text stringlengths 198 433k | conversation_id int64 0 109k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
#!/usr/bin/env python3
import sys
[n, m, s1, t1] = map(int, sys.stdin.readline().strip().split())
s, t = s1 - 1, t1 - 1
tos = [[] for _ in range(n)]
for _ in range(m):
[u1, v1] = map(int, sys.stdin.readline().strip().split())
v, u = v1 - 1, u1 - 1
tos[v].append(u)
tos[u].append(v)
def S_l(v0, tos):
n = len(tos)
visited = [False for _ in range(n)]
visited[v0] = True
queue = [v0]
hq = 0
l = [0 for _ in range(n)]
while len(queue) > hq:
u = queue[hq]
hq += 1
for v in tos[u]:
if not visited[v]:
l[v] = l[u] + 1
queue.append(v)
visited[v] = True
return l
lt = S_l(t, tos)
ls = S_l(s, tos)
d = lt[s]
count = 0
for u in range(n):
for v in range(u):
if (v not in tos[u]) and (min(lt[u] + ls[v], lt[v] + ls[u]) + 1 >= d):
count += 1
print (count)
```
| 3,900 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
from collections import deque, namedtuple
from heapq import *
from sys import stdin
inf = float('inf')
Edge = namedtuple('Edge', 'start, end, cost')
def make_edge(start, end, cost=1):
return Edge(start, end, cost)
class Graph:
def __init__(self, edges, bi=True):
wrong_edges = [i for i in edges if len(i) not in [2, 3]]
if wrong_edges:
raise ValueError('Wrong edges data: {}'.format(wrong_edges))
self.edges = [make_edge(*edge) for edge in edges]
self.vertices = set(
sum(
([edge.start, edge.end] for edge in self.edges), []
))
self.neighbors = {vertex: set() for vertex in self.vertices}
for edge in self.edges:
self.neighbors[edge.start].add(edge.end)
def get_node_pairs(self, n1, n2, both_ends=True):
if both_ends:
node_pairs = [[n1, n2], [n2, n1]]
else:
node_pairs = [[n1, n2]]
return node_pairs
def remove_edge(self, n1, n2, both_ends=True):
node_pairs = self.get_node_pairs(n1, n2, both_ends)
edges = self.edges[:]
for edge in edges:
if[edge.start, edge.end] in node_pairs:
self.edges.remove(edge)
def add_edge(self, n1, n2, cost=1, both_ends=True):
node_pairs = self.get_node_pairs(n1, n2, both_ends)
for edge in self.edges:
if [edge.start, edge.end] in node_pairs:
return ValueError('Edge {} {} already exists'.format(n1, n2))
self.edges.append(Edge(start=n1, end=n2, cost=cost))
if both_ends:
self.edges.append(Edge(start=n2, end=n1, cost=cost))
def dijkstra(self, source, dest):
assert source in self.vertices, 'Such source node doesn\'t exist'
distances = {vertex: inf for vertex in self.vertices}
distances[source] = 0
q, seen = [(0, source)], set()
while q:
(curr_cost, current_vertex) = heappop(q)
if current_vertex in seen:
continue
seen.add(current_vertex)
for neighbor in self.neighbors[current_vertex]:
cost = 1
if neighbor in seen:
continue
alternative_route = distances[current_vertex] + cost
if alternative_route < distances[neighbor]:
distances[neighbor] = alternative_route
heappush(q, (alternative_route, neighbor))
return distances
n, m, s, t = [int(x) for x in stdin.readline().rstrip().split()]
verts = []
for i in range(m):
verts.append(tuple([int(x) for x in stdin.readline().rstrip().split()]))
rev_verts = []
for i in verts:
rev_verts.append((i[1], i[0]))
for i in rev_verts:
verts.append(i)
graph = Graph(verts)
s_dist = graph.dijkstra(s, t)
t_dist = graph.dijkstra(t, s)
SHORTEST_DIST = s_dist[t]
count = 0
for i in range(1, n + 1):
for j in range(i + 1, n + 1):
if j not in graph.neighbors[i] and \
i not in graph.neighbors[j] and \
s_dist[i] + t_dist[j] + 1 >= SHORTEST_DIST and \
s_dist[j] + t_dist[i] + 1 >= SHORTEST_DIST:
count = count + 1
print(count)
```
| 3,901 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
class Node:
def __init__(self, label):
self.label, self.ways, self.distances = label, set(), dict()
def connect(self, other):
self.ways.add(other)
class Graph:
def __init__(self, count):
self.nodes = [Node(i) for i in range(count)]
def connect(self, a, b):
node_a, node_b = self.nodes[a], self.nodes[b]
node_a.connect(node_b)
node_b.connect(node_a)
def fill_distances(self, start):
looked, queue, next_queue = {start}, {start}, set()
start.distances[start] = 0
while len(queue) + len(next_queue) > 0:
for node in queue:
for neighbour in node.ways:
if neighbour not in looked:
looked.add(neighbour)
next_queue.add(neighbour)
neighbour.distances[start] = node.distances[start] + 1
queue, next_queue = next_queue, set()
def repeat(elem):
while True:
yield elem
def readline(types=repeat(int)):
return map(lambda x: x[0](x[1]), zip(types, input().split()))
def readinput():
nodes, edges, start, finish = readline()
start, finish = start - 1, finish - 1
graph = Graph(nodes)
for i in range(edges):
a, b = readline()
a, b = a - 1, b - 1
graph.connect(a, b)
return graph, graph.nodes[start], graph.nodes[finish]
def main():
graph, start, finish = readinput()
graph.fill_distances(start)
graph.fill_distances(finish)
distance = start.distances[finish]
answer = 0
for i in range(len(graph.nodes)):
for j in range(i + 1, len(graph.nodes)):
node_a, node_b = graph.nodes[i], graph.nodes[j]
if node_a in node_b.ways:
continue
if node_a.distances[start] + 1 + node_b.distances[finish] < distance:
continue
if node_a.distances[finish] + 1 + node_b.distances[start] < distance:
continue
answer += 1
print(answer)
main()
```
| 3,902 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
import sys
from collections import deque
readline = sys.stdin.readline
readlines = sys.stdin.readlines
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')
def bfs(G, s, t=None):
"""
G[v] = [x1, x2,...]
"""
dist = [-1]*len(G)
dist[s] = 0
q = deque([s])
while q:
v = q.popleft()
for x in G[v]:
if dist[x] < 0 or dist[x] > dist[v] + 1:
dist[x] = dist[v] + 1
q.append(x)
if t is None:
return dist
else:
return dist[t]
def solve():
n, m, s, t = nm()
s -= 1; t -= 1
G = [list() for _ in range(n)]
for _ in range(m):
u, v = nm()
u -= 1; v -= 1
G[u].append(v)
G[v].append(u)
dss = bfs(G, s)
dst = bfs(G, t)
ans = 0
sp = dss[t]
for i in range(n-1):
for j in range(i+1, n):
if dss[i] + dst[j] + 1 >= sp and dss[j] + dst[i] + 1 >= sp:
ans += 1
print(ans - m)
return
solve()
# T = ni()
# for _ in range(T):
# solve()
```
| 3,903 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
if __name__ == '__main__':
n, m, u, v = map(int, input().split())
graph = [[] for _ in range(n)]
mtx = [[0] * n for _ in range(n)]
for _ in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
graph[a].append(b)
graph[b].append(a)
mtx[a][b] = 1
mtx[b][a] = 1
q = [u - 1]
vec1 = [0] * n
vec1[u - 1] = 1
while q:
c = q.pop(0)
for i in graph[c]:
if not vec1[i]:
vec1[i] = vec1[c] + 1
q.append(i)
q = [v - 1]
vec2 = [0] * n
vec2[v - 1] = 1
while q:
c = q.pop(0)
for i in graph[c]:
if not vec2[i]:
vec2[i] = vec2[c] + 1
q.append(i)
for i in range(n):
vec1[i] -= 1
vec2[i] -= 1
res = 0
for i in range(n):
for j in range(i + 1, n):
if not mtx[i][j] and min(vec2[j] + vec1[i] + 1, vec2[i] + vec1[j] + 1) >= vec1[v - 1]:
res += 1
print(res)
```
| 3,904 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
import math,sys
#from itertools import permutations, combinations;import heapq,random;
from collections import defaultdict,deque
import bisect as bi
def yes():print('YES')
def no():print('NO')
#sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(sys.stdin.readline()))
def In():return(map(int,sys.stdin.readline().split()))
def Sn():return sys.stdin.readline().strip()
#sys.setrecursionlimit(1500)
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def find_gt(a, x):
'Find leftmost value greater than x'
i = bi.bisect_left(a, x)
if i != len(a):
return i
else:
return -1
def bfs(n,gp,st):
q=deque([st])
dist=[-1]*(n+1)
dist[st]=0
while q:
node=q.popleft()
for x in gp[node]:
if dist[x]==-1:
dist[x]=dist[node]+1
q.append(x)
return dist
def main():
try:
n,m,st,end=In()
gp=defaultdict(set)
for i in range(m):
a,b=In()
gp[a].add(b)
gp[b].add(a)
bfs1=bfs(n,gp,st)
bfs2=bfs(n,gp,end)
length=bfs1[end]
cnt=0
for i in range(1,n+1):
for j in range(i+1,n+1):
if j in gp[i]:
continue
else:
if bfs1[i]+bfs2[j]+1<length:
continue
elif bfs1[j]+bfs2[i]+1<length:
continue
else:
cnt+=1
print(cnt)
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
# for _ in range(I()):main()
for _ in range(1):main()
#End#
# ******************* All The Best ******************* #
```
| 3,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
n,m,s,t = map(int, input().split())
G = [[] for _ in range(n+1)]
for i in range(m):
a,b = map(int, input().split())
G[a].append(b)
G[b].append(a)
dists = [n+1]*(n+1)
distt = [n+1]*(n+1)
# BFS find distance
T = [s]
count = 0
while T:
newT = []
for i in T:
dists[i] = min(dists[i],count)
for j in G[i]:
if dists[j] == n+1:
newT.append(j)
count+=1
T = newT
T = [t]
count = 0
while T:
newT = []
for i in T:
distt[i] = min(distt[i],count)
for j in G[i]:
if distt[j] == n+1:
newT.append(j)
count+=1
T = newT
count = 0
mm = dists[t]
for i in range(1,n+1):
for j in range(i+1,n+1):
if j in G[i]:
continue
if min(dists[i] + distt[j] + 1, dists[j] + distt[i] + 1) < mm:
continue
count += 1
print(count)
```
| 3,906 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
n,m,s,e = map(int,input().split())
s -= 1
e -= 1
md = []
g = []
for i in range(n):
md.append([-1,-1])
g.append([])
for i in range(m):
u,v = map(int,input().split())
g[u-1].append(v-1)
g[v-1].append(u-1)
#bfs
md[s][0] = 0
d = 0
st = [s]
c = 1
#print('s')
while c < n:
#print(c)
d += 1
st2 = []
for node in st:
for p in g[node]:
if md[p][0] == -1:
st2.append(p)
md[p][0] = d
c+=len(st2)
st = st2[:]
#bfs
md[e][1] = 0
d = 0
st = [e]
c = 1
#print('s')
while c < n:
#print(c)
d += 1
st2 = []
for node in st:
for p in g[node]:
if md[p][1] == -1:
st2.append(p)
md[p][1] = d
c+=len(st2)
st = st2[:]
#print(md)
dis = md[e][0]
t = 0
for i in range(n):
for j in range(i):
if j in g[i]:continue
if dis <= min(md[i][0] + md[j][1],md[j][0] + md[i][1])+1:
t += 1
print(t)
```
| 3,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Submitted Solution:
```
def get_amount_variants(routes, n, s, t):
"""
:param List[List[int]] routes:
:param int n:
:param int s:
:param int t:
:return: int
"""
connects = [[] for _ in range(n + 1)]
for st, end in routes:
connects[st].append(end)
connects[end].append(st)
w_h = get_shortest_ways(connects, s)
w_j = get_shortest_ways(connects, t)
amount = 0
for i in range(1, n + 1):
for j in range(i + 1, n + 1):
if min(w_h[i] + w_j[j], w_h[j] + w_j[i]) + 1 >= w_h[t] and i not in connects[j]:
amount += 1
return amount
def get_shortest_ways(connects, st):
res = [int(1e9) if _ != st else 0 for _ in range(len(connects) + 1)]
available = [(x, 0) for x in connects[st]]
