_num.sort();
List<int> _answer = [];
for (int i = 1; i <= _num.last; i++) {
// help me pls
if (_num % i == 0) {
}
i'm trying to do Greatest common divisor code but i don't know how to check if int i can divide all items in the list at once and remainder == 0
If there is a better way to do this please tell me, thank you
Related
I was solving a question prime path
I am using bfs to solve this question
Here is my solution https://ideone.com/GMOyWX
when i am using this function to check for prime ,I am getting correct answer as 6
bool isprime(int number) {
for (int i = 2; i < sqrt(number); i++) {
if (number % i == 0 && i != number) return false;
}
return true;
}
But when I am using sieves I am getting wrong answer as 5
void sieves(int n) {
isPrime[0] = false;
isPrime[1] = false;
for (int i = 2; i <= n; i++) {
if (isPrime[i]) {
for (int j = 2*i; j <= n; j+=i) {
isPrime[j] = false;
}
}
}
}
can anyone tell me whats wrong with the sieves?
I think you sieve implementation is correct, so you are doing something wrong in any other place. from your code i see that you are calling your sieve function for every test case. This should not be necessary. In problem statement there is clearly stated that input will be 4 digit two number, so in maximum case, 100000 should be enough limit to generate checking vector if every number is prime or not.
So just call once your sieve with 100000 once. Prime numbers are fixed number.
So I have this code that I have written that correctly finds the optimal value for the knapsack problem.
int mat[2][size + 1];
memset(mat, 0, sizeof(mat));
int i = 0;
while(i < nItems)
{
int j = 0;
if(i % 2 != 0)
{
while(++j <= size)
{
if(weights[i] <= j) mat[1][j] = max(values[i] + mat[0][j - weights[i]], mat[0][j]);
else mat[1][j] = mat[0][j];
}
}
else
{
while(++j <= size)
{
if(weights[i] <= j) mat[0][j] = max(values[i] + mat[1][j - weights[i]], mat[1][j]);
else mat[0][j] = mat[1][j];
}
}
i++;
}
int val = (nItems % 2 != 0)? mat[0][size] : mat[1][size];
cout << val << endl;
return 0;
This part I udnerstand. However I am trying to keep the same memory space, i.e. O(W), but also now compute the optimal solution using backtracking. This is where I am finding trouble. The hints I have been given is this
Now suppose that we also want the optimal set of items. Recall that the goal
in finding the optimal solution in part 1 is to find the optimal path from
entry K(0,0) to entry K(W,n). The optimal path must pass through an
intermediate node (k,n/2) for some k; this k corresponds to the remaining
capacity in the knapsack of the optimal solution after items n/2 + 1,...n
have been considered
The question asked is this.
Implement a modified version of the algorithm from part 2 that returns not
only the optimal value, but also the remaining capacity of the optimal
solution after the last half of items have been considered
Any help would be apprecaited to get me started. Thanks
Working on a USACO programming problem, I got stuck when using a brute-force approach.
From a list of N elements, I need to compute all distinct pair-configurations.
My problem is twofold.
How do I express such a configuration in, lets say, an array?
How do I go about computing all distinct combinations?
I only resorted to the brute-force approach after I gave up solving it analytically. Although this is context-specific, I came as far as noting that one can quickly rule out the rows where there is only a single, so called, "wormhole" --- it isn't effectively in an infinite cycle.
Update
I'll express them with a tree structure. Set N = 6; {A,B,C,D,E,F}.
By constructing the following trees chronologically, all combinations are listed.
A --> B,C,D,E,F;
B --> C,D,E,F;
C --> D,E,F;
D --> E,F;
E --> F.
Check: in total there are 6 over 2 = 6!/(2!*4!) = 15 combinations.
Note. Once a lower node is selected, it should be discarded as a top node; it can only exist in one single pair.
Next, selecting them and looping over all configurations.
