So I have this code. Not sure if it works because the runtime for the program is still continuing.
void permute(std::vector<std::string>& wordsVector, std::string prefix, int length, std::string alphabet) {
if (length == 0) {
//end the recursion
wordsVector.push_back(prefix);
}
else {
for (int i = 0; i < alphabet.length(); ++i) {
permute(wordsVector, prefix + alphabet.at(i), length - 1, alphabet);
}
}}
where I'm trying to get all combinations of characters in the English alphabet of a given length. I'm not sure if the approach is correct at the moment.
Alphabet consists of A-Z in a string of length 26. WordsVectors holds all the different combinations of words. prefix is meant to pass through recursively until a word is made and length is self explanatory.
Example, if I give the length of 7 to the function, I expect a size of 26 x 25 x 24 x 23 x 22 x 21 x 20 = 3315312000 if I'm correct, following the formula for permutations.
I don't think programs are meant to run this long so either I'm hitting an infinite loop or something is wrong with my approach. Please advise. Thanks.
Surely the stack would overflow but concentrating on your question even if you write an iterative program it will take a long time ( not an infinite loop just very long )
[26L, 650L, 15600L, 358800L, 7893600L, 165765600L, 3315312000L, 62990928000L, 1133836704000L, 19275223968000L, 308403583488000L, 4626053752320000L, 64764752532480000L, 841941782922240000L, 10103301395066880000L, 111136315345735680000L, 1111363153457356800000L, 10002268381116211200000L, 80018147048929689600000L, 560127029342507827200000L, 3360762176055046963200000L, 16803810880275234816000000L, 67215243521100939264000000L, 201645730563302817792000000L, 403291461126605635584000000L, 403291461126605635584000000L]
The above list is the number of possibilities for 1<=n<=26. You can see as n increases number of possibilities increases tremendously. Say you have 1GHz processor that does 10^9 operations per second. Say consider number of possibilities for n=26 its 403291461126605635584000000L. Its evident that if you sit down to list all possibilities its so so long ( so so many years ) that
you will feel it has hit an infinite loop. Finally I have not looked that closely into your code , but in nutshell even if you write it correctly,iteratively and don't store (again can't have this much memory) and just print all possibilities its going to take long time for larger values of n.
EDIT
As jaromax and others said if you just want to write it for smaller values of n,
say less than 10-12 you can write an iterative program to list/print them. It will run quite fast for small values. But if you also want to store them them then n will have to be say less than 5 say. (Really depends on how much RAM is available or you could find some permutations write them to disk, then depends on how much disk memory you can spare, again refer the number of possibilities list I posted above. It gives a rough idea of both time and space complexity).
I think there could be quite a problem that you do this on stack. A large part of the calculation you do recursively and this means every time allocated space for function.
Try to reformulate it linearly. I think I had such a problem before.
Your question implies you think there are 26x25x24x ... permutations
Your code doesn't have anything I can see to avoid "AAAAAAA" being a permutation, in which case there are 26x26x26x ...
So in addition to being a very complicated way of counting in base 26, I think it's also giving bad answers?
Related
#include <iostream>
using namespace std;
int main() {
int n;
cin>>n;
int *arr=new int [n];
for(int k=0;n>k;k++)
{
cin>>*(arr+k);
}
long long sum1=0,sum2=0,sum3=0;
for(int k=0;n>k;k++)
{
sum1=sum1+*(arr+k);
if(*(arr+k)%2==0)
sum2++;
else
sum3++;
}
cout<<sum1<<" ";
cout<<sum3<<" ";
cout<<sum2;
return 0;
}
You're given a sequence of N integers, your task is to print sum of them, number of odd integers, and number of even integers respectively.
Input
The first line of input contains an integer N (1≤N≤10⁵).
The second line of input contains N integers separated by a single space (1≤Ai≤10⁵).
Output
Print the sum of them, number of odd integers, and number of even integers respectively, separated by a space.
Examples
input
5
1 2 3 4 5
output
15 3 2
Is there a better algorithm for this code? I need it to take less Execution Time.
