I need to find x from the below equation
h = pow(g,x,p)
I have the values of
h = 400 digit number
g = 400 digit number
p = 400 digit number
I need to find the value of x.
All the values are very big.
I have incomplete x with 10 missing digits in it and they are not continuous missing digits.Index of all the missing digits is different.
I am using 10 for loops
to replace that value every time to get H.It is taking too much time in python to do that.Is there anyway in which I can do it faster.
Don't call pow() ten billion times. Instead, use it to calculate the first value of h, and then calculate each successive value by multiplying the previous value by g (or g raised to the appropriate power of 10 in the case where the missing digits are not at the end of x).
Here's a worked example that finds 7 missing digits in just a few seconds. It should give an answer for ten missing digits in a few hours. But as others have noted, you will get better performance by porting this code to another language like C.
h = 27029080272084173153635398406622455117429159185281491773519587480106092289627
g = 26376362132555166607008315364046996472951702119314003469217622667073785183917
p = 81593331324697655999776287731387256090095437693547587289127550085992860325943
x = 31952256647378403805884540140134925446045889939629675712565170000000381638216
# The missing digits (3897858): ^^^^^^^
z = pow(g, x, p) # Initial value of h
d = pow(g, 10**9, p) # Multiplier to get next value of h
while z != h:
z = (z * d) % p
x += 10**9 # Increment x at position of last unknown digit
print(x)
EDIT:
If the missing digits are not consecutive, then you can still use a similar approach, but instead of incrementing x by the same amount each time, you'll need to keep tabs on which digits are changing at each step. For example given a number .0.....0...0.., where 0 represents a missing digit and . represents a known digit, you'll need to add 100 when incrementing the last unknown digit, 9999100 when the last unknown digit has reached 9, and 999990999100 when the last two unknown digits are both 9. It shouldn't be too hard to figure out the correct increments for any pattern of unknown digits.
.0.....0...0.. + 100
.0.....0...1.. + 100
.0.....0...2.. + 100
: : :
.0.....0...8.. + 100
.0.....0...9.. + 9999100
.0.....1...0.. + 100
.0.....1...1.. + 100
: : :
.0.....1...9.. + 9999100
.0.....2...0.. + 100
: : :
.0.....9...9.. + 999990999100
.1.....0...0.. + 100
: : :
This should run in a feasible amount of time with 10 unknown digits, but for a much faster solution, use the meet-in-the-middle approach suggested by James K Polk in the comments below.
Related
from math import log10, exp
places = int(input('Enter the number of decimal places: '))
one = 10**places
extra = 10**4
n = 1
term = one * extra
eee = 0
count = 0
while term > 0:
eee += term
count += 1
term = term // n
n += 1
eee = eee // extra
intPart = eee // one
fracPart = eee % one
eee = str(intPart)+'.'+ '0'*(places - (int(log10(fracPart)) + 1))+str(fracPart)
print("""
Python's value of e is:\n%.15f\n
e to %d decimal places is:\n%s\n
The number of terms in the series is %d""" \
% (exp(1), places, eee, count))
Questions:
What's the reason behind using extra?
The first term of the series is 1 but why is term = one * extra?
Why is while condition term > 0 ?
Logic behind term = term // n
In short problem with understanding the whole code.
Thank you.
(1) What's the reason behind using extra?
To keep four extra, hidden decimal places for extra accuracy during the execution of the main loop. As soon as the main loop ends, the line eee = eee // extra throws the four extra places away.
(2) The first term of the series is 1 but why is term = one * extra?
This is a fixed-point calculation. What do you know about fixed point? It requires a scale, here represented by one.
(3) Why is while condition term > 0?
Because only the first several terms of the series are large enough to matter. All the late, very small terms can be ignored. Do you know what the // operator does?
Your fourth question should answer itself once the other three answers are understood.
I have a number and I want to find the sum of all of its possible substrings. Since the sum may be very large, I am taking modulo 1e9+7. Here is the code I wrote for it:
n = input()
total = 0
for i in range(len(n)):
for j in range(i, len(n)):
total = (total + int(n[i:j+1]))%(1e9+7)
print(int(total))
But this gives me Overflow error:
OverflowError: int too large to convert to float
Taking modulo inside also doesn't help:
total = (total%(1e9+7) + int(n[i:j+1])%(1e9+7))%(1e9+7)
Neither does converting total to int every step:
total = int((total%(1e9+7) + int(n[i:j+1])%(1e9+7))%(1e9+7))
I searched online, and many people were using decimal, so I tried that too:
import decimal
n = input()
total = 0
for i in range(len(n)):
for j in range(i, len(n)):
total = decimal.Decimal((int(total)%(1e9+7) + int(n[i:j+1])%(1e9+7))%(1e9+7))
print(int(total))
This also gave me the same error. So how can I fix it?
