Matchstick Equation Problem

Problem Description

Given n matchsticks, determine how many equations of the form "A+B=C" can be formed. A, B, and C are integers constructed using matchsticks (if non-zero, the highest digit cannot be 0). The matchstick requirements for digits 0-9 are provided in a reference diagram.

Constraints:

• The plus and equals signs each require two matchsticks
• If A ≠ B, then A+B=C and B+A=C are considered different equations (A, B, C ≥ 0)
• All n matchsticks must be used

Algorithm Analysis

The total number of matchsticks available for digits is n-4, accounting for the two operators. Since forming a number requires at least two matchsticks, neither addend can exceed 1111.

Implementation Steps

1. Input Processing

int total_sticks;
cin >> total_sticks;
int digit_sticks = total_sticks - 4;

2. Matchstick Count Initialization

int stick_count[2223] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};

3. Calculate Matchstick Requirements for Larger Numbers

for(int num = 10; num <= 2222; num++) {
    stick_count[num] = stick_count[num / 10] + stick_count[num % 10];
}

4. Equation Counting

Use nested loops to iterate through all possible pairs of addends (0 to 1111), calculate the required matchsticks for the sum, and verify if the total matches the available count.

Broken Necklace Problem

Problem Description

Given a necklace with N beads colored red (r), blue (b), or white (w), find the optimal breaking point that maximizes the number of consecutive beads collceted from both ends. White beads can be considered as either red or blue during collection.

Algorithm Analysis

• Handle circular necklace structure
• Special handling for white beads
• Find optimal breaking point for maximum consecutive beads

Implementation Steps

1. Data Structure Setup

char beads[750];
int necklace_length;
cin >> necklace_length;
for(int i = 0; i < necklace_length; i++) {
    cin >> beads[i];
    beads[i + necklace_length] = beads[i]; // Duplicate for circular processing
}

2. Main Algorithm

int max_beads = 0;
for(int break_point = 0; break_point < necklace_length; break_point++) {
    int left_count = 0, right_count = 0;
    char left_color = ' ', right_color = ' ';
    
    // Count left side
    for(int i = break_point; i >= 0; i--) {
        if(beads[i] == 'w') left_count++;
        else {
            if(left_color == ' ') left_color = beads[i];
            if(beads[i] == left_color || beads[i] == 'w') left_count++;
            else break;
        }
    }
    
    // Count right side
    for(int i = break_point + 1; i < break_point + necklace_length; i++) {
        if(beads[i] == 'w') right_count++;
        else {
            if(right_color == ' ') right_color = beads[i];
            if(beads[i] == right_color || beads[i] == 'w') right_count++;
            else break;
        }
    }
    
    max_beads = max(max_beads, left_count + right_count);
}

3. Edge Case Handling

Check for all-white necklaces and ensure the solution doesn't exceed the necklace length.