Very poorly indeed. I'm reading it for - like - the 6th time and I just don't understand the task.

T_T help I'm new in C++

The Problem tells you everything you need to know to answer the question. You just have to read closely.

Here is my solution with added comments. I've used a raw array of arrays, dynamically allocated and via pointers, so they must be manually deleted afterwards. I could have also used a std::vector, but the question does say 'array'.

#include <iostream>

using namespace std;

int main() {
	// get length of 2D array and number of queries
	int n, q;
	cin >> n >> q;

	// create 2D array
	int** a = new int*[n]();

	// fill 2D array with 1D subarrays
	for (int i = 0; i < n; i++) {
		// get the length of the 1D array at a[i]
		int k;
		cin >> k;

		// create the 1D subarray with the given length
		int* i_arr = new int[k]();

		// fill the subarray with k values
		for (int j = 0; j < k; j++) {
			cin >> i_arr[j];
		}

		// store the subarray in a
		a[i] = i_arr;
	}


	// run queries on a
	for (int q_num = 0; q_num < q; q_num++) {
		// get i, j as the 'query' to get a value from a
		int i, j;
		cin >> i >> j;
		cout << a[i][j] << endl;
	}

	// delete 2D array (each subarray must be deleted)
	for (int i = 0; i < n; i++) {
		delete[] a[i];
	}
	delete[] a;

	return 0;
}

OR with std::vector:

#include <iostream>
#include <vector>

using namespace std;

int main() {
	// get length of array 'a' and number of queries
	int n, q;
	cin >> n >> q;

	// create vector of vectors
	vector<vector<int>> a(n);

	// fill each 2D vector i with k_i values
	for (int i = 0; i < n; i++) {
		// get the length k of the vector at a[i]
		int k;
		cin >> k;

		// fill the vector with k values
		a[i].resize(k);
		for (int j = 0; j < k; j++) {
			cin >> a[i][j];
		}
	}

	// run queries on a
	for (int q_num = 0; q_num < q; q_num++) {
		// get i, j as the 'query' to get a value from a
		int i, j;
		cin >> i >> j;
		cout << a[i][j] << endl;
	}

	return 0;
}

indeed

exactly

ditto, this should be in C, otherwise they should atleast allow use of C++ features like vectors :/

I think it even discourages idiomatic C++ by forcing you to use manual memory management. Even though code using vector can be easily adapted to pointers (thanks to vector's awesome interface) but it might create unneccessary memory leaks.

agree

I can't do it without vector (I don't know how to use it), and I think this challenger is not suitable for Introduction

int n,q;
cin >> n >> q;
vector< vector<int> > a(n);

// input each array
for (int i=0;i<n;i++) {
    int k;
    cin >> k;

    for (int j=0;j<k;j++) {
        int data;
        cin >> data;
        a[i].push_back(data);
    }
}

// do the queries
for (int i=0;i<q;i++) {
    int x,y;
    cin >> x >> y;
    cout << a[x][y] << endl;
}

This was exactly my solution. It seems like a lot of these problems were written for the C track and moved to C++ without updating the solutions. Nice solution.

beautiful code!

Well coded!!

Beautiful!!! you could inicialize "nvec" whit "n" , so you save and remove the line " nvec.push_back(ivec);"

This is very interesting. I initially tried to create an array of arrays in the very same manner as you created the vector of vectors, but I guess C++ does not allow that?

Between your example and a brief Youtube on vectors, I see how this works. Thanks for sharing this solution.

It was not updated however your code was good enough

Exactly what I have done! But I get segmentation fault, could anyone help?

After reading some notes and your code, I finally understood what is vector.

Kudos

but size of the variable vectors will not be given.

what is the use of "ivec.push_back(value);" what does it do!! please provide a preference!!

I solve this challange three way... but this code looks best from me....

BEAUTIFUL!!!

This method is easy and very clear to understand !

Thank you :)

can anyone **explain this code **please!!!