EDIT:
This is the input value causing the error:
63007895494540748697130257211701132911672127113982917934957238363754144356260578781606111036085360074421257207207387198523322868167701948879591559261313655853869741936915896141380413900486094968371175676410640884374632431850709016596137311611050415651036226139008641575836931194661785509132233097446922537947515703465398622143008692876882535026564205087414814876285063338031809221849792393893433149494497789721825006237838192606565116804799001721467342272432009712314084904086422379596732252144585986596196750288347933150685033753522443039157103807332491116413966300613741751053919270582143539182365841138533836973023237702830069573886631046833437773543939255134613488502488921701075580406239690138018459241613732213353199068094195960086440933264712021949041470198397933937575146560984080148459288992517286710576766386500347467387730678249218006735385649387335208267283324236829570036194140660709429852497101058284815529517587641628078980207037631383236114879153070060203938791830375096696531139157470783772857017638497070485512459440717225126809870697837609054291292934449251338418323104001620741264801956189141105715135298492818411518148353495966691130971474426577393248709217076189389546980748699933003944761579539683492598357473775080656936009069500959707744011739717600412538480688678363966062155016279422282550330186675591906410237102542897020260947416709372597337702901936954248900890739981634426827110725142264244439429503167749662657491819253898730703219914172569228601526136901294517438068414448408652485288627071736842683249099077215845807526690476654299190397040746528470362309997516648677591695269107335587796459234023073152908830766246216873096608906233643638927751392979298950952796180665578245384773623638725905735891369383833201685178140752435935281322907250538065071169304917667919102159102138937264124738496487528733665070884612044627688227281074562269370933157831720664584804270606755748075851155331082464002064939435532230555080408656278406091954428953100831871658476246847962625412944027106283123969563281644107592096927875380676545163227869818230832531369338104630185063964880075428674709642025343042348947946680468148204427273799945049955546582316448003742222236732940414274528382294789931354040436608227971710151521614759476639229242951124359411795292331914156083470605030293573012447568075893343623149168527048295141160679023585092938002707648154463420344913624817717119873800348683837221793276968989888478424382685528609128835823933770712162540205085109082587070367630067313533064776696964861827502508459812330918599106584868738991756659193306367398956355438229525202113654400239026884347461343064397664025248463273627585729047677445867514652971446394066091970616323396951168504154562483771764698148733870822358926184753051413327111690674109191730327822065319240519599901011286438320648370316577747649941281847889063992145722797473369698092782688190504008577236785654455943124233206826773616384851418122764189601284051343171878932046232940010251668912231630645042775980083200386480623888072677846344562885143783746246053531215869425826330930224051778041164960804697442635688285354389497086198777434999457478133661733653282161024413460328143721896372700138331902360395339801002849019324370472928179634514835029042196634893085002987449024538653084769349649361780544448023900419085030996025293267821633224287187700999723244561243626140770409199181283529285321805520719869999019359632752206587151332438340443688561748988667594627862916961421297673484360219893809090908042382241699048289169441570762688018233512738556484549696872250253106450823679449014903702137375687542527806962174586086391239919664138841439946317125776119213371905889154041992606193725968405596076183819908160092384091833692412052633523651650011510824185528743026380948023336408400010564640963891087899442082042255224880906285306297131459863380833411401041892413336493399820935278521945245608933772102214407947819925650130234521834285827823524419976919232687982300093777249782891412337818506769860369083173352493958839602578474422541850108379964969139387326989233531541598506923936086930628743016172543394216721521126188175476215460711352124477581111228767588146320455790337298329662450219245997135624667608463321777810934443458141729689725088157058043698135069334292367985922828854113138954089429957666350752537071506611029281378035862848313230313174298317644570058264296631872189631662359826821867318972251771854913270152954052083892228397794195634661858474972482726585256885550036123081470648813461106884531635459879919973084520662462817291748845738645963318596934145967253237900796485048018611047753698931624813763287089062288176545539923325893133941049487204878644538080562085371696536278985940315314490818357538393931021727970513972156001956807121425599497440857264884182862594848418387223228176775493691062756510368292537574241902481598785658616637811580124317555926904680461026827106118762447230822380546660264294026004268281275625671458696016538340789809948700289444406217883939127200226299337494154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53
Convert it into Decimal
from decimal import Decimal
