Given an integer 1 ≤ N ≤ 1,000,000 as input, output the last non-zero digit of N!, where ! is the factorial (the product of all numbers from 1 to N, inclusive). This is OEIS sequence A008904.
Your program needs finish within 10 seconds on a reasonable machine for any valid input.
1 => 1
2 => 2
3 => 6
4 => 4
5 => 2
6 => 2
7 => 4
8 => 2
9 => 8
10 => 8
100 => 4
1000 => 2
10000 => 8
100000 => 6
1000000 => 4
This is a code-golf so the shortest code in bytes wins!
f=->n{n<2?1:6*[1,1,2,6,4,4,4,8,4,6][n%10]*3**(n/5%4)*f[n/5]%10}
Source - http://oeis.org/A008904
Handles f upto a thousand digits under a second.
irb(main):014:0> for n in 2..6
irb(main):015:1> puts f[10**n]
irb(main):016:1> end
4
2
8
6
4
Last@Select[IntegerDigits[#!],#>0&]&
Very readable for a winning answer. :) (Then again, there's no GolfScript & Co. submission yet.)
This handles input 1,000,000 in about 5 seconds on my machine.
n=input()
g=1
while n:
g*=n
while g%10<1:g/=10
g%=10**9
n-=1
print g%10
This trades speed for size -- the testcase takes a long time (~6 seconds).
n->n!/10^valuation(n!,5)%10
This version is much faster (~15 microseconds) but takes 81 bytes:
n->r=1;while(n,r*=Mod(4,10)^(n\10%2)*[1,2,6,4,2,2,4,2,8][max(n%10,1)];n\=5);lift(r)
You can use this (non-golfed) code to test either:
[%(10^n) | n <- [1..6]]
($a=1).."$input"|%{$a="$($a*$_)".trim('0')%1e7}
$a%10
Notes:
History:
$input to a string is enough; the cast to int is applied implicitly.All whitespace can be removed, but I left it in for "readability". Note: not using some silly explicit formula from Sloane.
sub f {
$_ = $1 * ++$n || 1, /(.{1,7}?)0*$/ while $n < $_[0];
$1 % 10
}
Calculates f(10^6) in 8.7 seconds on my machine.
Update: OP wanted it to be a whole program:
$_ = $1 * ++$n || 1, /(.{1,7}?)0*$/ while $n < $ARGV[0];
print $1 % 10
That makes it 55 characters.
1ri{I)*_AbW%{}#A\#/1e7%}fIA%
You can try it at http://cjam.aditsu.net/ for values up to 10000 or so; for larger numbers you should use the java interpreter. 1000000 runs in about 3 seconds on my laptop.
Explanation:
Unfortunately the straightforward solution is too slow, so I'm keeping only the last 7 digits (before the trailing zeros) after each multiplication.
1 push 1 on the stack
ri read a token and convert to integer
{ loop (for I from 0 to N - 1)
I) push I and increment
* multiply with the previous value (initially 1)
_Ab duplicate and convert to array of digits
W% reverse array
{}# find the position of the first non-zero digit
A\# raise 10 to that power
/ divide, thus removing all trailing zeros
1e7% keep the remainder modulo 10000000
}fI end for loop
A% get the last digit
Note: this language is a lot newer than the question.
Mod[#!/10^IntegerExponent[#!],10]&
!0м¤
!0 # Push the factorial of the input and 0
м # Remove the occurences of 0 in the factorial
¤ # Push the last element, implicit display
!Ṛȯ/
Explanation
Uses the fact that when Ṛ (reverse a list; does not vectorize) is applied to an integer it automatically takes D (digits) first.
With input 8:
!Ṛȯ/
! Factorial: 8! = 40320
Ṛ Reverse: [0,2,3,0,4]
/ Reduce by...
ȯ ...logical OR: ((((0ȯ2)ȯ3)ȯ0)ȯ4) = first truthy element = 2
I don't think there is a one-byte "first truthy element" (which ȯ/ acts as) but if there is then this can be shortened to three bytes total.
n->{long f=n;for(;n>1||f%10==0;)f=n>1?f*--n:f/10;return f%10;}
Similar to @Kevin Cruijssen but saves 5 bytes by combining the loops.
|| with |, and an additional byte by replacing ==0 with <1. Enjoy your stay!
\$\endgroup\$
– Kevin Cruijssen
Feb 1 '18 at 12:19
r,d;f(k,x){r=x<5?3:f(k+1,x/5);return(d=x%5)?r*"33436"[d]*(1<<d*k%4)%5:r;}main(int c,char**v){c=atoi(*++v);printf("%d",c<2?1:2*f(0,c));}
This is the version for ASCII systems; replace the string 33436 with 11214 for an EBCDIC system, or with \1\1\2\1\4 for a portable program.
C solutions are a bit hampered by the requirement to provide a full program; however, this does answer the question fully.
