I am working on the following problem.
Write a program that continually prompts for positive integers and stops when the sum of numbers entered exceeds 1000.
But my code stop early if a negative integer is entered.
The numbers will not sum.
My code:
x = int(input("Enter an integer:"))
total = 0
sum = 0
while (0 <= x):
if sum <= 1000:
x += 1
sum += (int(input("Enter an integer:")))
elif sum >= 1000:
break
x = 0
total = 0
sum = 0
while sum <= 1000:
x = int(input("Enter an integer:"))
if x<0:
print("Invalid negative value given")
break
sum += x
First:
if sum >= 1000:
...
elif sum < 1000:
...
is redundant, because if you chek for sum >= 1000, and the elif is reached, you aready know, that the condition was False and therefore sum < 1000 has to be true. So you could replace it with
if sum >= 1000:
...
else:
...
Second:
You want to use x to check, whether the input is negative. Until now, you were simpy increasing it by one each time. Instead, you should first assing the input to x, and then add this to sum. So do it like this:
x = int(input("Enter an integer:"))
if x<0:
break
sum += x
In case you want to stop as soon as you encounter the negative number.
x = 0
total = 0
sum = 0
while (sum <= 1000):
x = (int(input("Enter an integer:"))
if (x < 0):
break
else:
sum += x
Related
I want the sum of all prime numbers till 2 million. My program logic is correct but it takes too much time with 2 million. How can I make it faster?
num_l=[]
y = int(input("enter till what number do you want the sum"))
for count in range(0,y):
num_l.append(count)
total = 0
for counter in range(0,y):
num = num_l[counter]
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
total = total + num
print(total)
just use a function to check that it is prime or not and if it is then add it to a counter (for total). The code below takes about 20 seconds to find what you want.
One more thing is that your algorithm for testing if a number is prime or not is having a long range of numbers to go through you don't need to test to the last number you just need to go through for a range of square root of the number.
import time
def is_prime(n):
if n == 1:
return False
elif n == 2:
return True
else:
for i in range(2, int(n**0.5)+1): # this range is just enough
if n % i == 0:
return False
return True
counter = 0
start = time.time()
for i in range(1, 2*10**6): # you can change this range for taking it as input
if is_prime(i):
counter += i
print(counter)
print(time.time() - start)
I am writing a small program to find all prime number from a range from 0 to whatever number the user want. How should I fix my code?
I tried to swap the loop position and the algorithm but it didn't work
g=int(input("type"))
def check_prime(g):
if(g<2):
return False
for x in range(2, g+1):
v=int(math.sqrt(x))
if(x%v==0):
return False
return True
if check_prime(g):
print(x)
I expect when the user input g they will get all the prime numbers from 0 to g and also the program will tell them how many prime numbers found
It's pretty simple, this code below will return you both the count of primes and their values.
g=int(input("type"))
def check_prime(g):
count = 0
primes = []
for number in range(g+1): #considering 'g' is inclusive
primes.append(number)
count += 1
if(number < 2 or number%2 == 0):
primes = primes[:-1]
count -= 1
continue
for x in range(3, number, 2):
if (number % x) == 0:
primes = primes[:-1]
count -= 1
break
return count, primes
print (check_prime(g))
I've to create a program that computes the average of a collection of values entered by the user. The user will enter 0 as a sentinel value to indicate that no further values will be provided. The program should display an appropriate error message if the first value entered by the user is 0.
Note: Number of inputs by the user can vary. Also, 0 marks the end of the
input it should not be included in the average
x = int(input("Enter Values\n"))
num = 1
count = 0
sum = 0.0
if x == 0:
print("Program exits")
exit()
while (x>0):
sum += num
count += 1
avg = (sum/(count-1))
print("Average: {}".format(avg))
You were not taking input inside while loop. You were taking input on the first line for once. So your program was not taking input repeatedly.
You may be looking for this -
sum = 0.0
count = 0
while(1):
x=int(input("Enter Values: "))
if x == 0:
print("End of input.")
break;
sum+=x;
count+=1;
if count == 0:
print("No input given")
else:
avg = sum/count;
print("Average is - ",avg)
Your code does not work because int function expect only one number.
