find the groups in which you and your crush are placed .

use ceil function .

As You want that you and your crush will be in same group , find out Group number for both $$$G_a$$$ = ceil(a/x) (group number where you are placed) , $$$G_b$$$ = ceil(b/x) (group number where your crush is placed) . there are total of n diffrent value of x(group size) which is equiprobable. Hence Probability = P/Q . P = Number of group size's in which $$$G_a == G_b$$$ , Q = n .

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define pb push_back
#define ppb pop_back
#define MOD 998244353

const int N=1e5;

int binexp(int a, int b) {
    int res = 1;
    while (b > 0) {
        if(b&1)
        res = ((res%MOD)*(a%MOD))%MOD;
        a = ((a%MOD)*(a%MOD))%MOD;
        b >>= 1;
    }
    return res;
}

void solve(){
    int n;
    string a,b;
    cin >> n >> a >> b;
    int an = stoi(a.substr(7,3));
    int bn = stoi(b.substr(7,3));
    int ans = 0;
    for(int i=1;i<=n;i++){
        if((an-1)/i == (bn-1)/i){
            ans++;
        }
    }
    
    ans *= binexp(n,MOD-2);
    ans %= MOD;

    cout << ans << endl;

}

int32_t main()
{
    auto begin = std::chrono::high_resolution_clock::now();
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

#ifndef ONLINE_JUDGE
    freopen("test0.in", "r", stdin);
    freopen("test0.out", "w", stdout);
#endif

    int t = 1;
    cin >> t;
    while (t--)
        solve();
        
    return 0;
}

Try doing something with the lowest common ancestor of $$$i$$$ and $$$i + 1$$$.

Prefix sums on tree?

Since we have to answer the problem offline, let's try to do something similar to difference arrays.

First root the tree on an arbitrary vertex — say $$$1$$$.

Say the lowest common ancestor of $$$i$$$ and $$$i + 1$$$ be $$$LCA$$$. We will add $$$1$$$ to $$$score[parent_i]$$$ and $$$score[i + 1]$$$, and subtract $$$1$$$ from $$$score[LCA]$$$ and $$$score[parent_{LCA}]$$$. A Now if we add up scores going from bottom to top, we will get the number of times each vertex is visited.

#include<bits/stdc++.h>
using namespace std;

#define int long long

// LCA code source - usaco.guide
int depth[200010];
int up[200010][20];
int score[200010];
vector<int> adj[200010];

void dfs(int v) {
    for(int i = 1; i < 20; i++) { up[v][i] = up[up[v][i - 1]][i - 1]; }
    for (int x : adj[v]) {
        if (x != up[v][0]) {
            depth[x] = depth[up[x][0] = v] + 1;
            dfs(x);
        }
    }
}

int jump(int x, int d) {
    for(int i = 0; i < 20; i++) {
        if ((d >> i) & 1) x = up[x][i];
    }
    return x;
}

int LCA(int a, int b) {
    if (depth[a] < depth[b]) swap(a, b);

    a = jump(a, depth[a] - depth[b]);
    if (a == b) return a;

    for(int i = 19; i >= 0; i--) {
        int aT = up[a][i], bT = up[b][i];
        if (aT != bT) a = aT, b = bT;
    }

    return up[a][0];
}

void clear(int n) {
    for(int i = 0; i < n; i++) {
        score[i] = 0;
        depth[i] = 0;
        adj[i].clear();
        for(int j = 0; j < 20; j++) 
            up[i][j] = 0;
    }   
}

int32_t main() {
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int t; cin >> t;
    while(t--) {
        int n; cin >> n;
        clear(n + 5);

        for(int i = 0; i < n - 1; i++) {
            int a, b; cin >> a >> b;
            adj[a].push_back(b);
            adj[b].push_back(a);
        }
        dfs(1);

        for(int i = 1; i < n; i++) {
            int par = LCA(i, i + 1);
            score[up[i][0]]++, score[i + 1]++;
            score[par]--, score[up[par][0]]--;
        }
        
        vector<pair<int,int>> vp;
        for(int i = 2; i <= n; i++) {
            vp.push_back({-depth[i], i});
        }
        sort(vp.begin(), vp.end());

        for(int i = 0; i < vp.size(); i++) {
            score[up[vp[i].second][0]] += score[vp[i].second];
        }

        score[1]++; // initial stand
        for(int i = 1; i <= n; i++) cout << score[i] << " ";
        cout << "\n";
    }
    return 0;
}