//amt = amount of cents to get change for
//onhand = array of available coins . Ex: {3, 0, 1, 0} = 3 quart, 0 dime, 1 nickel, 0 pennies
//denoms = {25, 10, 5, 1}
//ndenoms = 4 ; i.e the number of different denominations
// thechange = array of change. Ex: if amt = 80, then one possible thechange = {3, 0, 1, 0} b/c 3*25 + 1*5 = 80 cents
int i = 0;
void makechange(int amt, int *onhand, int *denoms, int ndenoms, int *thechange)
{
if ( (denoms[i] * onhand[i]) > amt)
{
onhand[i]--; // # of coins is too much, decrement and try again
makechange(amt, onhand, denoms, ndenoms, thechange); // try agan
}
thechange[i] = onhand[i]; //found #of coins
amt = amt - denoms[i]*onhand[i]; // get remaining amount from change
i++;
if (amt != 0) // we're not done with change so move on to next denomination
{
makechange(amt, onhand, denoms, ndenoms, thechange);
}
else if (amt == 0) // we're done with the change so all the other # coins = 0
{
for (int j = i; j < amt; j++)
{
thechange[j] = 0;
}
}
}
Now, down in main when I actually call the function prototype and print out the result
//
makechange(amt, onhand, denoms, ndenoms, thechange);
for (int k = 0; k < ndenoms; k++)
{
cout << thechange[i] << " ";
}
//
I get an error.
This algorithm seems seems sensible to me, does anyone know why it keeps crashing, though?
Have I properly used recursion here?
If you call makechange twice, the second time it won't work because the global variable i will be wrong.
Also what should happen if you try to makechange and don't have enough change on hand to make it?
Similarly what happens if you have 3 quarters and 3 dimes, and are asked to make 80 cents in change? Your algorithm will use all 3 quarters and then get stuck.
did you mean
for (int k = 0; k < ndenoms; k++) { cout << thechange[k] << " "; }
a little typo made possible by the use of global variable i.
also
for (int j = i; j < amt; j++) { thechange[j] = 0; }
I think you meant
for (int j = i; j < ndenoms; j++)
depending on the final value of amt, this will cause you to run off the end of the array, resulting in a crash.
you can solve this more easily without recursion. are you required to use recursion for an assignment? if not, try this:
int makechange(int amt, int *onhand, int *denoms, int ndenoms, int *thechange)
{
for (int i=0; i < ndenoms && amt > 0; i++)
{
while (onhand[i] > 0 && denoms[i] <= amt)
{
onhand[i]--; // use one coin
thechange[i]++;
amt -= denoms[i];
}
}
return amt; // this is the amount you owe if you dont have enough on hand
}
I made the changes as mentioned by shsmith and here is the modified and complete c++ program.
#include <iostream>
using namespace std;
int i = 0;
void makechange(int amt, int *onhand, int *denoms, int ndenoms, int *thechange)
{
if ( (denoms[i] * onhand[i]) > amt)
{
onhand[i]--; // # of coins is too much, decrement and try again
makechange(amt, onhand, denoms, ndenoms, thechange); // try agan
}
thechange[i] = onhand[i]; //found #of coins
amt = amt - denoms[i]*onhand[i]; // get remaining amount from change
i++;
if (amt != 0) // we're not done with change so move on to next denomination
{
makechange(amt, onhand, denoms, ndenoms, thechange);
}
else if (amt == 0) // we're done with the change so all the other # coins = 0
{
for (int j = i; j < ndenoms; j++)
{
thechange[j] = 0;
}
}}
int main(){
//Now, down in main when I actually call the function prototype and print out the result
int amt = 80, onhand[] = {3, 0, 1, 0}, denoms[] = {25, 10, 5, 1}, ndenoms = 4, thechange[4];
makechange(amt, onhand, denoms, ndenoms, thechange);
for (int k = 0; k < ndenoms; k++)
{
cout << thechange[k] << " ";
}
cout << "\n";
return 0;}
This code is running perfectly on my machine. I compiled it using Cygwin.
