I'm struggling with this problem I've found in a competitive programming book, but without a solution how to do it. For given two integers A and B (can fit in 64-bit integer type), where A is odd, find a pair of numbers X and Y such that A = X*Y and B = X xor Y.
My approach was to list all divisors of A and try pairing numbers under sqrt(A) with numbers over sqrt(A) that multiply up to A and see if their xor is equal to B. But I don't know if that's efficient enough.
What would be a good solution/algorithm to this problem?
You know that at least one factor is <= sqrt(A). Let's make that one X.
The length of X in bits will be about half the length of A.
The upper bits of X, therefore -- the ones higher in value than sqrt(A) -- are all 0, and the corresponding bits in B must have the same value as the corresponding bits in Y.
Knowing the upper bits of Y gives you a pretty small range for the corresponding factor X = A/Y. Calculate Xmin and Xmax corresponding to the largest and smallest possible values for Y, respectively. Remember that Xmax must also be <= sqrt(A).
Then just try all the possible Xs between Xmin and Xmax. There won't be too many, so it won't take very long.
The other straightforward way to solve this problem relies on the fact that the lower n bits of XY and X xor Y depend only on the lower n bits of X and Y. Therefore, you can use the possible answers for the lower n bits to restrict the possible answers for the lower n+1 bits, until you're done.
I've worked out that, unfortunately, there can be more than one possibility for a single n. I don't know how often there will be a lot of possibilities, but it's probably not too often if at all, so this may be fine in a competitive context. Probabilistically, there will only be a few possibilities, since a solution for n bits will provide either 0 or two solutions for n+1 bits, with equal probability.
It seems to work out pretty well for random input. Here's the code I used to test it:
public static void solve(long A, long B)
{
List<Long> sols = new ArrayList<>();
List<Long> prevSols = new ArrayList<>();
sols.add(0L);
long tests=0;
System.out.print("Solving "+A+","+B+"... ");
for (long bit=1; (A/bit)>=bit; bit<<=1)
{
tests += sols.size();
{
List<Long> t = prevSols;
prevSols = sols;
sols = t;
}
final long mask = bit|(bit-1);
sols.clear();
for (long prevx : prevSols)
{
long prevy = (prevx^B) & mask;
if ((((prevx*prevy)^A)&mask) == 0)
{
sols.add(prevx);
}
long x = prevx | bit;
long y = (x^B)&mask;
if ((((x*y)^A)&mask) == 0)
{
sols.add(x);
}
}
}
tests += sols.size();
{
List<Long> t = prevSols;
prevSols = sols;
sols = t;
}
sols.clear();
for (long testx: prevSols)
{
if (A/testx >= testx)
{
long testy = B^testx;
if (testx * testy == A)
{
sols.add(testx);
}
}
}
System.out.println("" + tests + " checks -> X=" + sols);
}
public static void main(String[] args)
{
Random rand = new Random();
for (int range=Integer.MAX_VALUE; range > 32; range -= (range>>5))
{
long A = rand.nextLong() & Long.MAX_VALUE;
long X = (rand.nextInt(range)) + 2L;
X|=1;
long Y = A/X;
if (Y==0)
{
Y = rand.nextInt(65536);
}
Y|=1;
solve(X*Y, X^Y);
}
}
You can see the results here: https://ideone.com/cEuHkQ
Looks like it usually only takes a couple thousand checks.
Here's a simple recursion that observes the rules we know: (1) the least significant bits of both X and Y are set since only odd multiplicands yield an odd multiple; (2) if we set X to have the highest set bit of B, Y cannot be greater than sqrt(A); and (3) set bits in X or Y according to the current bit in B.