Try it online (requires Javascript):
It's based on the algorithm outlined in Least Significant Non-Zero Digit of n!, turned around so that we recurse in to find the highest power of five, and do the calculation on the way out. The tables of constants were too big, so I reduced them by finding a relationship between the previous residue r, the current digit d and the recursion depth k:
0 1 2 3 4 =d
0 0 3×2^k 1×2^2k 3×2^3k 2
1 1 1×2^k 2×2^2k 1×2^3k 4
r 2 2 2×2^k 4×2^2k 2×2^3k 3
3 3 3×2^k 3×2^2k 3×2^3k 2
4 4 4×2^k 4×2^2k 4×2^3k 1
For r>0, this resolves to a constant times r times 2^dk (mod 5); the constants are in a[] below (inlined in the golfed code). We also observe that (2^4)%5 is 1, so we can reduce the exponent to avoid overflowing the range of int.
const int a[] = { 1, 1, 2, 1, 4 };
int f(int k, int x){
int r = x<5 ? 3 : f(k+1,x/5); /* residue - from recursing to higher-order quinary digits */
int d = x%5;
if (!d)
return r;
return r * a[d] * (1<<d*k%4) % 5;
}
int main(int c, char **v)
{
c = atoi(*++v);
printf("%d",
c<2
? 1 /* special-case 0 & 1 */
: 2*f(0,c)); /* otherwise, it's 2 times r */
}
$ for i in 100 1000 10000 100000; do echo $i: `./694 $i`; done
100: 4
1000: 2
10000: 8
100000: 6
1000000: 4
Performance is respectable, too. Here's a maximum input for a system with 32-bit int:
$ time ./694 2147483647
8
real 0m0.001s
user 0m0.000s
sys 0m0.000s
I got the same timings with a maximal 64-bit int, too.
2147483647! has over 19 billion digits, and (2^63-1)! over 170,000,000,000,000,000,000 digits, so this is a big win over calculating the factorials. 1000000! as specified in the question is feasible to calculate on current hardware; that's only 5½ million digits. :-)
\$\endgroup\$
– Toby Speight
May 17 '16 at 9:49
<?foreach(explode("\n",`cat`)as$n)if($n){$f=rtrim(gmp_strval(gmp_fact($n)),'0');echo substr($f,-1)."\n";}
Runs under 10 seconds with the given testcase.
Can do 10000 in ~3.2 seconds (Ideone cuts me off at 8 seconds, i'm sure it'll take longer than 10secs though :( )
from functools import *
N=100
r=range
s=(p for p in r(2,N)if all(p%n>0for n in r(2,p)))
f=lambda n,x:n//x+(n//x>0and f(n//x,x)or 0)
e=list([p,f(N,p)]for p in s)
e[0][1]-=e[2][1]
e[2][1]=0
print(reduce(lambda x,y:x*y,map(lambda x:x[0]**x[1],e))%10)
from itertools import *
N=100000
r=xrange
def s(c=count(2)):
while 1:p=c.next();c=ifilter(p.__rmod__,c);yield p
f=lambda n,x:n//x+(n//x>0and f(n//x,x)or 0)
e=[[p,f(N,p)]for p in takewhile(lambda x:x<N,s())]
e[0][1]-=e[2][1]
e[2][1]=0
print(reduce(lambda x,y:x*y,map(lambda x:pow(x[0],x[1],10),e))%10)
f n=head$dropWhile(=='0')$reverse$show$product[1..n]
main=interact(show.f.read)
(Would probably need to be compiled to compute 1,000,000! in 10 secs).
foldl1 with product (cf codegolf.stackexchange.com/questions/607/find-the-factorial/…). But have you actually tried with 1000000! ?
\$\endgroup\$
– J B
Feb 8 '11 at 14:48
A whole program. Save this program in a file and run with jconsole script.ijs 1234. Notice that this program does not exit the interpreter after printing a result. Type ^D or exit]0 to exit the interpreter.
echo([:{:@(#~*)10&#.inv@*)/1+i.".>{:ARGV
Here is an explanation:
x #. y interprets the integer vector y as a base x number; for example, 10 #. 1 2 3 4 yields 1234.u inv yields the inverse of a verb u. In particular, x #. inv y represents y as a base x number; for example, 10 #. 1234 yields 1 2 3 4. Notice that inv is defined as ^:_1, that is, u applied -1 time.x * y is the product of x and y, thus x 10&#.inv@* y yields a base-10 representation of the product of x and y.x # y copies the n-th item of y as often as the n-th item of x; when x is a vector of booleans, x selects which items of y to take. For instance, 1 0 1 0 # 1 2 3 4 yields 1 3.* y yields the signum of y.x u~ y is the reflexive of u, that is, the same as y u x.y #~ * y yields a vector of all items of y that are positive. In tacit notation, this can written with a hook as (#~ *).{: y yields the last item in y.([:{:@(#~*)10&#.inv@*).u/ y is the reduction of y, that is, the dyadic verb u inserted between elements of y. For instance, +/1 2 3 4 is like 1 + 2 + 3 + 4 and yields 10.([:{:@(#~*)10&#.inv@*)/ y yields the last digit of the product of the items of y.ARGV is a boxed vector of the command line arguments.".>{:ARGV is the last argument unboxed and interpreted as a number.i. y computes natural numbers from 0 to y - 1.1+i. y yields natural numbers from 1 to y. I could have also used >: increment here, but 1+ is clearer at the same cost of characters.1+i.".>{:ARGV (the vector of 1 to the number in the last command line argument) to the verb ([:{:@(#~*)10&#.inv@*)/ and prints the result with echo.Computes factorial, extracts the last non-zero digit. Thanks to Giuseppe for providing the basic structure.