Where can I find better algorithms for any code?
Unless you need to re-use the N integers that you have stored in the array, there's no point in storing them. You can get the sum as well as number of odd/even integers as you input them.
Additionally, you don't need long long as the input will never get that big, unless you mean 10^5?
Further, whenever you are thinking about improving performance you should take a look at the big O which in this case is O(N) where N is the number of integers that you have. From an algorithm point of view with N input there's generally very little that you can do to improve this. Maybe if we're talking streams, you can do some statistics but otherwise this implementation is as good as it gets. In some other situations, while the worst case can't be improved, we can improve the average case, which I don't think is applicable here.
Then you should look at profiling the code. That way you have a clear understanding of where bottlenecks are. For your code, there's probably not too much that can be done reasonably.
If we're trying to squeeze every ounce of performance possible, adjusting the compiler flags can bring some performance gains. You should research these but I would not prioritize this over the above.
I would also improve how you name your variables, but this has no impact on performance.
Actually C++ by default synchronizes cin/cout with C way of doing I/O - printf/scanf, which slows down I/O by quite a lot.
Switching to printf/scanf or adding something like ios::sync_with_stdio(0); at the start of main should speed this up a few times.
I am wondering how to make an array with values that start at 1111 and go all the way up to 8888. I am asking this because I need to generate a list of 4 digit numbers with each digit ranging from 1-8. I would like to have this in a loop form. Also, I am lost on my functions trim, methodicalEliminate, guessAndEliminate, and guessThreeThenEliminate in my following program. Here are the directions:
This assignment focuses on the use of arrays in a program, including using one as a parameter to a function.
PROBLEM DESCRIPTION
In the game of Mastermind, a player is only given a finite number of guesses to try to identify the hidden combination (such as twelve guesses). Often that is the only constraint in playing the game.
But some players might make a competition with each other to see who can guess the other's combination in fewer tries. In this case, the problem is not only to come up with a strategy that can find the answer within a specified limit, but to find one that is likely to require the minimal number of guesses.
Here is where the computer comes in -- one can write a program that would try out different guessing strategies, and see how they work out. Since a computer can do the analysis and computations more rapidly than a person, it could just pretend to play Mastermind on our behalf using any strategy we choose, and tell us how long it took to do that.
OVERALL SOLUTION
Of course, it would be extremely difficult to teach the computer to reason along the same lines as a person. For example, if we guessed a combination 1111 and got one black peg, we would make a mental note that the answer has exactly one 1 in it, and then proceed to make other guesses with that one fact in mind. If we next guesses 1222 and got one white peg, we would know there were no 2's, and that the single 1 is not in the first position. But how to keep track of such information after a series of guesses would be rather hard.
Fortunately, for a computer simulation with an array, we can record all of our known facts in a different way. We just maintain a list of all possible answers that there could be, and then remove numbers from the list that could no longer be the solution. If our first guess tells us there is exactly one 1 digit, we would remove all the numbers that do not have that feature. When we find out there are no 2's, we eliminate all the values that contain 2's. Eventually, the only number left would be the correct answer.
SOME SIMPLE STRATEGIES
This is a strategy that many players use, resembling what was described above. Just methodically got through the possibilities in a straightforward fashion. The first guess of 1111 would answer how many 1's are in the solution; the next guess would answer how many 2's are in the solution, and also say something about where any 1's might be, and so on.
With our list approach, which contain a whole lot of possibilities in order beginning with 1111, 1112, 1113, 1114, etc., our next guess would always be the first in the list.
The next strategy is for those who like a little more excitement. The guesses appear to be more or less random, with the hopes that a lot more information can be discovered. Simulating this approach is surprisingly simple -- if you have a list of numbers, just pick one at random. If you have 837 possibilities to choose from in an array, just pick a random subscript in the range of 0 to 836.
This third strategy considers the possibility that answers that give similar results to a given guess are in a sense similar to each other. So to try to get a little more information, it will still pick some numbers at random without regard to how they were evaluated, and then only start thinking about the results.
