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Let N be an integer. If N = 2536, the reversed N is 6352. If N = 1000000, the reversed N is 1.
We are given an integer M, where 1 <= M <= 10^(100000).
We need to find whether an integer N exists, where N + reversed(N) = M.
Any ideas, besides brute force ?
Here I will describe briefly an algorithm. It should be noted that many details needed to be filled in.
The basic idea is to look at the first and last digit of M to determine the sum of the first and last digit of N, and then subtract this quantity from M to reduce to the case of a shorter number.
Let us call a number good if it can be written as N + reverse(N).
(EDIT: in implementation, one will probably need a function IsGood(M, k) which judges whether M can be written as N + reverse(N) for some N < 10^k. But let's skip this detail for the moment.)
The algorithm for determining whether a given number M is good goes as follows:
Let c and d be the first and last digit of M, and let R be the middle part. That is, M has digital expression cRd.
There are two cases:
c is not equal to 1
c is equal to 1
In the case where c is not equal to 1, the digit c cannot be a carry. This is the normal case. Now look at d.
If d is equal to c, then M is good if and only if R is good.
If d is equal to c - 1, then there is a carry from R to c, so M is good if and only if 1R is good in the carry case (see below).
If d is equal to anything else, then M is not good.
In the case where c is equal to 1, there is the additional possibility that c is a carry.
Let e be the first digit of R, and write M as 1eTd.
If d = 9 or e < d, then the carry case is not possible.
(EDIT: this is wrong, the case d = 9 is possible if e = 0.)
Otherwise, the carry case is possible if and only if (e - d)(T - 1) is good.
If either the carry case hold, or the normal case hold, then M is good.
Example:
Let us start with M = 12001.
Since c = 1, there is the normal case and the carry case.
In the normal case, we have d = 1, so we need to test whether 200 is good. For M = 200, we have c = 2 and d = 0, so the number 200 is not good, hence the normal case for M = 12001 fails.
In the carry case, we need to test whether (12001 - 11000 - 11) / 10 = 99 is good. For M = 99, we have c = 9 and d = 9, so this again reduces to whether 0 is good, which obviously is true. Hence the carry case holds.
The conclusion is then M is good.
Time complexity:
With some detailed arguments (which I don't want to present here), it can be proved that the algorithm runs in O(log_10(M)) time.
I studied math, and I come up with this problem.
There are two permutations A and B and a integer M.
We say A almost equal to B if we can make from A to B doing following operations.
-1 Choose a M-length segment of the permutation A.
-2 Perform a cyclic shift on it towards the right.(So,if sub segment is "1 2 3 4 5"(m=5), then after this operation , it will be "5 1 2 3 4".)
Question : Does A almost equal to B?
I think it is typical , but I couldn't find the answer.
How to solve it?(not brute force!)
number of elements in the permutation<=10^5
example
A "1 2 3"
B "2 3 1"
m=2
answer YES
explain "1 2 3"->"2 1 3"->"2 3 1"
Here's a proof of my conjecture. Let n be the length of the permutations and m be the length of the windows that we are allowed to rotate, where 1 ≤ m ≤ n. Permutations P and Q are almost equal if there exists a sequence of window rotations that transforms P into Q. Almost equality is an equivalence relation. Here's the claimed characterization of the equivalence classes.
(1) m = 1: P almost equals Q if and only if P = Q
(2) m = n: P almost equals Q if and only if they're rotations of each other
(3) 1 < m < n, m odd: P almost equals Q if and only if they have the same parity
(4) 1 < m < n, n even: P almost equals Q
The first two claims are obvious. As for (3), the necessity of the parity condition follows from the fact that rotating a window of odd length is an even permutation.
The meat of the argument here is to find an algorithm for when n = m + 1 ≥ 4, since in general, we can use an algorithm similar to insertion sort to transform P so that all but the last m + 1 elements match Q, and specifically, the case (n, m) = (3, 2) can be solved by inspection. In case m is even, we ensure further that the transformation matches the parity of Q, by rotating the last m elements once if necessary. (For m odd, we just assume equal parities.)
We need a technique for moving fewer than m elements at once. Suppose that the state is as follows.
