First, calculate the total number of trees without worrying about visibility.
Layer 1 has 5 trees.
Layer 2 has 10 trees.
.......
Layer $$$k$$$ has $$$5k$$$ trees.
What is the sum of $$$5k$$$ for $$$k=1$$$ to $$$n$$$?

A tree is "visible" if no other tree blocks the line of sight from the center. Think of a tree's position as a fraction along one side of the pentagon. A tree at layer $$$k$$$ and position $$$m$$$ (where $$$0 \le m \le k$$$) corresponds to the fraction $$$m/k$$$. When is the fraction $$$m/k$$$ "blocked" by a previous layer? It is blocked if the fraction can be simplified (e.g., $$$2/4$$$ is blocked by $$$1/2$$$).

A tree is visible only if its position fraction $$$m/k$$$ cannot be simplified. This happens when $$$\gcd(m, k) = 1$$$. The number of integers $$$m \lt k$$$ coprime to $$$k$$$ is given by Euler's Totient Function, $$$\phi(k)$$$. The vertices of the pentagon are special cases; handle them carefully. Total Hidden = Total Trees — Visible Trees.

We need to find the number of "hidden" trees in an orchard formed by concentric pentagons. Let's break the counting into two parts:

1. Calculate the Total number of trees.
2. Calculate the Visible number of trees.

1. Counting Total Trees
Layer $$$k$$$ is a regular pentagon. The problem states it has exactly 5 vertices and $$$5(k-1)$$$ additional trees on the edges.
Total trees in layer $$$k = 5 + 5(k-1) = 5k$$$.
Summing this from $$$1$$$ to $$$n$$$:

2. Counting Visible Trees

Consider one side of the pentagon on layer $$$k$$$. The trees on this side divide it into $$$k$$$ equal segments. We can assign each tree an index $$$m$$$ ($$$0 \le m \le k$$$) representing its position along the side. The relative position of a tree can be described by the fraction $$$\frac{m}{k}$$$.

A tree at layer $$$k$$$ with index $$$m$$$ is visible from the center if and only if there is no tree at the same relative position in any previous layer $$$k' \lt k$$$. Mathematically, this means the fraction $$$\frac{m}{k}$$$ must be irreducible.

Let's analyze the counts for each layer $$$k$$$:

• Vertices ($$$m=0$$$ and $$$m=k$$$): For $$$k=1$$$, the vertices are visible. For any $$$k \gt 1$$$, the vertices are hidden by the vertices of layer 1 (since $$$\gcd(k,k) \neq 1$$$ and $$$\gcd(0,k) \neq 1$$$).
• Edge Trees ($$$1 \le m \lt k$$$): The number of visible trees strictly on the edge of one side is the count of integers $$$m \in [1, k-1]$$$ such that $$$\gcd(m, k) = 1$$$. This is exactly the definition of Euler's Totient Function, $$$\phi(k)$$$.

Conclusion
The number of visible trees up to layer $$$n$$$ is $$$\sum_{k=1}^n 5 \phi(k)$$$.
The number of hidden trees is:

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

#define int long long

const int MAXN = 100005;
int phi[MAXN];
int prefix_phi[MAXN];

void precompute() {
    for (int i = 0; i < MAXN; i++) phi[i] = i;
    for (int i = 2; i < MAXN; i++) {
        if (phi[i] == i) {
            for (int j = i; j < MAXN; j += i)
                phi[j] -= phi[j] / i;
        }
    }
    prefix_phi[0] = 0;
    for (int i = 1; i < MAXN; i++) {
        prefix_phi[i] = prefix_phi[i-1] + phi[i];
    }
}

void solve() {
    int n;
    cin >> n;
    int total_trees = 5 * n * (n + 1) / 2;
    int visible_trees = 5 * prefix_phi[n];
    cout << total_trees - visible_trees << endl;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    precompute();
    int t;
    cin >> t;
    while(t--) {
        solve();
    }
}

The core formula remains the same as the Easy Version:

However, since $$$n \le 10^9$$$, a simple linear loop is too slow.
You need an algorithm that runs faster than $$$O(n)$$$.