Mine is below. Yeah, the trick and grist of this one here is figuring out the 'nested' variable-sized array

int main() {
    int n, q;  
    cin >> n >> q; 
    //NOTE: C++'s variable-sized array: vector
    vector<vector<int>> a; 
    for (int i = 0; i < n; i++){
        int k;
        cin >> k; 
        vector<int> vect_j;
        for (int j = 0; j < k; j++){
            int val; 
            cin >> val;
            vect_j.push_back(val); 
        }
        a.push_back(vect_j);
    }
    for (int i = 0; i < q; i++){
        int vect_id, id;  
        cin >> vect_id >> id; 
        cout << a.at(vect_id).at(id) << "\n";
    }
    return 0;
}

nice going

Nice! My question is that how does the compiler know at this line:

nvec.push_back(ivec);

that it is a new line in the vector and not just concatenates the two vector after each other ?

int main() { int i,j,r,q,k,n=0,a[10][20]; cin>> r>> q; for(i=0;i> k; for(j=0;j> a[i][j]; } } do { cin>> i>> j; cout<

}while(n

I also did almost the same code.

elegant

vector ivec; You are initializing the vector with same name again and again , won't that override the previous array. Sorry if it makes no sense , can you explain ?

Cool code :)

Do you generate a k-element array by arr[i].resize(k)?

nice code.....

So clean clean! Thanks for the tip on the resize!

where can i learn vector from?

I don't know vector and dynamic arrays.... How to proceed without it... I know the logic, it is only that runtime error comes...

int main() { long n,q; int k[100000],a[100000][300000]; cin >> n >> q; for(int i = 0; i < n; i++) { cin >> k[i]; for(int j = 0; j < k[i]; j++) { cin >> a[i][j]; } } long x[100000],y[100000]; for(int i = 0; i < q; i++) { cin >> x[i] >> y[i]; cout << a[x[i]][y[i]] << endl; } return 0; }

int main() { long n,q; int k[100000],a[100000][300000]; cin >> n >> q; for(int i = 0; i < n; i++) { cin >> k[i]; for(int j = 0; j < k[i]; j++) { cin >> a[i][j]; } } long x[100000],y[100000]; for(int i = 0; i < q; i++) { cin >> x[i] >> y[i]; cout << a[x[i]][y[i]] << endl; } return 0; }

you need a fixed size array in which each element point to a variable array (whose size can be change )like vector v[2];

v[0] is pointing to a variable size array v[1] is pointing to a variable size array

if you want to push data into these vector : v[0].push_back(10); // put the 10 in at the end of the variable array

and you can access the element in array like: cout<

Can you explain your solution, you have interesting solution, we can learn lot from your explanation.

Beginner-friendly revision (also moved allocation for query, later on in the code)

int numSequence;
int numQueries;
int size;

// ask number of sequences and queries respectively
cin >> numSequence >> numQueries;

// where to store the sequence elements, dynamically allocate
int** arr = new int*[numSequence];
// where to store numbers for the queries, dynamically allocate
int**   dummy = new int*[numQueries];

// now enter sequences
for ( int i = 0; i < numSequence; i++){
    cin >> size;
    arr[i] = new int[size];             // allocate memory for integer array
    for( int j = 0; j < size; j++ ){    // enter numbers in the sequence
        cin >> arr[i][j];
    }
}

// now allocate memory for query elements
for( int i = 0; i < numQueries; i++){
    dummy[i] = new int[2];
}

// now enter queries
for( int i = 0; i < numQueries; i++){
    cin >> dummy[i][0] >> dummy[i][1];
}

// now output those queries
for( int i = 0; i < numQueries; i++){
    cout << arr[ dummy[i][0] ][ dummy[i][1] ] << endl;
}
return 0;

}

i did exactly the same ,but my all cases did not pass >_< here it is:

include

include

include

include

include

using namespace std;

int main() {
int **arr;
int n,a,x,y,T;
int b;
cin>>n>>T;
arr=(int**) new int*[n];
for(int i=0;i<n;i++)
{
    cin>>a;
    arr[i]=new int[a];
    for(int j=1;j<=a;j++)
    {
        cin>>b;
        arr[i][j]=b;
    }
}

int **arr2;
arr2=(int**) new int*[T];
for(int i=0;i<T;i++)
{
    arr2[i]=new int[2];
}
for(int i=0;i<T;i++)
{
    cin>>x>>y;
    arr2[i][0]=x;
    arr2[i][1]=y;
}
for(int i=0;i<T;i++)
{
    cout<<arr[arr2[i][0]][arr2[i][1]+1]<<endl;

}
return 0;

}

first line cosist number of array and queries p=2 and q=2 \n p number of arrays, line: 3 1 5 4, first element represents number of elements in an array = 3, array: 1 5 4, same for "5 1 2 8 9 3" . q number of queries line: 0 1 , 0 indexed array ("1 5 4"), 1 indexed element("5") answer : 5, same goes for "1 3" answer: 9