(y=(gamma(scan()+1))%/%10^(0:1e5)%%10)[!!y][1]
Alternatively, my old 51-byte answer:
Calculates factorial, converts to character, removes all 0s, and then takes the final character. Saved 2 bytes thanks to Giuseppe.
substring(x<-gsub("0","",gamma(scan())+1),nchar(x))
gamma(x+1) is shorter than factorial(x)
\$\endgroup\$
– Giuseppe
Jan 18 '18 at 18:44
(y=(x<-gamma(scan()+1))%/%10^(0:nchar(x))%%10)[!!y][1] at 54 bytes.
\$\endgroup\$
– Giuseppe
Jan 18 '18 at 18:49
nchar(x) with 1e5 for a 46-byte solution! Nice going.
\$\endgroup\$
– rturnbull
Jan 18 '18 at 21:54
v:a%:?n-a,!
1
>$::?!.1-}*
Handles 0! correctly as well. Value passed in through the -v flag.
1000 doesn't produce any output on TIO - what's wrong?
\$\endgroup\$
– Toby Speight
Jun 5 '18 at 10:27
{[*](1..$_)~~/.*<(<-[0]>/}
As a full program:
put [*](1..@*ARGS[0])~~/.*<(<-[0]>/
{
[*]( 1..$_ ) # reduce using &infix:« * »
~~ # match with
/
.* # any number of values (so it matches from the end)
<( # only capture the following
<-[0]> # any value but 0 (negated character class)
/
}
f(n,d)long long n,d;{for(d=1;n;d%=10000)for(d*=n--;d%10<1;d/=10);d%=10;}
main(){long long n,d=1;for(scanf("%lld",&n);n;d%=10000)for(d*=n--;d%10<1;d/=10);printf("%d",d%10);}
This question is just shy of 8 years old so "reasonable machine" is not the same as back then, but I'm getting times of ~.01 seconds on my computer when doing all test cases together, so unless computers have increased in speed by a factor of 1000 this last decade, it should be fine.
Last@`\&:(All@V)@Digits@`!
Last@`\&:(All@V)@Digits@`!
This is a composition of 4 functions:
`! - this is a functional version of the factorial operatorDigits - this gets the digits of the factorial\&:(All@V) - This is a selection function. It works by left-bonding (&:) the function All@V to \, which is select. In turn, All@V is a short way of testing if a number is not 0. It works by casting its input to a vector 0 -> [0] then querying if all those members are truthy (i.e. not 0). This gives the digits of the number without 0s.Last - this simply obtains the last member of this array.
{⊢/⍵/⍨0≠⍎¨⍵}⍕∘!
Tacit prefix function. Returns the correct digit for a single test case, or a string of digits for multiple test cases.
Thanks to @Adám and @ErikTheOutgolfer for 3 bytes each.
{⊢/⍵/⍨0≠⍎¨⍵}⍕∘! ⍝ Main function. Argument is a number following the !.
! ⍝ Factorial
∘ ⍝ then
⍕ ⍝ Format (stringify)
⍎¨⍵} ⍝ Execute (turn to number) each digit of the argument
0≠ ⍝ Check if each is ≠0. This returns a boolean vector
⍨ ⍝ Swap arguments for the following fn/op
⍵/ ⍝ Replicate. This takes a boolean vector as left arg and returns the truthy elements of the right arg. E.g.: 1 1 0/1 2 3 → 1 2.
{⊢/ ⍝ Reduce. This returns the rightmost (last) element of a vector argument.
{↑≠v/v←⌽⍎¨⍕!⍵}
I don't know why but this pass the test:
q←{↑≠v/v←⌽⍎¨⍕!⍵}
q¨1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 2 6 4 2 2 4 2 8 8 8 6 8 2
{for(p=$1;--$1;p=(p*$1)%1e4)while(!(p%10))p/=10;$0=p%10}1
Original solution did not handle "large" input values very well. Could add -M to force it to work, but that also requires a lot more processing time.
inf doesn't % very well. :(
\$\endgroup\$
– Robert Benson
Feb 20 '18 at 16:05
Came up with a few different 6-byters but I liked this one best. I'm convinced there must be a way to do it in 5, though.
Êsw ìv
Ê calculates the factorial of the input, s converts it to a string and back to an integer after w has reversed it, ì converts the result to an array of digits and v returns the first element.
Êì w æ
ÊìÈf Ì
Êì f o
Êsw sg
Êìf ìo
Êìf ìÌ
f (should be 4) and 100 ⇒ 7 (should be 4)
\$\endgroup\$
– Toby Speight
Feb 20 '18 at 14:22
