Input Format
The first line contains two space-separated integers denoting the respective values of (the number of variable-length arrays) and (the number of queries).
Each line of the subsequent lines contains a space-separated sequence in the format k a[i]0 a[i]1 … a[i]k-1 describing the -element array located at.
Each of the subsequent lines contains two space-separated integers describing the respective values of (an index in the array ) and (an index in the array referenced by ) for a query.
Output Format-
For each pair of and values (i.e., for each query), print a single integer denoting the element located at an index of the array referenced by. There should be a total of lines of output.
Sample Input
2 2
3 1 5 4
5 1 2 8 9 3
0 1
1 3
Sample Output
5
9
Somebody has solved this problem by -
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n,q; //n number of variable lenght arrays
// q no of queries asked
cin >>n >>q;
int ** Vectors = new int *[n];//no of length of var. arrays
int j;
for (int i=0;i<n;i++)
{
cin>>j;
Vectors[i] = new int [j];
for (int y=0;y<j;y++)
cin>>Vectors[i][y];
}
int q1,q2;
for (int i=0;i<q;i++)
{
cin >>q1 >> q2;
cout<<Vectors[q1][q2]<<endl;
}
return 0;
}
Can somebody explain me this code? Or if anyone has a better approach to solve this problem. Then please explain it in detail.
This shouldn't be hard to understand, that code is basically initializing dynamic 2D array at run time then inserting values to the 2D array and then accessing it by giving index:
int ** Vectors = new int *[n];//no of length of var. arrays
int j;
for (int i=0;i<n;i++)
{
cin>>j;
Vectors[i] = new int [j]; // initialzing inner array.. consider it as 2D array with n rows and j columns
for (int y=0;y<j;y++)
cin>>Vectors[i][y]; // insert element at specified index
}
cout<<Vectors[q1][q2]<<endl; // access element from 2D array
What you might want to use is a Matrix class.
Using
vector<vector<int>>
should do it.
Alternatively the snipet code should be refactored into a Matrix class with a constructor and a destructor.
The example you give present a memory leak since the allocated memory is not freed.
Related
I have a string say toioyhpknmtlghk.I want to create a 2-d character array having n rows such that it contains first,second,third sub strings of length n as it's rows.
For example,here n=5 the sub strings of length 5 are, toioy hpknm and tlghk
so array arr should look like
t o i o y
h p k n m
t l g h k
Now this could have been easier if I copy the characters by looping through the array but here I am trying to use memcpy as,
int main()
{
long n;
cin>>n;
char a[3][n+1];char str[20]; //I have taken n+1 columns as n for substring and 1 for '\0'
scanf("%s",str);
char *p=str;
memcpy(a[0],p,n);strcat(a[0],"\0");
p=p+n;
memcpy(a[1],p,n);strcat(a[1],"\0");
p=p+n;
memcpy(a[2],p,n);strcat(a[2],"\0");
for(int i=0;i<3;i++)
{
printf("%s\n",a[i]);
}
}
But on outputting the array following results are obtained,
//input
5
toioyhpknmtlghk
//output
toioy
hpknm{tlghk‼â
tlghk‼â
The command strcat(a[0],"\0"); is working on strings which are already terminated by \0. Otherwise it doesn't know where to append the second string. In your case a[0] is not terminated, so the function will induce undefined behavior. You can do the following instead:
a[0][n] = '\0';
(the same is for the rest of a elements)
I'm trying to teach myself programming by attempting problems from codeabbey.com.
I'm not getting the correct output on this question.
Question:
Here is an array of length M with numbers in the range 1 ... N, where N is less than or equal to 20. You are to go through it and count how many times each number is encountered.
Input data contain M and N in the first line.
The second (rather long) line will contain M numbers separated by spaces.
Answer should contain exactly N values, separated by spaces. First should give amount of 1-s, second - amount of 2-s and so on.
Data input:
10 3
1 2 3 2 3 1 1 1 1 3
Correct Output:
5 2 3
My Output:
7 3 4
You can check here
My Code:
#include <iostream>
using namespace std;
int main()
{
int arrayLength,range,a;
cin>>arrayLength>>range;
int array[20];
array[20]={0};
for(int i=0; i<arrayLength; i++)
{
cin>>a;
++array[a-1];
}
for(a=0; a<range; a++)
{
cout<<array[a]<<" ";
}
return 0;
}
There aren't any error messages or warnings. Also, if you have any suggestions for improving the code, that'd be nice.
int array[20];
array[20]={0};
is wrong, since it leave the array un-initialized and tries to initialize the 21st element (which is undefined behaviour btw, since your array has only 20 elements, remember that indexing starts from 0). Use
int array[20] = {0}; // this will initialize all elements to 0
and your code will work as expected. See here for more details regarding aggregate initialization in C++.
array[20]={0}; initializes the 21st element(non-existing) to 0.