1, 2, 3, 4, ..., m, m + 1
Rotate the second window m - 1 times (i.e., once in reverse).
1, 3, 4, ..., m, m + 1, 2
Rotate the first window m - 1 times.
3, 4, ..., m, m + 1, 1, 2
Second, twice.
3, 2, 4, ..., m, m + 1, 1
3, 1, 2, 4, ..., m, m + 1
We've succeeded in rotating the first three elements. This suffices in combination with conjugation by rotations to "insertion sort" the first m - 1 elements of Q into place. The other two are in the right order by the parity match.
I studied math, and I come up with this problem.
There are two permutations A and B and a integer M.
We say A almost equal to B if we can make from A to B doing following operations.
-1 Choose a M-length segment of the permutation A.
-2 Perform a cyclic shift on it towards the right.(So,if sub segment is "1 2 3 4 5"(m=5), then after this operation , it will be "5 1 2 3 4".)
Question : Does A almost equal to B?
I think it is typical , but I couldn't find the answer.
How to solve it?(not brute force!)
number of elements in the permutation<=10^5
example
A "1 2 3"
B "2 3 1"
m=2
answer YES
explain "1 2 3"->"2 1 3"->"2 3 1"
Here's a proof of my conjecture. Let n be the length of the permutations and m be the length of the windows that we are allowed to rotate, where 1 ≤ m ≤ n. Permutations P and Q are almost equal if there exists a sequence of window rotations that transforms P into Q. Almost equality is an equivalence relation. Here's the claimed characterization of the equivalence classes.
(1) m = 1: P almost equals Q if and only if P = Q
(2) m = n: P almost equals Q if and only if they're rotations of each other
(3) 1 < m < n, m odd: P almost equals Q if and only if they have the same parity
(4) 1 < m < n, n even: P almost equals Q
The first two claims are obvious. As for (3), the necessity of the parity condition follows from the fact that rotating a window of odd length is an even permutation.
The meat of the argument here is to find an algorithm for when n = m + 1 ≥ 4, since in general, we can use an algorithm similar to insertion sort to transform P so that all but the last m + 1 elements match Q, and specifically, the case (n, m) = (3, 2) can be solved by inspection. In case m is even, we ensure further that the transformation matches the parity of Q, by rotating the last m elements once if necessary. (For m odd, we just assume equal parities.)
We need a technique for moving fewer than m elements at once. Suppose that the state is as follows.
1, 2, 3, 4, ..., m, m + 1
Rotate the second window m - 1 times (i.e., once in reverse).
1, 3, 4, ..., m, m + 1, 2
Rotate the first window m - 1 times.
3, 4, ..., m, m + 1, 1, 2
Second, twice.
3, 2, 4, ..., m, m + 1, 1
3, 1, 2, 4, ..., m, m + 1
We've succeeded in rotating the first three elements. This suffices in combination with conjugation by rotations to "insertion sort" the first m - 1 elements of Q into place. The other two are in the right order by the parity match.
My input are three numbers - a number s and the beginning b and end e of a range with 0 <= s,b,e <= 10^1000. The task is to find the minimal Levenstein distance between s and all numbers in range [b, e]. It is not necessary to find the number minimizing the distance, the minimal distance is sufficient.
Obviously I have to read the numbers as string, because standard C++ type will not handle such large numbers. Calculating the Levenstein distance for every number in the possibly huge range is not feasible.
Any ideas?
[EDIT 10/8/2013: Some cases considered in the DP algorithm actually don't need to be considered after all, though considering them does not lead to incorrectness :)]
In the following I describe an algorithm that takes O(N^2) time, where N is the largest number of digits in any of b, e, or s. Since all these numbers are limited to 1000 digits, this means at most a few million basic operations, which will take milliseconds on any modern CPU.
Suppose s has n digits. In the following, "between" means "inclusive"; I will say "strictly between" if I mean "excluding its endpoints". Indices are 1-based. x[i] means the ith digit of x, so e.g. x[1] is its first digit.