Here is a sample code (in C/C++):
int c[N];
void LoopOverAll(int n)
{
if (n == N)
{
// output, the array c now contains a configuration
// do anything you want here
return;
}
if (c[n] != - 1)
{
// this warmhole is already paired with someone
LoopOverAll(n + 1);
return;
}
for (int i = n + 1; i < N; i ++)
{
if (c[i] != - 1)
{
// this warmhole is already paired with someone
continue;
}
c[i] = n; c[n] = i; LoopOverAll(n + 1);
c[i] = - 1;
}
c[n] = - 1;
}
int main()
{
for (int i = 0; i < N; i ++)
c[i] = - 1;
LoopOverAll(0);
return 1;
}
The program runs but it also spews out some other stuff and I am not too sure why. The very first output is correct but from there I am not sure what happens. Here is my code:
#include <iostream>
using namespace std;
const int MAX = 10;
int sum(int arrayNum[], int n)
{
int total = 0;
if (n <= 0)
return 0;
else
for(int i = 0; i < MAX; i ++)
{
if(arrayNum[i] % 2 != 0)
total += arrayNum[i];
}
cout << "Sum of odd integers in the array: " << total << endl;
return arrayNum[0] + sum(arrayNum+1,n-1);
}
int main()
{
int x[MAX] = {13,14,8,7,45,89,22,18,6,10};
sum(x,MAX);
system("pause");
return 0;
}
The term recursion means (in the simplest variation) solving a problem by reducing it to a simpler version of the same problem until becomes trivial. In your example...
To compute the num of the odd values in an array of n elements we have these cases:
the array is empty: the result is trivially 0
the first element is even: the result will be the sum of odd elements of the rest of the array
the first element is odd: the result will be this element added to the sum of odd elements of the rest of the array
In this problem the trivial case is computing the result for an empty array and the simpler version of the problem is working on a smaller array. It is important to understand that the simpler version must be "closer" to a trivial case for recursion to work.
Once the algorithm is clear translation to code is simple:
// Returns the sums of all odd numbers in
// the sequence of n elements pointed by p
int oddSum(int *p, int n) {
if (n == 0) {
// case 1
return 0;
} else if (p[0] % 2 == 0) {
// case 2
return oddSum(p + 1, n - 1);
} else {
// case 3
return p[0] + oddSum(p + 1, n - 1);
}
}
Recursion is a powerful tool to know and you should try to understand this example until it's 100% clear how it works. Try starting rewriting it from scratch (I'm not saying you should memorize it, just try rewriting it once you read and you think you understood the solution) and then try to solve small variations of this problem.
No amount of reading can compensate for writing code.
You are passing updated n to recursive function as argument but not using it inside.
change MAX to n in this statement
for(int i = 0; i < n; i ++)
so this doesnt really answer your question but it should help.
So, your code is not really recursive. If we run through your function
int total = 0; //Start a tally, good.
if (n <= 0)
return 0; //Check that we are not violating the array, good.
else
for(int i = 0; i < MAX; i ++)
{
if(arrayNum[i] % 2 != 0) //THIS PART IS WIERD
total += arrayNum[i];
}
And the reason it is wierd is because you are solving the problem right there. That for loop will run through the list and add all the odd numbers up anyway.
What you are doing by recursing could be to do this:
What is the sum of odd numbers in:
13,14,8,7,45,89,22,18,6,10
+
14,8,7,45,89,22,18,6
+
8,7,45,89,22,18
+
7,45,89,22 ... etc
And if so then you only need to change:
for(int i = 0; i < MAX; i ++)
to
for(int i = 0; i < n; i ++)
But otherwise you really need to rethink your approach to this problem.
It's not recursion if you use a loop.
It's also generally a good idea to separate computation and output.
int sum(int arrayNum[], int n)
{
if (n <= 0) // Base case: the sum of an empty array is 0.
return 0;
// Recursive case: If the first number is odd, add it to the sum of the rest of the array.
// Otherwise just return the sum of the rest of the array.
if(arrayNum[0] % 2 != 0)
return arrayNum[0] + sum(arrayNum + 1, n - 1);
else
return sum(arrayNum + 1, n - 1);
}
int main()
{
int x[MAX] = {13,14,8,7,45,89,22,18,6,10};
cout << sum(x,MAX);
}
I am attempting to make a maze-solver using a Breadth-first search, and mark the shortest path using a character '*'
The maze is actually just a bunch of text. The maze consists of an n x n grid, consisting of "#" symbols that are walls, and periods "." representing the walkable area/paths. An 'S' denotes start, 'F' is finish. Right now, this function does not seem to be finding the solution (it thinks it has the solution even when one is impossible). I am checking the four neighbors, and if they are 'unfound' (-1) they are added to the queue to be processed.
