here I am trying to convert my recursive solution into a dynamic solution but I am getting problem in converting. I am trying to count minimum possible coin to make value. what I am doing is I am taking all possible coin and putting inside a vector and finally in main function I will find and minimum of my vector and that will be my answer
int rec(vector<int>coins,int n,int sum,int counter)
{
if(sum==0)
{
return 1;
}
if(n==0)
{
return 0;
}
int total=rec(coins,n-1,sum,counter);
if(sum-coins[n-1]>=0)
{
total+=rec(coins,n,sum-coins[n-1],counter+1);
if(sum-coins[n-1]==0)
{
vec.push_back(counter+1);
}
}
return total;
}
you should first try to solve this type of problems by yourself.
BTW:
#define BIG 2147483647
int min_coin(int *coins, int m, int desire_value)
{
int dp[desire_value+1];
dp[0] = 0;
for (int i=1; i<=desire_value; i++)
dp[i] = BIG;
for (int i=1; i<=desire_value; i++)
{
for (int j=0; j<m; j++)
if (coins[j] <= i)
{
int diff = dp[i-coins[j]];
if (diff != BIG && diff + 1 < dp[i])
dp[i] = diff + 1;
}
}
if(dp[desire_value]==BIG)
return -1;
return dp[desire_value];
}
you can easily convert dp and coins to vector. vector is like array.(beware for allocate vector for dp you should reserve space in vector see here.)
So i am having difficulty understand a rather simple algorithm. If i want to compare all the elements within one array to another how would it work? For instance
For ( i = 0; i < size; i++ )
{
For ( k = 0; k < size; k++)
{
Do something if arrays are the same or not
}
}
Both i and k loops would just carry the same slot in the array. If i add a - 1 or + 1 it would just check the last or next slot... But what if i wanted to check 1 to 3 or the whole array preferably.
It really depends on what you consider "the same". If you want them to be in the same order, it can be a little simpler.
for(int i =0; i < size; i++){
if (arrayOne[i] != arrayTwo[i])
return false;
}
}
If the order doesn't matter, it can get a little more complicated.
for(int i = 0; i < size; i++){
int valueFound = 0;
for(int k = 0; k < size; k++){
if (arrayOne[i] == arrayTwo[k]){
valueFound = 1;
}
}
if (valueFound == 0)
return false;
}
return true;
But this assumes that the arrays are the same size. And it really only checks to see that all the values from arrayOne are inside arrayTwo so it is not exactly what you're looking for. I do hope that this provides some direction to you.
This is incorrect algorithm.
You can write your own function, it will be like this.
Note that a[] in function arguments will be replaced by compiler with *a
template<typename T>
bool arraycmp(T *a, size_t a_size, T *b, size_t b_size)
{
if(a_size != b_size) // If sizes is not same, we know that arrays not equal
return false;
for(size_t i = 0; i<a_size; ++i)
{
if(a[i]!=b[i]) // Just do check, and if array element mismatch return false
return false;
}
return true;
}
or like this, with one size_t
template<typename T>
bool arraycmp(T *a, T *b, size_t size)
{
for(size_t i = 0; i<size; ++i)
{
if(a[i]!=b[i]) // Just do check, and if array element mismatch return false
return false;
}
return true;
}
Or just use library function memcmp
I cant find out whats wrong with this part of my program, i want to find out most occuring number in my structure(array), but it finds only the last number :/
void Daugiausiai(int n)
{
int max = 0;
int sk;
for(int i = 0; i < n; i++){
int kiek = 0;
for(int j=0; j < n; j++){
if(A[i].datamet == A[j].datamet){
kiek++;
if(kiek > max){
max = kiek;
sk = A[i].datamet;
}
}
}
}
}
ps. its only a part of my code
You haven't shown us enough of your code, but it is likely that you are not looking at the real result of your function. The result, sk is local to the function and you don't return it. If you have global variable that is also named sk, it will not be touched by Daugiausiai.
In the same way, you pass the number of elements in your struct array, but work on a global struct. It is good practice to "encapsulate" functions so that they receive the data they work on as arguments and return a result. Your function should therefore pass both array length and array and return the result.
(Such an encapsulation doesn't work in all cases, but here, it has the benefit that you can use the same function for many different arrays of the same structure tape.)
It is also enough to test whether the current number of elements is more than the maximum so far after your counting loop.