The following Python code resulted in under 300 iterations for all but one of the random pairs I picked from Matt Timmermans' example code. But the first one took 231,199 iterations :)
from math import sqrt
def f(A, B):
i = 64
while not ((1<<i) & B):
i = i - 1
X = 1 | (1 << i)
sqrtA = int(sqrt(A))
j = 64
while not ((1<<j) & sqrtA):
j = j - 1
if (j > i):
i = j + 1
memo = {"it": 0, "stop": False, "solution": []}
def g(b, x, y):
memo["it"] = memo["it"] + 1
if memo["stop"]:
return []
if y > sqrtA or y * x > A:
return []
if b == 0:
if x * y == A:
memo["solution"].append((x, y))
memo["stop"] = True
return [(x, y)]
else:
return []
bit = 1 << b
if B & bit:
return g(b - 1, x, y | bit) + g(b - 1, x | bit, y)
else:
return g(b - 1, x | bit, y | bit) + g(b - 1, x, y)
g(i - 1, X, 1)
return memo
vals = [
(6872997084689100999, 2637233646), # 1048 checks with Matt's code
(3461781732514363153, 262193934464), # 8756 checks with Matt's code
(931590259044275343, 5343859294), # 4628 checks with Matt's code
(2390503072583010999, 22219728382), # 5188 checks with Matt's code
(412975927819062465, 9399702487040), # 8324 checks with Matt's code
(9105477787064988985, 211755297373604352), # 3204 checks with Matt's code
(4978113409908739575,67966612030), # 5232 checks with Matt's code
(6175356111962773143,1264664368613886), # 3756 checks with Matt's code
(648518352783802375, 6) # B smaller than sqrt(A)
]
for A, B in vals:
memo = f(A, B)
[(x, y)] = memo["solution"]
print "x, y: %s, %s" % (x, y)
print "A: %s" % A
print "x*y: %s" % (x * y)
print "B: %s" % B
print "x^y: %s" % (x ^ y)
print "%s iterations" % memo["it"]
print ""
Output:
x, y: 4251585939, 1616572541
A: 6872997084689100999
x*y: 6872997084689100999
B: 2637233646
x^y: 2637233646
231199 iterations
x, y: 262180735447, 13203799
A: 3461781732514363153
x*y: 3461781732514363153
B: 262193934464
x^y: 262193934464
73 iterations
x, y: 5171068311, 180154313
A: 931590259044275343
x*y: 931590259044275343
B: 5343859294
x^y: 5343859294
257 iterations
x, y: 22180179939, 107776541
A: 2390503072583010999
x*y: 2390503072583010999
B: 22219728382
x^y: 22219728382
67 iterations
x, y: 9399702465439, 43935
A: 412975927819062465
x*y: 412975927819062465
B: 9399702487040
x^y: 9399702487040
85 iterations
x, y: 211755297373604395, 43
A: 9105477787064988985
x*y: 9105477787064988985
B: 211755297373604352
x^y: 211755297373604352
113 iterations
x, y: 68039759325, 73164771
A: 4978113409908739575
x*y: 4978113409908739575
B: 67966612030
x^y: 67966612030
69 iterations
x, y: 1264664368618221, 4883
A: 6175356111962773143
x*y: 6175356111962773143
B: 1264664368613886
x^y: 1264664368613886
99 iterations
x, y: 805306375, 805306369
A: 648518352783802375
x*y: 648518352783802375
B: 6
x^y: 6
59 iterations
I have a formula of a sequence of double numbers k = a + d * n, where a and d are constant double values, n is an integer number, k >= 0, a >= 0. For example:
..., 300, 301.6, 303.2, 304.8, 306.4, ...
I want to round a given number c to a nearest value from this sequence which is lower than c.
Currently I use something like this:
double someFunc(double c) {
static double a = 1;
static double d = 2;
int n = 0;
double a1 = a;
if (c >= a) {
while (a1 < c) {
a1 += d;
}
a1 -= d;
} else {
while (a1 > c) {
a1 -= d;
}
}
return a1;
}
Is it possible to do the same without these awful cycles? I ask because the following situation may appear:
abs(a - c) >> abs(d) (the first number is much more then the second one and so a lot of iterations possible)
My question is similar to the following one. But in my case I also have a a variable which has influence on the final result. It means that a sequence may haven't number 0.
Suppose c is a number in your sequence. Then you have n = (c - a) / d.
Since you want an integer <= c, then take n = floor((c - a) / d).
Then you can round c to: a + d * floor((c - a) / d)
Suppose k = 3 + 5 * n and you round c=21.
And 3 + 5 * floor((21 - 3) / 5) = 3 + 5 * 3 = 18
I would like to ask:
how I can add expressions in Maxima? i.e. I have:
A = x + y;
B = 2*x + 2*y;
How to get Maxima to give me (A + B)?
how I can do numerical calculation in Maxima? I want to assign
x = 1;
b = 2;
How to get the numerical value of (A + B)?
