When Problem-Setting Syndrome Strikes, Start with the Strange Core

Content Covered#

• Problem-Setting Syndrome Struck

Original Problem-Setting Notes Organization#

Putting Coins Twice Becomes Three Times#

According to the description and solution process in the story, the operating mechanism of the magic bag can be modeled through mathematical functions. Define function $f(x)$ to represent the number of coins taken out after putting $x$ coins into the bag. The key condition is: for any initial quantity $y$, after putting $y$ coins into the bag and taking out $f(y)$, then putting $f(y)$ coins into the bag, the quantity taken out should be $f(f(y)) = 3y$. At the same time, function $f$ is strictly increasing (i.e., if $x_1 < x_2$, then $f(x_1) < f(x_2)$), and the number of coins must be integers.

In the story, the values of $f(x)$ are gradually derived by constructing a sequence, starting from small $x$, ensuring that the conditions $f(f(y)) = 3y$ and strict increase are satisfied. The following are the core steps of the derivation process:

• Starting from $y = 1$: Put 1 coin into the bag, take out $f(1)$; then put $f(1)$ into the bag, take out $f(f(1)) = 3 \times 1 = 3$. Since magic increases coins and coins are integers, $f(1) > 1$ and $f(f(1)) = 3 > f(1)$, so the only possibility is $f(1) = 2$, and thus $f(2) = 3$.
• For $y = 2$: Put 2 coins into the bag, take out $f(2) = 3$; then put 3 into the bag, take out $f(f(2)) = f(3) = 3 \times 2 = 6$, so $f(3) = 6$.
• For $y = 3$: Put 3 coins into the bag, take out $f(3) = 6$; then put 6 into the bag, take out $f(f(3)) = f(6) = 3 \times 3 = 9$, so $f(6) = 9$.
• For $y = 4$: Put 4 coins into the bag, take out $f(4)$; then put $f(4)$ into the bag, take out $f(f(4)) = 3 \times 4 = 12$. Since $f$ is strictly increasing, and $f(3) = 6$ and $f(6) = 9$, we have $6 < f(4) < 9$ (because $3 < 4 < 6$), so $f(4) = 7$ or $8$. Similarly, for $y = 5$, $f(5)$ satisfies $f(4) < f(5) < f(6) = 9$. By ensuring sequence consistency, we get $f(4) = 7$ and $f(5) = 8$, and thus $f(7) = 12$ (because $f(f(4)) = f(7) = 12$).
• Similarly, derive more values: $f(8) = 15$, $f(9) = 18$, etc.

By gradually constructing the sequence, we get the function value table as follows (partial key values):

For $x = 13$:

• Since $f$ is strictly increasing, and $f(12) = 21$ and $f(15) = 24$ (because when $y = 5$, $f(f(5)) = f(8) = 15$, and thus when $y = 8$, $f(f(8)) = f(15) = 24$), we have $21 < f(13) < 24$, so $f(13) = 22$ or $23$.
• If $f(13) = 22$, then $f(14) = 23$ (because $f(13) < f(14) < f(15) = 24$), and $f(f(13)) = f(22) = 39$ (because when $y = 13$, $f(f(13)) = 3 \times 13 = 39$).
• If $f(13) = 23$, then $f(14)$ cannot be an integer (because $f(13) = 23 < f(14) < 24$, no integer solution), contradiction.
• Therefore, the only possibility is $f(13) = 22$.

In the story, after the old man puts 13 coins into the bag and takes out coins, the number of coins in his hand is $f(13) = 22$. So, guess that he has 22 coins in his hand.

Problem: Magic Bag#

Problem Description#

There is a magic bag. When you put coins into the bag, the bag will cast magic, causing the number of coins taken out to change. The magic of the bag satisfies the following rules:

1. The function $f(x)$ implemented by the bag is strictly increasing: if $x_1 < x_2$, then $f(x_1) < f(x_2)$.
2. The magic of the bag has a composite effect: $f(f(x)) = 3x$ (i.e., after putting $x$ coins in and taking out $f(x)$ coins, putting $f(x)$ coins in again will take out $3x$ coins).
3. It is known that $f(1) = 2$.