Let $$$S(n) = \sum_{i=1}^n \phi(i)$$$. We need to calculate $$$S(n)$$$ efficiently.
Recall the property of the Euler Totient function: $$$\sum_{d|n} \phi(d) = n$$$.
If we sum this identity over all $$$i$$$ from $$$1$$$ to $$$n$$$, we get:

Can you rearrange the left side to express $$$S(n)$$$ in terms of smaller values of $$$S$$$?

By changing the order of summation, we derive the recurrence relation known as the Du Jiao Sieve:

You can compute this using recursion with memoization.
1. For small $$$n$$$ (up to $$$\approx 2 \cdot 10^6$$$), precompute $$$S(n)$$$ using a linear sieve.
2. For large $$$n$$$, use the recursive formula and a hash map (or std::map).

Difference from Easy Version
In the hard version, $$$N$$$ can be up to $$$10^9$$$. The formula derived in the easy version is:

Explanation: Du Jiao Sieve

Let $$$S(n) = \sum_{i=1}^n \phi(i)$$$.
We know the property of Euler's totient function involving Dirichlet convolution:

Algorithm:

1. Precompute $$$\phi(i)$$$ and its prefix sums for small values (up to $$$K \approx N^{2/3}$$$, e.g., $$$2 \cdot 10^6$$$) using a linear sieve.
2. Recursion with Memoization: For values larger than $$$K$$$, use the formula above.
   • The summation $$$\sum_{g=2}^n S(\lfloor n/g \rfloor)$$$ can be computed efficiently using harmonic ranges (since $$$\lfloor n/g \rfloor$$$ takes limited distinct values).
   • Store computed results in a hash map (map or unordered_map) to avoid re-calculation.

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

#define int long long
const int MAXK = 2000005;

int phi[MAXK];
int prefix_phi[MAXK];
map<int, int> memo;

void precompute() {
    phi[1] = 1;
    vector<int> primes;
    static bool is_prime[MAXK];
    memset(is_prime, true, sizeof(is_prime));
    for (int i = 2; i < MAXK; i++) {
        if (is_prime[i]) {
            primes.push_back(i);
            phi[i] = i - 1;
        }
        for (int p : primes) {
            if (i * p >= MAXK) break;
            is_prime[i * p] = false;
            if (i % p == 0) {
                phi[i * p] = phi[i] * p;
                break;
            } else {
                phi[i * p] = phi[i] * (p - 1);
            }
        }
    }
    prefix_phi[0] = 0;
    for (int i = 1; i < MAXK; i++) {
        prefix_phi[i] = prefix_phi[i-1] + phi[i];
    }
}

int get_sum_phi(int n) {
    if (n < MAXK) return prefix_phi[n];
    if (memo.count(n)) return memo[n];
    int total = n * (n + 1) / 2;
    for (int l = 2, r; l <= n; l = r + 1) {
        r = n / (n / l);
        int count = (r - l + 1);
        total -= count * get_sum_phi(n / l);
    }
    return memo[n] = total;
}

void solve() {
    int n;
    cin >> n;
    int total_trees = 5 * n * (n + 1) / 2;
    int visible_trees = 5 * get_sum_phi(n);    
    cout << total_trees - visible_trees << endl;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    precompute();

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

The answer is monotonic: if capacity $$$T$$$ works, any $$$T' \gt T$$$ works. Use Binary Search on the Answer.

Check Function: Greedily add people to the current trip. Start a new trip only when adding the next person would exceed the limit $$$T$$$.

Update Logic:

• If check(mid) is true, then mid is possible. Try to find a smaller answer: ans = mid, r = mid — 1.
• If check(mid) is false, we need more capacity: l = mid + 1.

