Python en línea a C Converter
console.log("Hello World!");
Good Grasshopper
console.log("Hello World!");#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')
eval("10")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)print("Hellow world")python("Hello")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
# 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")
a=input()
c=0
for i in a:
if(i>='0'and i<='9'):
if int(i)%2==0:
c+=1
print(c)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