Note: This algorithm would work only if you have the denomination coins more or correctly onhand. If there are insufficient number of coins onhand, then there is no exit for the recursive method because 'amt' would never become zero. At the same time, you never check for the 'i' value whether it is in bounds of the 'ndenoms'. This could also result in out of boundary errors which could cause ur program to exit incorrectly.
Related
In my situation, a lorry has a capacity of 30, while a van has a capacity of 10. I need to find the number of vans/lorries needed to transport a given amount of cargo, say 100. I need to find all possible combinations of lorries + vans that will add up to 100.
The basic math calculation would be: (30*lorrycount) + (10*vancount) = n, where n is number of cargo.
Output Example
Cargo to be transported: 100
Number of Lorry: 0 3 2 1
Number of Van: 10 1 4 7
For example, the 2nd combination is 3 lorries, 1 van. Considering that lorries have capacity = 30 and van capacity = 10, (30*3)+(10*1) = 100 = n.
For now, we only have this code, which finds literally all combinations of numbers that add up to given number n, without considering the formula given above.
#include <iostream>
#include <vector>
using namespace std;
void findCombinationsUtil(int arr[], int index,
int num, int reducedNum)
{
int lorry_capacity = 30;
int van_capacity = 10;
// Base condition
if (reducedNum < 0)
return;
// If combination is found, print it
if (reducedNum == 0)
{
for (int i = 0; i < index; i++)
cout << arr[i] << " ";
cout << endl;
return;
}
// Find the previous number stored in arr[]
// It helps in maintaining increasing order
int prev = (index == 0) ? 1 : arr[index - 1];
// note loop starts from previous number
// i.e. at array location index - 1
for (int k = prev; k <= num; k++)
{
// next element of array is k
arr[index] = k;
// call recursively with reduced number
findCombinationsUtil(arr, index + 1, num,
reducedNum - k);
}
}
void findCombinations(int n)
{
// array to store the combinations
// It can contain max n elements
std::vector<int> arr(n); // allocate n elements
//find all combinations
findCombinationsUtil(&*arr.begin(), 0, n, n);
}
int main()
{
int n;
cout << "Enter the amount of cargo you want to transport: ";
cin >> n;
cout << endl;
//const int n = 10;
findCombinations(n);
return 0;
}
Do let me know if you have any solution to this, thank you.
An iterative way of finding all possible combinations
#include <iostream>
#include <vector>
int main()
{
int cw = 100;
int lw = 30, vw = 10;
int maxl = cw/lw; // maximum no. of lorries that can be there
std::vector<std::pair<int,int>> solutions;
// for the inclusive range of 0 to maxl, find the corresponding no. of vans for each variant of no of lorries
for(int l = 0; l<= maxl; ++l){
bool is_integer = (cw - l*lw)%vw == 0; // only if this is true, then there is an integer which satisfies for given l
if(is_integer){
int v = (cw-l*lw)/vw; // no of vans
solutions.push_back(std::make_pair(l,v));
}
}
for( auto& solution : solutions){
std::cout<<solution.first<<" lorries and "<< solution.second<<" vans" <<std::endl;
}
return 0;
}
We will create a recursive function that walks a global capacities array left to right and tries to load cargo into the various vehicle types. We keep track of how much we still have to load and pass that on to any recursive call. If we reach the end of the array, we produce a solution only if the remaining cargo is zero.
