:
v = nxts.popleft()
for w in Adj[v]:
ed = tuple(sorted([v, w]))
if cols[w] is None:
if edges[ed] == 1:
cols[w] = cols[v]
else:
cols[w] = 1 - cols[v]
nxts.append(w)
else:
if edges[ed] == 1 and cols[w] != cols[v]:
return False
elif edges[ed] == 0 and cols[w] == cols[v]:
return False
return True
if __name__ == '__main__':
solve()
``` | output | 1 | 38,207 | 8 | 76,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked. | instruction | 0 | 38,208 | 8 | 76,416 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import ceil
def prod(a, mod=10 ** 9 + 7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def gcd(x, y):
while y:
x, y = y, x % y
return x
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def find_SCC(graph):
SCC, S, P = [], [], []
depth = [0] * len(graph)
stack = list(range(len(graph)))
while stack:
node = stack.pop()
if node < 0:
d = depth[~node] - 1
if P[-1] > d:
SCC.append(S[d:])
del S[d:], P[-1]
for node in SCC[-1]:
depth[node] = -1
elif depth[node] > 0:
while P[-1] > depth[node]:
P.pop()
elif depth[node] == 0:
S.append(node)
P.append(len(S))
depth[node] = len(S)
stack.append(~node)
stack += graph[node]
return SCC[::-1]
for _ in range(int(input()) if not True else 1):
#n = int(input())
n, m = map(int, input().split())
# a, b = map(int, input().split())
# c, d = map(int, input().split())
r = list(map(int, input().split()))
# b = list(map(int, input().split()))
# s = input()
doors = [[] for __ in range(n + 1)]
for i in range(m):
a = list(map(int, input().split()))
for j in range(1, len(a)):
doors[a[j]] += [i + 1]
graph = [[] for __ in range(2 * m + 1)]
def add_edge(x, y):
x2 = (x - m) if x > m else x + m
y2 = (y - m) if y > m else y + m
graph[x2] += [y]
graph[y2] += [x]
for i in range(1, n + 1):
x, y = doors[i]
x2 = (x - m) if x > m else x + m
y2 = (y - m) if y > m else y + m
if r[i-1]:
# maxterm - (!a + b) * (a + !b)
add_edge(x2, y)
add_edge(x, y2)
else:
# maxterm - (!a + !b) * (a + b)
add_edge(x2, y2)
add_edge(x, y)
scc = find_SCC(graph)
cx = [-1] * (2 * m + 1)
pos = True
for i in range(len(scc)):
for j in scc[i]:
cx[j] = i
for i in range(1, m + 1):
if cx[i] == cx[i + m]:
pos = False
break
print("YES" if pos else "NO")
``` | output | 1 | 38,208 | 8 | 76,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked. | instruction | 0 | 38,209 | 8 | 76,418 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
#!/usr/bin/env python3
def _vertex(lit):
if lit > 0:
return 2 * (lit - 1)
else:
return 2 * (-lit - 1) + 1
def tarjan(graph):
n = len(graph)
dfs_num = [None] * n
dfs_min = [n] * n
waiting = []
waits = [False] * n
sccp = []
dfs_time = 0
times_seen = [-1] * n
for start in range(n):
if times_seen[start] == -1:
times_seen[start] = 0
to_visit = [start]
while to_visit:
node = to_visit[-1]
if times_seen[node] == 0:
dfs_num[node] = dfs_time
dfs_min[node] = dfs_time
dfs_time += 1
waiting.append(node)
waits[node] = True
children = graph[node]
if times_seen[node] == len(children):
to_visit.pop()
dfs_min[node] = dfs_num[node]
for child in children:
if waits[child] and dfs_min[child] < dfs_min[node]:
dfs_min[node] = dfs_min[child]
if dfs_min[node] == dfs_num[node]:
component = []
while True:
u = waiting.pop()
waits[u] = False
component.append(u)
if u == node:
break
sccp.append(component)
else:
child = children[times_seen[node]]
times_seen[node] += 1
if times_seen[child] == -1:
times_seen[child] = 0
to_visit.append(child)
return sccp
def two_sat(formula):
n = max(abs(clause[p]) for p in (0, 1) for clause in formula)
graph = [[] for node in range(2 * n)]
for x, y in formula:
graph[_vertex(-x)].append(_vertex(y))
graph[_vertex(-y)].append(_vertex(x))
sccp = tarjan(graph)
comp_id = [None] * (2 * n)
#assignment = [None] * (2 * n)
for component in sccp:
rep = min(component)
for vtx in component:
comp_id[vtx] = rep
#if assignment[vtx] is None:
# assignment[vtx] = True
# assignment[vtx ^ 1] = False
for i in range(n):
if comp_id[2 * i] == comp_id[2 * i + 1]:
return "NO"
return "YES"
#return assignment[::2]
n, m = [int(x) for x in input().split()]
doors_status = [int(x) for x in input().split()]
switches = [list(map(int, input().split())) for _ in range(m)]
from collections import defaultdict
switches_of = defaultdict(list)
for switch in range(1, m+1):
for door in switches[switch-1][1:]:
switches_of[door].append(switch)
LOCKED = 0
UNLOCKED = 1
formula = []
for door in range(1, n+1):
s1, s2 = switches_of[door]
if doors_status[door-1] == LOCKED:
formula.append((s1, s2))
formula.append((-s1, -s2))
else:
formula.append((s1, -s2))
formula.append((-s1, s2))
print(two_sat(formula))
``` | output | 1 | 38,209 | 8 | 76,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked. | instruction | 0 | 38,210 | 8 | 76,420 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
import sys
import collections
n, m = map(int, input().split())
r = tuple(map(int, input().split()))
controls = [tuple(map(int, input().split()))[1:] for i in range(m)]
class DSU:
def __init__(self):
self.parent = None
self.has_zero = False
self.has_one = False
self.size = 1