while len(available) > 0:
point, weight = available.pop(0)
if res[point] != 0 and res[point] < weight + 1:
continue
res[point] = weight + 1
available += [(x, weight + 1) for x in connects[point] if res[x] == int(1e9) and x != st]
return res
n, m, s, t = map(int, input().strip().split())
routes = []
for i in range(m):
routes.append(map(int, input().split()))
print(get_amount_variants(routes, n, s, t))
```
Yes
| 3,908 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Submitted Solution:
```
import sys
input=sys.stdin.readline
from collections import deque
n,m,s,t=map(int,input().split())
s-=1;t-=1
g=[[] for i in range(n)]
gg=[set() for i in range(n)]
for _ in range(m):
u,v=map(int,input().split())
g[u-1].append(v-1)
gg[u-1].add(v-1)
g[v-1].append(u-1)
gg[v-1].add(u-1)
def bfs01(x,d):
dq=deque([x])
d[x]=0
while dq:
ver=dq.popleft()
for to in g[ver]:
if d[to]==-1:
dq.append(to)
d[to]=d[ver]+1
return d
cnt=0
d_s=bfs01(s,[-1]*n)
d_t=bfs01(t,[-1]*n)
l=d_s[t]
for i in range(n):
for j in range(i+1,n):
if j in gg[i] or i in gg[j]:
continue
if l<=min(l,d_s[i]+1+d_t[j],d_s[j]+1+d_t[i]):
cnt+=1
print(cnt)
```
Yes
| 3,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Submitted Solution:
```
from collections import defaultdict, deque
from sys import stdin, stdout
def In():return(map(int,stdin.readline().split()))
def bfs(n, graph, start):
q = deque([start])
dist = [-1]*(n+1)
dist[start] = 0
while q:
current = q.popleft()
for neigh in graph[current]:
if dist[neigh] == -1:
dist[neigh] = dist[current] + 1
q.append(neigh)
return dist
def main():
n, m, s, t = In()
graph = defaultdict(set)
for i in range(m):
a, b = In()
graph[a].add(b)
graph[b].add(a)
ds = bfs(n, graph, s)
dt = bfs(n, graph, t)
res = 0
aim = ds[t]
for i in range(1, n+1):
for j in range(i+1, n+1):
if j not in graph[i]:
if ds[i]+dt[j]+1 < aim or ds[j]+dt[i] + 1 < aim:
continue
else:
res += 1
print(res)
if __name__ == '__main__':
main()
```
Yes
| 3,910 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Submitted Solution:
```
import sys
import heapq
nnode,nedge,source,dest=map(int,sys.stdin.readline().split())
source-=1;dest-=1
adjl=[[] for _ in range(nnode)]
for _ in range(nedge):
u,v=map(int,sys.stdin.readline().split())
u-=1;v-=1
adjl[u].append(v)
adjl[v].append(u)
# why not BFS? >.<
def dijkstra_from(source):
pq=[]
dist=[10**9]*nnode
dist[source]=0
heapq.heappush(pq,(dist[source],source))
while pq:
dist_node,node=heapq.heappop(pq)
if dist[node]!=dist_node: continue
for adj in adjl[node]:
if dist[adj]>dist_node+1:
dist[adj]=dist_node+1
heapq.heappush(pq,(dist[adj],adj))
return dist
d_src=dijkstra_from(source)
d_dst=dijkstra_from(dest)
assert d_dst[source] == d_src[dest]
ans=0
for u in range(nnode):
adj_to_u=[False]*u
for v in adjl[u]:
if v<u:
adj_to_u[v]=True
for v in range(u):
if not adj_to_u[v]:
if min(
d_dst[u]+d_src[v]+1,
d_src[u]+d_dst[v]+1
)<d_src[dest]: continue
ans+=1
print(ans)
```
Yes
| 3,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Submitted Solution:
```
n, m, s, t = map(int, input().split())
G = [[] for i in range(n + 1)]
for i in range(m):
u, v = map(int, input().split())
G[u].append(v)
G[v].append(u)
def bfs(x, d):
q = [x]
while q:
c = q.pop()
for y in G[c]:
if not d[y]:
d[y] = d[c] + 1
q.insert(0, y)
d[x] = 0
ds = [0 for i in range(n + 1)]
dt = [0 for i in range(n + 1)]
bfs(s, ds)
bfs(t, dt)
D = ds[t]
pairs = 0
for u in range(n - 1):
for v in range(u + 1, n):
min_long = min(ds[u] + dt[v] + 1, ds[v] + dt[u] + 1)
if min_long >= D:
pairs += 1
print(pairs - m)
```
No
| 3,912 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Submitted Solution:
```
from collections import defaultdict
nodes,edges,start,end=map(int,input().split())
shortest=[-1]*(nodes+1)
shortest[start]=0
dic=defaultdict(set)
for i in range(edges):
x,y=map(int,input().split())
dic[x].add(y)
dic[y].add(x)
queue=[start]
def bfs(queue):
j=0
while(j<len(queue)):
node=queue[j]
for i in dic[node]:
if shortest[i]==-1:
shortest[i]=shortest[node]+1
queue.append(i)
j+=1
bfs(queue)
shorteststart=shortest[:]
shortest=[-1]*(nodes+1)
shortest[end]=0
queue=[end]
bfs(queue)
shortestend=shortest[:]
output=0
mindistance=shorteststart[end]
for i in range(1,nodes+1):
for j in range(i+1,nodes+1):
if j not in dic[i]:
if shorteststart[i]+1+shortestend[j]>=mindistance:
output+=1
print(output)
```
No
| 3,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Submitted Solution:
```
n, m, s, t = map(int, input().split())
G = [[] for i in range(n + 1)]
for i in range(m):
u, v = map(int, input().split())
G[u].append(v)
G[v].append(u)
def bfs(x, d):
q = [x]
while q:
c = q.pop()
for y in G[c]:
if not d[y]:
d[y] = d[c] + 1
q.insert(0, y)
d[x] = 0
ds = [0 for i in range(n + 1)]
dt = [0 for i in range(n + 1)]
bfs(s, ds)
bfs(t, dt)
D = ds[t]
pairs = 0
for u in range(n - 1):
for v in range(u + 1, n):
min_long = min(ds[u] + dt[v] + 1, ds[v] + dt[u] + 1)
if min_long >= D:
pairs += 1
if pairs == 0:
print(pairs)
else:
print(pairs - m)
```
No
| 3,914 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.
In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease.
You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.
Input
The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
Output
Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.
Examples
Input
5 4 1 5
1 2
2 3
3 4
4 5
Output
0
Input
5 4 3 5
1 2
2 3
3 4
4 5
Output
5
Input
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
Output
3
Submitted Solution:
```
n,m,s,t = map(int,input().split())
dic = {}
hasEdge = []
for i in range(n):
dic[i] = []
for i in range(n):
hasEdge.append([])
for j in range(n):
hasEdge[-1].append(0)
for i in range(m):
a,b = map(int,input().split())
dic[a-1].append(b-1)
dic[b-1].append(a-1)
hasEdge[a-1][b-1] = 1
hasEdge[b-1][a-1] = 1
distS = {}
distD = {}
def shortestDist(source,n):
lis = []
sol = {}
def bfs(source):
global dic
nonlocal lis
nonlocal sol
lis.append(source)
sol[source] = 0
while(len(lis)!=0):
temp = lis.pop()
for i in dic[temp]:
if sol.get(i)==None:
sol[i] = sol[temp]+1
lis.append(i)
bfs(source)
for i in range(n):
if sol.get(i)==None:
sol[i] = float("Inf")
return sol
distS = shortestDist(s-1,n)
distD = shortestDist(t-1,n)
if distS[j]==float("Inf"):
print("Hai")
# for checking if this is the case
else:
count = 0
for i in range(n):
for j in range(i+1,n):
if hasEdge[i][j]==0 and not(distS[i]+1+distD[j]<distS[t-1] or distS[j]+1+distD[i]<distS[t-1]):
count += 1
print(count)
```
No
| 3,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Tags: games, greedy
Correct Solution:
```
n, k = map(int, input().strip().split())
data = map(int, input().strip().split())
sol = []
mapping = [(-1,1000)]*256
for x in data:
if mapping[x][0] == -1:
for i in range(max(x-k+1,0), x+1):
if mapping[i][0] == -1:
if i > 0 and mapping[i-1][1]+(x-i+1) <= k:
p = mapping[i-1][1]+1
for j in range(i, x+1):
mapping[j] = (mapping[i-1][0], p)
p += 1
else:
p = 1
for j in range(i, x+1):
mapping[j] = (i, p)
p += 1
break
sol.append(mapping[x][0])
print(' '.join(map(str, sol)))
```
| 3,916 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Tags: games, greedy
Correct Solution:
```
R = lambda: map(int, input().split())
n, k = R()
a = list(range(0, 257)); v = [1]*257
# print(a)
# print(v)
for p in R():
if v[p]:
t = p
# print(t)
while t >= 0 and p-a[t]<=k-1:
# print(t)
t -= 1
t += 1
for i in range(t, p+1): a[i] = a[t]; v[i] = 0
print(a[p], end=' ')
```
| 3,917 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Tags: games, greedy
Correct Solution:
```
import atexit
import io
import sys
# Buffering IO
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
def main():
n, k = [int(x) for x in input().split()]
p = [int(x) for x in input().split()]
t = []
g={}
for x in p:
if x in g:
t.append(g[x])
continue
kk = x - 1
while True:
if kk in g:
if x - g[kk] < k:
ttt = g[kk]
else:
ttt= kk + 1
for i in range(kk +1 , x + 1):
g[i] = ttt
t.append(g[x])
break
elif kk<0 or x - kk == k:
for i in range(kk +1 , x + 1):
g[i] = kk + 1
t.append(g[x])
break
kk -= 1
print(' '.join((str(x) for x in t)))
if __name__ == '__main__':
main()
```
| 3,918 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Tags: games, greedy
Correct Solution:
```
N, K = input().split()
N, K = int(N), int(K)
P = [int(x) for x in input().split()]
A = [None]*256
A[0] = 0
for i in range(N):
pn = P[i]
if A[pn] is None:
for j in range(K-1, -1, -1):
if pn < j: continue
if A[pn-j] is None:
A[pn-j] = pn-j
break
else:
if A[pn-j] + K - 1 >= pn:
break
for jj in range(j, -1, -1):
A[pn-jj] = A[pn-j]
print(*[A[P[i]] for i in range(N)])
```
| 3,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Tags: games, greedy
Correct Solution:
```
n, k = map(int, input().split())
a = map(int, input().split())
ans = [0]*n
f = [-1]*256
i = -1
for x in a:
i += 1
if f[x] < 0:
l = x
while l > 0 > f[l - 1] and x - l < k - 1:
l -= 1
if l > 0 and x - f[l - 1] < k and f[l - 1] >= 0:
f[x] = f[l - 1]
else:
f[x] = l
for j in range(l, x):
f[j] = f[x]
ans[i] = f[x]
for x in ans:
print(x, end=' ')
```
| 3,920 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Tags: games, greedy
Correct Solution:
```
n, k = list(map(int, input().split()))
p = list(map(int, input().split()))
processed = set()
color = {}
length = {}
ans = []
def exists(p, elt, d):
for e in p:
if e > elt:
if e <= elt + d:
return True
elif e - d <= elt + d:
return False
return False
def exists2(p, elt, d):
for e in p:
if e > elt:
if e <= elt + d:
return False
elif e - d <= elt + d:
return [True, e - d]
return False
for i in range(n):
elt = p[i]
if elt in processed:
ans.append(color[elt])
else:
processed.add(elt)
new = 1
run = True
for j in range(1, k):
if elt - j < 0:
break
elif (elt - j) not in processed:
processed.add(elt - j)
new += 1
elif length[elt - j] + new <= k:
for i2 in range(length[elt - j] + new):
color[elt - i2] = color[elt - j]
length[elt] = length[elt - j] + new
run = False
break
else:
break
if run:
for j in range(new):
color[elt - j] = elt - new + 1
length[elt] = new
s = str(color[p[0]])
for elt in p[1:]:
s += ' ' + str(color[elt])
print(s)
```
| 3,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Tags: games, greedy
Correct Solution:
```
R = lambda: map(int, input().split())
n, k = R()
a = list(range(0, 257)); v = [1]*257
for p in R():
if v[p]:
t = p
while t >= 0 and p-a[t]<=k-1: t -= 1
t += 1
for i in range(t, p+1): a[i] = a[t]; v[i] = 0
print(a[p], end=' ')
```
| 3,922 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Tags: games, greedy
Correct Solution:
```
n,k = list(map(int,input().split()))
arr = list(map(int,input().split()))
val = [-1]*256