The problem asks us to partition the array of people $$$m_1, m_2, \dots, m_N$$$ into at most $$$K$$$ contiguous subarrays (trips) such that the maximum tension required for any single trip is minimized. The tension for a group with total mass $$$M_{load}$$$ is given by $$$T = M_{load} \cdot G_{eff}$$$.

This is a classic "Minimize the Maximum" problem, which can often be solved using Binary Search on the Answer.

1. Monotonicity
Let's define a predicate function check(T) which returns true if it is possible to rescue all people in $$$\le K$$$ trips using a cable with tension rating $$$T$$$, and false otherwise.
If a tension rating $$$T$$$ is sufficient to rescue everyone, then any rating $$$T' \gt T$$$ is definitely sufficient as well. This monotonicity allows us to binary search for the minimum valid $$$T$$$.

2. The Check Function (Greedy Strategy)
For a fixed tension capacity $$$T_{limit}$$$, how do we determine the minimum number of trips required? Since the order of people is fixed, we can use a greedy approach:

• Start the first trip with the first person.
• Continue adding the next person in the queue to the current trip as long as the total tension does not exceed $$$T_{limit}$$$.
• If adding the next person would exceed $$$T_{limit}$$$, we must finalize the current trip and start a new one with that person.
• If any single individual requires a tension greater than $$$T_{limit}$$$ on their own ($$$m_i \cdot G_{eff} \gt T_{limit}$$$), then this $$$T_{limit}$$$ is immediately invalid.

After iterating through all people, if the number of trips used is less than or equal to $$$K$$$, then $$$T_{limit}$$$ is feasible.
We perform binary search within the range $$$[L, R]$$$. In each step, if check(mid) is true, we store mid as a potential answer and try smaller values ($$$R = mid - 1$$$). Otherwise, we look for larger values ($$$L = mid + 1$$$).

3. Time Complexity
The check function iterates through the array once, taking $$$O(N)$$$ time. The binary search runs for $$$O(\log(\sum m_i \cdot G_{eff}))$$$ iterations.
The total time complexity is $$$O(N \cdot \log(\text{Total Mass} \cdot G_{eff}))$$$, which fits comfortably within the time limit for $$$N \le 2 \cdot 10^5$$$.

#include<bits/stdc++.h>
#define int long long
#define pb push_back
using namespace std;
 
 
bool ok(int n,int k,int g,vector<int>&a,int m){
    int c=1,cur=0;
    for(int i=0;i<n;i++){
        int w=a[i]*g;
        if(w>m) return 0;
        if(cur+w<=m){
            cur+=w;
        }else{
            c++;
            cur=w;
        }
    }
    return c<=k;
}
 
 
int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 
 
    int t;
    cin>>t;
    while(t--){
        int n,k,g;
        cin>>n>>k>>g;
 
 
        vector<int>a(n);
        int mx=0,sum=0;
        for(int i=0;i<n;i++){
            cin>>a[i];
            mx=max(mx,a[i]);
            sum+=a[i];
        }
 
 
        int l=mx*g,r=sum*g,ans=r;
        while(l<=r){
            int m=l+(r-l)/2;
            if(ok(n,k,g,a,m)){
                ans=m;
                r=m-1;
            }else{
                l=m+1;
            }
        }
        cout<<ans<<"\n";
    }
}

Analyze the bitwise difference for small numbers manually.
$$$1 (01) \to 2 (10)$$$: Difference is 2 bits. (Ends in 1 one)
$$$3 (011) \to 4 (100)$$$: Difference is 3 bits. (Ends in 2 ones)
$$$7 (0111) \to 8 (1000)$$$: Difference is 4 bits. (Ends in 3 ones)

Do you see the relationship between the number of trailing ones and the bitwise difference?

If a number ends in exactly $$$p$$$ ones, adding 1 will flip those $$$p$$$ ones to zeros and flip the next zero to a one. The total change is $$$p+1$$$ bits.
To get a difference of exactly $$$k$$$, the number must end in exactly $$$k-1$$$ ones.

