Answers for "online python to c converter"

online python to c converter

import random
from collections import defaultdict

def main_roll():
    dice_amount = int(input("Enter the number of dice: "))                  # Total Number of Dice Being Rolled
    sides_of_dice = int(input("Enter the number of sides: "))               # Total Number of Sides per Die
    rolls_of_dice = int(input("Enter the number of rolls to simulate: "))   # Total Number of Times Each Die Rolled
    result = roll(dice_amount, sides_of_dice, rolls_of_dice)                # This stores the results
    maxH = 0                                                                # Used for formulating

    for i in range(dice_amount, dice_amount * sides_of_dice + 1):
        if result[i] / rolls_of_dice > maxH: maxH = result[i] / rolls_of_dice
    for i in range(dice_amount, dice_amount * sides_of_dice + 1):
        print('{:2d}{:10d}{:8.2%} {}'.format(i, result[i], result[i] / rolls_of_dice, '#' * int(result[i] / rolls_of_dice / maxH * 40)))


def roll(dice_amount, sides_of_dice, rolls):
    d = defaultdict(int)
    for _ in range(rolls):
        d[sum(random.randint(1, sides_of_dice) for _ in range(dice_amount))] += 1
    return d

main_roll()

online python to 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")

online python to c converter

print(7)

online python to c converter

Python convert to c

online python to c converter

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])

online python to c converter

print("Hellow world")

online python to 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

online python to c converter

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

online python to c converter

lista[2:10] = [7]

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