Books Algorithm Dealing with Square Digit Sums

Question

A little context:

• Book has 411 Pages

• Read a random number of pages on the first day which is unknown

• Number of pages to read next day is the square of the sum of the digits of the page currently on

  I.e. on Page 36: read 81 pages next day

• On day 6 The square answer directly adds to current page number to make 411

• Finish Book

How many pages did she read on the first day and how many pages did she read on the 6 days?

This is what I have so far:

from itertools import accumulate, count

def read_book( start ):
    for i in range( 6 ): yield start; start += sum( int( digit ) for digit in str( start ) ) ** 2

def find_start( pages ):
    for start in count( 1 ):
        for read in read_book( start ): pass
        if read == pages: return start

previous_answer, start = 0, find_start( 411 )
for i, read in enumerate( read_book( start ), 1 ):
    print( "Pages read on day", i ,"was", read-previous_answer ,"\t(Now on page", str(read) + ")\n")
    previous_answer += read-previous_answer

Output:

Pages read on day 1 was 61 (Now on page 61)

Pages read on day 2 was 49 (Now on page 110)

Pages read on day 3 was 4 (Now on page 114)

Pages read on day 4 was 36 (Now on page 150)

Pages read on day 5 was 36 (Now on page 186)

Pages read on day 6 was 225 (Now on page 411)

Is there a more efficient way? Who can do it in the least no. chars

Answer 1

gcc - 176 chars

I don't grok Python (it is Python right?) so I don't understand how your example works. Is it actually treating the page number as a string?

I'm not sure but (at least in C) string handling uses more characters than integers. So to sum the digits of an integer between 0 and 1000 I use:

for(s=0,i=1;i<1001;i*=10)s+=(x%i*10)/i;

Where x is the input and s is the sum.

If you want the sum from some x for 7 days you could:

for(c=2;c<7;x+=s*s,c++)
  for(s=0,i=1;i<1001;i*=10)s+=(x%i*10)/i;

Where x is the input and the output.

To find the sum for 7 days which matches the number of pages in the book (p):

for(y=1;y<p&&x!=p;y++)
  for(x=y,c=2;c<7;x+=s*s,c++)
    for(s=0,i=1;i<1001;i*=10)s+=(x%i*10)/i;

Where y-1 is the calculated starting page when the sum adds up to p.

The whole program with which reads from the command line and prints to stdout:

main(_,a)char**a;{
  int p=atoi(a[1]),s,i,y,x,c;
  for(y=1;y<p&&x!=p;y++)
    for(x=y,c=2;c<7;x+=s*s,c++)
      for(s=0,i=1;i<1001;i*=10)s+=(x%i*10)/i;
  if(x==p)printf("%d",y-1);else puts("n/a");
}

It surely is possible to golf it some more. If you see something that would help please comment.