Given integer $n$ ($1 \leq n \leq 10^6$), find the value of $f(n)$.

Input Format#

One integer $n$.

Output Format#

One integer, representing the value of $f(n)$.

Example#

Input:
13
Output:
22

Data Range#

• For 30% of the data, $n \leq 1000$
• For 100% of the data, $n \leq 10^6$

Algorithm Approach#

1. Function properties: $f(f(x)) = 3x$ and $f$ is strictly increasing, $f(1) = 2$.
2. Dynamic construction: Calculate $f(x)$ in order from $x = 1$ to $n$:
   • Case 1: If $x$ is a function value from a previous calculation (i.e., $x = f(y)$), then $f(x) = 3y$.
   • Case 2: Otherwise, assign the smallest available integer $v$ greater than the current maximum function value and greater than $x$ as $f(x)$.
3. Maintain information:
   • f[x]: Store function values.
   • origin[v]: Record which $x$ produced $v$ (i.e., $v = f(x)$).
   • used[v]: Mark whether $v$ has been used as a function value.
   • Dynamically update the current maximum function value max_f and the next available integer next_avail.

C++ Code Implementation#

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

const int maxN = 1000000;   // Maximum value of input n
const int maxV = 3000000;   // Maximum possible range of function values

int f[maxN + 10];           // f[x] stores function values
int origin[maxV + 10];      // origin[v] = x means v is produced by x
bool used[maxV + 10];       // used[v] marks whether v has been used

int main() {
    int n;
    cin >> n;

    // Initialize
    for (int i = 1; i <= maxV; i++) {
        used[i] = false;
        origin[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
        f[i] = 0;
    }

    long long max_f = 0;    // Current maximum function value
    long long next_avail = 1; // Next available integer

    for (int x = 1; x <= n; x++) {
        if (f[x] != 0) continue;  // Skip if already calculated

        if (origin[x] != 0) {
            // Case 1: x is a function value from a previous calculation
            int y = origin[x];
            long long val = 3 * (long long)y;
            f[x] = val;
            max_f = max(max_f, val);  // Update maximum function value

            if (val <= maxV) {
                used[val] = true;     // Mark this function value as used
            }
        } else {
            // Case 2: Assign new function value
            long long v0 = max(max_f + 1, (long long)x + 1);
            long long v = next_avail;
            if (v < v0) v = v0;

            // Find the first unused v
            while (v <= maxV && used[v]) {
                v++;
            }
            if (v > maxV) {
                cerr << "No valid solution!" << endl;
                return 1;
            }

            f[x] = v;
            used[v] = true;
            max_f = max(max_f, v);    // Update maximum function value
            next_avail = v + 1;       // Update next available integer
        }

        // Set origin: record that f[x] is produced by x
        if (f[x] <= maxV) {
            origin[f[x]] = x;
        }
    }

    cout << f[n] << endl;
    return 0;
}

Complexity Analysis#

• Time complexity: $O(n + \text{maxV})$, where $\text{maxV} = 3 \times 10^6$. Loop $n$ times, each operation is constant or linear time (in the free allocation branch, searching for available integers, but amortized cost is low).
• Space complexity: $O(\text{maxV})$, mainly determined by the origin and used arrays.

Problem Extension Directions#

1. Larger data range: When $n \leq 10^7$, need to optimize the search efficiency of the used array (can use balanced tree or skip list).
2. Query multiple values: Query $f(n_i)$ multiple times, can preprocess the entire function table.
3. Inverse function solving: Given $k$, find the solution $x$ of $f(x) = k$.
4. Function property proof: Require proof of uniqueness or specific mathematical properties of $f$.

This problem cleverly transforms the logic in the story into a function construction problem, combining mathematical reasoning and algorithm design, suitable as a programming competition or algorithm training problem.

V1#

Directly give a certain form, output a certain formatted form according to the format (sign-in problem)

V2#

If coins can be torn (i.e., can be described as a non-integer geometric sequence), touching coins twice can become three times the original, then given a number of coins, output the number of coins (sign-in problem)

V3#

This is the current problem. If coins can only appear as integers, touching coins twice can become three times the original, then given an integer, how many coins will it become after touching once?