If you insert the numbers one by one, the following code works:
num = int(input("Enter a value: "))
count = 0
sum = 0.0
if num <= 0:
print("Program exits")
exit()
while (num>=0):
sum += num
count += 1
num = int(input("Enter a value: "))
avg = (sum/count)
print(f"Average: {avg}")
Im a newbie at python and i have a task. Given a number as input, i have to print the prime that belongs in the number/position on a list of primes starting from position 1 and not 0, until the input is 'END'. For example, if the input is 1, output should be the first prime which is 2, if the input is 5, output should be the 5th prime which is 11 and so. It works fine but after 3/4-digit numbers the output has a delay until i get the Error: Time limit exceeded. How can i make it run faster? Here's the code:
def primes_function(n):
primes = []
num = 2
while len(primes) <= n:
x = num // 2
while x > 1:
if num % x == 0:
break
x -= 1
else:
primes.append(num)
num += 1
return primes[n - 1]
#main
while True:
n = input()
if n == 'END':
break
elif n > '0':
n = int(n)
value = primes_function(n)
print(value)
Sorry if i made any mistakes in the description
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I combined this answer (1) and this answer (2) to speed up the function. The two key ideas are: When testing primality of a candidate ...
... do not divide by every number (2, 3, 4, 5, 6, ...) but only by the preceding prime numbers (2, 3, 5, ...). Every non-prime number > 2 is has to have some prime factor.
... divide only by numbers that are ≤ sqrt(candidate).
import math
def nth_prime(n):
prime_list = [2]
candidate = 3
while len(prime_list) < n:
max_factor = math.sqrt(candidate)
is_prime = True
for p in prime_list:
if p > max_factor:
break
elif candidate % p == 0:
is_prime = False
break
if is_prime:
prime_list.append(candidate)
candidate += 2
return prime_list[-1]
Benchmark of different solutions:
n=9000 n=15000 n=25000 n=75000
your solution 1m38.455s - - -
linked answer (1) 0m 2.954s 8.291s 22.482s -
linked answer (2) 0m 0.352s 0.776s 1.685s 9.567s
this answer 0m 0.120s 0.228s 0.410s 1.857s
Brij's answer 0m 0.315s 0.340s 0.317s 0.318s
For every n the programs where started from scratch.
As we can see, Brij's Sieve Of Eratosthenes takes a fairly low constant amount of time. If you want to find big prime numbers below a fixed limit then that's the best solution (here n < 78499, as the 78499-th prime number is 1 000 003 which is bigger than the sieve list).
If you also want to find a lot of smaller or medium sized prime numbers or cannot accept a fixed limit then go with this solution.
def SieveOfEratosthenes():
n = 1000000
prime = [True for i in range(n+1)]
p = 2
count = 0
while (p * p <= n):
if (prime[p] == True):
count = count + 1
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
seive = []
for p in range(2, n):
if prime[p]:
seive.append(p)
return seive
def primes_function(n , seive):
return seive[n - 1]
#main
seive = SieveOfEratosthenes()
while True:
n = input()
if n == 'END':
break
elif n > '0':
n = int(n)
value = primes_function(n,seive)
print(value)
Full working : https://ide.geeksforgeeks.org/QTSGQfhFV3
I have precomputed the primes below 10^6 and made a list of primes and accessed the nth prime number by the index.
I am failed to print only the nth fibonacci number.
In my code, when the user said to print nth trem it print the series upto nth term but i want to get the output only the nth term
e.g
if I say num=4
out put should be 2
please guide
here is the code:
N= int(input("How many terms? "))
N1 = 0
N2 = 1
sum = 2
if N <= 0:
print("Plese enter a positive integer")
elif N == 1:
print("Fibonacci sequence:")
print(N1)
else:
print("Fibonacci sequence:")
print(N1,",",N2,end=' , ')
while sum < N:
Nth = N1 + N2
print(Nth,end=' , ')
N1 = N2
N2 = Nth
sum += 1
The print stmt should be outside the loop
N= int(input("How many terms? "))
N1 = 0
N2 = 1
sum = 2
if N <= 0:
print("Plese enter a positive integer")
elif N == 1:
print("Fibonacci sequence:")
print(N1)
else:
print("Fibonacci sequence:")
print(N1,",",N2,end=' , ')
while sum < N:
Nth = N1 + N2
N1 = N2
N2 = Nth
sum += 1
print(Nth,end=' , ')
Simpler code, from the "How to Think Like a Comptuer Scientist: Python" book,
def fibonacci (n):
if n == 0:
return 0
elif n == 1:
return 1
else:
return fibonacci(n-1) + fibonacci(n-2)
Just call fibonacci passing your nth term as the argument.
to achieve that output you can simply decrease the value of n by 1 and then carry on all the computation.
For example:
def fib(n):
n = n-1
a, b = 0, 1
count = 1
while count <= abs(n):
next = a + b
a = b
b = next
count += 1
return a