The maze works on several mazes, but not on this one:
...###.#....
##.#...####.
...#.#.#....
#.####.####.
#F..#..#.##.
###.#....#S.
#.#.####.##.
....#.#...#.
.####.#.#.#.
........#...
What could be missing in my logic?
int mazeSolver(char *maze, int rows, int cols)
{
int start = 0;
int finish = 0;
for (int i=0;i<rows*cols;i++) {
if (maze[i] == 'S') { start=i; }
if (maze[i] == 'F') { finish=i; }
}
if (finish==0 || start==0) { return -1; }
char* bfsq;
bfsq = new char[rows*cols]; //initialize queue array
int head = 0;
int tail = 0;
bool solved = false;
char* prd;
prd = new char[rows*cols]; //initialize predecessor array
for (int i=0;i<rows*cols;i++) {
prd[i] = -1;
}
prd[start] = -2; //set the start location
bfsq[tail] = start;
tail++;
int delta[] = {-cols,-1,cols,+1}; // North, West, South, East neighbors
while(tail>head) {
int front = bfsq[head];
head++;
for (int i=0; i<4; i++) {
int neighbor = front+delta[i];
if (neighbor/cols < 0 || neighbor/cols >= rows || neighbor%cols < 0 || neighbor%cols >= cols) {
continue;
}
if (prd[neighbor] == -1 && maze[neighbor]!='#') {
prd[neighbor] = front;
bfsq[tail] = neighbor;
tail++;
if (maze[neighbor] == 'F') { solved = true; }
}
}
}
if (solved == true) {
int previous = finish;
while (previous != start) {
maze[previous] = '*';
previous = prd[previous];
}
maze[finish] = 'F';
return 1;
}
else { return 0; }
delete [] prd;
delete [] bfsq;
}
Iterating through neighbours can be significantly simplified(I know this is somewhat similar to what kobra suggests but it can be improved further). I use a moves array defining the x and y delta of the given move like so:
int moves[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
Please note that not only tis lists all the possible moves from a given cell but it also lists them in clockwise direction which is useful for some problems.
Now to traverse the array I use a std::queue<pair<int,int> > This way the current position is defined by the pair of coordinates corresponding to it. Here is how I cycle through the neighbours of a gien cell c:
pair<int,int> c;
for (int l = 0;l < 4/*size of moves*/;++l){
int ti = c.first + moves[l][0];
int tj = c.second + moves[l][1];
if (ti < 0 || ti >= n || tj < 0 || tj >= m) {
// This move goes out of the field
continue;
}
// Do something.
}
I know this code is not really related to your code, but as I am teaching this kind of problems trust me a lot of students were really thankful when I showed them this approach.
Now back to your question - you need to start from the end position and use prd array to find its parent, then find its parent's parent and so on until you reach a cell with negative parent. What you do instead considers all the visited cells and some of them are not on the shortest path from S to F.
You can break once you set solved = true this will optimize the algorithm a bit.
I personally think you always find a solution because you have no checks for falling off the field. (the if (ti < 0 || ti >= n || tj < 0 || tj >= m) bit in my code).
Hope this helps you and gives you some tips how to improve your coding.
A few comments:
You can use queue container in c++, its much more easier in use
In this task you can write something like that:
int delta[] = {-1, cols, 1 -cols};
And then you simple can iterate through all four sides, you shouldn't copy-paste the same code.
You will have problems with boundaries of your array. Because you are not checking it.
When you have founded finish you should break from cycle
And in last cycle you have an error. It will print * in all cells in which you have been (not only in the optimal way). It should look:
while (finish != start)
{
maze[finish] = '*';
finish = prd[finish];
}
maze[start] = '*';
And of course this cycle should in the last if, because you don't know at that moment have you reach end or not
PS And its better to clear memory which you have allocate in function