self.doors = []
def get_root(self):
if self.parent is None:
return self
self.parent = self.parent.get_root()
return self.parent
def unite(self, s):
r1 = self.get_root()
r2 = s.get_root()
if r1 is r2:
return r1
if r1.size < r2.size:
r1, r2 = r2, r1
r2.parent = r1
r1.size += r2.size
r1.has_zero = r1.has_zero or r2.has_zero
r1.has_one = r1.has_one or r2.has_one
return r1
door_dsus = [[] for i in range(n)]
for doors in controls:
n = DSU()
for door in doors:
n.doors.append(door - 1)
door_dsus[door - 1].append(n)
if r[door - 1]:
n.has_one = True
if not r[door - 1]:
n.has_zero = True
for door, is_open in enumerate(r):
n1, n2 = door_dsus[door]
if is_open:
n1.unite(n2)
G = {}
for door, is_open in enumerate(r):
if is_open:
continue
n1, n2 = door_dsus[door]
if n1.get_root() is n2.get_root():
print("NO")
sys.exit(0)
G.setdefault(n1.get_root(), set()).add(n2.get_root())
G.setdefault(n2.get_root(), set()).add(n1.get_root())
color = {}
for v in G.keys():
if v in color:
continue
color[v] = False
q = collections.deque([v])
while q:
v = q.popleft()
c = color[v]
for adj_v in G[v]:
if adj_v in color:
if color[adj_v] != (not c):
print("NO")
sys.exit(0)
else:
color[adj_v] = not c
q.append(adj_v)
print("YES")
``` | output | 1 | 38,210 | 8 | 76,421 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked. | instruction | 0 | 38,211 | 8 | 76,422 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
#!/usr/bin/env python3
from collections import deque
def bfs(adj, edge, col, u):
nxt = deque([u])
col[u] = 0
while nxt:
v = nxt.popleft()
for w in adj[v]:
a = [w,v]
a.sort()
a = tuple(a)
if col[w] is None:
col[w] = col[v]^edge[a]
nxt.append(w)
else:
if col[w]^col[v] != edge[a]:
return -1
return 0
def ri():
return map(int, input().split())
n, m = ri()
e = [1-i for i in ri()]
r = [[] for i in range(n)]
for i in range(m):
s = list(ri())
for j in range(1,len(s)):
r[s[j]-1].append(i)
edge = dict()
adj = [[] for i in range(m)]
for i in range(n):
a = r[i]
a.sort()
a = tuple(a)
if not a in edge:
edge[a] = e[i]
a1, a2 = a
adj[a1].append(a2)
adj[a2].append(a1)
else:
if edge[a] != e[i]:
print("No")
exit()
col = [None] * m
for u in range(m):
if col[u] is None:
if bfs(adj, edge, col, u):
print("No")
exit()
else:
pass
print("Yes")
``` | output | 1 | 38,211 | 8 | 76,423 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked. | instruction | 0 | 38,212 | 8 | 76,424 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
#####################################
import atexit, io, sys, collections, math, heapq, fractions,copy, os, functools
import sys
import random
import collections
from io import BytesIO, IOBase
##################################### python 3 START
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##################################### python 3 END
n,m = map(int, input().split())
ais = [0] + list(map(int, input().split()))
doors = collections.defaultdict(list)
for i in range(m):
#switch i+1
dis = list(map(int, input().split()))[1:]
for di in dis: doors[di].append(i+1)
adj = collections.defaultdict(set)
for key in doors:
u,v = doors[key]
w = 1-ais[key]
adj[u].add((v,w))
adj[v].add((u,w))
b = True
vis = {}
for cur in range(1, m+1):
if cur not in vis:
q = collections.deque([cur])
vis[cur] = 0
while(q):
u= q.popleft()
for v,w in adj[u]:
if v not in vis:
vis[v] = (1- vis[u]) if w == 1 else vis[u]
q.append(v)
elif v in vis:
if w == 0:
if vis[v] != vis[u]:
b = False
if w == 1:
if vis[v] == vis[u]:
b = False
print ('YES' if b else 'NO')
'''
3 3
1 0 1
1: 1 2 3
2: 2
3: 1 3
1 door1 : s1 ,s3
0 door2 : s2, s1
1 door3 : s3, s1
1
si ------- sj
2
di 1
si[0] si[1] 0 1
1 0
di 0
si[0] si[1] 0 0
0 1
'''
``` | output | 1 | 38,212 | 8 | 76,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked. | instruction | 0 | 38,213 | 8 | 76,426 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
import sys
n, m = map(int, input().split())
door = list(map(int, input().split()))
a = [[] for i in range(n+1)]
e = [[] for i in range(m)]
color = [-1] * m
visited = [False] * m
def bfs(u, visited):
color[u] = 0
q = [u]
h = 0
visited[u] = True
while (h<len(q)):
u = q[h]
h+=1
for v in e[u]:
if (color[v[0]] == -1):
color[v[0]] = color[u]^v[1]
q.append(v[0])
visited[v[0]] = True
else:
if color[v[0]] != color[u]^v[1]: return False
return True
for i in range(m):
b = list(map(int, input().split()))
for j in range(1, len(b)):
k = b[j]
a[k].append(i)
if len(a[k])==2:
e[a[k][0]].append((a[k][1],1-door[k-1]))
e[a[k][1]].append((a[k][0],1-door[k-1]))
flag = True
for i in range(m):
if color[i] == -1:
if bfs(i, visited) == False:
flag = False
break
if flag: print("YES")
else: print("NO")
``` | output | 1 | 38,213 | 8 | 76,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked. | instruction | 0 | 38,214 | 8 | 76,428 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
import sys
from math import inf
time = 0
cc = 0
l = {}
d = {}
f = {}
conn_comp = {}
on_stack = {}
stack = []
color = []
def tarjan(graph):
global l
global d
global f
global pi
global stack
global on_stack
l = {key : inf for key in graph}
