Project Euler #145: How many reversible numbers are there below one-billion? Discussions |

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6 months ago+ 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, me solution contains near 600 lines and 17 helper functions.

3 years ago+ 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)

4 years ago+ 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.. !!

4 years ago+ 0 comments

After many, many, MANY attempts, I finally did it!

6 years ago+ 0 comments

now why time out i even reduced the complexity before by checking if either of first or last digit odd and even

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int isdigitsodd(unsigned long long num);
int main() {
    unsigned long long number,reverse=0,sum=0,i,count=0,k,powervalue,logvalue,fd;
    int T,j,dig;
    scanf("%d",&T);
    for(j=1;j<=T;j++)
     {  count=0;
        scanf("%llu",&number);
        for(i=11;i<number;i++)
          {    k=i;reverse=0;
               logvalue=log10(i);
               powervalue=pow(10,logvalue);
               fd=i/powervalue;
               if(i%10==0)
                { ;}
               else if(((i%2)==0&&fd%2==1)||((i%2)==1&&fd%2==0))
                {
                   while(k!=0)
                     {
                        dig=k%10;
                        reverse=reverse*10+dig;
                        k/=10;
                     }
                    sum=reverse+i;
                    if(isdigitsodd(sum)==1)
                       { count++;}
                   
                } 
          }
        printf("%llu\n",count);
     }
    
    
    
    return 0;
}
int isdigitsodd(unsigned long long num)
{
    int flag=0,dig=0;
    while(num!=0)
    {
       dig=num%10;
       if(dig%2==1)
          flag=1;
       else 
          {flag=0;break;}
       num/=10;
    }
    if(flag==1)
       return 1;
    else return 0;
}

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