online python to c converter
from collections import deque
def BFS(a, b, target):
# Map is used to store the states, every
# state is hashed to binary value to
# indicate either that state is visited
# before or not
m = {}
isSolvable = False
path = []
# Queue to maintain states
q = deque()
# Initialing with initial state
q.append((0, 0))
while (len(q) > 0):
# Current state
u = q.popleft()
#q.pop() #pop off used state
# If this state is already visited
if ((u[0], u[1]) in m):
continue
# Doesn't met jug constraints
if ((u[0] > a or u[1] > b or
u[0] < 0 or u[1] < 0)):
continue
# Filling the vector for constructing
# the solution path
path.append([u[0], u[1]])
# Marking current state as visited
m[(u[0], u[1])] = 1
# If we reach solution state, put ans=1
if (u[0] == target or u[1] == target):
isSolvable = True
if (u[0] == target):
if (u[1] != 0):
# Fill final state
path.append([u[0], 0])
else:
if (u[0] != 0):
# Fill final state
path.append([0, u[1]])
# Print the solution path
sz = len(path)
for i in range(sz):
print("(", path[i][0], ",",
path[i][1], ")")
break
# If we have not reached final state
# then, start developing intermediate
# states to reach solution state
q.append([u[0], b]) # Fill Jug2
q.append([a, u[1]]) # Fill Jug1
for ap in range(max(a, b) + 1):
# Pour amount ap from Jug2 to Jug1
c = u[0] + ap
d = u[1] - ap
# Check if this state is possible or not
if (c == a or (d == 0 and d >= 0)):
q.append([c, d])
# Pour amount ap from Jug 1 to Jug2
c = u[0] - ap
d = u[1] + ap
# Check if this state is possible or not
if ((c == 0 and c >= 0) or d == b):
q.append([c, d])
# Empty Jug2
q.append([a, 0])
# Empty Jug1
q.append([0, b])
# No, solution exists if ans=0
if (not isSolvable):
print ("No solution")
# Driver code
if __name__ == '__main__':
Jug1, Jug2, target = 4, 3, 2
print("Path from initial state "
"to solution state ::")
BFS(Jug1, Jug2, target)
# This code is contributed by mohit kumar 29
3.8
5
for i in range(gradient1b.shape[0]):
for j in range(gradient1b.shape[1]):
if gradient1b[i, j] > thresholdHi:
gradient2b[i, j] = 0
elif ((gradient1b[i, j] <= thresholdHi) and (gradient1b[i, j] > thresholdLo)):
#gradient2b[i, j] = gradient1b[i, j] #255 #Binary thresholding
gradient2b[i, j] = 255 #Binary thresholding
else:
gradient2b[i, j] = 0
Thank you!
5
3.8
(5 Votes)
0
4.08
8
# A brute force approach based
# implementation to find if a number
# can be written as sum of two squares.
# function to check if there exist two
# numbers sum of whose squares is n.
def sumSquare( n) :
i = 1
while i * i <= n :
j = 1
while(j * j <= n) :
if (i * i + j * j == n) :
print(i, "^2 + ", j , "^2" )
return True
j = j + 1
i = i + 1
return False
# driver Program
n = 25
if (sumSquare(n)) :
print("Yes")
else :
print( "No")
Thank you!
8
4.08
(26 Votes)
0
0
0
n=int(input())
for i in range (n):
a,b,k=(map(int,input().split()))
if a>=b:
print(k//b)
else:
print(k//a)
Thank you!
0
0
0
0
0
if number == 2 : prime_con = True
if number>2 and number%2==0 : prime_con = False
stopper = math.floor(math.sqrt(number))
for j in range(3,100,2):
if number%j==0:
prime_con = False
break
Thank you!
0
0
0
0
0
def djikstra(graph, initial):
visited_weight_map = {initial: 0}
nodes = set(graph.nodes)
# Haven't visited every node
while nodes:
next_node = min(
node for node in nodes if node in visited
)
if next_node is None:
# If we've gone through them all
break
nodes.remove(next_node)
current_weight = visited_weight_map[next_node]
for edge in graph.edges[next_node]:
# Go over every edge connected to the node
weight = current_weight + graph.distances[(next_node, edge)]
if edge not in visited_weight_map or weight < visited_weight_map[edge]:
visited_weight_map[edge] = weight
return visited
Thank you!
0
0
0
4.1
10
a=[]
def knapsack(pro,wt,c,n,ans):
global a
if n==0 or c==0:
a+=ans,
elif wt[n-1]>c:
knapsack(pro,wt,c,n-1,ans)
else:
knapsack(pro,wt,c-wt[n-1],n-1,ans+pro[n-1])
knapsack(pro,wt,c,n-1,ans)
n=int(input())
profit=list(map(int,input().split()))
weights=list(map(int,input().split()))
capacity=int(input())
knapsack(profit,weights,capacity,n,0)
a.sort(reverse=True)
print(a[1 if a[0]<=10 and a[0]%3 else 0])
Thank you!
10
4.1
(10 Votes)
0
3
1
def djikstra(graph, initial):
visited_weight_map = {initial: 0}
nodes = set(graph.nodes)
# Haven't visited every node
while nodes:
next_node = min(
node for node in nodes if node in visited
)
if next_node is None:
# If we've gone through them all
break
nodes.remove(next_node)
current_weight = visited_weight_map[next_node]
for edge in graph.edges[next_node]:
# Go over every edge connected to the node
weight = current_weight + graph.distances[(next_node, edge)]
if edge not in visited_weight_map or weight < visited_weight_map[edge]:
visited_weight_map[edge] = weight
return visited
Thank you!
1
3
(1 Votes)
0
Are there any code examples left?
New code examples in category C
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C
2022-03-27 22:30:45
Problem Statement Print the following output: \ Input Format IN Output Format \
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