(1) assignment in Maxima uses the colon symbol (i.e., ":") not the equal sign ("=").
(2) there are a couple of ways to evaluate with specific values.
(2a) subst([x = ..., y = ...], foo) where foo is some expression such as foo : A + B.
(2b) ev(foo, x = ..., y = ...)
So:
(%i1) A : x + y;
(%o1) y + x
(%i2) B : 2*x + 2*y;
(%o2) 2 y + 2 x
(%i3) foo : A + B;
(%o3) 3 y + 3 x
(%i4) subst ([x = 1, y = 2], foo);
(%o4) 9
(%i5) ev (foo, x = 1, y = 2);
(%o5) 9
Yet another way to substitute values into a formula is with the '' operator as follows:
(%i57) A : 2*a+b ; B : a-b;
(%o57) b + 2 a
(%o58) a - b
(%i59) a : 4; b : 10;
(%o59) 4
(%o60) 10
(%i61) A;
(%o61) b + 2 a
(%i62) ''A;
(%o62) 18
(%i63) ''B;
(%o64) - 6
(%i65) ''A + ''B;
(%o65) 12
(%i66) ''(A+B);
(%o66) 12
I have three numbers x, y , z.
For a range between numbers x and y.
How can i find the total numbers whose % with z is 0 i.e. how many numbers between x and y are divisible by z ?
It can be done in O(1): find the first one, find the last one, find the count of all other.
I'm assuming the range is inclusive. If your ranges are exclusive, adjust the bounds by one:
find the first value after x that is divisible by z. You can discard x:
x_mod = x % z;
if(x_mod != 0)
x += (z - x_mod);
find the last value before y that is divisible by y. You can discard y:
y -= y % z;
find the size of this range:
if(x > y)
return 0;
else
return (y - x) / z + 1;
If mathematical floor and ceil functions are available, the first two parts can be written more readably. Also the last part can be compressed using math functions:
x = ceil (x, z);
y = floor (y, z);
return max((y - x) / z + 1, 0);
if the input is guaranteed to be a valid range (x >= y), the last test or max is unneccessary:
x = ceil (x, z);
y = floor (y, z);
return (y - x) / z + 1;
(2017, answer rewritten thanks to comments)
The number of multiples of z in a number n is simply n / z
/ being the integer division, meaning decimals that could result from the division are simply ignored (for instance 17/5 => 3 and not 3.4).
Now, in a range from x to y, how many multiples of z are there?
Let see how many multiples m we have up to y
0----------------------------------x------------------------y
-m---m---m---m---m---m---m---m---m---m---m---m---m---m---m---
You see where I'm going... to get the number of multiples in the range [ x, y ], get the number of multiples of y then subtract the number of multiples before x, (x-1) / z
Solution: ( y / z ) - (( x - 1 ) / z )
Programmatically, you could make a function numberOfMultiples
function numberOfMultiples(n, z) {
return n / z;
}
to get the number of multiples in a range [x, y]
numberOfMultiples(y) - numberOfMultiples(x-1)
The function is O(1), there is no need of a loop to get the number of multiples.
Examples of results you should find
[30, 90] ÷ 13 => 4
[1, 1000] ÷ 6 => 166
[100, 1000000] ÷ 7 => 142843
[777, 777777777] ÷ 7 => 111111001
For the first example, 90 / 13 = 6, (30-1) / 13 = 2, and 6-2 = 4
---26---39---52---65---78---91--
^ ^
30<---(4 multiples)-->90
I also encountered this on Codility. It took me much longer than I'd like to admit to come up with a good solution, so I figured I would share what I think is an elegant solution!
Straightforward Approach 1/2:
O(N) time solution with a loop and counter, unrealistic when N = 2 billion.
Awesome Approach 3:
We want the number of digits in some range that are divisible by K.
Simple case: assume range [0 .. n*K], N = n*K
N/K represents the number of digits in [0,N) that are divisible by K, given N%K = 0 (aka. N is divisible by K)
ex. N = 9, K = 3, Num digits = |{0 3 6}| = 3 = 9/3
Similarly,
N/K + 1 represents the number of digits in [0,N] divisible by K
ex. N = 9, K = 3, Num digits = |{0 3 6 9}| = 4 = 9/3 + 1
I think really understanding the above fact is the trickiest part of this question, I cannot explain exactly why it works.