val[0] = 0
for i in range(0,n):
if(val[arr[i]] > -1):
print(val[arr[i]],end = " ")
else:
flag = 0
for j in range(1,k):
if( val[arr[i]-j] > -1 ):
if(arr[i] - val[arr[i]-j] < k):
for p in range(arr[i] - j + 1,arr[i]+1):
val[p] = val[arr[i]-j]
else:
for t in range(1,j+1):
val[arr[i]- t + 1] = arr[i] - j + 1
# print(val[arr[i]],"OOO")
#print(val[:i],i)
print(val[arr[i]],end = " ")
flag = 1
break
if(flag == 0):
for j in range(0,k):
val[arr[i] - j] = arr[i] -k + 1
print(val[arr[i]],end = " ")
```
| 3,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n,k = [int(s) for s in input().split()]
p = [int(s) for s in input().split()]
map = {}
res = []
for pi in p:
if map.get(pi) is None:
key = pi
for j in range(pi, pi-k, -1):
if j < 0: break
if map.get(j) is None:
key = j
else:
if map[j] >= pi-k+1: key = map[j]
break
for j in range(pi, key-1, -1):
if map.get(j):
break
map[j] = key
res.append(map[pi])
print(*res, sep=" ")
```
Yes
| 3,924 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
read=lambda:map(int,input().split())
n,k=read()
p=list(read())
r=[-1]*256
for i in p:
for u in range(i,max(i-k,-1),-1):
if r[u]>=0: u+=1;break
if u<=i:
if u>0 and r[u-1]+k>i: r[i]=r[u-1]
else:
for j in range(u,i+1): r[j]=u
print(r[i])
```
Yes
| 3,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n,k=map(int,input().split())
arr=list(map(int,input().split()))
par=[i for i in range(260)]
path=[-1 for i in range(260)]
for i in range(n):
j=arr[i]
if path[j] >=0:
par[j] =par[path[j]]
continue
jump=1
while j>0 and path[j] ==-1 and jump <k:
path[j] =arr[i]
j-=1
jump +=1
if arr[i] -par[j] +1 <=k:
par[arr[i]] =par[j]
path[j] =arr[i]
else:
par[arr[i]] =par[j+1]
for i in range(n):
print(par[arr[i]],end=' ')
print()
```
Yes
| 3,926 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
def main():
n, k = map(int, input().split())
p = list(map(int, input().split()))
solve(n, k, p)
def solve(n, k, p):
group = 256 * [None]
r = p[:]
for i, pi in enumerate(p):
# print([(i, gi) for i, gi in enumerate(group)if gi is not None])
if group[pi] is not None:
r[i] = group[pi][0]
else:
lo = pi
while lo >= 0 and pi - lo < k and group[lo] is None:
lo -= 1
if lo < 0 or pi - lo == k:
lo += 1
hi = pi + 1
else: # group[lo] is not None
if pi - group[lo][0] < k:
lo = group[lo][0]
hi = pi + 1
else:
lo += 1
hi = pi + 1
lohi = (lo, hi)
for j in range(lo, hi):
group[j] = lohi
r[i] = group[pi][0]
print(" ".join(map(str, r)))
main()
```
Yes
| 3,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
from sys import stdin, stdout
a = stdin.readline()
p=stdin.readline()
n, k = [int(x) for x in a.split()]
b=p.split()
# print(b)
m=int(256/k)
out=[]
for i in b :
j = 0
for j in range(0, int(256/k)):
if int(i)>=(k*j) and int(i)<= ((k*j)+k-1):
out.append(str(k * j))
new_numbers = [];
for n in out:
new_numbers.append(int(n));
out = new_numbers;
print(out)
```
No
| 3,928 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n, k = map(int, input().split())
p = list(map(int, input().split()))
ranged = [None] * 256
for i in range(256):
ranged[i] = [0, None]
result = ""
for i, pi in enumerate(p):
if ranged[pi][0] == 2:
idx = pi
while ranged[idx][0] == 2:
ranged[idx][0] = 1
idx -= 1
elif ranged[pi][0] == 0:
# find smallest lexicographically possible value for pi
idx = pi
count = 0
_min = pi
while idx > 0 and ranged[idx - 1][0] != 1 and count < k - 1:
idx -= 1
count += 1
if str(idx) < str(_min):
_min = idx
count = 0
lowest = _min
# print("Min for {} is {}".format(pi, _min))
while count < k and idx < 256:
if idx <= pi:
ranged[idx][0] = 1
else:
ranged[idx][0] = 2
ranged[idx][1] = lowest
idx += 1
count += 1
# print(count, k - 1)
# print(ranged)
result += str(ranged[pi][1]) + " "
print(result[:-1])
```
No
| 3,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n,k=map(int,input().split())
ar=list(map(int,input().split()))
tmp=""
for j in ar:
if (j-k+1)>=0:
print(j-k+1,end=' ')
else:
print(0,end=' ')
```
No
| 3,930 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n,k=map(int,input().split())
par=[-1 for i in range(512)]
ls=map(int,input().split())
for v in ls:
v+=256
ans=v
if par[v]==-1:
for j in range(v-k+1,v+1):
if par[j]==-1:
ans=j
break
par[ans]=ans
#print('ans:%d'%(ans))
for j in range(ans+1,min(511,ans+k)):
if par[j]==-1:
par[j]=ans
else:
break
else:
ans=par[v]
print(max(0,ans-256),end=' ')
```
No
| 3,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Tags: brute force, implementation, math
Correct Solution:
```
#http://codeforces.com/problemset/problem/9/C
#solved
#finding lists
#
# from itertools import permutations
#
# print("a = ", sorted(list(map(int, map("".join, map(list, set(permutations("111111111000000000", 2))))))), "\n\n")
# print("b = ", sorted(list(map(int, map("".join, map(list, set(permutations("111111111000000000", 3))))))), "\n\n")
# print("c = ", sorted(list(map(int, map("".join, map(list, set(permutations("111111111000000000", 4))))))), "\n\n")
# print("d = ", sorted(list(map(int, map("".join, map(list, set(permutations("111111111000000000", 5))))))), "\n\n")
# print("e = ", sorted(list(map(int, map("".join, map(list, set(permutations("111111111000000000", 6))))))), "\n\n")
# print("f = ", sorted(list(map(int, map("".join, map(list, set(permutations("111111111000000000", 7))))))), "\n\n")
# print("g = ", sorted(list(map(int, map("".join, map(list, set(permutations("111111111000000000", 8))))))), "\n\n")
# print("h = ", sorted(list(map(int, map("".join, map(list, set(permutations("111111111000000000", 9))))))), "\n\n")
a = [1, 10, 11]
b = [1, 10, 11, 100, 101, 110, 111]
c = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111]
d = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111]
e = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111]
f = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001000, 1001001, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1010000, 1010001, 1010010, 1010011, 1010100, 1010101, 1010110, 1010111, 1011000, 1011001, 1011010, 1011011, 1011100, 1011101, 1011110, 1011111, 1100000, 1100001, 1100010, 1100011, 1100100, 1100101, 1100110, 1100111, 1101000, 1101001, 1101010, 1101011, 1101100, 1101101, 1101110, 1101111, 1110000, 1110001, 1110010, 1110011, 1110100, 1110101, 1110110, 1110111, 1111000, 1111001, 1111010, 1111011, 1111100, 1111101, 1111110, 1111111]
g = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001000, 1001001, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1010000, 1010001, 1010010, 1010011, 1010100, 1010101, 1010110, 1010111, 1011000, 1011001, 1011010, 1011011, 1011100, 1011101, 1011110, 1011111, 1100000, 1100001, 1100010, 1100011, 1100100, 1100101, 1100110, 1100111, 1101000, 1101001, 1101010, 1101011, 1101100, 1101101, 1101110, 1101111, 1110000, 1110001, 1110010, 1110011, 1110100, 1110101, 1110110, 1110111, 1111000, 1111001, 1111010, 1111011, 1111100, 1111101, 1111110, 1111111, 10000000, 10000001, 10000010, 10000011, 10000100, 10000101, 10000110, 10000111, 10001000, 10001001, 10001010, 10001011, 10001100, 10001101, 10001110, 10001111, 10010000, 10010001, 10010010, 10010011, 10010100, 10010101, 10010110, 10010111, 10011000, 10011001, 10011010, 10011011, 10011100, 10011101, 10011110, 10011111, 10100000, 10100001, 10100010, 10100011, 10100100, 10100101, 10100110, 10100111, 10101000, 10101001, 10101010, 10101011, 10101100, 10101101, 10101110, 10101111, 10110000, 10110001, 10110010, 10110011, 10110100, 10110101, 10110110, 10110111, 10111000, 10111001, 10111010, 10111011, 10111100, 10111101, 10111110, 10111111, 11000000, 11000001, 11000010, 11000011, 11000100, 11000101, 11000110, 11000111, 11001000, 11001001, 11001010, 11001011, 11001100, 11001101, 11001110, 11001111, 11010000, 11010001, 11010010, 11010011, 11010100, 11010101, 11010110, 11010111, 11011000, 11011001, 11011010, 11011011, 11011100, 11011101, 11011110, 11011111, 11100000, 11100001, 11100010, 11100011, 11100100, 11100101, 11100110, 11100111, 11101000, 11101001, 11101010, 11101011, 11101100, 11101101, 11101110, 11101111, 11110000, 11110001, 11110010, 11110011, 11110100, 11110101, 11110110, 11110111, 11111000, 11111001, 11111010, 11111011, 11111100, 11111101, 11111110, 11111111]
h = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001000, 1001001, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1010000, 1010001, 1010010, 1010011, 1010100, 1010101, 1010110, 1010111, 1011000, 1011001, 1011010, 1011011, 1011100, 1011101, 1011110, 1011111, 1100000, 1100001, 1100010, 1100011, 1100100, 1100101, 1100110, 1100111, 1101000, 1101001, 1101010, 1101011, 1101100, 1101101, 1101110, 1101111, 1110000, 1110001, 1110010, 1110011, 1110100, 1110101, 1110110, 1110111, 1111000, 1111001, 1111010, 1111011, 1111100, 1111101, 1111110, 1111111, 10000000, 10000001, 10000010, 10000011, 10000100, 10000101, 10000110, 10000111, 10001000, 10001001, 10001010, 10001011, 10001100, 10001101, 10001110, 10001111, 10010000, 10010001, 10010010, 10010011, 10010100, 10010101, 10010110, 10010111, 10011000, 10011001, 10011010, 10011011, 10011100, 10011101, 10011110, 10011111, 10100000, 10100001, 10100010, 10100011, 10100100, 10100101, 10100110, 10100111, 10101000, 10101001, 10101010, 10101011, 10101100, 10101101, 10101110, 10101111, 10110000, 10110001, 10110010, 10110011, 10110100, 10110101, 10110110, 10110111, 10111000, 10111001, 10111010, 10111011, 10111100, 10111101, 10111110, 10111111, 11000000, 11000001, 11000010, 11000011, 11000100, 11000101, 11000110, 11000111, 11001000, 11001001, 11001010, 11001011, 11001100, 11001101, 11001110, 11001111, 11010000, 11010001, 11010010, 11010011, 11010100, 11010101, 11010110, 11010111, 11011000, 11011001, 11011010, 11011011, 11011100, 11011101, 11011110, 11011111, 11100000, 11100001, 11100010, 11100011, 11100100, 11100101, 11100110, 11100111, 11101000, 11101001, 11101010, 11101011, 11101100, 11101101, 11101110, 11101111, 11110000, 11110001, 11110010, 11110011, 11110100, 11110101, 11110110, 11110111, 11111000, 11111001, 11111010, 11111011, 11111100, 11111101, 11111110, 11111111, 100000000, 100000001, 100000010, 100000011, 100000100, 100000101, 100000110, 100000111, 100001000, 100001001, 100001010, 100001011, 100001100, 100001101, 100001110, 100001111, 100010000, 100010001, 100010010, 100010011, 100010100, 100010101, 100010110, 100010111, 100011000, 100011001, 100011010, 100011011, 100011100, 100011101, 100011110, 100011111, 100100000, 100100001, 100100010, 100100011, 100100100, 100100101, 100100110, 100100111, 100101000, 100101001, 100101010, 100101011, 100101100, 100101101, 100101110, 100101111, 100110000, 100110001, 100110010, 100110011, 