Decimal(630078954945407486971302572117011329116721271139829179349572383637541443562605787816061110360853600744212572072073871985233228681677019488795915592613136558538697419369158961413804139004860949683711756764106408843746324318507090165961373116110504156510362261390086415758369311946617855091322330974469225379475157034653986221430086928768825350265642050874148148762850633380318092218497923938934331494944977897218250062378381926065651168047990017214673422724320097123140849040864223795967322521445859865961967502883479331506850337535224430391571038073324911164139663006137417510539192705821435391823658411385338369730232377028300695738866310468334377735439392551346134885024889217010755804062396901380184592416137322133531990680941959600864409332647120219490414701983979339375751465609840801484592889925172867105767663865003474673877306782492180067353856493873352082672833242368295700361941406607094298524971010582848155295175876416280789802070376313832361148791530700602039387918303750966965311391574707837728570176384970704855124594407172251268098706978376090542912929344492513384183231040016207412648019561891411057151352984928184115181483534959666911309714744265773932487092170761893895469807486999330039447615795396834925983574737750806569360090695009597077440117397176004125384806886783639660621550162794222825503301866755919064102371025428970202609474167093725973377029019369542489008907399816344268271107251422642444394295031677496626574918192538987307032199141725692286015261369012945174380684144484086524852886270717368426832490990772158458075266904766542991903970407465284703623099975166486775916952691073355877964592340230731529088307662462168730966089062336436389277513929792989509527961806655782453847736236387259057358913693838332016851781407524359352813229072505380650711693049176679191021591021389372641247384964875287336650708846120446276882272810745622693709331578317206645848042706067557480758511553310824640020649394355322305550804086562784060919544289531008318716584762468479626254129440271062831239695632816441075920969278753806765451632278698182308325313693381046301850639648800754286747096420253430423489479466804681482044272737999450499555465823164480037422222367329404142745283822947899313540404366082279717101515216147594766392292429511243594117952923319141560834706050302935730124475680758933436231491685270482951411606790235850929380027076481544634203449136248177171198738003486838372217932769689898884784243826855286091288358239337707121625402050851090825870703676300673135330647766969648618275025084598123309185991065848687389917566591933063673989563554382295252021136544002390268843474613430643976640252484632736275857290476774458675146529714463940660919706163233969511685041545624837717646981487338708223589261847530514133271116906741091917303278220653192405195999010112864383206483703165777476499412818478890639921457227974733696980927826881905040085772367856544559431242332068267736163848514181227641896012840513431718789320462329400102516689122316306450427759800832003864806238880726778463445628851437837462460535312158694258263309302240517780411649608046974426356882853543894970861987774349994574781336617336532821610244134603281437218963727001383319023603953398010028490193243704729281796345148350290421966348930850029874490245386530847693496493617805444480239004190850309960252932678216332242871877009997232445612436261407704091991812835292853218055207198699990193596327522065871513324383404436885617489886675946278629169614212976734843602198938090909080423822416990482891694415707626880182335127385564845496968722502531064508236794490149037021373756875425278069621745860863912399196641388414399463171257761192133719058891540419926061937259684055960761838199081600923840918336924120526335236516500115108241855287430263809480233364084000105646409638910878994420820422552248809062853062971314598633808334114010418924133364933998209352785219452456089337721022144079478199256501302345218342858278235244199769192326879823000937772497828914123378185067698603690831733524939588396025784744225418501083799649691393873269892335315415985069239360869306287430161725433942167215211261881754762154607113521244775811112287675881463204557903372983296624502192459971356246676084633217778109344434581417296897250881570580436981350693342923679859228288541131389540894299576663507525370715066110292813780358628483132303131742983176445700582642966318721896316623598268218673189722517718549132701529540520838922283977941956346618584749724827265852568855500361230814706488134611068845316354598799199730845206624628172917488457386459633185969341459672532379007964850480186110477536989316248137632870890622881765455399233258931339410494872048786445380805620853716965362789859403153144908183575383939310217279705139721560019568071214255994974408572648841828625948484183872232281767754936910627565103682925375742419024815987856586166378115801243175559269046804610268271061187624472308223805466602642940260042682812756256714586960165383407898099487002894444062178839391272002262993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1523212253)
The best solution to this question runs in linear time. Here's the idea:
let's look at a shorter example: 1234
consider the total sum of the substrings which end on the ith digit (0..3)
substringsum[0]: 1 (we have only a single substring)
substringsum[1]: 2 + 12 (two substrings)
substringsum[2]: 3+23+123 (three substrings)
substringsum[3]: 4 + 34+234+1234
see a pattern?