The rest boils down to prefix sums and handling special cases.
Now we don't always have a range that begins with 0, and we cannot assume the two bounds will be divisible by K.
But wait! We can fix this by calculating our own nice upper and lower bounds and using some subtraction magic :)
First find the closest upper and lower in the range [A,B] that are divisible by K.
Upper bound (easier): ex. B = 10, K = 3, new_B = 9... the pattern is B - B%K
Lower bound: ex. A = 10, K = 3, new_A = 12... try a few more and you will see the pattern is A - A%K + K
Then calculate the following using the above technique:
Determine the total number of digits X between [0,B] that are divisible by K
Determine the total number of digits Y between [0,A) that are divisible by K
Calculate the number of digits between [A,B] that are divisible by K in constant time by the expression X - Y
Website: https://codility.com/demo/take-sample-test/count_div/
class CountDiv {
public int solution(int A, int B, int K) {
int firstDivisible = A%K == 0 ? A : A + (K - A%K);
int lastDivisible = B%K == 0 ? B : B - B%K; //B/K behaves this way by default.
return (lastDivisible - firstDivisible)/K + 1;
}
}
This is my first time explaining an approach like this. Feedback is very much appreciated :)
This is one of the Codility Lesson 3 questions. For this question, the input is guaranteed to be in a valid range. I answered it using Javascript:
function solution(x, y, z) {
var totalDivisibles = Math.floor(y / z),
excludeDivisibles = Math.floor((x - 1) / z),
divisiblesInArray = totalDivisibles - excludeDivisibles;
return divisiblesInArray;
}
https://codility.com/demo/results/demoQX3MJC-8AP/
(I actually wanted to ask about some of the other comments on this page but I don't have enough rep points yet).
Divide y-x by z, rounding down. Add one if y%z < x%z or if x%z == 0.
No mathematical proof, unless someone cares to provide one, but test cases, in Perl:
#!perl
use strict;
use warnings;
use Test::More;
sub multiples_in_range {
my ($x, $y, $z) = #_;
return 0 if $x > $y;
my $ret = int( ($y - $x) / $z);
$ret++ if $y%$z < $x%$z or $x%$z == 0;
return $ret;
}
for my $z (2 .. 10) {
for my $x (0 .. 2*$z) {
for my $y (0 .. 4*$z) {
is multiples_in_range($x, $y, $z),
scalar(grep { $_ % $z == 0 } $x..$y),
"[$x..$y] mod $z";
}
}
}
done_testing;
Output:
$ prove divrange.pl
divrange.pl .. ok
All tests successful.
Files=1, Tests=3405, 0 wallclock secs ( 0.20 usr 0.02 sys + 0.26 cusr 0.01 csys = 0.49 CPU)
Result: PASS
Let [A;B] be an interval of positive integers including A and B such that 0 <= A <= B, K be the divisor.
It is easy to see that there are N(A) = ⌊A / K⌋ = floor(A / K) factors of K in interval [0;A]:
1K 2K 3K 4K 5K
●········x········x··●·····x········x········x···>
0 A
Similarly, there are N(B) = ⌊B / K⌋ = floor(B / K) factors of K in interval [0;B]:
1K 2K 3K 4K 5K
●········x········x········x········x···●····x···>
0 B
Then N = N(B) - N(A) equals to the number of K's (the number of integers divisible by K) in range (A;B]. The point A is not included, because the subtracted N(A) includes this point. Therefore, the result should be incremented by one, if A mod K is zero:
N := N(B) - N(A)
if (A mod K = 0)
N := N + 1
Implementation in PHP
function solution($A, $B, $K) {
if ($K < 1)
return 0;
$c = floor($B / $K) - floor($A / $K);
if ($A % $K == 0)
$c++;
return (int)$c;
}
In PHP, the effect of the floor function can be achieved by casting to the integer type:
$c = (int)($B / $K) - (int)($A / $K);
which, I think, is faster.