Putting all this together:
struct Data {
int datamet;
};
int Daugiausiai(const struct Data A[], int n)
{
int max = 0;
int sk;
for (int i = 0; i < n; i++){
int kiek = 0;
// Count occurrences
for(int j = 0; j < n; j++){
if(A[i].datamet == A[j].datamet) kiek++;
}
// Check for maximum
if (kiek > max) {
max = kiek;
sk = A[i].datamet;
}
}
return sk;
}
And you call it like this:
struct Data A[6] = {{1}, {2}, {1}, {4}, {1}, {2}};
int n = Daugiausiai(A, 6);
printf("%d\n", n); // 1
It would be nice if you had english variable names, so I could read them a bit better ^^. What should your paramter n do? Is that the array-length? And what should yout funtion do? It has no return value or something.
int getMostOccuring(int array[], int length)
{
int current_number;
int current_count = 0;
int most_occuring_number;
int most_occuring_count = 0;
for (int i = 0; i < length; i++)
{
current_number = array[i];
current_count = 0;
for (int j = i; j < length; j++)
{
int test_number = array[j];
if (test_number == current_number)
{
current_count ++;
if (current_count > most_occuring_count)
{
most_occuring_number = current_number;
most_occuring_count = current_count;
}
}
}
}
return most_occuring_number;
}
this should work and return the most occuring number in the given array (it has a bad runtime, but is very simple and good to understand).
I am supposed to get the output 8, 6. But I get 8,9 when this code is run. Why am I getting the out put 8,6 and how can I fix the code to make the output become 8,9.
int inputarray[]={9,8,9,9,9,9,6};
int length = 7;
int value = 9;
void arrayShift(int arr[], int length, int value)
{
for(int i = 0; i<length; i++)
{
if(arr[i] == value)
{
for (int k = i; k<length ; k++)
{
arr[k] = arr[k+1];
}
arr[length-1] = 0;
}
}
}
When shifting array, you may replace first element (containing number equal to value) with the same value from other element. In that case, you need to restart iteration on this element again, e.g.:
void arrayShift(int arr[], int length, int value)
{
for(int i = 0; i<length; i++)
{
if(arr[i] == value)
{
for (int k = i; k<length-1 ; k++)
{
arr[k] = arr[k+1];
}
arr[length-1] = 0;
i--; // <-- this
}
}
}
Your algorithm for shifting is wrong: you fail to adjust i on removal. In addition, it is rather inefficient: you can do this in a single loop with two indexes - r for reading and w for writing. When you see the value you want to keep, adjust both the reading and the writing index. Otherwise, increment only the reading index.
Once the reading index reaches the count, the writing index indicates how many items you have left. You need to return it to the caller somehow, otherwise he wouldn't know where the actual data ends. You can return the new length as the return value of your function, or take length as a pointer, and adjust it in place.
int arrayShift(int arr[], int length, int value) {
int r = 0, w = 0;
for (; r != length ; r++) {
if (arr[r] != value) {
arr[w++] = arr[r];
}
}
return w;
}
Here is how you call it:
int inputarray[]={9,8,9,9,9,9,6};
int length = 7;
int value = 9;
int newLen = arrayShift(inputarray, length, value);
for (int i = 0 ; i != newLen ; i++) {
printf("%d ", inputarray[i]);
}
printf("\n");
Demo.
I am currently reading "Programming: Principles and Practice Using C++", in Chapter 4 there is an exercise in which:
I need to make a program to calculate prime numbers between 1 and 100 using the Sieve of Eratosthenes algorithm.
This is the program I came up with:
#include <vector>
#include <iostream>
using namespace std;
//finds prime numbers using Sieve of Eratosthenes algorithm
vector<int> calc_primes(const int max);
int main()
{
const int max = 100;
vector<int> primes = calc_primes(max);
for(int i = 0; i < primes.size(); i++)
{
if(primes[i] != 0)
cout<<primes[i]<<endl;
}
return 0;
}
vector<int> calc_primes(const int max)
{
vector<int> primes;
for(int i = 2; i < max; i++)
{
primes.push_back(i);
}
for(int i = 0; i < primes.size(); i++)
{
if(!(primes[i] % 2) && primes[i] != 2)
primes[i] = 0;
else if(!(primes[i] % 3) && primes[i] != 3)
primes[i]= 0;
else if(!(primes[i] % 5) && primes[i] != 5)
primes[i]= 0;
else if(!(primes[i] % 7) && primes[i] != 7)
primes[i]= 0;
}
return primes;
}
Not the best or fastest, but I am still early in the book and don't know much about C++.
Now the problem, until max is not bigger than 500 all the values print on the console, if max > 500 not everything gets printed.
Am I doing something wrong?
P.S.: Also any constructive criticism would be greatly appreciated.
I have no idea why you're not getting all the output, as it looks like you should get everything. What output are you missing?