d = {key : -1 for key in graph}
f = {key : -1 for key in graph}
conn_comp = {key : -1 for key in graph}
on_stack = {key : False for key in graph}
for i in graph.keys():
if d[i] == -1:
strongconnect(graph, i)
def strongconnect(graph, v):
global time
global cc
stack.append(v)
on_stack[v] = True
time += 1
d[v] = time
for i in graph[v]:
if d[i] == -1:
strongconnect(graph, i)
l[v] = min(l[v], l[i])
if on_stack[i]:
l[v] = min(l[v], d[i])
if l[v] == d[v]:
cc += 1
w = stack.pop()
while w != v:
conn_comp[w] = cc
on_stack[w] = False
w = stack.pop()
conn_comp[v] = cc
on_stack[v] = False
def read():
n,m = map(int, sys.stdin.readline().split())
status = list(map(int, sys.stdin.readline().split()))
doors_switch = [[] for i in range(n)]
for i in range(m):
temp = list(map(int, sys.stdin.readline().split()))
for j in range (1,len(temp)):
door = temp[j]
doors_switch[door-1].append(i)
return m, status, doors_switch
def build_graph_scc():
m, status, doors_switch = read()
graph = {i : set() for i in range(2*m)}
for i in range (len(doors_switch)):
switch_1, switch_2 = tuple(doors_switch[i])
if status[i]:
graph[2*switch_1].add(2*switch_2)
graph[2*switch_2].add(2*switch_1)
graph[2*switch_1 + 1].add(2*switch_2 + 1)
graph[2*switch_2 + 1].add(2*switch_1 + 1)
else:
graph[2*switch_1].add(2*switch_2 + 1)
graph[2*switch_2].add(2*switch_1 + 1)
graph[2*switch_1 + 1].add(2*switch_2)
graph[2*switch_2 + 1].add(2*switch_1)
return graph
def build_graph_bfs():
m, status, doors_switch = read()
g = [[] for i in range(m)]
global color
color = [-1] * m
for i in range(len(status)):
switch_1, switch_2 = tuple(doors_switch[i])
g[switch_1].append((switch_2, 1 - status[i]))
g[switch_2].append((switch_1, 1 - status[i]))
return g
def bfs_bipartite(graph, v):
color[v] = 0
q = [v]
j = 0
while j < len(q):
v = q[j]
for w,st in graph[v]:
if color[w] == -1:
color[w] = color[v]^st
q.append(w)
else:
if color[w] != color[v]^st:
return False
j+=1
return True
def main():
# graph = build_graph_scc()
# tarjan(graph)
# for i in range(0, len(conn_comp), 2):
# if conn_comp[i] == conn_comp[i+1]:
# print("NO")
# return
# print("YES")
graph = build_graph_bfs()
for i in range(len(graph)):
if color[i] == -1 and not bfs_bipartite(graph, i):
print("NO")
return
print("YES")
main()
``` | output | 1 | 38,214 | 8 | 76,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked.
Submitted Solution:
```
from collections import defaultdict, deque
n, m = list(map(int, input().split()))
state = list(map(int, input().split()))
graph = defaultdict(list)
connected = [list() for _ in range(n + 1)] #Each node connected to which 2 edges
for i in range(m):
for j in list(map(int, input().split()))[1:]:
connected[j].append(i)
for index, i in enumerate(connected[1:]):
graph[i[0]].append((i[1], state[index]))
graph[i[1]].append((i[0], state[index]))
switch_color = [False] * m
switch_visited = [False] * m
# print(connected)
# print(graph)
queue = deque([])
yes = True
for nodes in graph:
if yes and (not switch_visited[nodes]):
# queue = deque([nodes])
# yes = True
queue.append(nodes)
while queue and yes:
now = queue.popleft()
switch_visited[now] = True
for i in graph[now]:
if not switch_visited[i[0]]:
switch_visited[i[0]] = True
queue.append(i[0])
if i[1] == 1:
switch_color[i[0]] = switch_color[now]
else:
switch_color[i[0]] = (not switch_color[now])
else:
if (i[1] == 1) and (switch_color[i[0]] != switch_color[now]):
print("NO")
yes = False
break
elif i[1] == 0 and switch_color[i[0]] == switch_color[now]:
print("NO")
yes = False
break
if yes:
print("YES")
# print(queue)
# print(switch_visited)
# print(switch_color)
``` | instruction | 0 | 38,215 | 8 | 76,430 |
Yes | output | 1 | 38,215 | 8 | 76,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked.
Submitted Solution:
```
import sys
from collections import deque
def debug(x, table):
for name, val in table.items():
if x is val:
print('DEBUG:{} -> {}'.format(name, val), file=sys.stderr)
return None
def solve():
n, m = map(int, sys.stdin.readline().split())
r = [1 - int(i) for i in sys.stdin.readline().split()]
es = [[] for i in range(n)]
for i in range(m):
line = [int(j) for j in sys.stdin.readline().split()]
for u in line[1:]:
es[u - 1].append(i)
Adj = [[] for i in range(m)]
for u, v in es:
Adj[u].append(v)
Adj[v].append(u)
edges = dict()
for i, e in enumerate(es):
e.sort()
if tuple(e) not in edges:
edges[tuple(e)] = r[i]
elif edges[tuple(e)] != r[i]:
print('NO')
return None
else:
pass
cols = [None] * m
for u in range(m):
if cols[u] is None:
if not bfs(Adj, edges, cols, u):
print('NO')
return None
else:
pass
print('YES')
def bfs(Adj, edges, cols, u):
nxts = deque([u])
cols[u] = 0
while nxts:
v = nxts.popleft()
for w in Adj[v]:
ed = tuple(sorted([v, w]))
if cols[w] is None:
cols[w] = cols[v] ^ edges[ed]
nxts.append(w)
else:
if cols[w] ^ cols[v] != edges[ed]:
return False
return True
if __name__ == '__main__':
solve()
``` | instruction | 0 | 38,216 | 8 | 76,432 |
Yes | output | 1 | 38,216 | 8 | 76,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked.