So you have to use int array[20] = {0}; which will initialize all 20 elements to zero.
Also from your code, you are not storing the elements to an array. You are just incrementing the corresponding count when an input is read. If so, what is the need of initializing an array to max limit. Just declare the array as you need it. In your case,
int array[range] = {0};
It will initialize an array of three (range =3 here) elements.
Okay so I have;
int grid_x = 5
int * grid;
grid = new int[grid_x];
*grid = 34;
cout << grid[0];
Should line 3 create an array with 5 elements? Or fill the first element with the number 5?
Line 4 fills the first element, how do I fill the rest?
Without line 4, line 5 reads "-842150451".
I don't understand what is going on, I'm trying to create a 2 dimensional array using x and y values specified by the user, and then fill each element one by one with numeric values also specified by the user. My above code was an attempt to try it out with a 1 dimensional array first.
The default C++ way of creating a dynamic(ally resizable) array of int is:
std::vector<int> grid;
Don't play around with unsafe pointers and manual dynamic allocation when the standard library already encapsulates this for you.
To create a vector of 5 elements, do this:
std::vector<int> grid(5);
You can then access its individual elements using []:
grid[0] = 34;
grid[1] = 42;
You can add new elements to the back:
// grid.size() is 5
grid.push_back(-42);
// grid.size() now returns 6
Consult reference docs to see all operations available on std::vector.
Should line 3 create an array with 5 elements?
Yes. It won't initialise them though, which is why you see a weird value.
Or fill the first element with the number 5?
new int(grid_x), with round brackets, would create a single object, not an array, and specify the initial value.
There's no way to allocate an array with new and initialise them with a (non-zero) value. You'll have to assign the values after allocation.
Line 4 fills the first element, how do I fill the rest?
You can use the subscript operator [] to access elements:
grid[0] = 34; // Equivalent to: *(grid) = 34
grid[1] = 42; // Equivalent to: *(grid+1) = 42
// ...
grid[4] = 77; // That's the last one: 5 elements from 0 to 4.
However, you usually don't want to juggle raw pointers like this; the burden of having to delete[] the array when you've finished with it can be difficult to fulfill. Instead, use the standard library. Here's one way to make a two-dimensional grid:
#include <vector>
std::vector<std::vector<int>> grid(grid_x, std::vector<int>(grid_y));
grid[x][y] = 42; // for any x is between 0 and grid_x-1, y between 0 and grid_y-1
Or might be more efficient to use a single contiguous array; you'll need your own little functions to access that as a two-dimenionsal grid. Something like this might be a good starting point:
template <typename T>
class Grid {
public:
Grid(size_t x, size_t y) : size_x(x), size_y(y), v(x*y) {}
T & operator()(size_t x, size_t y) {return v[y*size_x + x];}
T const & operator()(size_t x, size_t y) const {return v[y*size_x + x];}
private:
size_t size_x, size_y;
std::vector<T> v;
};
Grid grid(grid_x,grid_y);
grid(x,y) = 42;
Should line 3 create an array with 5 elements? Or fill the first element with the number 5?
Create an array with 5 elements.
Line 4 fills the first element, how do I fill the rest?
grid[n] = x;
Where n is the index of the element you want to set and x is the value.
Line 3 allocates memory for 5 integers side by side in memory so that they can be accessed and modified by...
The bracket operator, x[y] is exactly equivalent to *(x+y), so you could change Line 4 to grid[0] = 34; to make it more readable (this is why grid[2] will do the same thing as 2[grid]!)
An array is simply a contiguous block of memory. Therefore it has a starting address.
int * grid;
Is the C representation of the address of an integer, you can read the * as 'pointer'. Since your array is an array of integers, the address of the first element in the array is effectively the same as the address of the array. Hence line 3
grid = new int[grid_x];
allocates enough memory (on the heap) to hold the array and places its address in the grid variable. At this point the content of that memory is whatever it was when the physical silicon was last used. Reading from uninitialised memory will result in unpredictable values, hence your observation that leaving out line 4 results in strange output.