std::vector<int> capacities = { 30, 10 };
using Solution = std::vector<int>;
using Solutions = std::vector<Solution>;
void tryLoad(int remaining_cargo, int vehicle_index, Solution so_far, std::back_insert_iterator<Solutions>& solutions) {
if (vehicle_index == capacities.size()) {
if (remaining_cargo == 0) // we have a solution
*solutions++ = so_far;
return;
}
int capacity = capacities[vehicle_index];
for (int vehicles = 0; vehicles <= remaining_cargo / capacity; vehicles++) {
Solution new_solution = so_far;
new_solution.push_back(vehicles);
tryLoad(remaining_cargo - vehicles * capacity, vehicle_index + 1, new_solution, solutions);
}
}
Calling this as follows should produce the desired output in all_solutions:
Solutions all_solutions;
auto inserter = std::back_inserter(all_solutions)
tryLoad(100, 0, Solution{}, inserter);
I'm trying to find all the prime numbers between two integers and place them in an integer array.
The catch is that i have to use a specific method of doing so (divide each subsequent integer by all the primes in my array). So I can't use the sieve of Eratosthanes or any other 'easier' methods.
My code successfully prompts the user for two integers, but for now I do not use either of them. First I want to make sure the program works for values between 0 and whatever, in this case 200 just to test it.
Problem is, when I run the program and print the first 20 or so values in the array, I'm getting
2, 3, 5, 7, 11, 200, 0, 0, 0, 0, 0, 0 ...... more zeroes.
The first 5 values are correct because they start in the array, but after that the whole thing goes haywire.
I've worked through my nested loop by hand for a couple values and it SEEMS like it should work. I feel like there's a specific array property that I'm overlooking.
Here's my code:
#include "stdafx.h"
#include "iostream"
#include "climits"
#include "cmath"
#include "array"
using namespace std;
int main()
{
// declare variables to store user input
int lowerBound, upperBound;
// prompt user for lesser and greater integers and store them
cout << "Program to find all primes between two integers." << endl;
cout << "Enter lesser integer: " << endl;
cin >> lowerBound;
cout << "Enter greater integer: " << endl;
cin >> upperBound;
// if statement to switch the input variables if the user accidentally enters them backwards
if (lowerBound > upperBound) {
int temp = lowerBound;
lowerBound = upperBound;
upperBound = temp;
}
// initialize int array with the first 5 primes
int primes[100] = { 2, 3, 5, 7, 11 };
// loop to find primes between 12 and 200 (since we already have primes from 1-11 in the array)
for (int i = 12; i <= 200; i++) {
// the maximum divisor needed to determine if the current integer being tested is prime
double maxDivisor = sqrt(i);
// variable for the current size of the array
int size = 5;
// boolean variable is set to true by default
bool isPrime = true;
for (int j = 0; j < size; j++) { // changed "j<=size" to "j<size"
int remainder = (i % primes[j]);
// once the maximum divisor is reached, there is no need to continue testing for the current integer
if (primes[j] > maxDivisor) {
break;
}
// if the remainder of divison by a prime is 0, the number is not prime, so set the boolean variable to false
if (remainder = 0) {
isPrime = false;
}
}
// if isPrime is still true after the nested loop, the integer value being tested will be placed in the next element of the array
if (isPrime == true) {
primes[size] = i;
// since we added to the array, increment size by 1
size++;
}
}
// display the first 20 values in the array for debugging
for (int k = 0; k < 20; k++) {
cout << primes[k] << ", ";
}
system("pause");
return 0;
}
This here
if (remainder = 0) {
isPrime = false;
}
Needs to be changed to
if (remainder == 0) {
isPrime = false;
}
Because = does assignment, not comparison. So what remainder = 0 does it setting remainder to 0, and then it returns that 0, which gets casted to false, which is on of the reasons why it's not finding any primes.
Also, as Fantastic Mr Fox pointed out, for (int j = 0; j <= size; j++) needs to be changed to for (int j = 0; j < size; j++).
Also, did your compiler issue any warnings? If not, try to see if you can set it to be more strict with warnings. I figure most modern compilers will give you a hint at if (remainder = 0). Getting useful warnings from the compiler helps a lot with preventing bugs.
Edit:
As Karsten Koop pointed out, you need to move the int size = 5; out of the loop, to before the for (int i = 12;. With those changes, it's now working on my machine.
