The minimum position in vector of the number x

[swingby]

The program reads from the keyboard, on the first line, the natural number n. On the next line reads the string of n natural numbers, sorted in ascending order. From the third line we read a natural number m, and from the fourth we read m natural numbers, representing possible values for x.
Displays, for each of the m numbers read, the minimum position in the initial string.

Input :
10
1 2 3 4 4 4 5 6 6 7
4
2 4 5 6
Output:
2 4 7 8

Change my code please.

 #include <iostream>

using namespace std;

int main() {
    int n, m ,i, j, v[10002], w[102], ok = 0;
    cin >> n;
    for (i = 1; i <=n; i++) {
        cin >> v[i];
    }
    cin >> m;
    for (j = 1; j <= m; j++) {
        cin >> w[j];

       for (i = 1; i <= n; i++) {
        if ( v[i] == w[j]) {
            cout << i << " ";
        }

    }
    }
    return 0;

}

[dutch]

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> v;

    int n;
    cin >> n;
    for (int i = 0, x; i < n && cin >> x; ++i)
        v.push_back(x);

    int m;
    cin >> m;
    for (int i = 0, j = 0, x; i < m && cin >> x; ++i) {
        while (j < n && x > v[j]) ++j;
        cout << j + 1 << ' ';
    }
    cout << '\n';
}

[swingby]

And if you need to find the maximum position of x?
Input:
10
1 2 3 4 4 4 5 6 6 7
4
2 6 5 4

Output :
2 9 7 6

[keskiverto]

Note that dutch's solution does optimize on assumption that both input sequences have ascending order.

More generically, you must find x from v and report the position, if found.

For reporting the last position you should find from v in reverse order; start search from last, rather than first element.