To implement this one, let us just pick any three possible answers and guess them, temporarily ignoring how many black pegs and white pegs they earn us. Only after making those guesses will we trim the list of possibilities, then proceeding as the second strategy above.
SAMPLE INTERFACE
These are the sample results from the current implementation: Please enter a combination to try for, or 0 for a random value: 0
Guessing at 2475
Guessing 1111...
Guessing 2222...
Guessing 2333...
Guessing 2444...
Guessing 2455...
Guessing 2456...
Guessing 2475...
Methodical Eliminate required 7 tries.
Guessing 6452...
Guessing 2416...
Guessing 2485...
Guessing 2445...
Guessing 2435...
Guessing 2425...
Guessing 2475...
Guess and Eliminate required 7 tries.
Guessing 7872...
Guessing 6472...
Guessing 1784...
Guessing 2475...
Guess Three then Eliminate required 4 tries.
Play another game? (y/n) y
Please enter a combination to try for, or 0 for a random value: 0
Guessing at 4474
Guessing 1111...
Guessing 2222...
Guessing 3333...
Guessing 4444...
Guessing 4445...
Guessing 4464...
Guessing 4474...
Methodical Eliminate required 7 tries.
Guessing 3585...
Guessing 7162...
Guessing 4474...
Guess and Eliminate required 3 tries.
Guessing 8587...
Guessing 1342...
Guessing 1555...
Guessing 7464...
Guessing 6764...
Guessing 4468...
Guessing 4474...
Guess Three then Eliminate required 7 tries.
NOTE: This program allows each digit to go up to 8 instead of 6. Even though there are 4096 possible answers, it still finds them rather rapidly.
PROGRAM SPECIFICATIONS
The assigned program must implement all of the following functions. Additional ones are permitted as desired -- these below are required. Future assignments will not detail the functions as below -- but will instead require the students to design their own function descriptions in advance to writing the program. main:
Simply governs the overall behavior of the program. A number will be
chosen as the target combination, and then each strategy will attempt
to find it.
Calls: generateAnswer, (to compare all three must have the same answer)
methodicalEliminate, guessAndEliminate, guessThreeThenEliminate
generateAnswer:
Either lets the user at the keyboard choose the mystery combination,
or gives the option to have the computer generate a random combination.
(For a competitive game, it might be interesting to know what sorts
of combinations would be the hardest to guess!)
Parameters: none!
Returns: a 4-digit combination, each digit in the range 1 to 8
generateSearchSpace:
Populates an array with all possible combinations of four-digit
values in the range 1 to 8.
Parameters:
guesses (modified int array) list of guesses
length (output int) number of values in list
Pre-condition:
The array must be allocated to no fewer than 4096 elements.
trim:
Analyzes the response to a particular guess and then eliminates
any values from the list of possibilities that are no longer
possible answers. In each case, it assumes that a value in the
list is an answer, and evaluates the guess accordingly. If the
number of black and white pegs is not the same as those specified,
then it cannot be the correct answer.
Parameters:
guesses (modified int array) list of guesses
length (modified int) number of values in list
guess (input int) a guess that has been evaluated
black (input int) how many black pegs that guess earned
white (input int) how many white pegs that guess earned
Pre-condition:
black and white actually do contain the results of comparing
the guess with the actual answer
Post-condition:
length has been reduced (we learned something)
the viable answers occupy the first 'length' positions
in the guesses array (so the list is shorter)
Calls: evaluate
methodicalEliminate:
beginning with a list of all possible candidate answers
continually guesses the first element in the list, and
trim answers accordingly, until an answer is found
Parameter:
answer (input int) the actual answer
(necessary to get black/white pegs)
Returns: number of guesses required to find the answer
Calls: generateSearchSpace, evaluate, trim
gusssAndEliminate:
beginning with a list of all possible candidate answers
continually guesses a random element in the list, and
trim answers accordingly, until an answer is found
Parameter:
answer (input int) the actual answer
Returns: number of guesses required to find the answer
Calls: generateSearchSpace, evaluate, trim
gusssThreeThenEliminate:
beginning with a list of all possible candidate answers
first guesses three answers at random before trimming
the list of possibilites, and then narrows on the answer
one random guess at a time
Parameter:
answer (input int) the actual answer
Returns: number of guesses required to find the answer
Calls: generateSearchSpace, evaluate, trim
NOTE: These last functions use the correct answer to evaluate
each guess and then use the black/white pegs for the guessing
strategy. NOne of these strategies may peek at the answer to
decide what to do next!