let's look at substringsum[2]:
3 + 23 + 123 = 3 + 20+3 + 120+3 = 3*3 + 20+120 = 3*3 + 10*(2+12) = 3*3 +10*substringsum[1]
in general:
substringsum[k] = (k+1)*digit[k] + 10 * substringsum[k-1]
This you can compute in linear time.
This type of ideas is called "Dynamic Programming"
The overflow error suggest that there's an undesired conversion to float, which I assume you did not intend. The conversion happens because the type of 1e9 is float. To fix this, use % int(1e9+7) instead of % 1e9+7
Given two numbers a and b, we have to find the nth number which is divisible by a or b.
The format looks like below:
Input :
First line consists of an integer T, denoting the number of test cases.
Second line contains three integers a, b and N
Output :
For each test case, print the Nth
number in a new line.
Constraints :
1≤t≤105
1≤a,b≤104
1≤N≤10
Sample Input
1
2 3 10
Sample Output
15
Explanation
The numbers which are divisible by 2
or 3 are: 2,3,4,6,8,9,10,12,14,15 and the 10th number is 15
My code
test_case=input()
if int(test_case)<=100000 and int(test_case)>=1:
for p in range(int(test_case)):
count=1
j=1
inp=list(map(int,input().strip('').split()))
if inp[0]<=10000 and inp[0]>=1 and inp[1]<=10000 and inp[1]>=1 and inp[1]<=1000000000 and inp[1]>=1:
while(True ):
if count<=inp[2] :
k=j
if j%inp[0]==0 or j%inp[1] ==0:
count=count+1
j=j+1
else :
j=j+1
else:
break
print(k)
else:
break
Problem Statement:
For single test case input 2000 3000 100000 it is taking more than one second to complete.I want if i can get the results in less than 1 second. Is there a time efficient approach to this problem,may be if we can use some data structure and algorithms here??
For every two numbers there will be number k such that k=a*b. There will only be so many multiples of a and b under k. This set can be created like so:
s = set(a*1, b*1, ... a*(b-1), b*(a-1), a*b)
Say we take the values a=2, b=3 then s = (2,3,4,6). These are the possible values of c:
[1 - 4] => (2,3,4,6)
[5 - 8] => 6 + (2,3,4,6)
[9 - 12] => 6*2 + (2,3,4,6)
...
Notice that the values repeat with a predictable pattern. To get the row you can take the value of c and divide by length of the set s (call it n). The set index is the mod of c by n. Subtract 1 for 1 indexing used in the problem.
row = floor((c-1)/n)
column = `(c-1) % n`
result = (a*b)*row + s(column)
Python impl:
a = 2000
b = 3000
c = 100000
s = list(set([a*i for i in range(1, b+1)] + [b*i for i in range(1, a+1)]))
print((((c-1)//len(s)) * (a*b)) + s[(c - 1)%len(s)])
I'm not certain to grasp exactly what you're trying to accomplish. But if I get it right, isn't the answer simply b*(N/2)? since you are listing the multiples of both numbers the Nth will always be the second you list times N/2.
In your initial example that would be 3*10/2=15.
In the code example, it would be 3000*100000/2=150'000'000
Update:
Code to compute the desired values using set's and lists to speed up the calculation process. I'm still wondering what the recurrence for the odd indexes could be if anyone happens to stumble upon it...
a = 2000
b = 3000
c = 100000
a_list = [a*x for x in range(1, c)]
b_list = [b*x for x in range(1, c)]
nums = set(a_list)
nums.update(b_list)
nums = sorted(nums)
print(nums[c-1])
This code runs in 0.14s on my laptop. Which is significantly below the requested threshold. Nonetheless, this values will depend on the machine the code is run on.
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Closed 10 years ago.