100110100, 100110101, 100110110, 100110111, 100111000, 100111001, 100111010, 100111011, 100111100, 100111101, 100111110, 100111111, 101000000, 101000001, 101000010, 101000011, 101000100, 101000101, 101000110, 101000111, 101001000, 101001001, 101001010, 101001011, 101001100, 101001101, 101001110, 101001111, 101010000, 101010001, 101010010, 101010011, 101010100, 101010101, 101010110, 101010111, 101011000, 101011001, 101011010, 101011011, 101011100, 101011101, 101011110, 101011111, 101100000, 101100001, 101100010, 101100011, 101100100, 101100101, 101100110, 101100111, 101101000, 101101001, 101101010, 101101011, 101101100, 101101101, 101101110, 101101111, 101110000, 101110001, 101110010, 101110011, 101110100, 101110101, 101110110, 101110111, 101111000, 101111001, 101111010, 101111011, 101111100, 101111101, 101111110, 101111111, 110000000, 110000001, 110000010, 110000011, 110000100, 110000101, 110000110, 110000111, 110001000, 110001001, 110001010, 110001011, 110001100, 110001101, 110001110, 110001111, 110010000, 110010001, 110010010, 110010011, 110010100, 110010101, 110010110, 110010111, 110011000, 110011001, 110011010, 110011011, 110011100, 110011101, 110011110, 110011111, 110100000, 110100001, 110100010, 110100011, 110100100, 110100101, 110100110, 110100111, 110101000, 110101001, 110101010, 110101011, 110101100, 110101101, 110101110, 110101111, 110110000, 110110001, 110110010, 110110011, 110110100, 110110101, 110110110, 110110111, 110111000, 110111001, 110111010, 110111011, 110111100, 110111101, 110111110, 110111111, 111000000, 111000001, 111000010, 111000011, 111000100, 111000101, 111000110, 111000111, 111001000, 111001001, 111001010, 111001011, 111001100, 111001101, 111001110, 111001111, 111010000, 111010001, 111010010, 111010011, 111010100, 111010101, 111010110, 111010111, 111011000, 111011001, 111011010, 111011011, 111011100, 111011101, 111011110, 111011111, 111100000, 111100001, 111100010, 111100011, 111100100, 111100101, 111100110, 111100111, 111101000, 111101001, 111101010, 111101011, 111101100, 111101101, 111101110, 111101111, 111110000, 111110001, 111110010, 111110011, 111110100, 111110101, 111110110, 111110111, 111111000, 111111001, 111111010, 111111011, 111111100, 111111101, 111111110, 111111111]
n = int(input())
dlugosc = len(str(n))
if dlugosc == 1:
print(1)
elif dlugosc == 2:
for i in range(len(a)):
if a[i] > n:
print(len(a[:i]))
exit()
print(len(a))
elif dlugosc == 3:
for i in range(len(b)):
if b[i] > n:
print(len(b[:i]))
exit()
print(len(b))
elif dlugosc == 4:
for i in range(len(c)):
if c[i] > n:
print(len(c[:i]))
exit()
print(len(c))
elif dlugosc == 5:
for i in range(len(d)):
if d[i] > n:
print(len(d[:i]))
exit()
print(len(d))
elif dlugosc == 6:
for i in range(len(e)):
if e[i] > n:
print(len(e[:i]))
exit()
print(len(e))
elif dlugosc == 7:
for i in range(len(f)):
if f[i] > n:
print(len(f[:i]))
exit()
print(len(f))
elif dlugosc == 8:
for i in range(len(g)):
if g[i] > n:
print(len(g[:i]))
exit()
print(len(g))
elif dlugosc == 9:
for i in range(len(h)):
if h[i] > n:
print(len(h[:i]))
exit()
print(len(h))
elif dlugosc == 10:
print(len(h) + 1)
```
| 3,932 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Tags: brute force, implementation, math
Correct Solution:
```
n=int(input())
k=1
z=1
s=1
if n<10 :
print(1)
exit()
for i in range(1,len(str(n))-1) :
z*=2
k+=z
n=str(n)
d=0
if int(n[0])>1 :
d=1
for i in range(1,len(n)) :
if int(n[i])>1 :
d=1
if n[i]!='0' or d==1 :
k+=2**(len(n)-i-1)
print(k+1)
```
| 3,933 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Tags: brute force, implementation, math
Correct Solution:
```
def main():
alpha = 'abcdefghijklmnopqrstuvwxyz'
ALPHA = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
inf = 1e17
mod = 10 ** 9 + 7
# Max = 10 ** 1
# primes = []
# prime = [True for i in range(Max + 1)]
# p = 2
# while (p * p <= Max + 1):
#
# # If prime[p] is not
# # changed, then it is a prime
# if (prime[p] == True):
#
# # Update all multiples of p
# for i in range(p * p, Max + 1, p):
# prime[i] = False
# p += 1
#
# for p in range(2, Max + 1):
# if prime[p]:
# primes.append(p)
#
# print(primes)
def factorial(n):
f = 1
for i in range(1, n + 1):
f = (f * i) % mod # Now f never can
# exceed 10^9+7
return f
def ncr(n, r):
# initialize numerator
# and denominator
num = den = 1
for i in range(r):
num = (num * (n - i)) % mod
den = (den * (i + 1)) % mod
return (num * pow(den,
mod - 2, mod)) % mod
def solve(n):
cnt = 0
for i in range(1,pow(2,len(str(n)))):
if int(bin(i)[2:]) <= n:
cnt += 1
print(cnt)
pass
t = 1#int(input())
ans = []
for _ in range(t):
n = int(input())
#n,m = map(int, input().split())
# s = input()
#arr = [int(x) for x in input().split()]
# a2 = [int(x) for x in input().split()]
# grid = []
# for i in range(n):
# grid.append([int(x) for x in input().split()][::-1])
ans.append(solve(n))
# for answer in ans:
# print(answer)
if __name__ == "__main__":
import sys, threading
import bisect
import math
import itertools
from sys import stdout
# Sorted Containers
import heapq
from queue import PriorityQueue
# Tree Problems
#sys.setrecursionlimit(2 ** 32 // 2 - 1)
#threading.stack_size(1 << 27)
# fast io
input = sys.stdin.readline
thread = threading.Thread(target=main)
thread.start()
thread.join()
```
| 3,934 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Tags: brute force, implementation, math
Correct Solution:
```
n=int(input())
a=[]
c=int
for i in range(515):
a.append(c(bin(i)[2:]))
a.remove(0)
ans=0
for i in a:
if i<=n:
ans+=1
print(ans)
```
| 3,935 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Tags: brute force, implementation, math
Correct Solution:
```
def answer(n):
k=len(str(n))
count=2**(k-1)-1
arr=["1"]
for i in range(k-1):
temp=[]
for x in range(len(arr)):
temp.append(arr[x]+"1")
temp.append(arr[x]+"0")
arr=temp[:]
for i in arr:
if int(i)<=n:
count+=1
return count
t=int(input())
print(answer(t))
```
| 3,936 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Tags: brute force, implementation, math
Correct Solution:
```
n=int(input())
x=1
c=0
while(int(bin(x)[2:])<=n):
x+=1
c+=1
print(c)
```
| 3,937 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Tags: brute force, implementation, math
Correct Solution:
```
num=int(input())
li=[]
for i in range(1,520) :
li.append(int((bin(i)).replace("0b","")) )
count=0
for i in li :
if i<=num :
count+=1
print(count)
```
| 3,938 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Tags: brute force, implementation, math
Correct Solution:
```
import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
# Hex nums
#flop [not conv to bin exact :))]
n,=I()
x=[]
for i in range(1,2**10+1):
k=format(i,'b');p=0
o=len(k)
for j in range(o):
if k[o-j-1]=='1':
p+=10**(j)
x.append(p)
an=0
for j in range(len(x)):
if x[j]>n:
an=j;break
print(an)
```
| 3,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Submitted Solution:
```
n = input()
result = 2**len(n)-1
for i in range(len(n)):
if n[i]=='0':
result-=2**(len(n)-i-1)
elif n[i]=='1':
continue
else:
break
print(result)
```
Yes
| 3,940 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Submitted Solution:
```
import sys
sys_input = sys.stdin.readline
def inp():
return int(sys_input())
def minp():
return map(int,sys_input().split())
def linp():
return list(map(int,sys_input().split()))
def inpsl():
return list(input().split())
def write(s):
sys.stdout.write(s+" ")
def main():
n = inp()
cnt =0
for i in range(1,2**10):
if int(bin(i)[2:]) <= n:
cnt += 1
else:
break
print(cnt)
main()
```
Yes
| 3,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Submitted Solution:
```
n = 0
res = 0
def dfs(x):
global n
global res
if x > n:
return
res += 1
dfs(x * 10)
dfs(x * 10 + 1)
n = int(input())
dfs(1)
print(res)
```
Yes
| 3,942 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Submitted Solution:
```
n=int(input())
def jinzhi(a):
k=bin(a)
return int(k[2:])
i=1
while jinzhi(i)<=n:
i=i+1
print(i-1)
```
Yes
| 3,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Submitted Solution:
```
#http://codeforces.com/problemset/problem/9/C
#solved
a = [1, 10, 11]
b = [1, 10, 11, 100, 101, 110, 111]
c = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111]
d = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111]
e = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111]
f = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001000, 1001001, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1010000, 1010001, 1010010, 1010011, 1010100, 1010101, 1010110, 1010111, 1011000, 1011001, 1011010, 1011011, 1011100, 1011101, 1011110, 1011111, 1100000, 1100001, 1100010, 1100011, 1100100, 1100101, 1100110, 1100111, 1101000, 1101001, 1101010, 1101011, 1101100, 1101101, 1101110, 1101111, 1110000, 1110001, 1110010, 1110011, 1110100, 1110101, 1110110, 1110111, 1111000, 1111001, 1111010, 1111011, 1111100, 1111101, 1111110, 1111111]
g = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001000, 1001001, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1010000, 1010001, 1010010, 1010011, 1010100, 1010101, 1010110, 1010111, 1011000, 1011001, 1011010, 1011011, 1011100, 1011101, 1011110, 1011111, 1100000, 1100001, 1100010, 1100011, 1100100, 1100101, 1100110, 1100111, 1101000, 1101001, 1101010, 1101011, 1101100, 1101101, 1101110, 1101111, 1110000, 1110001, 1110010, 1110011, 1110100, 1110101, 1110110, 1110111, 1111000, 1111001, 1111010, 1111011, 1111100, 1111101, 1111110, 1111111, 10000000, 10000001, 10000010, 10000011, 10000100, 10000101, 10000110, 10000111, 10001000, 10001001, 10001010, 10001011, 10001100, 10001101, 10001110, 10001111, 10010000, 10010001, 10010010, 10010011, 10010100, 10010101, 10010110, 10010111, 10011000, 10011001, 10011010, 10011011, 10011100, 10011101, 10011110, 10011111, 10100000, 10100001, 10100010, 10100011, 10100100, 10100101, 10100110, 10100111, 10101000, 10101001, 10101010, 10101011, 10101100, 10101101, 10101110, 10101111, 10110000, 10110001, 10110010, 10110011, 10110100, 10110101, 10110110, 10110111, 10111000, 10111001, 10111010, 10111011, 10111100, 10111101, 10111110, 10111111, 11000000, 11000001, 11000010, 11000011, 11000100, 11000101, 11000110, 11000111, 11001000, 11001001, 11001010, 11001011, 11001100, 11001101, 11001110, 11001111, 11010000, 11010001, 11010010, 11010011, 11010100, 11010101, 11010110, 11010111, 11011000, 11011001, 11011010, 11011011, 11011100, 11011101, 11011110, 11011111, 11100000, 11100001, 11100010, 11100011, 11100100, 11100101, 11100110, 11100111, 11101000, 11101001, 11101010, 11101011, 11101100, 11101101, 11101110, 11101111, 11110000, 11110001, 11110010, 11110011, 11110100, 11110101, 11110110, 11110111, 11111000, 11111001, 11111010, 11111011, 11111100, 11111101, 11111110, 11111111]