Submitted Solution:
```
from collections import defaultdict, deque
n, m = list(map(int, input().split()))
state = list(map(int, input().split()))
graph = defaultdict(list)
connected = [list() for _ in range(n + 1)] #Each node connected to which 2 edges
for i in range(m):
for j in list(map(int, input().split()))[1:]:
connected[j].append(i)
for index, i in enumerate(connected[1:]):
graph[i[0]].append((i[1], state[index]))
graph[i[1]].append((i[0], state[index]))
switch_color = [False] * m
switch_visited = [False] * m
# print(connected)
# print(graph)
queue = deque([])
yes = True
for nodes in graph:
if yes and (not switch_visited[nodes]):
# queue = deque([nodes])
# yes = True
queue.append(nodes)
while queue and yes:
now = queue.popleft()
switch_visited[now] = True
for i in graph[now]:
if not switch_visited[i[0]]:
switch_visited[i[0]] = True
queue.append(i[0])
if i[1] == 1:
switch_color[i[0]] = switch_color[now]
else:
switch_color[i[0]] = (not switch_color[now])
else:
if (i[1] == 1) and (switch_color[i[0]] != switch_color[now]):
print("NO")
yes = False
break
elif i[1] == 0 and switch_color[i[0]] == switch_color[now]:
print("NO")
yes = False
break
if yes:
print("YES")
``` | instruction | 0 | 38,217 | 8 | 76,434 |
Yes | output | 1 | 38,217 | 8 | 76,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked.
Submitted Solution:
```
def Console():
temp = input().split()
n = int((temp[0]))
m = int((temp[1]))
temp = input().split()
doors = []
check = []
i = 0
j = 0
while i < n:
doors.append([])
i+= 1
i = 0
while i < m:
temp1 = input().split()
j = 1
while j <= int(temp1[0]):
doors[int(temp1[j])-1].append(i)
check.append(temp1)
j += 1
i += 1
i = 0
j = 0
status = m*[False]
switches = {}
for i in range(m):
switches[i]= []
for i in range(len(doors)):
for j in range(len(doors[i])):
switches[doors[i][j]].append([doors[i][j-1] if(j > 0) else doors[i][j + 1], bool(int(temp[i]))])
if DoorProblemHasSol(switches, status):
print("YES")
else:
print("NO")
def DoorProblemHasSol(switches, status):
visited = len(switches)*[0];
for i in range(len(switches)):
if not visited[i]:
if not Auxiliar(i, switches, status, visited):
return False
return True
def Auxiliar(vertexU, switches, status, visited):
visited[vertexU] = True;
for i in range(len(switches[vertexU])):
vertexV = switches[vertexU][i]
if not visited[vertexV[0]]:
if (vertexV[1] and status[vertexU]) or (not vertexV[1] and not status[vertexU]):
status[vertexV[0]] = True
if not Auxiliar(vertexV[0], switches, status, visited):
return False
else:
if ((vertexV[1] and
not ((status[vertexU] and status[0]) or (not status[vertexU] and not status[0]))
or (not vertexV[1] and
not ((status[vertexU] and not status[0]) or (not status[vertexU] and status[0]))))):
return False
return True
def Cheking(check, doors, sol):
i = 0
while i < len(check):
j = 0
if sol[i]:
while j < len(check[i]):
doors[int(check[i][j])] = not doors[int(check[i][j])]
j += 1
i += 1
for door in doors:
if not door:
return False
return True
def Generator():
n = random.randrange(2, 10**5 + 1)
m = random.randrange(2, 10**5 + 1)
i = 0
check = []
while i < m:
j = 0
``` | instruction | 0 | 38,218 | 8 | 76,436 |
No | output | 1 | 38,218 | 8 | 76,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked.
Submitted Solution:
```
n, m = map(int, input().split())
r = tuple(map(int, input().split()))
controls = [tuple(map(int, input().split()))[1:] for i in range(m)]
class DSU:
def __init__(self):
self.parent = None
self.has_zero = False
self.has_one = False
self.size = 1
self.doors = []
def get_root(self):
if self.parent is None:
return self
self.parent = self.parent.get_root()
return self.parent
def unite(self, s):
r1 = self.get_root()
r2 = s.get_root()
if r1 is r2:
return r1
if r1.size < r2.size:
r1, r2 = r2, r1
r2.parent = r1
r1.size += r2.size
r1.has_zero = r1.has_zero or r2.has_zero
r1.has_one = r1.has_one or r2.has_one
return r1
door_dsus = [[] for i in range(n)]
for doors in controls:
n = DSU()
for door in doors:
n.doors.append(door - 1)
door_dsus[door - 1].append(n)
if r[door - 1]:
n.has_one = True
if not r[door - 1]:
n.has_zero = True
for door, is_open in enumerate(r):
n1, n2 = door_dsus[door]
if is_open:
n1.unite(n2)
for door, is_open in enumerate(r):
n1, n2 = door_dsus[door]
if not is_open:
if n1.get_root().has_one and n2.get_root().has_one:
print("NO")
break
else:
print("YES")
``` | instruction | 0 | 38,219 | 8 | 76,438 |
No | output | 1 | 38,219 | 8 | 76,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked.