Here is my short and simple solution in C++ which got 100/100 on codility. :)
Runs in O(1) time. I hope its not difficult to understand.
int solution(int A, int B, int K) {
// write your code in C++11
int cnt=0;
if( A%K==0 or B%K==0)
cnt++;
if(A>=K)
cnt+= (B - A)/K;
else
cnt+=B/K;
return cnt;
}
(floor)(high/d) - (floor)(low/d) - (high%d==0)
Explanation:
There are a/d numbers divisible by d from 0.0 to a. (d!=0)
Therefore (floor)(high/d) - (floor)(low/d) will give numbers divisible in the range (low,high] (Note that low is excluded and high is included in this range)
Now to remove high from the range just subtract (high%d==0)
Works for integers, floats or whatever (Use fmodf function for floats)
Won't strive for an o(1) solution, this leave for more clever person:) Just feel this is a perfect usage scenario for function programming. Simple and straightforward.
> x,y,z=1,1000,6
=> [1, 1000, 6]
> (x..y).select {|n| n%z==0}.size
=> 166
EDIT: after reading other's O(1) solution. I feel shamed. Programming made people lazy to think...
Division (a/b=c) by definition - taking a set of size a and forming groups of size b. The number of groups of this size that can be formed, c, is the quotient of a and b. - is nothing more than the number of integers within range/interval ]0..a] (not including zero, but including a) that are divisible by b.
so by definition:
Y/Z - number of integers within ]0..Y] that are divisible by Z
and
X/Z - number of integers within ]0..X] that are divisible by Z
thus:
result = [Y/Z] - [X/Z] + x (where x = 1 if and only if X is divisible by Y otherwise 0 - assuming the given range [X..Y] includes X)
example :
for (6, 12, 2) we have 12/2 - 6/2 + 1 (as 6%2 == 0) = 6 - 3 + 1 = 4 // {6, 8, 10, 12}
for (5, 12, 2) we have 12/2 - 5/2 + 0 (as 5%2 != 0) = 6 - 2 + 0 = 4 // {6, 8, 10, 12}
The time complexity of the solution will be linear.
Code Snippet :
int countDiv(int a, int b, int m)
{
int mod = (min(a, b)%m==0);
int cnt = abs(floor(b/m) - floor(a/m)) + mod;
return cnt;
}
here n will give you count of number and will print sum of all numbers that are divisible by k
int a = sc.nextInt();
int b = sc.nextInt();
int k = sc.nextInt();
int first = 0;
if (a > k) {
first = a + a/k;
} else {
first = k;
}
int last = b - b%k;
if (first > last) {
System.out.println(0);
} else {
int n = (last - first)/k+1;
System.out.println(n * (first + last)/2);
}
Here is the solution to the problem written in Swift Programming Language.
Step 1: Find the first number in the range divisible by z.
Step 2: Find the last number in the range divisible by z.
Step 3: Use a mathematical formula to find the number of divisible numbers by z in the range.
func solution(_ x : Int, _ y : Int, _ z : Int) -> Int {
var numberOfDivisible = 0
var firstNumber: Int
var lastNumber: Int
if y == x {
return x % z == 0 ? 1 : 0
}
//Find first number divisible by z
let moduloX = x % z
if moduloX == 0 {
firstNumber = x
} else {
firstNumber = x + (z - moduloX)
}
//Fist last number divisible by z
let moduloY = y % z
if moduloY == 0 {
lastNumber = y
} else {
lastNumber = y - moduloY
}
//Math formula
numberOfDivisible = Int(floor(Double((lastNumber - firstNumber) / z))) + 1
return numberOfDivisible
}
public static int Solution(int A, int B, int K)
{
int count = 0;
//If A is divisible by K
if(A % K == 0)
{
count = (B / K) - (A / K) + 1;
}
//If A is not divisible by K
else if(A % K != 0)
{
count = (B / K) - (A / K);
}
return count;
}
This can be done in O(1).
Here you are a solution in C++.
auto first{ x % z == 0 ? x : x + z - x % z };
auto last{ y % z == 0 ? y : y - y % z };
auto ans{ (last - first) / z + 1 };
Where first is the first number that ∈ [x; y] and is divisible by z, last is the last number that ∈ [x; y] and is divisible by z and ans is the answer that you are looking for.