V4#

Extensions#

1. Mathematical Properties and Closed-Form Formula#

Extension direction: Explore the mathematical properties of function $f$, derive closed-form expression Core properties:

• $f(f(x)) = 3x$
• $f(3^k \cdot x) = 3^k \cdot f(x)$ (numbers divisible by 3 can be recursively decomposed)
• Function values are linearly grouped within the interval $[3^k, 3^{k+1})$

Derivation process:

1. Orbit decomposition: Each orbit consists of a seed number $s$ (not divisible by 3) and a sequence:

   $$s \rightarrow f(s) \rightarrow 3s \rightarrow f(3s) \rightarrow 9s \rightarrow \cdots$$

   For example, the orbit of seed number 1: $1 \rightarrow 2 \rightarrow 3 \rightarrow 6 \rightarrow 9 \rightarrow 18 \rightarrow \cdots$

2. Function value distribution:

   • Within interval $[3^k, 2\cdot3^k)$: $f(x) = x + 3^k$
   • Within interval $[2\cdot3^k, 3^{k+1})$: $f(x) = 3x - 3^{k+1}$

3. Closed-form expression:

   $$f(x) = \begin{cases} x + 3^k & \text{if } 3^k \leq x < 2\cdot3^k \\ 3x - 3^{k+1} & \text{if } 2\cdot3^k \leq x < 3^{k+1} \end{cases}$$

   where $k = \lfloor \log_3 x \rfloor$

Example verification:

• $x=4$: $k=1$, $3^1 \leq 4 < 6$ → $f(4)=4+3=7$
• $x=7$: $k=1$, $6 \leq 7 < 9$ → $f(7)=3\times7-9=12$

2. Efficient Algorithm Optimization#

Extension direction: Handle large-scale data with $n \leq 10^7$ Optimization strategies:

1. Grouped calculation: Directly calculate function values using interval characteristics

   int f(int x) {
       if (x == 1) return 2;
       int k = 0, temp = x;
       while (temp % 3 == 0) { k++; temp /= 3; }
       if (k > 0) return pow(3, k) * f(temp);
       
       int exp = 0;
       while (pow(3, exp + 1) <= x) exp++;
       int lower = pow(3, exp);
       
       if (x < 2 * lower) return x + lower;
       return 3 * x - 3 * pow(3, exp + 1);
   }

   cpp

2. Preprocessing technique:

   • Block preprocessing: Every $10^6$ as a group, store interval boundary values
   • Binary search: Quickly locate the interval where $x$ is located

Complexity:

• Time complexity: $O(1)$ per query (after preprocessing)
• Space complexity: $O(\log n)$

3. Inverse Function Solving#

Extension direction: Given $k$, find the solution $x$ of $f(x) = k$ Mathematical derivation:

1. Determine the interval $[3^m, 3^{m+1})$ where $k$ is located
2. Solve in two cases: $$x = \begin{cases} k - 3^m & \text{if } 3^m \leq k < 2\cdot3^m \\ \frac{k + 3^{m+1}}{3} & \text{if } 2\cdot3^m \leq k < 3^{m+1} \end{cases}$$

Algorithm implementation:

int inverse_f(int k) {
    if (k == 2) return 1;
    int m = 0;
    while (pow(3, m + 1) <= k) m++;
    int lower = pow(3, m);
    
    if (k < 2 * lower) return k - lower;
    if ((k + pow(3, m + 1)) % 3 != 0) return -1; // No solution
    return (k + pow(3, m + 1)) / 3;
}

4. Dynamic Version Maintenance#

Extension direction: Support dynamic function modification and queries Operation types:

1. Modification operation: Update function value at a specific point
2. Query operation: Find $f^{(k)}(x)$ ($k$-th composite function)
3. Range query: Find $\sum_{i=l}^r f(i)$

Data structures:

• Segment tree + lazy propagation: Maintain function value range sum
• Skip list index: Accelerate composite function queries
• Version control: Use persistent data structures to support historical queries

