As the name suggest it is an easy problem all what is needed is to take a string s as input and print Mohamed Salah

#include<iostream>
using namespace std;
int main()
{
 string str;cin>>str;
 cout<<"Mohamed Salah";
}

If $$$a$$$%$$$b=0$$$ ($$$a$$$ is divisible by $$$b$$$), just print $$$0$$$. Otherwise, we need exactly $$$b − a$$$%$$$b$$$ moves to make zero remainder of $$$a$$$ modulo $$$b$$$. % is the modulo operation.

#include<iostream>
using namespace std;
int main(){
    int t = 1; cin >> t;
    while(t--){
        int a,b;
        cin >> a >> b;
        if(!(a%b)) cout<<"0\n";
        else cout<<b-(a%b)<<"\n";
    }
    return 0;
}

In this problem we want to find a number that is divisible by t and has n digits. But we found a series of problems:

1. n can be 100 -> there is no data type that handles such a big number (long long take up to pow(2,64))

2. How can I easily get a number divisible by t

The answer is t itself and

what about multiplying the number by 10 until I get the desired length? why 10: multiplying by 10 will keep the divisibility state (2x10 is divisible by 2 and 2x10x10 is still divisible by 2....)

so the answer will be the t multiplied by 10 until I reach the desired length. -> problem 2 solved

multiplying by 10 means adding 0 to the end of the number (2x10 = 20, 330x10 = 3300)

so we can print the t followed by zeros -> problem 1 solved

but when I will print -1? the answer is if the number of digits in t is less than n as the smallest number is divisble by t is t itself

#include <iostream>
#include <vector>
using namespace std ;
int32_t main() {
    int n,t;cin>>n>>t;
    if(t == 10 && n== 1){
        cout<<"-1";
    }else{
        cout<<t;
        n--;
        if(t == 10)n--;
        for (int i = 0; i < n; ++i) {
            cout<<"0";
        }
    }
}

The Solution is to check the sum for each cell. For each cell (i , j) we can iterate over all elements in its diagonals. We have four different directions to go to - top right each cell in this direction (i, j) the next one is (i-1, j+1) - bottom left each cell in this direction (i, j) the next one is (i+1, j-1) - top left each cell in this direction (i, j) the next one is (i-1, j-1) - bottom right each cell in this direction (i, j) the next one is (i+1, j+1)

#include<bits/stdc++.h>
using namespace std;
void fast(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);}

int main()
{
    fast();
    int t;
    cin >>t;
    while(t--){   // our test cases
        int n, m;
        cin >> n >> m;
        int a[n][m];
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                cin >> a[i][j];
            }
        }
        int mx = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                int now = 0;
                int ci = i, cj = j;
                while(ci >= 0 && ci < n && cj >= 0 && cj < m)
                {
                    now+=a[ci][cj];
                    ci--;
                    cj--;     // top left direction
                }
                ci = i, cj = j;
                while(ci >= 0 && ci < n && cj >= 0 && cj < m)
                {
                    now+=a[ci][cj];
                    ci++;
                    cj--;   // bottom left direction
                }
                ci = i, cj = j;
                while(ci >= 0 && ci < n && cj >= 0 && cj < m)
                {
                    now+=a[ci][cj];
                    ci--;
                    cj++;   // top right direction
                }
                ci = i, cj = j;
                while(ci >= 0 && ci < n && cj >= 0 && cj < m)
                {
                    now+=a[ci][cj];
                    ci++;
                    cj++;   // bottom right direction 
                }
                now-=a[i][j]*3;  // we have added a[i][j] to sum 4 times. we need to add it only one time, so we subtract 3*a[i][j]
                mx = max(mx, now); // save maximum sum in mx
            }
        }
        cout << mx << endl;
    }
    return 0;
}

Not finished yet

#include<bits/stdc++.h>
using namespace std;
void fast(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);}

int main()
{
    fast();
    int n, m;
    cin >> n >> m;
    int arr[n];
    for(int i = 0; i<n; i++){
        cin >> arr[i];
    }
    int freq[100005]={0},pref[n+1];
    pref[n] = 0;
    for(int i = n-1; i>=0; i--){
        if(freq[arr[i]] == 0)
            pref[i] = pref[i+1] + 1;
        else
            pref[i] = pref[i+1];
        freq[arr[i]]++;
    }
    int q;
    while(m--){
        cin >> q;
        cout<<pref[q-1]<<"\n";
    }
    return 0;
}