h = [0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001000, 1001001, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1010000, 1010001, 1010010, 1010011, 1010100, 1010101, 1010110, 1010111, 1011000, 1011001, 1011010, 1011011, 1011100, 1011101, 1011110, 1011111, 1100000, 1100001, 1100010, 1100011, 1100100, 1100101, 1100110, 1100111, 1101000, 1101001, 1101010, 1101011, 1101100, 1101101, 1101110, 1101111, 1110000, 1110001, 1110010, 1110011, 1110100, 1110101, 1110110, 1110111, 1111000, 1111001, 1111010, 1111011, 1111100, 1111101, 1111110, 1111111, 10000000, 10000001, 10000010, 10000011, 10000100, 10000101, 10000110, 10000111, 10001000, 10001001, 10001010, 10001011, 10001100, 10001101, 10001110, 10001111, 10010000, 10010001, 10010010, 10010011, 10010100, 10010101, 10010110, 10010111, 10011000, 10011001, 10011010, 10011011, 10011100, 10011101, 10011110, 10011111, 10100000, 10100001, 10100010, 10100011, 10100100, 10100101, 10100110, 10100111, 10101000, 10101001, 10101010, 10101011, 10101100, 10101101, 10101110, 10101111, 10110000, 10110001, 10110010, 10110011, 10110100, 10110101, 10110110, 10110111, 10111000, 10111001, 10111010, 10111011, 10111100, 10111101, 10111110, 10111111, 11000000, 11000001, 11000010, 11000011, 11000100, 11000101, 11000110, 11000111, 11001000, 11001001, 11001010, 11001011, 11001100, 11001101, 11001110, 11001111, 11010000, 11010001, 11010010, 11010011, 11010100, 11010101, 11010110, 11010111, 11011000, 11011001, 11011010, 11011011, 11011100, 11011101, 11011110, 11011111, 11100000, 11100001, 11100010, 11100011, 11100100, 11100101, 11100110, 11100111, 11101000, 11101001, 11101010, 11101011, 11101100, 11101101, 11101110, 11101111, 11110000, 11110001, 11110010, 11110011, 11110100, 11110101, 11110110, 11110111, 11111000, 11111001, 11111010, 11111011, 11111100, 11111101, 11111110, 11111111, 100000000, 100000001, 100000010, 100000011, 100000100, 100000101, 100000110, 100000111, 100001000, 100001001, 100001010, 100001011, 100001100, 100001101, 100001110, 100001111, 100010000, 100010001, 100010010, 100010011, 100010100, 100010101, 100010110, 100010111, 100011000, 100011001, 100011010, 100011011, 100011100, 100011101, 100011110, 100011111, 100100000, 100100001, 100100010, 100100011, 100100100, 100100101, 100100110, 100100111, 100101000, 100101001, 100101010, 100101011, 100101100, 100101101, 100101110, 100101111, 100110000, 100110001, 100110010, 100110011, 100110100, 100110101, 100110110, 100110111, 100111000, 100111001, 100111010, 100111011, 100111100, 100111101, 100111110, 100111111, 101000000, 101000001, 101000010, 101000011, 101000100, 101000101, 101000110, 101000111, 101001000, 101001001, 101001010, 101001011, 101001100, 101001101, 101001110, 101001111, 101010000, 101010001, 101010010, 101010011, 101010100, 101010101, 101010110, 101010111, 101011000, 101011001, 101011010, 101011011, 101011100, 101011101, 101011110, 101011111, 101100000, 101100001, 101100010, 101100011, 101100100, 101100101, 101100110, 101100111, 101101000, 101101001, 101101010, 101101011, 101101100, 101101101, 101101110, 101101111, 101110000, 101110001, 101110010, 101110011, 101110100, 101110101, 101110110, 101110111, 101111000, 101111001, 101111010, 101111011, 101111100, 101111101, 101111110, 101111111, 110000000, 110000001, 110000010, 110000011, 110000100, 110000101, 110000110, 110000111, 110001000, 110001001, 110001010, 110001011, 110001100, 110001101, 110001110, 110001111, 110010000, 110010001, 110010010, 110010011, 110010100, 110010101, 110010110, 110010111, 110011000, 110011001, 110011010, 110011011, 110011100, 110011101, 110011110, 110011111, 110100000, 110100001, 110100010, 110100011, 110100100, 110100101, 110100110, 110100111, 110101000, 110101001, 110101010, 110101011, 110101100, 110101101, 110101110, 110101111, 110110000, 110110001, 110110010, 110110011, 110110100, 110110101, 110110110, 110110111, 110111000, 110111001, 110111010, 110111011, 110111100, 110111101, 110111110, 110111111, 111000000, 111000001, 111000010, 111000011, 111000100, 111000101, 111000110, 111000111, 111001000, 111001001, 111001010, 111001011, 111001100, 111001101, 111001110, 111001111, 111010000, 111010001, 111010010, 111010011, 111010100, 111010101, 111010110, 111010111, 111011000, 111011001, 111011010, 111011011, 111011100, 111011101, 111011110, 111011111, 111100000, 111100001, 111100010, 111100011, 111100100, 111100101, 111100110, 111100111, 111101000, 111101001, 111101010, 111101011, 111101100, 111101101, 111101110, 111101111, 111110000, 111110001, 111110010, 111110011, 111110100, 111110101, 111110110, 111110111, 111111000, 111111001, 111111010, 111111011, 111111100, 111111101, 111111110, 111111111]
n = int(input())
dlugosc = len(str(n))
if dlugosc == 1:
print(1)
elif dlugosc == 2:
for i in range(len(a)):
if a[i] > n:
print(len(a[:i]))
exit()
print(len(a))
elif dlugosc == 3:
for i in range(len(b)):
if b[i] > n:
print(len(b[:i]))
exit()
print(len(b))
elif dlugosc == 4:
for i in range(len(c)):
if c[i] > n:
print(len(c[:i]))
exit()
print(len(c))
elif dlugosc == 5:
for i in range(len(d)):
if d[i] > n:
print(len(d[:i]))
exit()
print(len(d))
elif dlugosc == 6:
for i in range(len(e)):
if e[i] > n:
print(len(e[:i]))
exit()
print(len(e))
elif dlugosc == 7:
for i in range(len(f)):
if f[i] > n:
print(len(f[:i]))
exit()
print(len(f))
elif dlugosc == 8:
for i in range(len(g)):
if g[i] > n:
print(len(g[:i]))
exit()
print(len(g))
elif dlugosc == 2:
for i in range(len(h)):
if h[i] > n:
print(len(h[:i]))
exit()
print(len(h))
```
No
| 3,944 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Submitted Solution:
```
"""
from sys import stdin, stdout
import math
from functools import reduce
import statistics
import numpy as np
import itertools
import operator
"""
from math import sqrt
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")
def prog_name():
# n = int(input())
# if n < 10:
# print(1)
# else:
# cnt = 0
# check = 9
# t = 9
# flag = True
# while flag:
# if check <= n:
# cnt += 2**(len(str(check)) - 1)
# check = check*10 + t
# else:
# flag = False
# num = (int('1'*len(str(n))))
# start = int(str(check)[:-1])
#
# if (min(n,num)) == num and n != start:
# cnt += 2 ** (len(str(check)) - 1)
# else:
# for x in range(start + 1,min(n,num) + 1):
# j = str(x)
# if j.count('0') + j.count('1') == len(j):
# cnt += 1
# print(cnt)
x = 1
n = int(input())
flag = True
while flag:
if int(bin(x)[2:]) < n:
x +=1
else:
flag = False
print(x - 1)
def main():
t = 1
for unique in range(t):
# print("Case #"+str(unique+1)+":",end = " ")
prog_name()
if __name__ == "__main__":
main()
```
No
| 3,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Submitted Solution:
```
x=input("saisir un entier ")
tab=list(x)
entry=int(x)
aid=str()
if entry==0 :
print("0");
elif entry<=9 :
print("2")
for i in range(len(tab)) :
tab[i]=int(tab[i])
if 1<tab[0]:
x=(2**len(tab))-1
elif tab[0]==1:
if 1<tab[1]:
x=(2**len(tab))-1
else:
x=(2**(len(tab)-1))-1
A=list()
a=str()
for i in range(len(tab)) :
if i == 0 :
A.append(1)
a=str(A[i])
else :
A.append(0)
a=a+str(A[i])
number=int(a)
b=str(number)
b=int(b,2)
while number<=entry :
x=x+1
alter=int(bin(b),2)+int(bin(1),2)
help=bin(alter)
aid=help[2:]
s=str(aid)
number=int(s)
b=str(number)
b=int(b,2)
print(x)
```
No
| 3,946 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
Input
10
Output
2
Note
For n = 10 the answer includes numbers 1 and 10.
Submitted Solution:
```
n = input()
ans = 2 ** len(n) - 1
i = 1
while i < len(n):
if n[i] == '0':
ans -= 2**(len(n)-i-1)
i += 1
print(ans)
```
No
| 3,947 |
Provide a correct Python 3 solution for this coding contest problem.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
"Correct Solution:
```
s = str(input())
if(s[-1]=="s"):
print(s+"es")
else:
print(s+"s")
```
| 3,948 |
Provide a correct Python 3 solution for this coding contest problem.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
"Correct Solution:
```
S=input()
if S[-1] == 's':
print(S+"es")
else:
print(S+"s")
```
| 3,949 |
Provide a correct Python 3 solution for this coding contest problem.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
"Correct Solution:
```
s = input()
if s[-1] == "s":
print(s+"es")
else:
print(s+"s")
```
| 3,950 |
Provide a correct Python 3 solution for this coding contest problem.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
"Correct Solution:
```
S = input()
if S.endswith("s"): print(S + "es")
else: print(S + "s")
```
| 3,951 |
Provide a correct Python 3 solution for this coding contest problem.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
"Correct Solution:
```
s = input()
a = s[-1]
if(a == 's'):print(s+"es")
else:print(s+"s")
```
| 3,952 |
Provide a correct Python 3 solution for this coding contest problem.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
"Correct Solution:
```
a=input()
if(a[len(a)-1]=="s"):
a+="e"
print(a+"s")
```
| 3,953 |
Provide a correct Python 3 solution for this coding contest problem.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
"Correct Solution:
```
x=input()
if x[-1]=="s":
print(x+"es")
else:
print(x+"s")
```
| 3,954 |
Provide a correct Python 3 solution for this coding contest problem.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
"Correct Solution:
```
A = input()
A = A + "es" if A[-1] == "s" else A + "s"
print(A)
```
| 3,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
Submitted Solution:
```
s=input()
s+=(s[-1]=='s')*'e'
print(s+'s')
```
Yes
| 3,956 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
Submitted Solution:
```
n=input()
print([n+"s",n+"es"][n[-1]=="s"])
```
Yes
| 3,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
Submitted Solution:
```
s = input()
if s[-1] == "s":
s += "e"
print(s + "s")
```
Yes
| 3,958 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
Submitted Solution:
```
s=str(input())
print(s+"s" if s[-1]!="s" else s+"es")
```
Yes
| 3,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
Submitted Solution:
```
print(input()+'s')
```
No
| 3,960 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
Submitted Solution:
```
S = input()
if S[-1] == 'e':
print(S+'s')
else:
print(S+'es')
```
No
| 3,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
Submitted Solution:
```
from pattern.en import pluralize, singularize
s=input()
print(pluralize(s))
```
No
| 3,962 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`, append `es` to the end of the singular form.