Remember that * pointer? On line four you can read it as 'the content of the pointer' and therefore
*grid = 34;
means set the content of the memory pointed to by grid to the value 34. But line 3 gave grid the address of the first element of the array. So line 4 sets the first element of the array to be 34.
In C, arrays use a zero-based index, which means that the first element of the array is number 0 and the last is number-of-elements-in-the-array - 1. So one way of filling the array is to index each element in turn to set a value to it.
for(int index = 0; index < grid_x; index++)
{
grid[index] = 34;
}
Alternatively, you could continue to use a pointer to do the same job.
for(int* pointerToElement = grid; 0 < grid_x; grid_x-- )
{
// save 34 to the address held by the pointer
/// and post-increment the pointer to the next element.
*pointerToElement++ = 34;
}
Have fun with arrays and pointers, they consistently provide a huge range of opportunities to spend sleepless hours wondering why your code doesn't work, PC reboots, router catches fire, etc, etc.
int grid_x = 5
int * grid;
grid = new int[grid_x];
*grid = 34;
cout << grid[0];
Should line 3 create an array with 5 elements? Or fill the first
element with the number 5?
Definitely the former. With the operator "new" you are allocating memory
Line 4 fills the first element, how do I fill the rest?
Use operator [], e.g.:
for int (i=0; i < grid_x; i++) { //Reset the buffer
grid[i] = 0;
}
Without line 4, line 5 reads "-842150451".
You are just reading uninitialized memory, it could be any value.
I don't understand what is going on, I'm trying to create a 2
dimensional array using x and y values specified by the user, and then
fill each element one by one with numeric values also specified by the
user. My above code was an attempt to try it out with a 1 dimensional
array first.
Other users explained how to use vectors. If you have to set only once the size of your array, I usually prefer boost::scoped_array which takes care of deleting when the variable goes out of scope.
For a two dimensional array of size not known at compile time, you need something a little bit trickier, like a scoped_array of scoped_arrays. Creating it will require necessarily a for loop, though.
using boost::scoped_array;
int grid_x;
int grid_y;
///Reading values from user...
scoped_array<scoped_array<int> > grid(new scoped_array<int> [grid_x]);
for (int i = 0; i < grid_x; i++)
grid[i] = scoped_array<int>(new int[grid_y] );
You will be able then to access your grid elements as
grid[x][y];
Note:
It would work also taking scoped_array out of the game,
typedef int* p_int_t;
p_int_t* grid = new p_int_t [grid_x];
for (int i = 0; i < grid_x; i++)
grid[i] = new int[grid_y];
but then you would have to take care of deletion at the end of the array's life, of ALL sub arrays.
I want to add a pair to a vector of pairs.
When I print the vector elements, I don't get the integers I input.
Please help.
#include<iostream>
#include<vector>
#include<utility>
using namespace std;
int main()
{
vector<vector<unsigned long int> >matrix;
vector<pair<int,int> >pmat;
int L;
cin>>L;
int n1, n2;
pmat.resize(L);
for(int k = 0; k<L; k++)
{
cin>>n1>>n2;
pair<int,int> p = make_pair(n1,n2);
cout<<p.first<<p.second<<endl;
pmat.push_back(p);
}
for(int k = 0; k<L; k++)
{
cout<<pmat[k].first<<','<<pmat[k].second<<' ';
}
cout<<endl;
return 0;
}
Method 1:
Delete this line:
pmat.resize(L);
You don't need to resize it in the first place as you do push_back() when adding afterwards.
Method 2:
Change the following line
pmat.push_back(p);
to
pmat[k] = p;
You can do resize() in the first place, but after this, you should not use push_back() when adding, just use pmat[k] = p.
PS: you should not mix these two ways up. Always use one of them consistently.
Since you're using pmat.resize(L) and L times pmat.push_back(...), you're ending up having stored 2 * L entries in your vector. However you're printing just the first half, index 0 to L - 1. The values you want are stored from index L to 2 * L - 1.
Just change pmat.resize(L) to pmat.reserve(L).
Alternatively, you can use the resize(L), but to end up with L entries, you need to store each input pair to pmat[k], hence you write pmat[k] = p;.
As a rule of thumb, I recommend using the reserve + push_back approach if you know how many elements you're going to add. The reason is, that resize initializes the elements, while reserving just asserts that there will be enough space and no reallocation will be necessary with any following push_back.
You don't want to add more pairs after you allocated them. You can now directly access them.
Just use pmat[k] = p; instead of pmat.push_back(p);
If you print the size of the vector after reading the values, you will notice a small problem with your program:
./test
2
1 2
12
3 4
34
Size of the vector: 4
0,0 0,0
Huh? Even though I only entered 2 pairs, the size of the vector is 4. How did this happen?
