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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!
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.
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
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]]
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]&
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\$
Feb 1, 2018 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\$
May 17, 2016 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\$
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.
{[*](1..$_)~~/.*<(<-[0]>/}
As a full program:
put [*](1..@*ARGS[0])~~/.*<(<-[0]>/
{
[*]( 1..$_ ) # reduce using &infix:« * »
~~ # match with
/
.* # any number of values (so it matches from the end)
<( # only capture the following
<-[0]> # any value but 0 (negated character class)
/
}
f(n,d)long long n,d;{for(d=1;n;d%=10000)for(d*=n--;d%10<1;d/=10);d%=10;}
main(){long long n,d=1;for(scanf("%lld",&n);n;d%=10000)for(d*=n--;d%10<1;d/=10);printf("%d",d%10);}
This question is just shy of 8 years old so "reasonable machine" is not the same as back then, but I'm getting times of ~.01 seconds on my computer when doing all test cases together, so unless computers have increased in speed by a factor of 1000 this last decade, it should be fine.
{k.=Floor[_/5]If[_<2,1,6*Digits@$`U1sTny@(_%10)*3^(k%4)*$@k%10]}
Simple implementation of the OEIS formula. Digits@$`U1sTny compresses the hardcoded dictionary using Attache's compressed number feature. $ refers to the function itself within this answer.
{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\$
Feb 20, 2018 at 16:05
Based on this answer
f=n=>n>9?'6248'[(a=n/5|0)%4]*f(a)*f(n%5)%10:'1126422428'[n]
Try it:
f=n=>n>9?'6248'[(a=n/5|0)%4]*f(a)*f(n%5)%10:'1126422428'[n]
;[
0, // 1
1, // 1
2, // 2
3, // 6
4, // 4
5, // 2
6, // 2
7, // 4
8, // 2
9, // 8
10, // 8
20, // 4
100, // 4
1000, // 2
10000, // 8
100000, // 6
1000000, // 4
].forEach(n=>console.log(n, f(n)))Modified formula is:
$$ D(0) = 1 \\ D(1) = 1 \\ D(2) = 2 \\ D(3) = 6 \\ D(4) = 4 \\ D(5) = 2 \\ D(6) = 2 \\ D(7) = 3 \\ D(8) = 2 \\ D(9) = 8 \\ D(n) = (LastDigitOf(2^{\lfloor n/5 \rfloor}) * D(\lfloor n/5 \rfloor) * D(n \mod 5)) \mod 10, \;where \;n > 9 $$
The original formula is \$D(n) = (2^{[n/5]} * D([n/5]) * D(n \mod 5)) \mod 10\$
But there is a problem: we need to calculate \$2^{[n/5]}\$. For example if \$n = 10 000 \$ then we need to calculate \$2^{2000}\$ which is already too hard (in JS it will output Infinity). But I discovered that we don't need to calculate it instead of this we just need to calculate the last digit of \$2^{[n/5]}\$. Why? Because each time we just need to calculate the last digit of \$2^{[n/5]} * D([n/5]) * D(n \mod 5)\$
The last digit of multiplication is the last digit of multiplication of last digits of each multiplier
Let's say we have two numbers \$a\$ and \$b\$. We can represent each of them to the power of 10:
$$ a = x_n * 10^n + x_{n - 1} * 10^{n - 1} + \ldots + x_1 * 10^1 + x_0 \\ b = y_n * 10^m + y_{m - 1} * 10^{m - 1} + \ldots + y_1 * 10^1 + y_0 $$
or
$$ a = (x_n * 10^{n - 1} + x_{n - 1} * 10^{n - 2} + \ldots + x_1) * 10 + x_0 \\ b = (y_n * 10^{m - 1} + y_{m - 1} * 10^{m - 2} + \ldots + y_1) * 10 + y_0 $$
When we multiply them we will get:
$$ a * b = (...) * 10 + x_0 * y_0 $$
Now it is clear that the last digit of multiplication is the last digit of \$x_0 * y_0\$, where the \$x_0\$ and \$y_0\$ are the last digits of \$a\$ and \$b\$ respectively. So we proved the theorem
Now we need to find the last digits of powers of 2. Let's write some of them:
$$ 2^0 = 1 \\ 2^1 = 2 \\ 2^2 = 4 \\ 2^3 = 8 \\ 2^4 = 16 \\ 2^5 = 32 \\ 2^6 = 64 \\ 2^7 = 128 \\ 2^8 = 256 $$
Now we can see the pattern 1 (2 4 8 6) (2 4 8 6) and can write it \$(k = [n / 5])\$:
if (k == 0) return 1;
else return [6, 4, 8, 7][k % 4];
We can ignore the first case when k == 0 and here is why:
k % 4 == 0 when k is one of [0, 4, 8, 12, 16, ...] but we need to consider cases when k is one of [0, 4, 8] because for bigger values we will run recursion. The second multiplier in our formula is \$D(k)\$ and for k is on of [0, 4, 8] we get [1, 4, 2] respectively. We had to multiply these values by 1 but instead of this we will multiply them by 6 so we will get [1, 24, 12], the last digits of these values are [1, 4, 2]. Now we can see that it doesn't matter to multiply them by 1 or 6 the last digits will be the same
The last thing we need it is to prove that \$[n / 5] = \lfloor n / 5 \rfloor\$ for \$n \ge 0\$. I think it is obvious