ALSO: The following functions should also appear in this program
from the previous assignment, though they are not themselves
part of the grade for this one.
evaluate:
evaluates a combination by comparing it with the answer
Correctness is indicated by black pegs (correct digit in correct position)
and white pegs (correct digit in incorrect position)
Parameters:
answer (input int) the correct combination
guess (input int) the current guess
black (output int) number of black pegs
white (output int) number of white pegs
pre-conditions:
answer and guess are both 4-digit numbers with no zero digits
post-conditions:
black and white are both > 0 and their sum is <= 4
Calls: nthDigit, clearNthDigit
nthDigit:
identified the n'th digit of a combination
whether digits count from left to right or right to left is unspecified
Parameters:
combination (input int) combination to examine
position (input int) which digit to examine
(returned) (output int) the value of the actual digit
pre-conditions:
combination has the appropriate number of digits, and
0 < position <= number of digits
post-condition:
0 <= returned digit <= 9 (single digit)
clearNthDigit:
ears the n'th digit of a combination to zero, so it will no longer match
digits must be counted in the same manner as nthDigit above.
parameters:
combination (in/out int) combination to modify
position (input int) which digit to set to 0
pre-condition:
same as those for nthDigit above
post-condition:
corresponding digit is set to zero
Calls: nthDigit (optional, depending on the implementation)
Thank you for reading such this long question, and I hope you can help me on arrays!
Just because your guesses have the form of 1111 through to 8888, it does not mean they are numbers.
They are numbers if it makes sense to do arithmetic calculations on them. It does not make sense to define arithmetic calculations on the guesses: what would it mean to add a guess 4571 to a guess 6214?
If your guesses are not numbers, don't use a representation that is reserved for numbers. You can, however, use an array of four integers:
int guesses[4][4096];
int g = 0;
for (int i = 1; i <= 8; ++i)
for (int j = 1; j <= 8; ++j)
for (int k = 1; k <= 8; ++k)
for (int m = 1; m <= 8; ++m)
guesses[g++] = {i, j, k, m};
I am pretty convinced that putting all possible guesses into the array guesses like that is not a good idea either; the code mainly demonstrates how the guesses could be generated.
Go through the other functions, think of what high level operations should be performed on the remaining gueses (like "eliminate all guesses that has a certain number in the thrid position" etc). This should give you an idea about what would be a better data structure to replace the guesses array.
I removed the codes because it's homework. If you actually needs the help, you can either look at the discussion I had with George B (below), or PM me.
Hi guys. This is a homework assignment. I have tested it against other sorting algorithms, and Q.S. is the only one that is crashing on some random inputs.
The program is quit long (with other stuff), but input is randomly generated....
I spent a few hours tracing the codes and still couldn't figure out any error....
Q.S. is probably very easy for the professionals, so I hope to receive advices on this implementation....
Any input is appreciated!
Q: What is "random"?
A: A portion of generation is included.
void randomArray(unsigned long*& A, unsigned long size)
{
//Note that RAND_MAX is a little small for some compilers (2^16-1).
//In order to test our algorithms on large arrays without huge
//numbers of duplicates, we'll set the high-order and low-order
//parts of the return value with two random values.
A = new unsigned long[size];
for(unsigned long i=0; i<size; i++)
A[i] = (rand()<<16) | (rand());
//Another note: initially, if you want to test your program out with smaller
//arrays and small numbers, just reduce A[i] mod k for some small value k as in the following:
//A[i] = rand() % 16;
//this may help you debug at first.