Core operations:

class DynamicF {
    struct Node {
        int l, r, sum, lazy;
    } tree[4 * MAXN];
    
    void build(int id, int l, int r) {
        // Initialize segment tree
    }
    
    void update(int id, int pos, int val) {
        // Update single point function value
    }
    
    int query_composite(int x, int k) {
        // Query f^{(k)}(x)
        while (k > 0) {
            x = f(x); // Optimize using skip list
            k--;
        }
        return x;
    }
};

5. Multi-dimensional Extension#

Extension direction: Extend the problem to higher dimensions Problem definition:

• Define $d$-dimensional function $f: \mathbb{Z}^d \to \mathbb{Z}^d$
• Satisfies $f^{(m)}(x_1, \dots, x_d) = (a_1x_1, \dots, a_dx_d)$
• Constraint: $\prod_{i=1}^d a_i = 3^m$

Solution approach:

1. Coordinate separation: Solve each dimension independently
2. Tensor decomposition: Handle correlations between dimensions $$f(\mathbf{x}) = \bigotimes_{i=1}^d f_i(x_i)$$
3. Grid optimization: Use KD-Tree to accelerate multi-dimensional queries

6. Function Property Proof#

Extension direction: Prove existence and uniqueness of the function Proof framework:

1. Existence proof (constructive):

   • Base function: $f(1)=2$
   • Recursive definition: $$f(x) = \begin{cases} 3f(x/3) & x \equiv 0 \pmod{3} \\ \text{smallest available integer} & \text{otherwise} \end{cases}$$

2. Uniqueness proof:

   • Assume there exist $f_1 \neq f_2$ satisfying the conditions
   • Take the smallest difference point $x_0 = \min\{x \mid f_1(x) \neq f_2(x)\}$
   • Derive contradiction: $f_1(f_1(x_0)) = 3x_0 = f_2(f_2(x_0))$

3. Monotonicity proof:

   • Prove by induction $\forall x_1 < x_2, f(x_1) < f(x_2)$
   • Key step: Compare function values when crossing intervals

Extended Problem Example (ACM Style)#

Problem: The Mystery of the Magic Bag#

Problem Description
The king’s magic bag satisfies strange rules: after putting $x$ coins into the bag, you get $f(x)$ coins. Function $f$ satisfies:

1. $f(f(x)) = 3x$
2. $f(3x) = 3f(x)$
3. $f$ is strictly increasing
4. $f(1) = 2$

You need to handle three types of operations:

1. 1 x: Query the value of $f(x)$
2. 2 x k: Query $f^{(k)}(x)$ ($k$-th composite function)
3. 3 l r: Calculate $\sum_{i=l}^r f(i)$

Input Format
The first line contains the number of operations $q$ and maximum range $n$
The next $q$ lines, each line is an operation

Output Format
Output the result for each query

Data Range

• $1 \leq q \leq 10^5$
• $1 \leq n \leq 10^7$
• $1 \leq x, k, l, r \leq n$

Sample Input

3 20
1 13
2 4 2
3 1 5

Sample Output

22
12
36

Sample Explanation:

• $f(13) = 22$
• $f^{(2)}(4) = f(f(4)) = f(7) = 12$
• $\sum_{i=1}^5 f(i) = f(1)+\dots+f(5) = 2+3+6+7+8=26$

Summary#

Through extensions from multiple dimensions such as mathematical properties, algorithm optimization, dynamic maintenance, and multi-dimensional extension, the original interesting story can be developed into an algorithm competition problem with depth. Key extension points include:

1. Mathematical essence: Reveal the closed-form expression and orbit decomposition characteristics of the function
2. Efficient algorithms: Use interval characteristics to achieve $O(1)$ queries
3. Dynamic version: Support real-time modification and complex queries
4. Multi-dimensional extension: Generalize the problem to high-dimensional space
5. Rigorous proof: Train mathematical reasoning ability

These extensions both maintain the interest of the original problem and increase the depth and challenge of the algorithm, suitable as advanced algorithm competition problems.