The sieve is implemented wrongly. Something like
vector<int> sieve;
vector<int> primes;
for (int i = 1; i < max + 1; ++i)
sieve.push_back(i); // you'll learn more efficient ways to handle this later
sieve[0]=0;
for (int i = 2; i < max + 1; ++i) { // there are lots of brace styles, this is mine
if (sieve[i-1] != 0) {
primes.push_back(sieve[i-1]);
for (int j = 2 * sieve[i-1]; j < max + 1; j += sieve[i-1]) {
sieve[j-1] = 0;
}
}
}
would implement the sieve. (Code above written off the top of my head; not guaranteed to work or even compile. I don't think it's got anything not covered by the end of chapter 4.)
Return primes as usual, and print out the entire contents.
Think of the sieve as a set.
Go through the set in order. For each value in thesive remove all numbers that are divisable by it.
#include <set>
#include <algorithm>
#include <iterator>
#include <iostream>
typedef std::set<int> Sieve;
int main()
{
static int const max = 100;
Sieve sieve;
for(int loop=2;loop < max;++loop)
{
sieve.insert(loop);
}
// A set is ordered.
// So going from beginning to end will give all the values in order.
for(Sieve::iterator loop = sieve.begin();loop != sieve.end();++loop)
{
// prime is the next item in the set
// It has not been deleted so it must be prime.
int prime = *loop;
// deleter will iterate over all the items from
// here to the end of the sieve and remove any
// that are divisable be this prime.
Sieve::iterator deleter = loop;
++deleter;
while(deleter != sieve.end())
{
if (((*deleter) % prime) == 0)
{
// If it is exactly divasable then it is not a prime
// So delete it from the sieve. Note the use of post
// increment here. This increments deleter but returns
// the old value to be used in the erase method.
sieve.erase(deleter++);
}
else
{
// Otherwise just increment the deleter.
++deleter;
}
}
}
// This copies all the values left in the sieve to the output.
// i.e. It prints all the primes.
std::copy(sieve.begin(),sieve.end(),std::ostream_iterator<int>(std::cout,"\n"));
}
From Algorithms and Data Structures:
void runEratosthenesSieve(int upperBound) {
int upperBoundSquareRoot = (int)sqrt((double)upperBound);
bool *isComposite = new bool[upperBound + 1];
memset(isComposite, 0, sizeof(bool) * (upperBound + 1));
for (int m = 2; m <= upperBoundSquareRoot; m++) {
if (!isComposite[m]) {
cout << m << " ";
for (int k = m * m; k <= upperBound; k += m)
isComposite[k] = true;
}
}
for (int m = upperBoundSquareRoot; m <= upperBound; m++)
if (!isComposite[m])
cout << m << " ";
delete [] isComposite;
}
Interestingly, nobody seems to have answered your question about the output problem. I don't see anything in the code that should effect the output depending on the value of max.
For what it's worth, on my Mac, I get all the output. It's wrong of course, since the algorithm isn't correct, but I do get all the output. You don't mention what platform you're running on, which might be useful if you continue to have output problems.
Here's a version of your code, minimally modified to follow the actual Sieve algorithm.
#include <vector>
#include <iostream>
using namespace std;
//finds prime numbers using Sieve of Eratosthenes algorithm
vector<int> calc_primes(const int max);
int main()
{
const int max = 100;
vector<int> primes = calc_primes(max);
for(int i = 0; i < primes.size(); i++)
{
if(primes[i] != 0)
cout<<primes[i]<<endl;
}
return 0;
}
vector<int> calc_primes(const int max)
{
vector<int> primes;
// fill vector with candidates
for(int i = 2; i < max; i++)
{
primes.push_back(i);
}
// for each value in the vector...
for(int i = 0; i < primes.size(); i++)
{
//get the value
int v = primes[i];
if (v!=0) {
//remove all multiples of the value
int x = i+v;
while(x < primes.size()) {
primes[x]=0;
x = x+v;
}
}
}
return primes;
}
In the code fragment below, the numbers are filtered before they are inserted into the vector. The divisors come from the vector.