Submitted Solution:
```
import sys
from collections import deque
def debug(x, table):
for name, val in table.items():
if x is val:
print('DEBUG:{} -> {}'.format(name, val), file=sys.stderr)
return None
def solve():
n, m = map(int, sys.stdin.readline().split())
r = [int(i) for i in sys.stdin.readline().split()]
es = [[] for i in range(n)]
for i in range(m):
line = [int(j) for j in sys.stdin.readline().split()]
for u in line[1:]:
es[u - 1].append(i)
Adj = [[] for i in range(m)]
for u, v in es:
Adj[u].append(v)
Adj[v].append(u)
# debug(Adj, locals())
edges = dict()
for i, e in enumerate(es):
e.sort()
if tuple(e) not in edges:
edges[tuple(e)] = r[i]
elif edges[tuple(e)] != r[i]:
print('NO')
return None
else:
pass
# debug(edges, locals())
cols = [None] * m
for u in range(m):
if cols[u] is None:
if bfs(Adj, edges, cols, u):
print('NO')
return None
else:
pass
print('YES')
def bfs(Adj, edges, cols, u):
nxts = deque([u])
cols[u] = 0
while nxts:
v = nxts.popleft()
for w in Adj[v]:
ed = tuple(sorted([v, w]))
if cols[w] is None:
if edges[ed] == 1:
cols[w] = cols[v]
else:
cols[w] = 1 - cols[v]
nxts.append(w)
else:
if edges[ed] == 1 and cols[w] != cols[v]:
return False
elif edges[ed] == 0 and cols[w] == cols[v]:
return False
return True
if __name__ == '__main__':
solve()
``` | instruction | 0 | 38,220 | 8 | 76,440 |
No | output | 1 | 38,220 | 8 | 76,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.
You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.
You need to tell Sherlock, if there exists a way to unlock all doors at the same time.
Input
First line of input contains two integers n and m (2 β€ n β€ 105, 2 β€ m β€ 105) β the number of rooms and the number of switches.
Next line contains n space-separated integers r1, r2, ..., rn (0 β€ ri β€ 1) which tell the status of room doors. The i-th room is locked if ri = 0, otherwise it is unlocked.
The i-th of next m lines contains an integer xi (0 β€ xi β€ n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1 to n. It is guaranteed that each door is controlled by exactly two switches.
Output
Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.
Examples
Input
3 3
1 0 1
2 1 3
2 1 2
2 2 3
Output
NO
Input
3 3
1 0 1
3 1 2 3
1 2
2 1 3
Output
YES
Input
3 3
1 0 1
3 1 2 3
2 1 2
1 3
Output
NO
Note
In the second example input, the initial statuses of the doors are [1, 0, 1] (0 means locked, 1 β unlocked).
After toggling switch 3, we get [0, 0, 0] that means all doors are locked.
Then, after toggling switch 1, we get [1, 1, 1] that means all doors are unlocked.
It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked.
Submitted Solution:
```
class TwoSat(object):
def __init__(self,n):
self.n = n
self.G = {}
self.mark = [0] * 400010
self.S = [-1] * 400010
self.c = 0
for i in range(n*2):
self.G[i] = []
def dfs(self, x):
if (self.mark[x ^ 1]): return False
if (self.mark[x]): return True
self.mark[x] = True
self.c = self.c + 1
self.S[self.c] = x
for i in range(len(self.G[x])):
if (not self.dfs(self.G[x][i])): return False
return True
def add_clause(self, x, xval, y, yval):
x = x * 2 + xval
y = y * 2 + yval
self.G[x ^ 1].append(y)
self.G[y ^ 1].append(x)
def solve(self):
for i in range(0, self.n*2, 2):
if not self.mark[i] and not self.mark[i+1]:
self.c = 0
if not self.dfs(i):
while self.c > 0:
self.c -= 1
self.mark[self.S[self.c]] = False
if not self.dfs(i+1): return False
return True
if __name__ == '__main__':
g = {}
# n,m = (3,3)
n, m = list(map(int, input().split()))
# a = [1,0,1]
a = list(map(int, input().split()))
# data = [[1, 3], [1, 2], [2, 3]]
# g[0] = [0,1]
# g[1] = [0,1]
# g[2] = [0,2]
for i in range(n):
g[i] = []
for i in range(m):
temp = list(map(int, input().split()))
x = temp[0]
for j in range(x):
u = temp[j+1] - 1
g[u].append(i)
solver = TwoSat(m)
for i in range(n):
u = g[i][0]
v = g[i][1]
solver.add_clause(u, 0, v, 1 if a[1] == 1 else 0)
solver.add_clause(u, 1, v, 0 if a[1] == 1 else 1)
if solver.solve() == 1 : print("YES\n")
else: print("NO\n")
``` | instruction | 0 | 38,221 | 8 | 76,442 |
No | output | 1 | 38,221 | 8 | 76,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
AOR Ika got a water tank with a size of $ 1 $ in length and $ N $ in width. The aquarium is tall enough to hold water. The aquarium has $ N-1 $ partitions and is evenly spaced into $ N $ compartments. When water was poured here, the height of the water in each section became $ a_i $.
AOR Ika decided to remove some dividers and reduce the number of compartments to $ M $ or less. When you have removed the dividers, find the maximum sum of the water heights in each compartment.
The thickness of the partition can be ignored.
input
$ N \ M $
$ a_1 \ cdots a_N $
output
Output the maximum value of the total water height of each section in one line. Also, output a line break at the end. It is acceptable if the relative or absolute error is less than or equal to $ 10 ^ {-6} $.