Splitting up the problem
The first thing to do is to break up the problem into a series of subproblems in which each b and e have the same number of digits. Suppose e has k >= 0 more digits than s: break up the problem into k+1 subproblems. E.g. if b = 5 and e = 14032, create the following subproblems:
b = 5, e = 9
b = 10, e = 99
b = 100, e = 999
b = 1000, e = 9999
b = 10000, e = 14032
We can solve each of these subproblems, and take the minimum solution.
The easy cases: the middle
The easy cases are the ones in the middle. Whenever e has k >= 1 more digits than b, there will be k-1 subproblems (e.g. 3 above) in which b is a power of 10 and e is the next power of 10, minus 1. Suppose b is 10^m. Notice that choosing any digit between 1 and 9, followed by any m digits between 0 and 9, produces a number x that is in the range b <= x <= e. Furthermore there are no numbers in this range that cannot be produced this way. The minimum Levenshtein distance between s (or in fact any given length-n digit string that doesn't start with a 0) and any number x in the range 10^m <= x <= 10^(m+1)-1 is necessarily abs(m+1-n), since if m+1 >= n it's possible to simply choose the first n digits of x to be the same as those in s, and delete the remainder, and if m+1 < n then choose the first m+1 to be the same as those in s and insert the remainder.
In fact we can deal with all these subproblems in a single constant-time operation: if the smallest "easy" subproblem has b = 10^m and the largest "easy" subproblem has b = 10^u, then the minimum Levenshtein distance between s and any number in any of these ranges is m-n if n < m, n-u if n > u, and 0 otherwise.
The hard cases: the end(s)
The hard cases are when b and e are not restricted to have the form b = 10^m and e = 10^(m+1)-1 respectively. Any master problem can generate at most two subproblems like this: either two "ends" (resulting from a master problem in which b and e have different numbers of digits, such as the example at the top) or a single subproblem (i.e. the master problem itself, which didn't need to be subdivided at all because b and e already have the same number of digits). Note that due to the previous splitting of the problem, we can assume that the subproblem's b and e have the same number of digits, which we will call m.
Super-Levenshtein!
What we will do is design a variation of the Levenshtein DP matrix that calculates the minimum Levenshtein distance between a given digit string (s) and any number x in the range b <= x <= e. Despite this added "power", the algorithm will still run in O(n^2) time :)
First, observe that if b and e have the same number of digits and b != e, then it must be the case that they consist of some number q >= 0 of identical digits at the left, followed by a digit that is larger in e than in b. Now consider the following procedure for generating a random digit string x:
Set x to the first q digits of b.
Append a randomly-chosen digit d between b[i] and e[i] to x.
If d == b[i], we "hug" the lower bound:
For i from q+1 to m:
If b[i] == 9 then append b[i]. [EDIT 10/8/2013: Actually this can't happen, because we chose q so that e[i] will be larger then b[i], and there is no digit larger than 9!]
Otherwise, flip a coin:
Heads: Append b[i].
Tails: Append a randomly-chosen digit d > b[i], then goto 6.
Stop.
Else if d == e[i], we "hug" the upper bound:
For i from q+1 to m:
If e[i] == 0 then append e[i]. [EDIT 10/8/2013: Actually this can't happen, because we chose q so that b[i] will be smaller then e[i], and there is no digit smaller than 0!]
Otherwise, flip a coin:
Heads: Append e[i].
Tails: Append a randomly-chosen digit d < e[i], then goto 6.
Stop.
Otherwise (if d is strictly between b[i] and e[i]), drop through to step 6.
Keep appending randomly-chosen digits to x until it has m digits.
The basic idea is that after including all the digits that you must include, you can either "hug" the lower bound's digits for as long as you want, or "hug" the upper bound's digits for as long as you want, and as soon as you decide to stop "hugging", you can thereafter choose any digits you want. For suitable random choices, this procedure will generate all and only the numbers x such that b <= x <= e.