You are given the singular form S of a Taknese noun. Output its plural form.
Constraints
* S is a string of length 1 between 1000, inclusive.
* S contains only lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
Print the plural form of the given Taknese word.
Examples
Input
apple
Output
apples
Input
bus
Output
buses
Input
box
Output
boxs
Submitted Solution:
```
n = int(input())
count = 0
for i in range(n):
if i == 0:
count = 0
else:
if n % i == 0:
count += (n//i) -1
else:
count += (n//i)
print(count)
```
No
| 3,963 |
Provide a correct Python 3 solution for this coding contest problem.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
"Correct Solution:
```
from math import cos, sqrt, pi
A, B, H, M = map(int, input().split())
print(sqrt(A*A + B*B - 2*A*B*cos((30*H-5.5*M)*pi/180)))
```
| 3,964 |
Provide a correct Python 3 solution for this coding contest problem.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
"Correct Solution:
```
import math
a,b,h,m=map(int,input().split())
kaku=math.cos(math.radians(abs((h*60+m)/2-m*6)))
print((a**2+b**2-2*a*b*kaku)**(1/2))
```
| 3,965 |
Provide a correct Python 3 solution for this coding contest problem.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
"Correct Solution:
```
A,B,H,M=map(int,input().split())
k=abs(11/2*M-30*H)
import math
l=2*A*B*math.cos(k*math.pi/180)
print(math.sqrt(A**2+B**2-l))
```
| 3,966 |
Provide a correct Python 3 solution for this coding contest problem.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
"Correct Solution:
```
import math
A,B,H,M=map(int,input().split())
P=H*30-M*5.5
c = math.cos(math.radians(P))
L=A*A+B*B-2*A*B*c
x=math.sqrt(L)
print(x)
```
| 3,967 |
Provide a correct Python 3 solution for this coding contest problem.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
"Correct Solution:
```
import math
a,b,h,m = map(int,input().split())
s = abs(6*m-(h*30+m/2))
print((a**2+b**2-2*a*b*math.cos(math.radians(s)))**(1/2))
```
| 3,968 |
Provide a correct Python 3 solution for this coding contest problem.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
"Correct Solution:
```
import math
a,b,h,m=map(int,input().split())
c=math.radians(abs(30*(h+m/60)-6*m))
print(math.sqrt(a**2+b**2-2*a*b*math.cos(c)))
```
| 3,969 |
Provide a correct Python 3 solution for this coding contest problem.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
"Correct Solution:
```
from math import *
a,b,h,m=map(int,input().split())
print(sqrt(a*a+b*b-2*a*b*cos(radians(abs(30*h-5.5*m)))))
```
| 3,970 |
Provide a correct Python 3 solution for this coding contest problem.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
"Correct Solution:
```
import math
a, b, h, m = map(int, input().split())
print(math.sqrt(a**2 + b**2 - 2*a*b*math.cos(2*(h*60 - m*11)*math.pi/720)))
```
| 3,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
Submitted Solution:
```
from math import*
A,B,H,M = map(int, input().split())
print((A*A+B*B-2*A*B*cos((M*11/360-H/6)*pi))**.5)
```
Yes
| 3,972 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
Submitted Solution:
```
import math
A,B,H,M=map(int,input().split())
print(math.sqrt(A**2+B**2-2*A*B*math.cos((30*H-11*M/2)*math.pi /180)))
```
Yes
| 3,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
Submitted Solution:
```
import math
a,b,h,m = map(int,input().split())
c2 = a**2 + b**2 - 2 * a * b * math.cos((60*h-11*m)*math.pi/360)
print(math.sqrt(c2))
```
Yes
| 3,974 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
Submitted Solution:
```
import math
A,B,H,M=map(int,input().split())
X=H*30+M*0.5
Y=M*6
t=(X-Y)*math.pi/180
print(math.sqrt(A**2+B**2-2*A*B*math.cos(t)))
```
Yes
| 3,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
Submitted Solution:
```
from math import degrees, cos,sqrt ,radians
A, B, H, M = map(int, input().split())
if H == 6 and M == 0:
print(A+B)
exit()
kaku = abs((6*M-(0.5*M+H*30)))%180
# print(kaku)
if kaku > 90:
kaku = 180-kaku
# print(kaku)
# print(cos(radians(kaku)))
print(sqrt(A**2 + B**2 -2*A*B*abs(cos(radians(kaku)))))
# print(-A*B*cos(kaku))
```
No
| 3,976 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
Submitted Solution:
```
import math
a, b, h, m = map(int, input().split())
p1 = 30 * h + m // 2
p2 = 6 * m
s = min(abs(p1 - p2), abs(abs(p1 - p2) - 360))
ans = pow(a * math.sin(math.radians(s)), 2) + pow((b - a * math.cos(math.radians(s))), 2)
print((ans) ** (1 / 2))
```
No
| 3,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
Submitted Solution:
```
import sys
import math
from collections import Counter
import itertools
import fractions
#from functools import reduce
# 入力を整数に変換して受け取る
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
# 入力全てを整数に変換したものの配列を受け取る
def LI(): return list(map(int, sys.stdin.readline().split()))
# 入力全てを整数に変換して1引いたものの配列を受け取る
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SR(): return sys.stdin.readline().rstrip()
ascii_lowercase = 'abcdefghijklmnopqrstuvwxyz'
ascii_uppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
ascii_uppercase2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZ'
# リストを改行区切りで出力する
p2D = lambda x: print(*x, sep="\n")
p2E = lambda x: print(''.join(x))
p2S = lambda x: print(*x, sep=" ")
# ###########################################
A,B,H,M=MI()
kakuM = M*6
kakuH = H*30+0.5*M
kakudo = kakuH - kakuM
if kakudo > 180:
kakudo -= 180
ans = A**2 + B**2 - 2 * A * B * math.cos(math.radians(kakudo))
print(math.sqrt(ans))
```
No
| 3,978 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.
An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.
At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 1000
* 0 \leq H \leq 11
* 0 \leq M \leq 59
Input
Input is given from Standard Input in the following format:
A B H M
Output
Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.
Examples
Input
3 4 9 0
Output
5.00000000000000000000
Input
3 4 10 40
Output
4.56425719433005567605
Submitted Solution:
```
from math import sqrt, cos, pi
a, b, h, m = map(int, input().split())
rot_a = h*30 + m*0.5
rot_b = m*6
rot = rot_a-rot_b if rot_a>rot_b else rot_b-rot_a
if rot > 180:
rot = min(rot_a, rot_b) + 360 - max(rot_a, rot_b)
ans = sqrt(a**2 + b**2 - 2*a*b*cos(rot*pi/180))
print(rot)
print(ans)
```
No
| 3,979 |
Provide a correct Python 3 solution for this coding contest problem.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
"Correct Solution:
```
# 2点 or 3点
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def 外心(p1,p2,p3):
# https://qiita.com/tatesuke/items/59133758ec25146766c3
a,b = p1
c,d = p2
e,f = p3
aa = a * a
bb = b * b
cc = c * c
dd = d * d
ee = e * e
ff = f * f
# 直線判定がめんどくさい!!!
if (2 * (e - a)*(b - d) - 2 * (c - a) * (b - f)) == 0:
return 0, 0
py = ((e - a) * (aa + bb - cc - dd) - (c - a) * (aa + bb - ee- ff)) / (2 * (e - a)*(b - d) - 2 * (c - a) * (b - f))
if c == a:
px = (2 * (b - f) * py - aa - bb + ee + ff) / (2 * (e - a))
else:
px = (2 * (b - d) * py - aa - bb + cc + dd) / (2 * (c - a))
return px, py
n = int(input())
points = [list(map(int, input().split())) for i in range(n)]
from itertools import combinations
候補 = []
# two
for p1, p2 in combinations(points, 2):
px, py = (p1[0]+p2[0])/2, (p1[1]+p2[1])/2
候補.append((px, py))
# three
for p1, p2, p3 in combinations(points, 3):
候補.append(外心(p1,p2,p3))
from math import sqrt
ans = float("Inf")
for x,y in 候補:
tmp = 0
for px, py in points:
tmp = max(tmp, (px-x)**2 + (py-y)**2)
ans = min(ans, tmp)
ans = sqrt(ans)
print(ans)
```
| 3,980 |
Provide a correct Python 3 solution for this coding contest problem.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
"Correct Solution:
```
n=int(input())
def bai(x):
return int(x)*2.0
xy=[list(map(bai, input().split())) for _ in range(n)]
#nは50だから4重ループでも間に合う？
#2500*2500
if n==2:
print((((xy[0][0]-xy[1][0])**2 +(xy[0][1]-xy[1][1])**2) **0.5)*0.5*0.5)
exit()
ans=10**9
def gaishin(a,b,c,d,e,f):
if (a-c)*(d-f)==(c-e)*(b-d):
y=(max([b,d,f])-min([b,d,f]))/2.0+min([b,d,f])
x=(max([a,c,e])-min([a,c,e]))/2.0+min([a,c,e])
#print(a,b,c,d,e,f,"==>",x,y)
else:
y=1.0*((e-a)*(a**2+b**2-c**2-d**2)-(c-a)*(a**2+b**2-e**2-f**2))/(2.0*(e-a)*(b-d)-2.0*(c-a)*(b-f))
if e-a!=0:
x=(2.0*(b-f)*y-a**2-b**2+e**2+f**2)/(2.0*(e-a))
elif c-a!=0:
x=(2.0*(b-d)*y-a**2-b**2+c**2+d**2)/(2.0*(c-a))
else:
print(a,b,c,d,e,f)
exit()
return x,y
xy.sort()
for i in range(n-2):
for j in range(i+1,n-1):
for k in range(i+2,n):
#print(xy[i],xy[j],xy[k])
gx,gy=gaishin(xy[i][0],xy[i][1],xy[j][0],xy[j][1],xy[k][0],xy[k][1])
d=((1.0*(gx-xy[i][0]))**2+(1.0*(gy-xy[i][1]))**2)
flag=0
#print(gx,gy)
for (x,y) in xy:
if (x,y) != xy[i] and (x,y) != xy[j] and (x,y) != xy[k]:
if ((1.0*(gx-x))**2+(1.0*(gy-y))**2) >d+1:
flag=1
#print(((1.0*(gx-x))**2+(1.0*(gy-y))**2)**0.5,"-",d,"=",((1.0*(gx-x))**2+(1.0*(gy-y))**2)**0.5-d)
break
if flag==0:
ans=min(ans,d**0.5)
#print(">>>>",xy[i],xy[j],xy[k],d**0.5/2,gx,gy)
print(ans/2)
```
| 3,981 |
Provide a correct Python 3 solution for this coding contest problem.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
"Correct Solution:
```
import sys
import cmath
from itertools import combinations
inf = float('inf')
n, *xy = map(int, sys.stdin.read().split())
xy = zip(*[iter(xy)] * 2)
z = [x + y*1j for x, y in xy]
def circumcenter(z1, z2, z3):
a = z2 - z1
b = z3 - z1
numerator = a * b * (b.conjugate() - a.conjugate())
denominator = a * b.conjugate() - a.conjugate() * b
o = numerator / denominator + z1
return o
def center(z1, z2):
return (z1 + z2) / 2
def main():
cand = []
for comb in combinations(z, 3):
try:
cand.append(circumcenter(*comb))
except:
pass
for comb in combinations(z, 2):
cand.append(center(*comb))
res = inf
for o in cand:
r = 0
for i in z:
r = max(r, abs(o - i))
res = min(res, r)
return res
if __name__ == '__main__':
ans = main()
print(ans)
```
| 3,982 |
Provide a correct Python 3 solution for this coding contest problem.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
"Correct Solution:
```
from math import sqrt
N=int(input())
XY=[tuple(map(int,input().split())) for _ in range(N)]
def calc(r):
lst=[]
for i in range(N-1):
x1=XY[i][0]
y1=XY[i][1]
for j in range(i+1,N):
x2=XY[j][0]
y2=XY[j][1]
diff=sqrt((x1-x2)**2+(y1-y2)**2)
if diff>2*r:
return False
h=sqrt(r**2-diff**2/4)
if x1==x2:
ny=(y1+y2)/2
lst.append([x1-h,ny])
lst.append([x1+h,ny])
elif y1==y2:
nx=(x1+x2)/2
lst.append([nx,y1-h])
lst.append([nx,y1+h])
else:
a=(y2-y1)/(x2-x1)
b=-1/a
size=sqrt(1+b**2)
dx=h/size
dy=dx*b
nx=(x1+x2)/2
ny=(y1+y2)/2
lst.append([nx+dx,ny+dy])
lst.append([nx-dx,ny-dy])
nr=r+10**(-9)
for x,y in lst:
flag=True
for X,Y in XY:
tmp=(x-X)**2+(y-Y)**2
if tmp>nr**2:
flag=False
break
if flag:
return True
return False
l=0