Example
Input
5 3
9 1 2 3 9
Output
20.000000
Submitted Solution:
```
N,M = map(int,input().split())
src = list(map(int,input().split()))
INF = float('inf')
mem = [(src[i], 1) for i in range(N)]
for i in range(N-M):
minh = INF
mini = -1
minl = -1
for i,((h1,l1),(h2,l2)) in enumerate(zip(mem, mem[1:])):
w = (h1*l1 + h2*l2)
h = w / (l1+l2)
if h < minh:
minh = h
mini = i
minl = l1+l2
del mem[mini]
mem[mini] = (minh,minl)
#print(mem)
print(sum([h for h,l in mem]))
``` | instruction | 0 | 38,569 | 8 | 77,138 |
No | output | 1 | 38,569 | 8 | 77,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator. | instruction | 0 | 38,634 | 8 | 77,268 |
Tags: implementation
Correct Solution:
```
(x, y, z, t1, t2, t3) = map(int, input().split())
a1 = abs(x - y) * t1
a2 = abs(x - z) * t2 + abs(x - y) * t2 + t3 * 3
if (a2 <= a1):
print("YES")
else:
print("NO")
``` | output | 1 | 38,634 | 8 | 77,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator. | instruction | 0 | 38,635 | 8 | 77,270 |
Tags: implementation
Correct Solution:
```
x,y,z,t1,t2,t3 = map(int, input().rstrip().split())
distance = abs(x-y)
time_walk = distance*t1
distance_lyft = abs(z-x) + abs(x-y)
time_lyft = (distance_lyft*t2) + 3*t3
if(time_lyft>time_walk):
print('NO')
else:
print('YES')
``` | output | 1 | 38,635 | 8 | 77,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator. | instruction | 0 | 38,636 | 8 | 77,272 |
Tags: implementation
Correct Solution:
```
from sys import stdin
x,y,z,t1,t2,t3=map(int,stdin.readline().strip().split())
a=(abs(x-z)*t2)+(t3*3)+(abs(x-y)*t2)
e=(abs(x-y)*t1)
if a<=e:
print("YES")
else:
print("NO")
``` | output | 1 | 38,636 | 8 | 77,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator. | instruction | 0 | 38,637 | 8 | 77,274 |
Tags: implementation
Correct Solution:
```
x, y, z, t1, t2, t3 = map(int, input().split())
stairs = abs(y - x) * t1
movement = abs(z - x) + abs(x - y)
elevator = (movement * t2) + t3 * 3
if stairs >= elevator:
print('YES')
else:
print('NO')
``` | output | 1 | 38,637 | 8 | 77,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator. | instruction | 0 | 38,638 | 8 | 77,276 |
Tags: implementation
Correct Solution:
```
x, y, z, t1, t2, t3 = map(int, input().split())
ladder = abs(x - y) * t1
elevator = abs(x - z) * t2 + 3 * t3 + abs(x - y) * t2
if elevator > ladder:
print("NO")
else:
print("YES")
``` | output | 1 | 38,638 | 8 | 77,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator. | instruction | 0 | 38,639 | 8 | 77,278 |
Tags: implementation
Correct Solution:
```
(x , y, z, t1, t2, t3) = map(int, input().split())
if abs(z-x) * t2 + t3 + t3 + abs(x-y) * t2 + t3 <= abs(x-y) * t1:
print('YES')
else:
print('NO')
``` | output | 1 | 38,639 | 8 | 77,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator. | instruction | 0 | 38,640 | 8 | 77,280 |
Tags: implementation
Correct Solution:
```
x, y, z, t1, t2, t3 = map(int, input().split())
l = abs(x - y)
w = abs(z - x)
if t1 * l >= t2 * (w + l) + (3 * t3):
print("YES")
else:
print("NO")
``` | output | 1 | 38,640 | 8 | 77,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator. | instruction | 0 | 38,641 | 8 | 77,282 |
Tags: implementation
Correct Solution:
```
x, y, z, t1, t2, t3 = [int(i) for i in input().split()]
ts = abs(x-y) * t1
te = abs(x-y) * t2 + abs(x-z) * t2 + 3 * t3
if te <= ts:
print('YES')
else:
print('NO')
``` | output | 1 | 38,641 | 8 | 77,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator.
Submitted Solution:
```
"""
instagram : essipoortahmasb2018
telegram channel : essi_python
"""
i = input
List = [*map(int,i().split())]
print("YES") if (abs(List[2]-List[0])*List[4])+(abs(List[0]-List[1])*List[4])+(List[5]*3) <= List[3]*abs(List[0]-List[1]) else print("NO")
``` | instruction | 0 | 38,642 | 8 | 77,284 |
Yes | output | 1 | 38,642 | 8 | 77,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator.
Submitted Solution:
```
x, y, z, t1, t2, t3 = list(map(int, input().split()))
print("YES" if 3 * t3 + abs(x - z) * t2 + abs(x - y) * t2 <= abs(x - y) * t1 else "NO")
``` | instruction | 0 | 38,643 | 8 | 77,286 |
Yes | output | 1 | 38,643 | 8 | 77,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator.
Submitted Solution:
```
x, y, z, t2, t1, t3 = map(int,input().split())
l = abs(z - x) * t1 + 2 * t3 + abs(y - x) * t1 + t3
r = t2 * abs(x - y)
if l <= r:
print("YES")
else:
print("NO")
``` | instruction | 0 | 38,644 | 8 | 77,288 |
Yes | output | 1 | 38,644 | 8 | 77,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator.
Submitted Solution:
```
[x,y,z,t1,t2,t3]=[int(i) for i in input().split()]
e=t2*abs(x-y)+t2*abs(z-x)+3*t3
if(e<=abs(x-y)*t1):
print("YES")
else:
print("NO")
``` | instruction | 0 | 38,645 | 8 | 77,290 |
Yes | output | 1 | 38,645 | 8 | 77,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator.
Submitted Solution:
```
x,y,z,t1,t2,t3 = map(int, input().split())
d = abs(x-y)
tel = (abs(z-x) + d)*t2 + 3*t3
ts = d*t1
if tel < ts:
print("YES")
else:
print("NO")
``` | instruction | 0 | 38,646 | 8 | 77,292 |
No | output | 1 | 38,646 | 8 | 77,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator.
Submitted Solution:
```
x, y, z, t1, t2, t3 = map(int, input().split())
tt = abs(x - y) * t1
tt1 = t3 * 2 + abs(x - y) * t2 + abs(x - z) * t2
if tt >= tt1:
print('YES')
else:
print('NO')
``` | instruction | 0 | 38,647 | 8 | 77,294 |
No | output | 1 | 38,647 | 8 | 77,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator.
Submitted Solution:
```
x,y,z,t1,t2,t3=map(int,input().split())
a=abs(x-y)
st=a*t1
el=2*t3+abs(x-z)*t2+a*t2
if el<st:
print("YES")
else:
print("NO")
``` | instruction | 0 | 38,648 | 8 | 77,296 |
No | output | 1 | 38,648 | 8 | 77,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a multi-storey building, where floors are numbered with positive integers. Two floors are called adjacent if their numbers differ by one. Masha decided to visit Egor. Masha lives on the floor x, Egor on the floor y (not on the same floor with Masha).