SAME QUSETION HERE

use int** a

My way of doing this

int N,Q;
cin >> N >> Q;

int *array[N];    

for(int i = 0;i<N;++i)
{
    int size;
    cin >> size;
    array[i] = (int *)malloc(size * sizeof(int));
    for(int j = 0;j< size;++j)
    {
        int num;
        cin >> num;
        array[i][j] = num;
    }
}

for(int i=0 ; i<Q;++i)
{
    int nth,index; 
    cin >> nth >>index;
    cout << array[nth][index]<<endl;
}


return 0;

Indeed vector can be used.

#include <iostream>

int main() {
  void do_work();
  do_work();
}

#include <vector>

void do_work() {
  std::vector<int> vec(10);
  std::cout << "vec.size() -> " << vec.size() << '\n'; 
}

I think so!!! for every line in Q ,the length is not sure....

int N; int Q; cin >> N >> Q; int** arr = new int*[N]; for (int i = 0; i> l; arr[i] = new int[l]; int tmp; for (int j= 0; j> tmp; arr[i][j] = tmp; } } int** arr1 = new int*[Q]; for (int i = 0; i> tmp; arr1[i][j] = tmp; } } for (int i = 0; i < Q; i++) { cout << arr[arr1[i][0]][arr1[i][1]] << endl; }

You can do it with one dimensional array. Just need to keep track of each seq length and value in the array. Please refer my solution below:

int N,Q;
cin>>N>>Q;   //read no of seq and no of queries
int seq[500002];  //this will store all sequences, seq[0] to seq[N-1] will store the length of each sequence, the values of each seq will be stored seq[N] onwards
int csum[N]={0};  //it will store the cummulative sum of lengths of each seq till seq N i.e  csum[0] stores first seq length, csum[1] stores 0th + 1st seq length and so on
int k,inp,curr=N; //curr denotes the point from where we need to fill the current seq values in the array
for(int i=0;i<N;i++){
    cin>>k;
    seq[i]=k;  //store length of current seq
    for(int j=0;j<k;j++){
        cin>>inp;
        seq[curr+j] = inp;   //store values of current seq
    }
    if(i==0) csum[i] = k;
    else csum[i] = csum[i-1]+k;        
    curr +=k;   //adjust curr value for next seq     
}
for(int x=0;x<Q;x++){
    int a,b;
    cin>>a>>b;
    cout<<seq[N+b+csum[a]-seq[a]]<<endl;   //output the value of the element b from sequence a 
}

int **array = new int*[sequences];
for (int x = 0; x < sequences; x++) {
  array[x] = new int[sequence_width];
  for (int y = 0; y < sequence_width; y++) {
    cin >> *(array[x] + y);
  }
}

You need to allocate an array of pointer to int. Then you need to allocate each array with variable width. Finally you need to read in the int into the array offset by Y ints. The memory is contingous so *(array[x] + y) gets the reference of array[x] (which is a pointer to int) and offsets the memory address y ints. Then we dereference that to insert the int.

BTW this has nothing to do with C++. But it is useful to know. If you were to do C++ you would use

std::vector<std::vector<int> >

using pointer and multidimensional arrays

using pointer and multidimensional arrays

int n,q;
int k,i=0;
cin>>n;
cin>>q;
int size_of_n=n;
int size_of_q=q;
//declaring arraay of pointers to hold the array
int **array1;
//array to hold the query answer
int *array2;
array1=(int **)malloc(n*sizeof(int*));
array2=(int *)malloc(q*sizeof(int));
while(n!=0)
{

    cin>>k;
    array1[i]=(int *)malloc((k+1)*sizeof(int));
    array1[i][0]=k;
    for(int j=1;j<k+1;j++)
   {
      cin>>array1[i][j];
   }
    //function(array1[i],k+1);
    ++i;
    --n;

}

i=0;
while(q!=0)
{
    int a,b;
    cin>>a>>b;
    array2[i]=array1[a][b+1];
    --q;
    i++;
}

for(int i=0;i<size_of_q;i++)
{
    cout<<*(array2+i)<<endl;
}
for (int i = 0; i < size_of_n; ++i)
{
    free(array1[i]);
}
free(array2);

return 0;

}

Why did you comment using namespace std;?