We need to count numbers $$$i \lt n$$$ that end in the pattern ...011...1 (with $$$k-1$$$ ones).
This pattern repeats every $$$2^k$$$ numbers.

We need to count how many integers $$$i$$$ in the range $$$[0, n-1]$$$ change exactly $$$k$$$ bits when incremented to $$$i+1$$$.

Mathematical Derivation

1. The Bitwise Property
When we add $$$1$$$ to an integer $$$i$$$, the number of bits that flip is determined by the number of trailing ones in $$$i$$$.
Specifically, if $$$i$$$ ends with $$$z$$$ ones (e.g., $$$\dots 0\underbrace{11\dots1}_{z}$$$), adding $$$1$$$ will flip all those $$$z$$$ ones to zeros, and flip the zero immediately to their left to a one.
The total bits changed (the bitwise difference) is $$$z + 1$$$.

2. The Condition
We are given that the difference is $$$k$$$.

3. The Formula
This pattern repeats with a period of $$$2^k$$$. The first number satisfying this is $$$2^{k-1} - 1$$$.
To count how many such numbers exist strictly less than $$$n$$$, we can use the direct formula for counting arithmetic progressions:

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

void solve() {
    int n, k;
    cin >> n >> k;
    if (k == 0) {
        cout << 0 << endl;
        return;
    }
    int ans = (n + (1ll << (k - 1))) >> k;
    cout << ans << endl;
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

The problem asks us to minimize $$$\sum a_i^2$$$.
First, notice that the sign of $$$a_i$$$ does not affect its square ($$$(-x)^2 = x^2$$$). Since we can increment or decrement freely, we can treat all scores as their absolute values $$$|a_i|$$$ and always choose to decrement them towards 0.
The problem effectively becomes: Reduce the sum of squares of $$$|a_i|$$$ using $$$k$$$ operations.

The function $$$f(x) = x^2$$$ is convex. Reducing a larger number gives a bigger drop in the sum of squares than reducing a smaller number.
This implies a Greedy Strategy: We should always apply our operation to the student with the currently largest absolute score. We are essentially "leveling down" the peaks of the distribution from top to bottom.

Sort the array of absolute values in descending order.
Instead of simulating one by one (which is too slow), process them in groups. If the top $$$i$$$ elements are equal, we can try to reduce all of them together to match the next largest value ($$$a_{i+1}$$$).

Let $$$S$$$ be the sum of all absolute values. If $$$k \ge S$$$, we can reduce everyone to 0. If $$$(k-S)$$$ is odd, the answer is 1; otherwise, it is 0.

If $$$k \lt S$$$, we need to perform the operations greedily on the largest elements.

Sorting Algorithm

1. Convert all $$$a_i$$$ to $$$|a_i|$$$ and sort them in descending order. Append a $$$0$$$ at the end to handle the boundary.
2. Iterate through the sorted array. Let's say we are at index $$$i$$$ (1-based), meaning the first $$$i$$$ elements are currently the "largest" and have been flattened to the same value $$$a_i$$$.
3. Calculate the operations needed to bring these $$$i$$$ elements down to the level of the next element $$$a_{i+1}$$$.
   $$$ \text{needed} = i \times (a_i - a_{i+1}) $$$
4. If $$$k \ge \text{needed}$$$:
   We have enough operations. Perform them (subtract `needed` from $$$k$$$) and continue to the next level. Effectively, the first $$$i$$$ elements are now equal to $$$a_{i+1}$$$.
5. If $$$k \lt \text{needed}$$$:
   We cannot fully lower this group to $$$a_{i+1}$$$. We stop here.
   We distribute the remaining $$$k$$$ operations as evenly as possible among these $$$i$$$ elements.
   Each of the $$$i$$$ elements is lowered by $$$k / i$$$.
   The remaining $$$k \% i$$$ elements are lowered by an additional 1.
   Calculate the final sum of squares and print the answer.

