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  1. PO
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    <p>Here's my python solution that prints out the correct answer in a fraction of a second. I'm not a python programmer generally, so apologies for any egregious code style errors.</p> <pre><code>#!/usr/bin/env python import sys ONES=[ "", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen","eighteen", "nineteen", ] TENS=[ "zero", "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety", ] def to_s_h(i): if(i&lt;20): return(ONES[i]) return(TENS[i/10] + ONES[i%10]) def to_s_t(i): if(i&lt;100): return(to_s_h(i)) return(ONES[i/100] + "hundred" + to_s_h(i%100)) def to_s_m(i): if(i&lt;1000): return(to_s_t(i)) return(to_s_t(i/1000) + "thousand" + to_s_t(i%1000)) def to_s_b(i): if(i&lt;1000000): return(to_s_m(i)) return(to_s_m(i/1000000) + "million" + to_s_m(i%1000000)) def try_string(s,t): global letters_to_go,word_sum l=len(s) letters_to_go -= l word_sum += t if(letters_to_go == 0): print "solved: " + s print "sum is: " + str(word_sum) sys.exit(0) elif(letters_to_go &lt; 0): print "failed: " + s + " " + str(letters_to_go) sys.exit(-1) def solve(depth,prefix,prefix_num): global millions,thousands,ones,letters_to_go,onelen,thousandlen,word_sum src=[ millions,thousands,ones ][depth] for x in src: num=prefix + x[2] nn=prefix_num+x[1] try_string(num,nn) if(x[0] == 0): continue if(x[0] == 1): stl=(len(num) * 999) + onelen ss=(nn*999) + onesum else: stl=(len(num) * 999999) + thousandlen + onelen*999 ss=(nn*999999) + thousandsum if(stl &lt; letters_to_go): letters_to_go -= stl word_sum += ss else: solve(depth+1,num,nn) ones=[] thousands=[] millions=[] onelen=0 thousandlen=0 onesum=(999*1000)/2 thousandsum=(999999*1000000)/2 for x in range(1,1000): s=to_s_b(x) l=len(s) ones.append( (0,x,s) ) onelen += l thousands.append( (0,x,s) ) thousands.append( (1,x*1000,s + "thousand") ) thousandlen += l + (l+len("thousand"))*1000 millions.append( (0,x,s) ) millions.append( (1,x*1000,s + "thousand") ) millions.append( (2,x*1000000,s + "million") ) ones.sort(key=lambda x: x[2]) thousands.sort(key=lambda x: x[2]) millions.sort(key=lambda x: x[2]) letters_to_go=51000000000 word_sum=0 solve(0,"",0) </code></pre> <p>It works by precomputing the length of the numbers from 1..999 and 1..999999 so that it can skip entire subtrees unless it knows that the answer lies somewhere within them.</p>
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    1. POSorting numbers from 1 to 999,999,999 in words as strings
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    1. USElectric Wig
    1. USElectric Wig
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      1. VTUpMod
    1. COThat's insanely quick - finished in 6ms on my fairly run of the mill PC. Kudos, sir.
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      1. USDominic Rodger
    2. COThanks - I'd missed the part where it's supposed to compute the sum of the numbers to that point. Fixed now :)
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      1. USElectric Wig
 

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