The house has a staircase and an elevator. If Masha uses the stairs, it takes t_1 seconds for her to walk between adjacent floors (in each direction). The elevator passes between adjacent floors (in each way) in t_2 seconds. The elevator moves with doors closed. The elevator spends t_3 seconds to open or close the doors. We can assume that time is not spent on any action except moving between adjacent floors and waiting for the doors to open or close. If Masha uses the elevator, it immediately goes directly to the desired floor.
Coming out of the apartment on her floor, Masha noticed that the elevator is now on the floor z and has closed doors. Now she has to choose whether to use the stairs or use the elevator.
If the time that Masha needs to get to the Egor's floor by the stairs is strictly less than the time it will take her using the elevator, then she will use the stairs, otherwise she will choose the elevator.
Help Mary to understand whether to use the elevator or the stairs.
Input
The only line contains six integers x, y, z, t_1, t_2, t_3 (1 β€ x, y, z, t_1, t_2, t_3 β€ 1000) β the floor Masha is at, the floor Masha wants to get to, the floor the elevator is located on, the time it takes Masha to pass between two floors by stairs, the time it takes the elevator to pass between two floors and the time it takes for the elevator to close or open the doors.
It is guaranteed that x β y.
Output
If the time it will take to use the elevator is not greater than the time it will take to use the stairs, print Β«YESΒ» (without quotes), otherwise print Β«NO> (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 4 2 1
Output
YES
Input
1 6 6 2 1 1
Output
NO
Input
4 1 7 4 1 2
Output
YES
Note
In the first example:
If Masha goes by the stairs, the time she spends is 4 β
4 = 16, because she has to go 4 times between adjacent floors and each time she spends 4 seconds.
If she chooses the elevator, she will have to wait 2 seconds while the elevator leaves the 4-th floor and goes to the 5-th. After that the doors will be opening for another 1 second. Then Masha will enter the elevator, and she will have to wait for 1 second for the doors closing. Next, the elevator will spend 4 β
2 = 8 seconds going from the 5-th floor to the 1-st, because the elevator has to pass 4 times between adjacent floors and spends 2 seconds each time. And finally, it will take another 1 second before the doors are open and Masha can come out.
Thus, all the way by elevator will take 2 + 1 + 1 + 8 + 1 = 13 seconds, which is less than 16 seconds, so Masha has to choose the elevator.
In the second example, it is more profitable for Masha to use the stairs, because it will take 13 seconds to use the elevator, that is more than the 10 seconds it will takes to go by foot.
In the third example, the time it takes to use the elevator is equal to the time it takes to walk up by the stairs, and is equal to 12 seconds. That means Masha will take the elevator.
Submitted Solution:
```
Masha_Floor, Wants, elv_located, stair_passing, elv_passing, open_close = map(int, input().split())
"""Masha_Floor
Wants
elv_located
stair_passing
elv_passing
open_close"""
Difference = abs(Wants-Masha_Floor)
stair_time = Difference * stair_passing
elv_dfr = abs(Masha_Floor -elv_located)
difference_elv = abs(elv_located - Masha_Floor)
elv_time = 3+(Difference*elv_passing)+(difference_elv*elv_passing)
if stair_time >= elv_time:
print('YES')
else:
print('NO')
print(stair_time, elv_time)
``` | instruction | 0 | 38,649 | 8 | 77,298 |
No | output | 1 | 38,649 | 8 | 77,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1. | instruction | 0 | 38,853 | 8 | 77,706 |
Tags: greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
count_1 = 0
for i in range(n):
if(a[i] == 1):
count_1 += 1
temp = 1
for i in range(n):
if(a[i] == 1):
j = i+1
while(j < n):
if(a[j] == 1):
temp += 1
j += 1
else:
break
break
if(temp == count_1):
print(0)
continue
count_1_temp = 1
count = 0
one_found = False
for i in range(n):
if (one_found == True and count_1_temp != count_1):
if(a[i] == 0):
count += 1
else:
count_1_temp += 1
else:
if a[i] == 1:
one_found = True
print(count)
``` | output | 1 | 38,853 | 8 | 77,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1. | instruction | 0 | 38,854 | 8 | 77,708 |
Tags: greedy, implementation
Correct Solution:
```
t=int(input())
while t!=0:
t-=1
n=int(input())
l=list(map(int,input().split()))
if 1 not in l:
print(0)
else:
fi=l.index(1)
while(fi<n and l[fi]!=0):
fi+=1
ans=0
while(fi<n):
inc=0
fl=0
while (l[fi]!=1):
inc+=1
fi+=1
if fi>=n:
fl=1
break
if fl==1:
break
else:
ans+=inc
while(fi<n and l[fi]!=0):
fi+=1
print(ans)
``` | output | 1 | 38,854 | 8 | 77,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1. | instruction | 0 | 38,855 | 8 | 77,710 |
Tags: greedy, implementation
Correct Solution:
```
from collections import Counter
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
if l.count(1)==1:
print(0)
else:
f=l.index(1)
for i in range(f,len(l)):
if l[i]==1:
x=i
ind=x
k=l[f:ind+1].count(0)
print(k)
``` | output | 1 | 38,855 | 8 | 77,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1. | instruction | 0 | 38,856 | 8 | 77,712 |
Tags: greedy, implementation
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
flag=0
sum_of_ziro=0
ans=0
for i in a:
if i==1 and flag==0:
flag=1
if i==1 and flag==1:
ans+=sum_of_ziro
sum_of_ziro=0
if i==0 and flag==1:
sum_of_ziro+=1
print(ans)