Last but not least, a tip: instead of if (isPrime == true), you can just write if (isPrime).
I am using an arduino to read a sensor which stores 256 values into an array. I am trying to find local max's but some values being stored have repeating values to the left and right of itself causing the value to print multiple times. Is there a way to take all true values meaning they are a max value and store them in another array to process and reduce the repeated values to just 1 value...
OR is there a way to send the max values to another array where the repeated values get reduced to just 1? OR
IE:
Array1[] = {1,2,3,4,4,4,3,2,7,8,9,10}
max = 4 at index 3
max = 4 at index 4
max = 4 at index 5
since 4 is a peak point but repeats how can I reduce it so that the array looks like
Array2[] = {1,2,3,4,3,2,7,8,9,10}
max = 4 at index 3
I need the most basic breakdown if possible nothing on an expert level, thanks.
Code from Arduino:
int inp[20] = {24,100,13,155,154,157,156,140,14,175,158,102,169,160,190,100,200,164,143,20};
void setup()
{
Serial.begin(9600); // for debugging
}
void loop()
{
int i;
int count = 0;
for (i = 0; i < 20; i++)
{
Serial.println((String)inp[i]+" index at - "+i);
delay(100);
};
int N = 5; // loc max neighborhood size
for (int i = N-1; i < 19-N; i++)
{
bool loc = false;
for (int j = 1; j < N; j++) // look N-1 back and N-1 ahead
{
if (inp[i] > inp[i-j] && inp[i] > inp[i+j]) loc = true;
}
if (loc == true)
{
Serial.println((String)"max = "inp[i]+" at index "+i);
}
}
Serial.println("----------------------------------");
}
You can detect "local maxima" or peaks in a single loop without the need of copying something into another array. You just have to ignore repeating values, and you just have to keep track if the values considered are currently increasing or decreasing. Each value after which this status switches from increasing to decreasing is then a peak:
int main() {
int Array1[] = {1,2,3,4,4,4,3,2,7,8,9,10};
int prevVal = INT_MIN;
enum {
Ascending,
Descending
} direction = Ascending;
for (int i=0; i<sizeof(Array1)/sizeof(*Array1); i++) {
int curVal = Array1[i];
if (prevVal < curVal) { // (still) ascending?
direction = Ascending;
}
else if (prevVal > curVal) { // (still) descending?
if (direction != Descending) { // starts descending?
cout << "peak at index " << i-1 << ": " << prevVal << endl;
direction = Descending;
}
}
// prevVal == curVal is simply ignored...
prevVal = curVal;
}
}
I am new to functions and i am really trying to understand how they work, my teacher gave us a problem where by we were to pass a number to a function between the range of 1-12 and the function was then meant to do the times tales of that number so I asked the user to enter a number and if the number is less then 1 and greater then 12 exit, else pass the number to the function and then I used a for loop to do the multiplication for me (as far as I am aware) but nothing seems to happen? Νo doubt I am doing something really stupid, any help is much appreciated.
#include <iostream>
using namespace std;
int TimesTables (int num);
int main(int argc, const char * argv[]) {
int number;
cout << "enter a number to multiply by, with a range of 1-12: ";
cin >> number;
if (number < 1 && number > 12)
return EXIT_FAILURE;
else {
int tables = TimesTables(number);
cout << tables;
}
return 0;
}
int TimesTables (int num) {
for ( int i = 0; num <=12; i ++)
num = num * i;
return num;
}
Running i from 0 is going to set num to 0, and therefore any multiplication after that.
Your loop is also rather dubious. Why are you checking num <= 12 rather than i <= 12?
Shouldn't your loop take the form
for ( int i = 1; i <=12; i ++){
// Print num * i
cout << num * i;
}
// There's no need to return anything back to the caller
for ( int i = 0; num <=12; i ++)
num = num * i;
Here i starts from 0, so any multiplication you do afterwards doesn't affect the result (num). Moreover, you want to go from 1 to 12, so you should start from 0 and finish at 12 - 1, or start from 1 and finish at 12.
