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Project Euler #145: How many reversible numbers are there below one-billion?
Project Euler #145: How many reversible numbers are there below one-billion?
+ 0 comments Time Out Error
def is_reversible(n): if n % 10 == 0: # Avoid leading zeroes in the reverse return False sum_n_reverse = n + int(str(n)[::-1]) # Sum of n and its reverse for digit in str(sum_n_reverse): if int(digit) % 2 == 0: # Check for even digits in the sum return False return True def count_reversible_numbers(limit): count = 0 for i in range(1, limit): if is_reversible(i): count += 1 return count def main(): test_cases = int(input()) for _ in range(test_cases): limit = int(input()) reversible_count = count_reversible_numbers(limit) print(reversible_count) if __name__ == "__main__": main()
+ 0 comments For those who are struggling solving this. At one time it took me about a week of work to solve this problem, may be longer. This is no way an easy thing, and I doubt the solution can be written as one function. Particularly, my solution contains near 600 lines and 17 helper functions.
+ 0 comments Time out can anyone fix it
n = [int(input()) for i in range(0,int(input()))] def reverse(x): return int(str(x)[::-1]) def isInputOdd(x): lis = str(x) for i in lis: if int(i)%2==0: flag = False return flag else: flag = True return flag def main(iter): sum1=0 for i in range(iter): k = i + reverse(i) if i%10!=0 and isInputOdd(k): sum1+=1 print (sum1) for i in n: main(i)
+ 2 comments Facing timeout problem... !! def procedure(n): s=str(n) #print("s = ",s) s1 = s[::-1] if int(s1)>10: #print("s1 = ",s1) #print(int(s1)) s2=n+int(s1) #print("s2 = ",s2) if '2' not in str(s2) and '4' not in str(s2) and'6' not in str(s2) and'8' not in str(s2) and'0' not in str(s2): return True else: return False
t=int(input())#no of test cases while t>0: n=int(input()) c=0 for i in range(10,n): if procedure(i)==True: c=c+1
print(c) t=t-1
Time Complexity :--- O(no.of test cases* n).. Suggest me ...some alternatives.. !!
+ 0 comments After many, many, MANY attempts, I finally did it!
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