``` | output | 1 | 38,856 | 8 | 77,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1. | instruction | 0 | 38,857 | 8 | 77,714 |
Tags: greedy, implementation
Correct Solution:
```
for i in range(int(input())):
input()
line = input().replace(' ', '')
line = line[line.find('1'):][::-1]
line = line[line.find('1'):][::-1]
print(line.count('0'))
``` | output | 1 | 38,857 | 8 | 77,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1. | instruction | 0 | 38,858 | 8 | 77,716 |
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
ans = 0
l = -1
r = -1
lol = [int(n) for n in input().split()]
if(lol.count(1) == 1):
print(0)
else:
for i in range(n):
if(lol[i]==1):
l = i
break
for j in range(len(lol)-1,-1,-1):
if(lol[j]==1):
r = j
break
#print(l,r)
for z in range(l+1,r):
if(lol[z]==0):
ans += 1
print(ans)
``` | output | 1 | 38,858 | 8 | 77,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1. | instruction | 0 | 38,859 | 8 | 77,718 |
Tags: greedy, implementation
Correct Solution:
```
import re
import sys
def solve(nums: str) -> int:
nums = nums.replace(' ', '')
space_map = re.findall(r'0+', nums)
distance_from_left = (sum([int(len(i)) for i in space_map])
- (0 if nums[-1] == '1' else len(space_map[-1]))
)
distance_from_right = (sum([int(len(i)) for i in space_map])
- (0 if nums[0] == '1' else len(space_map[0]))
)
distance_from_center = (sum([int(len(i)) for i in space_map])
- (0 if nums[0] == '1' else len(space_map[0]))
- (0 if nums[-1] == '1' else len(space_map[-1]))
)
return min(distance_from_left, distance_from_right, distance_from_center)
if __name__ == '__main__':
for (count, line) in enumerate(sys.stdin):
if count == 0:
continue
if count % 2 == 1:
continue
print(solve(line.rstrip()))
``` | output | 1 | 38,859 | 8 | 77,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1. | instruction | 0 | 38,860 | 8 | 77,720 |
Tags: greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
# u,v = map(int,input().split())
l = list(map(int,input().split()))
arr = []
for i in range(n):
if l[i] == 1:
arr.append(i)
ans = 0
for i in range(1,len(arr)):
# if arr[i] - arr[i-1] > 1:
ans += arr[i] - arr[i-1] - 1
print(ans)
# if len(ans) == 0:
# print(0)
# else:
# print(max(max(ans),len(ans)))
``` | output | 1 | 38,860 | 8 | 77,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1.
Submitted Solution:
```
tc = int(input())
for case in range(tc):
input()
array = input().split(" ")
l = 0
while True:
if array[l] == '1':
break
else:
l += 1
r = len(array) - 1
while True:
if array[r] == '1':
break
else:
r -= 1
count = 0
for i in array[l:r]:
if i == '0':
count += 1
print(count)
``` | instruction | 0 | 38,861 | 8 | 77,722 |
Yes | output | 1 | 38,861 | 8 | 77,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
a = ''.join(str(i) for i in a)
a = a.strip("0")
print(a.count("0"))
``` | instruction | 0 | 38,862 | 8 | 77,724 |
Yes | output | 1 | 38,862 | 8 | 77,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1.
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
i = 0
b = []
while(i < len(a)):
if a[i] == 1:
j = i
while(j < len(a) and a[j] == 1):
j += 1
b.append([i, j])
i = j
i += 1
s = 0
for i in range(len(b) - 1):
s += b[i + 1][0] - b[i][1]
print(s)
``` | instruction | 0 | 38,863 | 8 | 77,726 |
Yes | output | 1 | 38,863 | 8 | 77,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1.
Submitted Solution:
```
import sys
input=sys.stdin.readline
T=int(input())
for _ in range(T):
n=int(input())
A=list(map(int,input().split()))
flag=0
v=A.index(1)
ans=0
t=0
for i in range(v,n):
if (A[i]==1 and flag==0):
flag=1
ans=ans+t
t=0
elif (A[i]==0):
t=t+1
flag=0
print(ans)
``` | instruction | 0 | 38,864 | 8 | 77,728 |
Yes | output | 1 | 38,864 | 8 | 77,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1.
Submitted Solution:
```
import sys
import math
input = sys.stdin.readline
def algo(n, arr):
xn = arr.count(1)
avg = (xn+1) // 2
t = 0
for i in range(n):
if arr[i] == 1:
t += 1
if t == avg:
t = i
break
ans = 0
j = 0
for i in range(t-1, -1, -1):
if arr[i] == 1:
j += 1
ans += t - i - j
j = 0
for i in range(t+1, n):
if arr[i] == 1:
j += 1
ans += i - t - j
return ans
for _ in range(int(input())):
n = int(input().strip())
arr = list(map(int, input().split()))
res = algo(n, arr)
print(res)
``` | instruction | 0 | 38,865 | 8 | 77,730 |
No | output | 1 | 38,865 | 8 | 77,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
l = list(map(int,input().split()))
mx = 0
if l.count(0) > 1:
ans = 1
for i in range(len(l)-1):
if l[i] == l[i+1] and l[i]==0:
ans = ans + 1
else:
if l[i+1] == 1:
mx = max(ans,mx)
ans = 1
print(mx)
``` | instruction | 0 | 38,866 | 8 | 77,732 |
No | output | 1 | 38,866 | 8 | 77,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1.
Submitted Solution:
```
n = int(input())
numberOfblock = []
blocks = []
for i in range(n):
numberOfblock.append(int(input()))
temp = input().split(" ")
blocks.append(temp)
for blc in range(len(blocks)):
iter = numberOfblock[blc]
ones = []
meet = []
for i in range(iter):
if(blocks[blc][i] == '1'):
ones.append(i)
counter = 0
i = 0
for x in ones:
if(len(ones) > 1):
while ((x + 1 not in ones) and ((x + 1) < iter)):
if((x + 1) not in meet):
meet.append(x+1)
counter += 1
x = x + 1
while ((x - 1 not in ones) and ((x - 1) > 0) ):
if ((x - 1) not in meet):
meet.append(x - 1)
counter += 1
x = x - 1
print(counter)
``` | instruction | 0 | 38,867 | 8 | 77,734 |
No | output | 1 | 38,867 | 8 | 77,735 |
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