If you look at the documentation for resize, it says
Resizes the container to contain count elements.
So even before you read any values, your vector will already contain 2 elements! Those will be default-constructed and therefore be 0. When you then push_pack the elements you read in, those will land at the indices 2 and 3 in the vector, so the end vector has twice as much elements as you wanted (4 in this case). You only print out the first half, which are the zero values.
The solution is to use reserve instead of resize, which doesn't create the elements but only reserves space for them, or just delete the call to resize. Using reserve is more efficient though, because then the vector will only need to allocate memory once.
pmat.resize(L);
if vector in empty its going to initialize a vector pmat with size L then assign default values to vector so now pmat size is L with default values
for(int k = 0; k<L; k++)
{
cin>>n1>>n2;
pair<int,int> p = make_pair(n1,n2);
cout<<p.first<<p.second<<endl;
pmat.push_back(p);
}
then u r pushing values to pmat L times so the final size is 2*L
for(int k = 0; k<L; k++)
{
cout<<pmat[k].first<<','<<pmat[k].second<<' ';
}
here u r going to read from 0 to L-1 , it contains default values you can see your values from L-1 to 2L-1.
so finally what i want say is use reserve instead of resize
or
pmat.resize(L); comment this line
i've been thinking for a long time and havent got anywhere with the program. i dont know where to begin. The assignment requires use of single function main and only iostream library to be used.
the task is to Declare a char array of 10 elements. Take input from user. Determine if array contains any values more than 1 times . do not show the characters that appears 1 time only.
Sample output:
a 2
b 4
..
a an b are characters. and 2 and 4 represents number of times they appear in the array B.
i tried to use nested loop to compare a character with all the character in array and incrementing a counter each time similer character id sound but unexpected results are occuring.
Here is the code
#include <iostream>
using namespace std;
void main()
{
char ara[10];
int counter=0;
cout<<"Enter 10 characters in an array\n";
for ( int a=0; a<10; a++)
cin>>ara[a];
for(int i=0; i<10; i++)
{
for(int j=i+1; j<10; j++)
{
if(ara[i] == ara[j])
{
counter++;
cout<<ara[i]<<"\t"<<counter<<endl;
}
}
}
}
Algorithm 2: std::map
Declare / define the container:
std::map<char, unsigned int> frequency;
Open the file
read a letter.
find the letter: frequency.find(letter)
If letter exists, increment the frequency: frequency[letter]++;
If letter no exists, insert into frequency: frequency[letter] = 1;
After all letters processed, iterate through the map displaying the letter and its frequency.
Here's one possible way you can solve this. I'm not giving you full code; it's considered bad to just give full implementations for other people's homework.
First, fill a new array with only unique characters. For example, if the input was:
abacdadeff
The new array should only have:
abcdef
That is, every character should appear only once in it. Do not forget to \0-terminate it, so that you can tell where it ends (since it can have a length smaller than 10).
Then create a new array of int (or unsigned, since you can't have negative occurrences) values that holds the frequency of occurence of every character from the unique array in the original input array. Every value should be initially 1. You can achieve this with a declaration like:
unsigned freq[10] = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
Now, iterate over the unique array and every time you find the current character in the original input array, increment the corresponding element of the frequencies array. So at the end, for the above input, you would have:
a b c d e f (unique array)
3 1 1 2 1 2 (frequencies array)
And you're done. You can now tell how many times each characters appears in the input.
Here, I'll tell you what you should do and you code it yourself:
include headers ( stdio libs )
define main ( entry point for your app )
declare input array A[amount_of_chars_in_your_input]
write output requesting user to input
collect input
now the main part:
declare another array of unsigned shorts B[]
declare counter int i = 0
declare counter int j = 0
loop through the array A[] ( in other words i < sizeof ( A ); or a[i] != '\0' )
now loop as much as there is different letters in the array A
store the amount of letters in the B[]
print it out
Now there are some tricks applying this but you can handle it
Try this:
unsigned int frequency[26] = {0};
char letters[10];
Algorithm:
Open file / read a letter.
Search for the letters array for the new letter.
If the new letter exists: increment the frequency slot for that
letter: frequency[toupper(new_letter) - 'A']++;
If the new letter is missing, add to array and set frequency to 1.
After all letters are processed, print out the frequency array:
`cout << 'A' + index << ": " << frequency[index] << endl;