I'm also passing the vector by reference. This means that the huge vector won't be copied from the function to the caller. (Large chunks of memory take long times to copy)
vector<unsigned int> primes;
void calc_primes(vector<unsigned int>& primes, const unsigned int MAX)
{
// If MAX is less than 2, return an empty vector
// because 2 is the first prime and can't be placed in the vector.
if (MAX < 2)
{
return;
}
// 2 is the initial and unusual prime, so enter it without calculations.
primes.push_back(2);
for (unsigned int number = 3; number < MAX; number += 2)
{
bool is_prime = true;
for (unsigned int index = 0; index < primes.size(); ++index)
{
if ((number % primes[k]) == 0)
{
is_prime = false;
break;
}
}
if (is_prime)
{
primes.push_back(number);
}
}
}
This not the most efficient algorithm, but it follows the Sieve algorithm.
below is my version which basically uses a bit vector of bool and then goes through the odd numbers and a fast add to find multiples to set to false. In the end a vector is constructed and returned to the client of the prime values.
std::vector<int> getSieveOfEratosthenes ( int max )
{
std::vector<bool> primes(max, true);
int sz = primes.size();
for ( int i = 3; i < sz ; i+=2 )
if ( primes[i] )
for ( int j = i * i; j < sz; j+=i)
primes[j] = false;
std::vector<int> ret;
ret.reserve(primes.size());
ret.push_back(2);
for ( int i = 3; i < sz; i+=2 )
if ( primes[i] )
ret.push_back(i);
return ret;
}
Here is a concise, well explained implementation using bool type:
#include <iostream>
#include <cmath>
void find_primes(bool[], unsigned int);
void print_primes(bool [], unsigned int);
//=========================================================================
int main()
{
const unsigned int max = 100;
bool sieve[max];
find_primes(sieve, max);
print_primes(sieve, max);
}
//=========================================================================
/*
Function: find_primes()
Use: find_primes(bool_array, size_of_array);
It marks all the prime numbers till the
number: size_of_array, in the form of the
indexes of the array with value: true.
It implemenets the Sieve of Eratosthenes,
consisted of:
a loop through the first "sqrt(size_of_array)"
numbers starting from the first prime (2).
a loop through all the indexes < size_of_array,
marking the ones satisfying the relation i^2 + n * i
as false, i.e. composite numbers, where i - known prime
number starting from 2.
*/
void find_primes(bool sieve[], unsigned int size)
{
// by definition 0 and 1 are not prime numbers
sieve[0] = false;
sieve[1] = false;
// all numbers <= max are potential candidates for primes
for (unsigned int i = 2; i <= size; ++i)
{
sieve[i] = true;
}
// loop through the first prime numbers < sqrt(max) (suggested by the algorithm)
unsigned int first_prime = 2;
for (unsigned int i = first_prime; i <= std::sqrt(double(size)); ++i)
{
// find multiples of primes till < max
if (sieve[i] = true)
{
// mark as composite: i^2 + n * i
for (unsigned int j = i * i; j <= size; j += i)
{
sieve[j] = false;
}
}
}
}
/*
Function: print_primes()
Use: print_primes(bool_array, size_of_array);
It prints all the prime numbers,
i.e. the indexes with value: true.
*/
void print_primes(bool sieve[], unsigned int size)
{
// all the indexes of the array marked as true are primes
for (unsigned int i = 0; i <= size; ++i)
{
if (sieve[i] == true)
{
std::cout << i <<" ";
}
}
}
covering the array case. A std::vector implementation will include minor changes such as reducing the functions to one parameter, through which the vector is passed by reference and the loops will use the vector size() member function instead of the reduced parameter.
Here is a more efficient version for Sieve of Eratosthenes algorithm that I implemented.
#include <iostream>
#include <cmath>
#include <set>
using namespace std;
void sieve(int n){
set<int> primes;
primes.insert(2);
for(int i=3; i<=n ; i+=2){
primes.insert(i);
}
int p=*primes.begin();
cout<<p<<"\n";
primes.erase(p);
int maxRoot = sqrt(*(primes.rbegin()));
while(primes.size()>0){
if(p>maxRoot){
while(primes.size()>0){
p=*primes.begin();
cout<<p<<"\n";
primes.erase(p);
}
break;
}
int i=p*p;
int temp = (*(primes.rbegin()));
while(i<=temp){
primes.erase(i);
i+=p;
i+=p;
}
p=*primes.begin();
cout<<p<<"\n";
primes.erase(p);
}
}
int main(){
int n;
n = 1000000;
sieve(n);
return 0;
}
Here's my implementation not sure if 100% correct though :
http://pastebin.com/M2R2J72d
#include<iostream>
#include <stdlib.h>
using namespace std;
void listPrimes(int x);
int main() {
listPrimes(5000);
}
void listPrimes(int x) {
bool *not_prime = new bool[x];
unsigned j = 0, i = 0;
for (i = 0; i <= x; i++) {
if (i < 2) {
not_prime[i] = true;
} else if (i % 2 == 0 && i != 2) {
not_prime[i] = true;
}
}
while (j <= x) {
for (i = j; i <= x; i++) {
if (!not_prime[i]) {
j = i;
break;
}
}
for (i = (j * 2); i <= x; i += j) {
not_prime[i] = true;
}
j++;
}
for ( i = 0; i <= x; i++) {
if (!not_prime[i])
cout << i << ' ';
}
return;
}
I am following the same book now. I have come up with the following implementation of the algorithm.