In the "usual" Levenshtein distance computation between two strings s and x, of lengths n and m respectively, we have a rectangular grid from (0, 0) to (n, m), and at each grid point (i, j) we record the Levenshtein distance between the prefix s[1..i] and the prefix x[1..j]. The score at (i, j) is calculated from the scores at (i-1, j), (i, j-1) and (i-1, j-1) using bottom-up dynamic programming. To adapt this to treat x as one of a set of possible strings (specifically, a digit string corresponding to a number between b and e) instead of a particular given string, what we need to do is record not one but two scores for each grid point: one for the case where we assume that the digit at position j was chosen to hug the lower bound, and one where we assume it was chosen to hug the upper bound. The 3rd possibility (step 5 above) doesn't actually require space in the DP matrix because we can work out the minimal Levenshtein distance for the entire rest of the input string immediately, very similar to the way we work it out for the "easy" subproblems in the first section.
Super-Levenshtein DP recursion
Call the overall minimal score at grid point (i, j) v(i, j). Let diff(a, b) = 1 if characters a and b are different, and 0 otherwise. Let inrange(a, b..c) be 1 if the character a is in the range b..c, and 0 otherwise. The calculations are:
# The best Lev distance overall between s[1..i] and x[1..j]
v(i, j) = min(hb(i, j), he(i, j))
# The best Lev distance between s[1..i] and x[1..j] obtainable by
# continuing to hug the lower bound
hb(i, j) = min(hb(i-1, j)+1, hb(i, j-1)+1, hb(i-1, j-1)+diff(s[i], b[j]))
# The best Lev distance between s[1..i] and x[1..j] obtainable by
# continuing to hug the upper bound
he(i, j) = min(he(i-1, j)+1, he(i, j-1)+1, he(i-1, j-1)+diff(s[i], e[j]))
At the point in time when v(i, j) is being calculated, we will also calculate the Levenshtein distance resulting from choosing to "stop hugging", i.e. by choosing a digit that is strictly in between b[j] and e[j] (if j == q) or (if j != q) is either above b[j] or below e[j], and thereafter freely choosing digits to make the suffix of x match the suffix of s as closely as possible:
# The best Lev distance possible between the ENTIRE STRINGS s and x, given that
# we choose to stop hugging at the jth digit of x, and have optimally aligned
# the first i digits of s to these j digits
sh(i, j) = if j >= q then shc(i, j)+abs(n-i-m+j)
else infinity
shc(i, j) = if j == q then
min(hb(i, j-1)+1, hb(i-1, j-1)+inrange(s[i], (b[j]+1)..(e[j]-1)))
else
min(hb(i, j-1)+1, hb(i-1, j-1)+inrange(s[i], (b[j]+1)..9),
he(i, j-1)+1, he(i-1, j-1)+inrange(s[i], (0..(e[j]-1)))
The formula for shc(i, j) doesn't need to consider "downward" moves, since such moves don't involve any digit choice for x.
The overall minimal Levenshtein distance is the minimum of v(n, m) and sh(i, j), for all 0 <= i <= n and 0 <= j <= m.
Complexity
Take N to be the largest number of digits in any of s, b or e. The original problem can be split in linear time into at most 1 set of easy problems that collectively takes O(1) time to solve and 2 hard subproblems that each take O(N^2) time to solve using the super-Levenshtein algorithm, so overall the problem can be solved in O(N^2) time, i.e. time proportional to the square of the number of digits.
A first idea to speed up the computation (works if |e-b| is not too large):
Question: how much can the Levestein distance change when we compare s with n and then with n+1?
Answer: not too much!
Let's see the dynamic-programming tables for s = 12007 and two consecutive n
n = 12296
0 1 2 3 4 5
1 0 1 2 3 4
2 1 0 1 2 3
3 2 1 1 2 3
4 3 2 2 2 3
5 4 3 3 3 3
and
n = 12297
0 1 2 3 4 5
1 0 1 2 3 4
2 1 0 1 2 3
3 2 1 1 2 3
4 3 2 2 2 3
5 4 3 3 3 2
As you can see, only the last column changes, since n and n+1 have the same digits, except for the last one.
If you have the dynamic-programming table for the edit-distance of s = 12001 and n = 12296, you already have the table for n = 12297, you just need to update the last column!
Obviously if n = 12299 then n+1 = 12300 and you need to update the last 3 columns of the previous table.. but this happens just once every 100 iteration.