I want to know what is the biggest number you can make from multiplying digits entered by the user like this:
5*6*7*2 OR 567*2 OR 67*25 ...etc
so 5/6/7/2 should be entered by the user as variables, but how do I tell python to form a number from two variables (putting the two digits next to each other and treating the outcome as a number by itself).
how do I tell python to form a number from two variables (putting the two digits next to each other and treating the outcome as a number by itself)
Provided the two digits are stored in integer variables, the following will do it:
In [1]: v1 = 5
In [2]: v2 = 6
In [3]: v1 * 10 + v2
Out[3]: 56
This can be generalized to a sequence of digits:
In [7]: l = (1, 2, 6, 3)
In [8]: reduce(lambda x,y: x * 10 + y, l)
Out[8]: 1263
I feel you have posted a mistake in your question. You ask for permutations? Are you sure?
If so, see #mbeckish's answer. It's pretty simple, and not a very good exercise in programming.
(it is, however, a good exercise in trick questions with riddle-like "gotchas" for solutions)
A better approach is to ditch the permutations requirement, so that the order of the input can actually affect the answer. For that approach, we get a much more interesting solution:
def largest_product(*args):
numbers = ''.join(map(str, args))
results = []
for i in range(1, len(numbers) - 1):
multicand = int(numbers[:i])
multiplier = int(numbers[i:])
m, n = multicand, multiplier
results.append(( m * n, "%s * %s" % (m, n)))
return max(results)
>>> largest_product(*range(8))
(827115, '12345 * 67')
so 5/6/7/2 should be entered by the user as variables, but how do I tell python to form a number from two variables (putting the two digits next to each other and treating the outcome as a number by itself).
Seems the root of your problem is capturing data from the user, combining it, and converting it:
>>> a = raw_input()
8
>>> b = raw_input()
3
>>> a
'8'
>>> b
'3'
>>> a + b
'83'
>>> int(a+b)
83
It's that easy.
Now as far as biggest number you can make from multiplying digits entered goes... we can prove that with math if you'd like so you don't have a pile of combinations to try:
We can sort the digits a >= b >= c >= d
First let's look at splitting the digits 3 and 1. We need to compare a * bcd, b * acd, c * abd, d * abc.
Comparing a * bcd = 100ab + 10ac + ad with b * acd = 100ab + 10bc + bd we see the former is larger because a >= b. A similar argument will show that a * bcd beats the others.
Similarly we can compare ac * bd = 100ab + 10(ad+bc) + bd with ad * bc = 100ab + 10(ac+bd) + cd. We would rather have more copies of the big a, so the second wins.
Finally we need to compare a * bcd = 100ab + 10ac + ad with ad * bc = 100ab + 10(ac+bd) + cd. The second is the winner.
You probably took the input in a loop as an array, so if you have:
(a) arr[0] = '5' arr[0] = '7'
(b) arr[1] = '6' sort em => arr[1] = '6'
(c) arr[2] = '7' arr[2] = '5'
(d) arr[3] = '2' arr[3] = '2'
The largest would be:
int(arr[0] + arr[3]) * int(arr[1] + arr[2]) = 4680
Any solution that has you trying all permutations of digits will be horribly inefficient, running in O(n!). Just 14 digits (and the multiply operator) would give around 1 trillion combinations!
An O(n lg n) solution would be:
Sort the digits from high to low.
Concatenate them into one string.
Print the string.
If you must multiply at least one digit, then
Sort.
Take the highest digit and multiply by the concatenation of the remaining digits.
Print the result.
If you must multiply at least one digit, then you might need to try all permutations (see #Mike's answer).
I assume you get the numbers as string, so you can simply strip them, join and translate to int:
string = "5*6*7*2"
value = int( "".join(string.split('*')) )
# value == 5672
This is the problem:
How many integers 0 ≤ n < 10^18 have the property that the sum of the digits of n equals the sum of digits of 137n?
This solution is grossly inefficient. What am I missing?
#!/usr/bin/env python
#coding: utf-8
import time
from timestrings import *
start = time.clock()
maxpower = 18
count = 0
for i in range(0, 10 ** maxpower - 1):
if i % 9 == 0:
result1 = list(str(i))
result2 = list(str(137 * i))
sum1 = 0
for j in result1:
sum1 += int(j)
sum2 = 0
for j in result2:
sum2 += int(j)
if sum1 == sum2:
print (i, sum1)
count += 1
finish = time.clock()
print ("Project Euler, Project 290")
print ()
print ("Answer:", count)
print ("Time:", stringifytime(finish - start))
First of all, you are to count, not to show the hits.