So change this:
for ( int i = 0; num <=12; i ++)
to this:
for ( int i = 1; i <=12; i ++)
since you want to stop when i reaches 12, not num, i is the counter of the for-loop!
I want to start by saying I am new to programming. I have a problem with writing a list of distinct numbers from another list in c++. Let's say I have a list l1 = {1, 12, 2, 4, 1, 3, 2} and I want to create a new list that looks like this l2 = {1, 12, 2, 4, 3}...
This is what I wrote:
#include <iostream>
using namespace std;
int main() {
int l1[100], l2[100], length, length1 = 0, i, j, a = 0;
cin >> length; //set the length
for (i = 0; i < length; i++) {
cin >> l1[i]; //add numbers to the list
}
l2[0] = l1[0]; //added the first number manually
for (i = 0; i < length; i++) {
length1++;
a = 0;
for (j = 0; j < length1; j++) {
if (l1[i] != l2[j]) //this checks numbers in the second list
a = 1; // and if they aren't found a gets the value
} //1 so after it's done checking if a is 1 it
if (a == 1) //will add the number to the list, but if the
l2[j] = l1[i]; //number is found then a is 0 and nothing happens,
} // SUPPOSEDLY
for (j = 0; j < length1; j++) {
cout << l2[j] << " ";
}
}
The output of this is 1 -858993460 12 2 4 1 3 so obviously I did something very wrong. I'd welcome any suggestion you might have, I don't necessarily need a solution to this, I just want to get unstuck.
Thanks a lot for taking time to reply to this.
std::sort(l1, l1 + 100);
int* end_uniques = std::unique(l1, l1 + 100);
std::copy(l1, end_uniques, l2);
size_t num_uniques = end_uniques - l1;
This is O(N log N) instead of your O(N^2) solution, so theoretically faster. It requires first sorting the array l1 (in-place) to let std::unique work. Then you get a pointer to the end of the unique elements, which you can use to copy to l2 and of course get the count (because it may be less than the full size of 100 of course).
Most Important : This solution assumes that we've to preserve order
Well....
try out this one....
I've changed identifiers a bit ( of course that's not gonna affect the execution )
It'll just help us to identify what is the sake of that variable.
Here's the code
#include <iostream>
using namespace std;
int main()
{
int Input[100], Unique[100], InSize, UniLength = 0;
cin >> InSize;
for (int ii = 0 ; ii < InSize ; ii++ )
{
cin >> Input[ii];
}
Unique[0] = Input[0];
UniLength++;
bool IsUnique;
for ( int ii = 1 ; ii < InSize ; ii++ )
{
IsUnique=true;
for (int jj = 0 ; jj < UniLength ; jj++ )
{
if ( Input[ii] == Unique[jj] )
{
IsUnique=false;
break;
}
}
if ( IsUnique )
{
Unique[UniLength] = Input[ii];
UniLength++;
}
}
for ( int jj = 0 ; jj < UniLength ; jj++ )
{
cout << Unique[jj] << " ";
}
}
You were inserting Unique element at it's original index in new array..... and in place of those elements which were duplicate.... you was not doing any kind of shifting.... i.e. they were uninitialized..... and were giving something weird like -858993460
I appreciate above mentioned two answers but again..... I think this question was placed on hackerrank.... and unqiue_array() doesn't work there.....
Added
Of course we can only add Unique elements to our Input array..... but... this solution works..... Moreover we have 2 seconds of execution time limit.... and just 100 elements..... Keeping in mind.... that Big Oh Notation works good for really large Inputs .... Which is not case here....So there's really no point looking at time complexity....... What I'll choose is the algorithm which is easy to understand.
I hope this is what you were looking for...
Have a nice day.