“Python en línea a C Converter” Código de respuesta

Python en línea a C Converter

console.log("Hello World!");
Good Grasshopper

Python en línea a C Converter

#Importting csv module
import csv
#csv file name
filename= "aapi.csv"
# intiliaze titles and rows
fields =0
rows = 0
# reading csv file
with open(filename,'r') as csvfile:
    # extract each data row one by one 
    csvreader=csv.reader(csvfile)
    rows.append(rows)
    # get total no of rwos
    print("Total no of rows:%d%(csvreader.line_num)")
    # print fields name
    print('Field names are:' + ', '.join(field for field in fields))
    # print first 5 rows
    print('\n First 5 rows are:\n')
    for row in rows[:5]:
    # parsing each column of rows
        for col in row:
            print("%10s"%col),
        print('\n')
Enchanting Echidna

Python en línea a C Converter

eval("10")
Cristielson Silva de Morais

Python en línea a C Converter

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)
Spotless Sable

Python en línea a C Converter

print("Hellow world")
Modern Mongoose

Python en línea a C Converter

python("Hello")
Spotless Sheep

Python en línea a C Converter

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
Amused Antelope

Python en línea a C Converter

# 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")
     
Yawning Yacare

Python en línea a C Converter

a=input()
c=0
for i in a:
        if(i>='0'and i<='9'):
                if int(i)%2==0:
                        c+=1
print(c)
Hurt Hyena

Python en línea a 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
Tough Teira

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