That is very important. All you have to do is to device an efficient way to count it. Like Jon Bentley wrote in Programming Pearls: "Any methond that considers all permutations of letters for a word is doomed to failure". In fact, I tried in python, as soon as "i" hit 10^9, the system already freezed. 1.5 G memory was consumed. Let alone 10^18. And this also tells us, cite Bentley again, "Defining the problem was about ninety percent of this battle."
And to solve this problem, I can't see a way without dynamic programming (dp). In fact, most of those ridiculously huge Euler problems all require some sort of dp. The theory of dp itself is rather academic and dry, but to implement the idea of dp to solve real problems is not, in fact, the practice is fun and colorful.
One solution to the problem is, we go from 0-9 then 10-99 then 100-999 and so on and extract the signatures of the numbers, summarize numbers with the same signature and deal with all of them as a piece, thus save space and time.
Observation:
3 * 137 = 411 and 13 * 137 = 1781. Let's break the the first result "411" down into two parts: the first two digits "41" and the last digit "1". The "1" is staying, but the "41" part is going to be "carried" to further calculations. Let's call "41" the carry, the first element of the signature. The "1" will stay as the rightest digit as we go on calculating 13 * 137, 23 * 137, 33 * 137 or 43 * 137. All these *3 numbers have a "3" as their rightest digit and the last digit of 137*n is always 1. That is, the difference between this "3" and "1" is +2, call this +2 the "diff" as the second element of the signature.
OK, if we are gonna find a two-digit number with 3 as its last digit, we have to find a digit "m" that satisfies
diff_of_digitsum (m, 137*m+carry) = -2 (1)
to neutralize our +2 diff accumulated earlier. If m could do that, then you know m * 10 + 3, on the paper you write: "m3", is a hit.
For example, in our case we tried digit 1. diff_of_digitsum (digit, 137*digit+carry) = diff_of_digitsum (1, 137*1+41) = -15. Which is not -2, so 13 is not a hit.
Let's see 99. 9 * 137 = 1233. The "diff" is 9 - 3 = +6. "Carry" is 123. In the second iteration when we try to add a digit 9 to 9 and make it 99, we have diff_of_digitsum (digit, 137*digit+carry) = diff_of_digitsum (9, 137*9+123) = diff_of_digitsum (9, 1356) = -6 and it neutralizes our surplus 6. So 99 is a hit!
In code, we just need 18 iteration. In the first round, we deal with the single digit numbers, 2nd round the 2-digit numbers, then 3-digit ... until we get to 18-digit numbers. Make a table before the iterations that with a structure like this:
table[(diff, carry)] = amount_of_numbers_with_the_same_diff_and_carry
Then the iteration begins, you need to keep updating the table as you go. Add new entries if you encounter a new signature, and always update amount_of_numbers_with_the_same_diff_and_carry. First round, the single digits, populate the table:
0: 0 * 137 = 0, diff: 0; carry: 0. table[(0, 0)] = 1
1: 1 * 137 = 137. diff: 1 - 7 = -6; carry: 13. table[(-6, 13)] = 1
2: 2 * 137 = 274. diff: 2 - 7 = -5; carry: 27. table[(-5, 27)] = 1
And so on.
Second iteration, the "10"th digit, we will go over the digit 0-9 as your "m" and use it in (1) to see if it can produce a result that neutralizes the "diff". If yes, it means this m is going to make all those amount_of_numbers_with_the_same_diff_and_carry into hits. Hence counting not showing. And then we can calculate the new diff and carry with this digit added, like in the example 9 has diff 6 and carry 123 but 99 has the diff 9 - 6 ( last digit from 1356) = 3 and carry 135, replace the old table using the new info.
Last comment, be careful the digit 0. It will appear a lot of times in the iteration and don't over count it because 0009 = 009 = 09 = 9. If you use c++, make sure the sum is in unsigned long long and that sort because it is big. Good luck.
You are trying to solve a Project Euler problem by brute force. That may work for the first few problems, but for most problems you need think of a more sophisticated approach.
Since it is IMHO not OK to give advice specific to this problem, take a look at the general advice in this answer.
This brute force Python solution of 7 digits ran for 19 seconds for me:
print sum(sum(map(int, str(n))) == sum(map(int, str(137 * n)))
for n in xrange(0, 10 ** 7, 9))
On the same machine, single core, same Python interpreter, same code, would take about 3170 years to compute for 18 digits (as the problem asked).
See dgg32's answer for an inspiration of a faster counting.