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
inline void keep_window_open() { char ch; cin>>ch; }
int main ()
{
int max_no = 100;
vector <int> numbers (max_no - 1);
iota(numbers.begin(), numbers.end(), 2);
for (unsigned int ind = 0; ind < numbers.size(); ++ind)
{
for (unsigned int index = ind+1; index < numbers.size(); ++index)
{
if (numbers[index] % numbers[ind] == 0)
{
numbers.erase(numbers.begin() + index);
}
}
}
cout << "The primes are\n";
for (int primes: numbers)
{
cout << primes << '\n';
}
}
Here is my version:
#include "std_lib_facilities.h"
//helper function:check an int prime, x assumed positive.
bool check_prime(int x) {
bool check_result = true;
for (int i = 2; i < x; ++i){
if (x%i == 0){
check_result = false;
break;
}
}
return check_result;
}
//helper function:return the largest prime smaller than n(>=2).
int near_prime(int n) {
for (int i = n; i > 0; --i) {
if (check_prime(i)) { return i; break; }
}
}
vector<int> sieve_primes(int max_limit) {
vector<int> num;
vector<int> primes;
int stop = near_prime(max_limit);
for (int i = 2; i < max_limit+1; ++i) { num.push_back(i); }
int step = 2;
primes.push_back(2);
//stop when finding the last prime
while (step!=stop){
for (int i = step; i < max_limit+1; i+=step) {num[i-2] = 0; }
//the multiples set to 0, the first none zero element is a prime also step
for (int j = step; j < max_limit+1; ++j) {
if (num[j-2] != 0) { step = num[j-2]; break; }
}
primes.push_back(step);
}
return primes;
}
int main() {
int max_limit = 1000000;
vector<int> primes = sieve_primes(max_limit);
for (int i = 0; i < primes.size(); ++i) {
cout << primes[i] << ',';
}
}
Here is a classic method for doing this,
int main()
{
int max = 500;
vector<int> array(max); // vector of max numbers, initialized to default value 0
for (int i = 2; i < array.size(); ++ i) // loop for rang of numbers from 2 to max
{
// initialize j as a composite number; increment in consecutive composite numbers
for (int j = i * i; j < array.size(); j +=i)
array[j] = 1; // assign j to array[index] with value 1
}
for (int i = 2; i < array.size(); ++ i) // loop for rang of numbers from 2 to max
if (array[i] == 0) // array[index] with value 0 is a prime number
cout << i << '\n'; // get array[index] with value 0
return 0;
}
I think im late to this party but im reading the same book as you, this is the solution in came up with! Feel free to make suggestions (you or any!), for what im seeing here a couple of us extracted the operation to know if a number is multiple of another to a function.
#include "../../std_lib_facilities.h"
bool numIsMultipleOf(int n, int m) {
return n%m == 0;
}
int main() {
vector<int> rawCollection = {};
vector<int> numsToCheck = {2,3,5,7};
// Prepare raw collection
for (int i=2;i<=100;++i) {
rawCollection.push_back(i);
}
// Check multiples
for (int m: numsToCheck) {
vector<int> _temp = {};
for (int n: rawCollection) {
if (!numIsMultipleOf(n,m)||n==m) _temp.push_back(n);
}
rawCollection = _temp;
}
for (int p: rawCollection) {
cout<<"N("<<p<<")"<<" is prime.\n";
}
return 0;
}
Try this code it will be useful to you by using java question bank
import java.io.*;
class Sieve
{
public static void main(String[] args) throws IOException
{
int n = 0, primeCounter = 0;
double sqrt = 0;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println(“Enter the n value : ”);
n = Integer.parseInt(br.readLine());
sqrt = Math.sqrt(n);
boolean[] prime = new boolean[n];
System.out.println(“\n\nThe primes upto ” + n + ” are : ”);
for (int i = 2; i<n; i++)
{
prime[i] = true;
}
for (int i = 2; i <= sqrt; i++)
{
for (int j = i * 2; j<n; j += i)
{
prime[j] = false;
}
}
for (int i = 0; i<prime.length; i++)
{
if (prime[i])
{
primeCounter++;
System.out.print(i + ” “);
}
}
prime = new boolean[0];
}
}