r=1000
for i in range(100):
mid=(l+r)/2
if calc(mid):
r=mid
else:
l=mid
print(r)
```
| 3,983 |
Provide a correct Python 3 solution for this coding contest problem.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
"Correct Solution:
```
#!/usr/bin/env python3
import sys
import math
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(2147483647)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
n = I()
xy = [LI() for _ in range(n)]
def d(x1,y1,x2,y2):
return math.sqrt((x1-x2)**2 + (y1-y2)**2)
def f(x,y):
res = 0
for x1,y1 in xy:
res = max(res,d(x,y,x1,y1))
return res
def g(x):
l,r = 0,1000
for _ in range(80):
a = (l*2+r)/3
b = (l+r*2)/3
if f(x,a) > f(x,b):
l = a
else:
r = b
return f(x,l)
l,r = 0,1000
for _ in range(80):
a = (l*2+r)/3
b = (l+r*2)/3
if g(a) > g(b):
l = a
else:
r = b
print(g(l))
```
| 3,984 |
Provide a correct Python 3 solution for this coding contest problem.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
"Correct Solution:
```
n = int(input())
l = [[0, 0] for _ in range(n)]
for i in range(n):
x, y = map(int, input().split())
l[i][0] = x
l[i][1] = y
import math
if n == 2:
r = math.sqrt((l[1][0]-l[0][0])**2+(l[1][1]-l[0][1])**2)/2
print(r)
exit()
ans = float('inf')
import itertools
combs = list(itertools.combinations(range(n), 3))
for comb in combs:
x0, y0 = l[comb[0]]
x1, y1 = l[comb[1]]
x2, y2 = l[comb[2]]
a = x1 - x0
b = y1 -y0
p = (x1**2-x0**2+y1**2-y0**2)/2
c = x2 - x0
d = y2 -y0
q = (x2**2-x0**2+y2**2-y0**2)/2
if a*d - b*c == 0:
continue
xc = (p*d-q*b)/(a*d-b*c)
yc = (-p*c+q*a)/(a*d-b*c)
r = math.sqrt((x0-xc)**2+(y0-yc)**2)
for i in range(n):
x, y = l[i]
if math.sqrt((x-xc)**2+(y-yc)**2) > r:
break
else:
ans = min(ans, r)
combs = list(itertools.combinations(range(n), 2))
for comb in combs:
x0, y0 = l[comb[0]]
x1, y1 = l[comb[1]]
xc = (x0+x1)/2
yc = (y0+y1)/2
r = math.sqrt((x0-xc)**2+(y0-yc)**2)
for i in range(n):
x, y = l[i]
if math.sqrt((x-xc)**2+(y-yc)**2) > r:
break
else:
ans = min(ans, r)
print(ans)
```
| 3,985 |
Provide a correct Python 3 solution for this coding contest problem.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
"Correct Solution:
```
def min_disc(points):
from random import sample
N = len(points)
if N == 1:
return points[0], 0
points = sample(points, N)
def cross(a, b):
return a.real * b.imag - a.imag * b.real
def norm2(a):
return a.real * a.real + a.imag * a.imag
def make_circle_3(a, b, c):
A, B, C = norm2(b-c), norm2(c-a), norm2(a-b)
S = cross(b-a, c-a)
p = (A*(B+C-A)*a + B*(C+A-B)*b + C*(A+B-C)*c) / (4*S*S)
radius = abs(p-a)
return p, radius
def make_circle_2(a, b):
c = (a+b) / 2
radius = abs(a-c)
return c, radius
def in_circle(point, circle):
return abs(point-circle[0]) <= circle[1]+1e-7
p0 = points[0]
circle = make_circle_2(p0, points[1])
for i, p_i in enumerate(points[2:], 2):
if not in_circle(p_i, circle):
circle = make_circle_2(p0, p_i)
for j, p_j in enumerate(points[1:i], 1):
if not in_circle(p_j, circle):
circle = make_circle_2(p_i, p_j)
for p_k in points[:j]:
if not in_circle(p_k, circle):
circle = make_circle_3(p_i, p_j, p_k)
return circle
N = int(input())
XY = []
for _ in range(N):
x, y = map(int, input().split())
XY.append(x + y*1j)
center, rad = min_disc(XY)
print(rad)
```
| 3,986 |
Provide a correct Python 3 solution for this coding contest problem.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
"Correct Solution:
```
from math import sqrt
N = int(input())
THRE = 0.0000001
ps = []
for _ in range(N):
x, y = map(int, input().split())
ps.append((x,y))
# 2点を直径とする円の中心と半径を求める
def circleThrough2Points(x1, y1, x2, y2):
if x1 == x2 and y1 == y2:
return None, None, None
px = (x1 + x2) / 2
py = (y1 + y2) / 2
r = (px - x1)**2 + (py - y1)**2
return px, py, r
# 3点を通る円の中心と半径を求める
def circleThrough3Points(x1, y1, x2, y2, x3, y3):
a, b, c, d, e, f = x1, y1, x2, y2, x3, y3
a2, b2, c2, d2, e2, f2 = a*a, b*b, c*c, d*d, e*e, f*f
py_1 = 2*(e-a)*(b-d) - 2*(c-a)*(b-f)
if py_1 == 0:
return None, None, None
py_2 = (e-a)*(a2+b2-c2-d2) - (c-a)*(a2+b2-e2-f2)
py = py_2 / py_1
if a != c:
px = (2*(b-d)*py-a2-b2+c2+d2) / (2*(c-a))
else:
px = (2*(b-f)*py-a2-b2+e2+f2) / (2*(e-a))
r = (px - a)**2 + (py - b)**2
return px, py, r
ans = float('inf')
for i in range(N):
x1, y1 = ps[i]
for j in range(i+1, N):
x2, y2 = ps[j]
px, py, r = circleThrough2Points(x1, y1, x2, y2)
if r is not None:
if all([(px-x)**2 + (py-y)**2 <= r + THRE for x,y in ps]):
ans = ans if ans < r else r
for k in range(j+1, N):
x3, y3 = ps[k]
px, py, r = circleThrough3Points(x1, y1, x2, y2, x3, y3)
if r is not None:
if all([(px-x)**2 + (py-y)**2 <= r + THRE for x,y in ps]):
ans = ans if ans < r else r
print(sqrt(ans))
```
| 3,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
Submitted Solution:
```
import math
N=int(input())
x=[0 for i in range(N)]
y=[0 for i in range(N)]
p=[]
for i in range(N):
x[i],y[i]=map(int,input().split())
p.append((x[i],y[i]))
def dist(p,q):
a,b=p
c,d=q
return math.sqrt((a-c)**2+(b-d)**2)
def check(R):
Q=[]
for i in range(N):
for j in range(i+1,N):
d=dist(p[i],p[j])/2
if R<d:
return False
h=math.sqrt(R*R-d*d)
#print(d,h,R)
nx,ny=(x[j]-x[i],y[j]-y[i])
nr=math.sqrt(nx**2+ny**2)
dx,dy=h*nx/nr,h*ny/nr
mx,my=(x[i]+x[j])/2,(y[i]+y[j])/2
#print(mx,my)
#print(dx,dy)
Q.append((mx+dy,my-dx))
Q.append((mx-dy,my+dx))
#print(Q,R)
for pa in Q:
flag=1
for i in range(N):
#print(pa,i,dist(pa,p[i]),R)
if dist(pa,p[i])>R+9e-7:
flag=0
break
if flag==1:
return True
return False
low=0
high=2000
for line in range(50):
mid=(low+high)/2
if check(mid):
high=mid
else:
low=mid
print(mid)
```
Yes
| 3,988 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
Submitted Solution:
```
from math import sqrt
N = int(input())
XY = [tuple(map(int, input().split())) for _ in range(N)]
def calc(r): #半径rが与えられたときに全てが重なるかを判定
lst = [] #交点を入れるリスト
for i in range(N - 1):
x1 = XY[i][0]
y1 = XY[i][1]
for j in range(i + 1, N):
x2 = XY[j][0]
y2 = XY[j][1]
diff = sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)
if diff > 2 * r:
return False
h = sqrt(r ** 2 - (diff / 2) ** 2)
if x1 == x2:
ny = (y1 + y2) / 2
lst.append([x1 - h, ny])
lst.append([x1 + h, ny])
elif y1 == y2:
nx = (x1 + x2) / 2
lst.append([nx, y1 - h])
lst.append([nx, y1 + h])
else:
a = (y2 - y1) / (x2 - x1) #2点を結ぶ直線の傾き
b = -1 / a
size = sqrt(1 + b ** 2)
nx = h / size
ny = nx * b
#もとの2点の中点
xc = (x1 + x2) / 2
yc = (y1 + y2) / 2
lst.append([xc + nx, yc + ny])
lst.append([xc - nx, yc - ny])
# print (r)
# print (lst)
nr = r + eps
for x, y in lst: #中心の点
flag = True
for X, Y in XY:
tmp = (x - X) ** 2 + (y - Y) ** 2
if tmp > nr ** 2:
flag = False
break
if flag:
return True
l = 0
r = 1000
eps = 10 ** (-9)
for i in range(100):
mid = (l + r) / 2
if calc(mid):
r = mid
else:
l = mid
print (r)
```
Yes
| 3,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
Submitted Solution:
```
from decimal import Decimal
import sys
input = sys.stdin.readline
def dist(x1, x2, y1, y2):
return (y1-y2)*(y1-y2)+(x1-x2)*(x1-x2)
def farthest(x, y):
ret = 0
point = (-1, -1)
for u, v in X:
d = dist(x, u, y, v)
if d > ret:
ret = d
point = u, v
return ret, point
N = int(input())
X = []
for _ in [0]*N:
x, y = map(int, input().split())
X.append((x, y))
x = (max(x for x, y in X) - min(x for x, y in X))/2
y = (max(y for x, y in X) - min(y for x, y in X))/2
x = Decimal(str(x))
y = Decimal(str(y))
step = Decimal('0.5')
o = 1
eps = Decimal('0.00000000000001')
while o > eps:
_, (u, v) = farthest(x, y)
x += (u-x)*step
y += (v-y)*step
step *= Decimal('0.999')
o = (abs(u-x)+abs(x-y))*step
ans, _ = farthest(x, y)
ans **= Decimal('0.5')
print(ans)
```
Yes
| 3,990 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
Submitted Solution:
```
import math
from itertools import combinations
eps = 10 ** -7
def midpoint(a, b):
return (a[0] + b[0]) / 2, (a[1] + b[1]) / 2
def distance(a, b):
return ((a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2) ** 0.5
def within(c, r, a):
return (a[0] - c[0]) ** 2 + (a[1] - c[1]) ** 2 <= r ** 2 + eps
def tcircle(t1, t2, t3):
x1, y1 = t1
x2, y2 = t2
x3, y3 = t3
d = 2 * ((y1 - y3) * (x1 - x2) - (y1 - y2) * (x1 - x3))
if d == 0:
return (0.0, 0.0), -1.0
x = ((y1 - y3) * (y1 ** 2 - y2 ** 2 + x1 ** 2 - x2 ** 2) - (y1 - y2) * (y1 ** 2 - y3 ** 2 + x1 ** 2 - x3 ** 2)) / d
y = ((x1 - x3) * (x1 ** 2 - x2 ** 2 + y1 ** 2 - y2 ** 2) - (x1 - x2) * (x1 ** 2 - x3 ** 2 + y1 ** 2 - y3 ** 2)) / -d
r = math.sqrt((x - x1) ** 2 + (y - y1) ** 2)
return (x, y), r
def main():
N = int(input())
L = [tuple(map(int, input().split())) for _ in range(N)]
br = float('inf')
for p, q in combinations(L, 2):
c, r = midpoint(p, q), distance(p, q) / 2
if all(within(c, r, i) for i in L):
return r
ac, ar = (0.0, 0.0), 0.0
for i in L:
if within(c, r, i):
continue
pc, pr = tcircle(p, q, i)
if pr == -1:
break
if ar < pr:
ac, ar = pc, pr
else:
if ar < br and all(within(ac, ar, i) for i in L):
br = ar
return br
print(main())
```
Yes
| 3,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
Submitted Solution:
```
from math import sqrt
n = int(input())
xy = [list(map(int, input().split())) for _ in range(n)]
mx = 0
for i in range(n):
for j in range(i + 1, n):
xi, yi = xy[i]
xj, yj = xy[j]
dist = sqrt((xi - xj) ** 2 + (yi - yj) ** 2)
mx = max(mx, dist)
r = mx / 2
print(r)
```
No
| 3,992 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
Submitted Solution:
```
import numpy as np
import math
def get_distance(xy1, xy2):
d = math.sqrt((xy2[0] - xy1[0]) ** 2 + (xy2[1] - xy1[1]) ** 2)
return d
N = int(input())
XY = [list(map(int, input().split())) for _ in range(N)]
ans = 10**9
for i in range(N):
for j in range(N):
flag = True
xy1 = np.array(XY[i])
xy2 = np.array(XY[j])
center = (xy1 + xy2) / 2
dis = get_distance(xy1, center)
# dis = np.linalg.norm(xy1 - center)
for k in range(N):
_dis = get_distance(XY[k], center)
# _dis = np.linalg.norm(np.array(XY[k]) - center)
if _dis > dis:
flag = False
break
if flag:
ans = min(ans, dis)
print(ans)
```
No
| 3,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
Submitted Solution:
```
#放送参照
n = int(input())
x = []
y = []
for i in range(n):
xi,yi = map(int, input().split( ))
x.append(xi)
y.append(yi)
s = sum(x)/n
t = sum(y)/n
tmp2 = -1
for _ in range(100000):###回数*ep >=1000くらいは要ると思う
ep = 0.001###要調整
tmp = 0
for i in range(n):
if tmp<(x[i]-s)**2 + (y[i]-t)**2:
tmp2 = i
tmp = (x[i]-s)**2 + (y[i]-t)**2
r = ((x[tmp2] - s)**2 + (y[tmp2] - t)**2)**0.5
dx = x[tmp2] - s
dy = y[tmp2] - t
s += dx/r * ep
t += dy/r * ep
ep *= 0.998###要調整
ans = 0
for i in range(n):
ans = max(ans, (x[i] - s)**2 + (y[i] - t)**2)
print(ans**0.5)
```
No
| 3,994 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are N points (x_i, y_i) in a two-dimensional plane.