}
Q: What kind of error?
Well, I am not getting compilation error. Without Q.S., I can ran other four sorting algorithm without problems (I can continuously running the sorting). When Q.S is activated, after running the program one or two or three times, or even at the first time of running, the program ends (I am using Eclipse, so the consoles ends).
enter the number of elements, or a
negative number to quit: 5 {some
arrays}
selection sort took 0 seconds. merge
sort took 0 seconds. quick sort took 0
seconds. heap sort took 0 seconds.
bucket sort took 0 seconds. {output of
5 sorted arrays}
enter the number of elements, or a
negative number to quit: 6 {some
arrays}
selection sort took 0 seconds. merge
sort took 0 seconds. quick sort took 0
seconds. heap sort took 0 seconds.
bucket sort took 0 seconds.
{output of 5 sorted arrays}
enter the number of elements, or a
negative number to quit: 8 {arrays}
--- console ends---
Again, the problem is that it crashes quite often, so this suggests that there is a high possibility of access violation,,, but doing 10+ tracings I don't see the problem.... (maybe I overloaded my brain stack -_- )
Thanks.
Hint:
q is unsigned (the result of the partition function)
so, q-1 is also unsigned
what if q is zero?
(It is homework so you have to figure it out I guess :) )
Trace your algorithm with the array {2,5,2}. Obviously your program crashes as soon as there are duplicate numbers in your list. The first call of Partition will return 2 as the index of r. Thus, the second call of quickSort(A,3,2) will access memory locations not within array boundaries. Its always a good idea to do the boundary checks for arrays manually and generate understandable output to more easily trace and debug your program.
If we have an array of all the numbers up to N (N < 10), what is the best way to find all the numbers that are missing.
Example:
N = 5
1 5 3 2 3
Output: 1 5 4 2 3
In the ex, the number 4 was the missing one and there were 2 3s, so we replaced the first one with 4 and now the array is complete - all the numbers up to 5 are there.
Is there any simple algorithm that can do this ?
Since N is really small, you can use F[i] = k if number i appears k times.
int F[10]; // make sure to initialize it to 0
for ( int i = 0; i < N; ++i )
++F[ numbers[i] ];
Now, to replace the duplicates, traverse your number array and if the current number appears more than once, decrement its count and replace it with a number that appears 0 times and increment that number's count. You can keep this O(N) if you keep a list of numbers that don't appear at all. I'll let you figure out what exactly needs to be done, as this sounds like homework.
Assume all numbers within the range 1 ≤ x ≤ N.
Keep 2 arrays of size N. output, used (as an associative array). Initialize them all to 0.
Scan from the right, fill in values to output unless it is used.
Check for unused values, and put them into the empty (zero) slots of output in order.
O(N) time complexity, O(N) space complexity.
You can use a set data structure - one for all the numbers up to N, one for the numbers you actually saw, and use a set difference.
One way to do this would be to look at each element of the array in sequence, and see whether that element has been seen before in elements that you've already checked. If so, then change that number to one you haven't seen before, and proceed.
Allow me to introduce you to my friend Schlemiel the Painter. Discovery of a more efficient method is left as a challenge for the reader.
This kind of looks like homework, please let us know if it isn't. I'll give you a small hint, and then I'll improve my answer if you confirm this isn't homework.
My tip for now is this: If you were to do this by hand, how would you do it? Would you write out an extra list of numbers of some time, would you read through the list (how many times?)? etc.
For simple problems, sometimes modelling your algorithm after an intuitive by-hand approach can work well.
Here's a link I read just today that may be helpful.
http://research.swtch.com/2008/03/using-uninitialized-memory-for-fun-and.html
I understand this is a classic programming problem and therefore I want to be clear I'm not looking for code as a solution, but would appreciate a push in the right direction. I'm learning C++ and as part of the learning process I'm attempting some programming problems. I'm attempting to write a program which deals with numbers up to factorial of 1billion. Obviously these are going to be enormous numbers and way too big to be dealing with using normal arithmetic operations. Any indication as to what direction I should go in trying to solve this type of problem would be appreciated.