In general, you have to
update the last column on every iterations (so, length(s) cells)
update the second-to-last too, once every 10 iterations
update the third-to-last, too, once every 100 iterations
so let L = length(s) and D = e-b. First you compute the edit-distance between s and b. Then you can find the minimum Levenstein distance over [b,e] looping over every integer in the interval. There are D of them, so the execution time is about:
Now since
we have an algorithm wich is
The question is Number of solutions to a1 x1+a2 x2+....+an xn=k with constraints: 1)ai>0 and ai<=15 2)n>0 and n<=15 3)xi>=0 I was able to formulate a Dynamic programming solution but it is running too long for n>10^10. Please guide me to get a more efficient soution.
The code
int dp[]=new int[16];
dp[0]=1;
BigInteger seen=new BigInteger("0");
while(true)
{
for(int i=0;i<arr[0];i++)
{
if(dp[0]==0)
break;
dp[arr[i+1]]=(dp[arr[i+1]]+dp[0])%1000000007;
}
for(int i=1;i<15;i++)
dp[i-1]=dp[i];
seen=seen.add(new BigInteger("1"));
if(seen.compareTo(n)==0)
break;
}
System.out.println(dp[0]);
arr is the array containing coefficients and answer should be mod 1000000007 as the number of ways donot fit into an int.
Update for real problem:
The actual problem is much simpler. However, it's hard to be helpful without spoiling it entirely.
Stripping it down to the bare essentials, the problem is
Given k distinct positive integers L1, ... , Lk and a nonnegative integer n, how many different finite sequences (a1, ..., ar) are there such that 1. for all i (1 <= i <= r), ai is one of the Lj, and 2. a1 + ... + ar = n. (In other words, the number of compositions of n using only the given Lj.)
For convenience, you are also told that all the Lj are <= 15 (and hence k <= 15), and n <= 10^18. And, so that the entire computation can be carried out using 64-bit integers (the number of sequences grows exponentially with n, you wouldn't have enough memory to store the exact number for large n), you should only calculate the remainder of the sequence count modulo 1000000007.
To solve such a problem, start by looking at the simplest cases first. The very simplest cases are when only one L is given, then evidently there is one admissible sequence if n is a multiple of L and no admissible sequence if n mod L != 0. That doesn't help yet. So consider the next simplest cases, two L values given. Suppose those are 1 and 2.
0 has one composition, the empty sequence: N(0) = 1
1 has one composition, (1): N(1) = 1
2 has two compositions, (1,1); (2): N(2) = 2
3 has three compositions, (1,1,1);(1,2);(2,1): N(3) = 3
4 has five compositions, (1,1,1,1);(1,1,2);(1,2,1);(2,1,1);(2,2): N(4) = 5
5 has eight compositions, (1,1,1,1,1);(1,1,1,2);(1,1,2,1);(1,2,1,1);(2,1,1,1);(1,2,2);(2,1,2);(2,2,1): N(5) = 8
You may see it now, or need a few more terms, but you'll notice that you get the Fibonacci sequence (shifted by one), N(n) = F(n+1), thus the sequence N(n) satisfies the recurrence relation
N(n) = N(n-1) + N(n-2) (for n >= 2; we have not yet proved that, so far it's a hypothesis based on pattern-spotting). Now, can we see that without calculating many values? Of course, there are two types of admissible sequences, those ending with 1 and those ending with 2. Since that partitioning of the admissible sequences restricts only the last element, the number of ad. seq. summing to n and ending with 1 is N(n-1) and the number of ad. seq. summing to n and ending with 2 is N(n-2).
That reasoning immediately generalises, given L1 < L2 < ... < Lk, for all n >= Lk, we have
N(n) = N(n-L1) + N(n-L2) + ... + N(n-Lk)
with the obvious interpretation if we're only interested in N(n) % m.
Umm, that linear recurrence still leaves calculating N(n) as an O(n) task?
Yes, but researching a few of the mentioned keywords quickly leads to an algorithm needing only O(log n) steps ;)
Algorithm for misinterpreted problem, no longer relevant, but may still be interesting:
The question looks a little SPOJish, so I won't give a complete algorithm (at least, not before I've googled around a bit to check if it's a contest question). I hope no restriction has been omitted in the description, such as that permutations of such representations should only contribute one to the count, that would considerably complicate the matter. So I count 1*3 + 2*4 = 11 and 2*4 + 1*3 = 11 as two different solutions.