Find the minimum radius of a circle such that all the points are inside or on it.
Constraints
* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum radius of a circle such that all the N points are inside or on it.
Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
2
0 0
1 0
Output
0.500000000000000000
Input
3
0 0
0 1
1 0
Output
0.707106781186497524
Input
10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9
Output
6.726812023536805158
Submitted Solution:
```
import math
from itertools import combinations
n = int(input())
p = [list(map(int, input().split())) for i in range(n)]
def enclose(points):
a = points[0]
b = points[1]
c = points[2]
d = points[3]
e = points[4]
f = points[5]
aa = a * a
bb = b * b
cc = c * c
dd = d * d
ee = e * e
ff = f * f
py = ((e - a) * (aa + bb - cc - dd) - (c - a) * (aa + bb - ee- ff)) / (2 * (e - a)*(b - d) - 2 * (c - a) * (b - f))
px = (2 * (b - f) * py - aa - bb + ee + ff) / (2 * (e - a)) \
if (c == a) else (2 * (b - d) * py - aa - bb + cc + dd) / (2 * (c - a))
r = math.hypot(px - a, py - b)
return (px, py, r)
ans = 1000000
if n >= 3:
for i, j, k in combinations(range(n), 3):
points = []
points.extend(p[i])
points.extend(p[j])
points.extend(p[k])
try:
px, py, r = enclose(points)
except:
continue
flag = 1
for l in range(n):
if (px-p[l][0])**2 + (py-p[l][1])**2 > r**2:
flag = 0
break
if flag == 1:
ans = min(ans, r)
"""
for i, j in combinations(range(n), 2):
px = (p[i][0]+p[j][0]) / 2
py = (p[i][1]+p[j][1]) / 2
r = ((p[i][0] - px)**2 + (p[i][1] - py) ** 2) ** (1/2)
flag = 1
for l in range(n):
if (px-p[l][0])**2 + (py-p[l][1])**2 > r**2:
flag = 0
break
if flag == 1:
ans = min(ans, r)
"""
print(ans)
```
No
| 3,995 |
Provide a correct Python 3 solution for this coding contest problem.
There are N positive integers arranged in a circle.
Now, the i-th number is A_i. Takahashi wants the i-th number to be B_i. For this objective, he will repeatedly perform the following operation:
* Choose an integer i such that 1 \leq i \leq N.
* Let a, b, c be the (i-1)-th, i-th, and (i+1)-th numbers, respectively. Replace the i-th number with a+b+c.
Here the 0-th number is the N-th number, and the (N+1)-th number is the 1-st number.
Determine if Takahashi can achieve his objective. If the answer is yes, find the minimum number of operations required.
Constraints
* 3 \leq N \leq 2 \times 10^5
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
B_1 B_2 ... B_N
Output
Print the minimum number of operations required, or `-1` if the objective cannot be achieved.
Examples
Input
3
1 1 1
13 5 7
Output
4
Input
4
1 2 3 4
2 3 4 5
Output
-1
Input
5
5 6 5 2 1
9817 1108 6890 4343 8704
Output
25
"Correct Solution:
```
import heapq as h
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
e6 = 10**6
q = [-b[i]*e6-i for i in range(n)]
h.heapify(q)
cnt = 0
while q:
w = -h.heappop(q)
v,i = w//e6, w%e6
u = b[(i-1)%n] + b[(i+1)%n]
if v < u or a[i] == v: continue
elif a[i] > v: print(-1); exit(0)
t = (v-a[i])//u
if not t: print(-1); exit(0)
cnt += t
b[i] -= t*u
if b[i] > a[i]:
h.heappush(q,-b[i]*e6-i)
print(cnt if a==b else -1)
```
| 3,996 |
Provide a correct Python 3 solution for this coding contest problem.
There are N positive integers arranged in a circle.
Now, the i-th number is A_i. Takahashi wants the i-th number to be B_i. For this objective, he will repeatedly perform the following operation:
* Choose an integer i such that 1 \leq i \leq N.
* Let a, b, c be the (i-1)-th, i-th, and (i+1)-th numbers, respectively. Replace the i-th number with a+b+c.
Here the 0-th number is the N-th number, and the (N+1)-th number is the 1-st number.
Determine if Takahashi can achieve his objective. If the answer is yes, find the minimum number of operations required.
Constraints
* 3 \leq N \leq 2 \times 10^5
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
B_1 B_2 ... B_N
Output
Print the minimum number of operations required, or `-1` if the objective cannot be achieved.
Examples
Input
3
1 1 1
13 5 7
Output
4
Input
4
1 2 3 4
2 3 4 5
Output
-1
Input
5
5 6 5 2 1
9817 1108 6890 4343 8704
Output
25
"Correct Solution:
```
def main():
from sys import exit
n = int(input())
a = [int(e) for e in input().split()]
b = [int(e) for e in input().split()]
result = 0
q = [i for i in range(n) if b[i] != a[i]]
while len(q) != 0:
nq = []
c = 0
for i in q:
if i == 0 or i == n - 1:
j = b[(n + i - 1) % n] + b[(n + i + 1) % n]
else:
j = b[i - 1] + b[i + 1]
if j > b[i] - a[i]:
nq.append(i)
continue
c += 1
k = (b[i] - a[i]) // j
result += k
b[i] -= j * k
if a[i] != b[i]:
nq.append(i)
if c == 0 and len(nq) != 0:
print(-1)
exit()
q = nq
print(result)
main()
```
| 3,997 |
Provide a correct Python 3 solution for this coding contest problem.
There are N positive integers arranged in a circle.
Now, the i-th number is A_i. Takahashi wants the i-th number to be B_i. For this objective, he will repeatedly perform the following operation:
* Choose an integer i such that 1 \leq i \leq N.
* Let a, b, c be the (i-1)-th, i-th, and (i+1)-th numbers, respectively. Replace the i-th number with a+b+c.
Here the 0-th number is the N-th number, and the (N+1)-th number is the 1-st number.
Determine if Takahashi can achieve his objective. If the answer is yes, find the minimum number of operations required.
Constraints
* 3 \leq N \leq 2 \times 10^5
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
B_1 B_2 ... B_N
Output
Print the minimum number of operations required, or `-1` if the objective cannot be achieved.
Examples
Input
3
1 1 1
13 5 7
Output
4
Input
4
1 2 3 4
2 3 4 5
Output
-1
Input
5
5 6 5 2 1
9817 1108 6890 4343 8704
Output
25
"Correct Solution:
```
from heapq import heappush,heappop
n,*t=map(int,open(0).read().split())
a=t[:n]
b=t[n:]
B=[]
for i,j in enumerate(b):heappush(B,(-j,i))
c=0
d=0
while B and d<2*10**6:
d+=1
y,i=heappop(B)
x,z=b[(i-1)%n],b[(i+1)%n]
t=(b[i]-a[i])//(x+z)
b[i]-=t*(x+z)
c+=t
if b[i]>a[i]:
heappush(B,(-b[i],i))
print([-1,c][a==b])
```
| 3,998 |
Provide a correct Python 3 solution for this coding contest problem.
There are N positive integers arranged in a circle.
Now, the i-th number is A_i. Takahashi wants the i-th number to be B_i. For this objective, he will repeatedly perform the following operation:
* Choose an integer i such that 1 \leq i \leq N.
* Let a, b, c be the (i-1)-th, i-th, and (i+1)-th numbers, respectively. Replace the i-th number with a+b+c.
Here the 0-th number is the N-th number, and the (N+1)-th number is the 1-st number.
Determine if Takahashi can achieve his objective. If the answer is yes, find the minimum number of operations required.
Constraints
* 3 \leq N \leq 2 \times 10^5
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
B_1 B_2 ... B_N
Output
Print the minimum number of operations required, or `-1` if the objective cannot be achieved.
Examples
Input
3
1 1 1
13 5 7
Output
4
Input
4
1 2 3 4
2 3 4 5
Output
-1
Input
5
5 6 5 2 1
9817 1108 6890 4343 8704
Output
25
"Correct Solution:
```
from heapq import heapify, heappop, heappush
def solve(n, aaa, bbb):
q = [(-b, i) for i, b in enumerate(bbb) if b != aaa[i]]
heapify(q)
ans = 0
while q:
b, i = heappop(q)
b = -b - aaa[i]
d, b = divmod(b, bbb[(i - 1) % n] + bbb[(i + 1) % n])
if d == 0:
return -1
b += aaa[i]
bbb[i] = b
ans += d
if b != aaa[i]:
heappush(q, (-b, i))
return ans
n = int(input())
aaa = list(map(int, input().split()))
bbb = list(map(int, input().split()))
print(solve(n, aaa, bbb))
```
| 3,999 |
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