I'd rather try to solve this without using additional libraries if possible
Thanks
PS - the problem is here http://www.codechef.com/problems/FCTRL
Here's the method I used to solve the problem, this was achieved by reading the comments below:
Solution -- The number 5 is a prime factor of any number ending in zero. Therefore, dividing the factorial number by 5, recursively, and adding the quotients, you get the number of trailing zeros in the factorial result
E.G. - Number of trailing zeros in 126! = 31
126/5 = 25 remainder 1
25/5 = 5 remainder 0
5/5 = 1 remainder 0
25 + 5 + 1 = 31
This works for any value, just keep dividing until the quotient is less
than 5
Skimmed this question, not sure if I really got it right but here's a deductive guess:
First question - how do you get a zero on the end of the number? By multiplying by 10.
How do you multiply by 10? either by multiplying by either a 10 or by 2 x 5...
So, for X! how many 10s and 2x5s do you have...?
(luckily 2 & 5 are prime numbers)
edit: Here's another hint - I don't think you need to do any multiplication. Let me know if you need another hint.
Hint: you may not need to calculate N! in order to find the number of zeros at the end of N!
To solve this question, as Chris Johnson said you have to look at number of 0's.
The factors of 10 will be 1,2,5,10 itself. So, you can go through each of the numbers of N! and write them in terms of 2^x * 5^y * 10^z. Discard other factors of the numbers.
Now the answer will be greaterof(x,y)+z.
One interesting thing I learn from this question is, its always better to store factorial of a number in terms of prime factors for easy comparisons.
To actually x^y, there is an easy method used in RSA algorithm, which don't remember. I will try to update the post if I find one.
This isn't a good answer to your question as you've modified it a bit from what I originally read. But I will leave it here anyway to demonstrate the impracticality of actually trying to do the calculations by main brute force.
One billion factorial is going to be out of reach of any bignum library. Such numbers will require more space to represent than almost anybody has in RAM. You are going to have to start paging the numbers in from storage as you work on them. There are ways to do this. The guy who recently calculated π out to 2700 billion places used such a library
Do not use the naive method. If you need to calculate the factorial, use a fast algorithm: http://www.luschny.de/math/factorial/FastFactorialFunctions.htm
I think that you should come up with a way to solve the problem in pseudo code before you begin to think about C++ or any other language for that matter. The nature of the question as some have pointed out is more of an algorithm problem than a C++ problem. Those who suggest searching for some obscure library are pointing you in the direction of a slippery slope, because learning to program is learning how to think, right? Find a good algorithm analysis text and it will serve you well. In our department we teach from the CLRS text.
You need a "big number" package - either one you use or one you write yourself.
I'd recommend doing some research into "large number algorithms". You'll want to implement the C++ equivalent of Java's BigDecimal.
Another way to look at it is using the gamma function. You don't need to multiply all those values to get the right answer.
To start you off, you should store the number in some sort of array like a std::vector (a digit for each position in the array) and you need to find a certain algorithm that will calculate a factorial (maybe in some sort of specialized class). ;)
//SIMPLE FUNCTION TO COMPUTE THE FACTORIAL OF A NUMBER
//THIS ONLY WORKS UPTO N = 65
//CAN YOU SUGGEST HOW WE CAN IMPROVE IT TO COMPUTE FACTORIAL OF 400 PLEASE?
#include <iostream>
#include <cmath>
using namespace std;
int factorial(int x); //function to compute factorial described below
int main()
{
int N; //= 150; //you can also get this as user input using cin.
cout<<"Enter intenger\n";
cin>>N;
factorial(N);
return 0;
}//end of main
int factorial(int x) //function to compute the factorial
{
int i, n;
long long unsigned results = 1;
for (i = 1; i<=x; i++)
{
results = results * i;
}
cout<<"Factorial of "<<x<<" is "<<results<<endl;
return results;
}