Some notations first. For m-tuples of numbers, let < | > denote the canonical bilinear pairing, i.e.
<a|x> = a_1*x_1 + ... + a_m*x_m. For a positive integer B, let A_B = {1, 2, ..., B} be the set of positive integers not exceeding B. Let N denote the set of natural numbers, i.e. of nonnegative integers.
For 0 <= m, k and B > 0, let C(B,m,k) = card { (a,x) \in A_B^m × N^m : <a|x> = k }.
Your problem is then to find \sum_{m = 1}^15 C(15,m,k) (modulo 1000000007).
For completeness, let us mention that C(B,0,k) = if k == 0 then 1 else 0, which can be helpful in theoretical considerations. For the case of a positive number of summands, we easily find the recursion formula
C(B,m+1,k) = \sum_{j = 0}^k C(B,1,j) * C(B,m,k-j)
By induction, C(B,m,_) is the convolution¹ of m factors C(B,1,_). Calculating the convolution of two known functions up to k is O(k^2), so if C(B,1,_) is known, that gives an O(n*k^2) algorithm to compute C(B,m,k), 1 <= m <= n. Okay for small k, but our galaxy won't live to see you calculating C(15,15,10^18) that way. So, can we do better? Well, if you're familiar with the Laplace-transformation, you'll know that an analogous transformation will convert the convolution product to a pointwise product, which is much easier to calculate. However, although the transformation is in this case easy to compute, the inverse is not. Any other idea? Why, yes, let's take a closer look at C(B,1,_).
C(B,1,k) = card { a \in A_B : (k/a) is an integer }
In other words, C(B,1,k) is the number of divisors of k not exceeding B. Let us denote that by d_B(k). It is immediately clear that 1 <= d_B(k) <= B. For B = 2, evidently d_2(k) = 1 if k is odd, 2 if k is even. d_3(k) = 3 if and only if k is divisible by 2 and by 3, hence iff k is a multiple of 6, d_3(k) = 2 if and only if one of 2, 3 divides k but not the other, that is, iff k % 6 \in {2,3,4} and finally, d_3(k) = 1 iff neither 2 nor 3 divides k, i.e. iff gcd(k,6) = 1, iff k % 6 \in {1,5}. So we've seen that d_2 is periodic with period 2, d_3 is periodic with period 6. Generally, like reasoning shows that d_B is periodic for all B, and the minimal positive period divides B!.
Given any positive period P of C(B,1,_) = d_B, we can split the sum in the convolution (k = q*P+r, 0 <= r < P):
C(B,m+1, q*P+r) = \sum_{c = 0}^{q-1} (\sum_{j = 0}^{P-1} d_B(j)*C(B,m,(q-c)*P + (r-j)))
+ \sum_{j = 0}^r d_B(j)*C(B,m,r-j)
The functions C(B,m,_) are no longer periodic for m >= 2, but there are simple formulae to obtain C(B,m,q*P+r) from C(B,m,r). Thus, with C(B,1,_) = d_B and C(B,m,_) known up to P, calculating C(B,m+1,_) up to P is an O(P^2) task², getting the data necessary for calculating C(B,m+1,k) for arbitrarily large k, needs m such convolutions, hence that's O(m*P^2).
Then finding C(B,m,k) for 1 <= m <= n and arbitrarily large k is O(n^2*P^2), in time and O(n^2*P) in space.
For B = 15, we have 15! = 1.307674368 * 10^12, so using that for P isn't feasible. Fortunately, the smallest positive period of d_15 is much smaller, so you get something workable. From a rough estimate, I would still expect the calculation of C(15,15,k) to take time more appropriately measured in hours than seconds, but it's an improvement over O(k) which would take years (for k in the region of 10^18).
¹ The convolution used here is (f \ast g)(k) = \sum_{j = 0}^k f(j)*g(k-j).
² Assuming all arithmetic operations are O(1); if, as in the OP, only the residue modulo some M > 0 is desired, that holds if all intermediate calculations are done modulo M.