Time Complexity

Sorting takes $$$O(N \log N)$$$. The linear pass takes $$$O(N)$$$.
Total: $$$O(N \log N)$$$.

In 1 step, we can move the center by any distance $$$x \in (0, 2r]$$$ by choosing an appropriate rotation angle $$$\theta = 2 \arcsin\left(\frac{x}{2r}\right)$$$.Therefore, if $$$d \% 2r \neq 0$$$, we take $$$\lfloor \frac{d}{2r} \rfloor$$$ steps of length $$$2r$$$ plus 1 smaller step of length $$$d \bmod 2r$$$, totaling $$$\lceil \frac{d}{2r} \rceil$$$ steps.

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define loop(i, n) for (int i = 0; i < n; i++)
#define print_vec(a) for (int i = 0; i < a.size(); i++) cout << a[i] << " ";
#define endl '\n'
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int r, x, y, xf, yf;
    cin >> r >> x >> y >> xf >> yf;
    int dx = xf - x;
    int dy = yf - y;
    
    int d2 = dx*dx + dy*dy;
    double d = sqrtl(d2);

    int ans = ceil(d/(2.0*r));
    
    cout << ans << "\n";


    return 0;
}

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

void solve() {
    vector<long long> x(4);
    for (int i = 0; i < 4; ++i) {
        cin >> x[i];
    }
    sort(x.begin(), x.end());

    // S = x[3] = a + b + c
    long long a = x[3] - x[2];
    long long b = x[3] - x[1];
    long long c = x[3] - x[0];

    cout << a << " " << b << " " << c << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    solve();
    
    return 0;
}

We construct the array starting with $$$P_1 = 1$$$. Each subsequent term $$$P_i$$$ ($$$1 \lt i \le N$$$) is formed by alternatingly applying step sizes (jumps) of $$$N$$$ and $$$N+2$$$:

$$$P_i = \vert{}N - P_{i-1}\vert{}$$$ if $$$i$$$ is odd,
$$$P_i = \vert{}(N + 2) - P_{i-1}\vert{}$$$ if $$$i$$$ is even.

Modulo $$$(N+1)$$$, notice that:
$$$N \equiv -1 \pmod{N+1}$$$
$$$N+2 \equiv +1 \pmod{N+1}$$$

2. Jump Sum Analysis ($$$L$$$):

For a contiguous block of length $$$k$$$, the jump sum $$$L_k$$$ alternates between terms of $$$N$$$ and $$$N+2$$$. The possible jump sums $$$L$$$ take two forms:

$$$\text{Type 1 (starts with } N \text{)}: N + (N+2) + N + (N+2) + \dots$$$
$$$\text{Type 2 (starts with } N+2 \text{)}: (N+2) + N + (N+2) + N + \dots$$$

For example, when $$$k = 3$$$:
$$$\text{Type 1: } N + (N+2) + N = 3N + 2$$$
$$$\text{Type 2: } (N+2) + N + (N+2) = 3N + 4$$$

Let L be the above value

3. Parity & Impossibility Proof:

Let's analyze $$$L$$$ and $$$N+1$$$ based on the parity of $$$N$$$:

It will be different in both the cases either n = even or odd which implies L%(N+1) can never be equals to zero

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define loop(i, n) for (int i = 0; i < n; i++)
#define print_vec(a) for (int i = 0; i < a.size(); i++) cout << a[i] << " ";
#define endl '\n'
signed main() {
    int t=1;
    // cin >> t;
    while (t--) {
        int n; cin >> n;
        int s = (n*(n+1))/2;
        if(s%(n+1)){
            vi ans(n);
            ans[0] = 1;
            for(int i=1;i<n;i++){
                if(i&1) ans[i] = abs(n-ans[i-1]);
                else ans[i] = abs(n+2-ans[i-1]);
            }
            print_vec(ans);
            cout<<endl;
        }else cout<<-1<<endl;

    }
    return 0;
}