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can someone look at this code and tell me if I am creating the pointer and object correctly please.
int main()
{
Square<int>* originalSquare = new Square<int>(3, 3);
for(int r = 0; r < originalSquare -> rowSize; r++)
{
for(int c = 0; c < originalSquare -> colSize; c++)
{
int num= 0;
originalSquare -> setElement(r, c, num);
}
}
return 0;
}
//quick_sort function
void quick_sort(Square<int>* square)
{
//nothing yet.
}
I keep getting a access violation error for somer reason... Program works fine before I changed this from stack to heap...
Any help will be greatful.
Thanks
This is not the code that shows your problem. Although, I would guess Square allocates a dynamically sized array, and setElement sets it? Can we see your constructor, and the code for setElement?
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Im trying to opptimise this peice of code as its a small section of a longer code for speed rather than memory. How best would I do that. I was thinking to use set the v_vtx vector to be able to just to define the chitemp array.
double chitemp[nvert1][2];
for (int i=0;i<nvert1;i++){
chitemp[i][1]=v_vtx[i];
chitemp[i][0]=chi2->at(v_vtx[i]);
}
for (int k = 0; k < nvert1; k++){
for( int p = k+1; p < nvert1; p++){
if( chitemp[k][0] > chitemp[p][0]){
swap(chitemp[k][0], chitemp[p][0]);
swap(chitemp[k][1], chitemp[p][1]);
}
}
}
edit:
Im trying to sort chi2 (double) into order and know which v_vtx (int) links to the chi2 value
You could instead store your values as pairs (using std::array is optional, but offers a richer interface than an inbuilt array):
std::array<std::pair<double>, nvert1> chitemp;
for (size_t i = 0; i < nvert1; ++i) {
chitemp[i].second = v_vtx[i];
chitemp[i].first = chi2->at(v_vtx[i]);
}
Then, use...
std::sort(chitemp.begin(), chitemp.end());
...instead of your (inefficient) home-grown bubble-sort.
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I want to create some variables in a loop,e.g.
for(int i = 0; i < _vector.size(); i++) //_vector is a vector struct
{
auto v = _vector.at(i);
auto xi = get_name(v); //how to create x0,x1,x2,x3.....dynamically in this loop
}
anyone knows how to do it like that?
thanks very much!
If you want to reference these variables as x1, x2, etc., it would be better to create a vector to store these.
The code below is written to support integers, however, this can be replaced with another data type.
vector<int> x;
for(int i = 0; i < _vector.size(); i++) //_vector is a vector struct
{
auto v = _vector.at(i);
x.push_back(get_name(v)); //sets the value of x.at(0), x.at(1)...
}
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I have written this recursion for the problem https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee
I want to know how can we memoize this solution (using a dp array).
or do we have to write recursion in a specific way to memoize it?
class Solution {
public:
int solve(vector<int>& prices,int fee,int i,int profit,int buy,int sell)
{
if(i==prices.size())
{
if(buy==sell)
return profit;
else
return 0;
}
int ans = 0;
ans = max(ans,solve(prices,fee,i+1,profit,buy,sell));
if(buy>sell)
{
ans = max(ans,solve(prices,fee,i+1,profit+prices[i]-fee,buy,sell+1));
}
else
{
ans = max(ans,solve(prices,fee,i+1,profit-prices[i],buy+1,sell));
}
return ans;
}
int maxProfit(vector<int>& prices, int fee) {
vector<int> diff;
int sum = 0;
sum = solve(prices,fee,0,0,0,0);
return sum;
}
};
You can just create an array where element i is equal to solve(i). Then, inside your function, you can pass this array through by reference to each call. You add an if/else structure in your function testing if the input you got was defined in the array, if so return arr[input] and if not, run through your normal function except just before you return, you initialize arr[input] to the value you will return.
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1st code: works fine gives success with time of 0sec
int main()
{
int n=100000;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{}
cout<<"ffdfdf";
}
2nd code: gives a time limit exceeded
int main()
{
int n=100000;
bool **a=new bool*[n];
for(int i=0;i<n;i++)
{
bool[i]=new bool[n];
for(int j=0;j<n;j++)
{
bool[i][j]=1;
}
}
cout<<"ffdfdf";
}
can anyone explain why the two code fragments have a vast time difference.I am not understanding it.
bool[i]=new bool[n]; is extremely expensive cf. the other statements.
A good compiler will optimise out your first program to cout << "ffdfdf";, since it will know that the loop doesn't do anything.
Once you've replaced your errant bools with as so the second snippet actually compiles, you'd be advised to pair your new[] calls with delete[].
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Can someone help me with my code ? I'm trying to use different algorithm but it returns way to big numbers, when i use algorithm between /* */ it works perfect, anyone can see whats wrong with my new code ? (same on java works)
int* czynnikiPierwsze(int n)throw (string){
if(n<0){
string wyjatek1="Nie mozna rozlozyc ujemnej liczby";
throw wyjatek1;
}
int b=0;
while(n>2){
n=n/tab[n-2];
b++;
}
dzielniki=new int[b]();
int j=0;
while(n>2){
dzielniki[j]=tab[n-2];
n=n/tab[n-2];
j++;
}
/* int a=n;
int*dzielniki=new int[30]();
for(int j=0;j<n+a;j++){
while(n>2){
dzielniki[j]=tab[n-2];
n=n/tab[n-2];
break;
}
}*/
return dzielniki;
}
There's no chance your second while(n>2) loop runs even once, since the first loop only exited when that same condition was no longer true.