Contest Overview

Total Score: 290
Ranking: 3
Problems Solved: 2

Problem Solutions

Problem 1: Interview Screening

Score: 100

**Analysis:**This is a relatively straightforward problem that resembles bucket sorting. The key is to properly handle the conditional checks without mixing them up.

Implementation:

// Store possible outcomes
string results[4] = {
    "hired",
    "failed",
    "offer",
    "special offer"
};

// Process input
for(int i = 0; i < input.size(); i++) {
    if(input[i] == 'A') qualifications[1]++;
    else if(input[i] == 'B') qualifications[2]++;
    else if(input[i] == 'C') qualifications[3]++;
    else qualifications[4]++; 
}

if(qualifications[4] != 0 || qualifications[3] >= 2) 
    cout << results[1] << endl;
else if(qualifications[4] == 0 && qualifications[1] >= 3) 
    cout << results[3] << endl;
else 
    cout << results[2] << endl;

**Potential Pitfalls:**1. Remember to reset the counter variables between test cases 2. Ensure conditional statements are properly ordered

Problem 2: Excel Column Labeling

Score: 50

**Analysis:**This problem is essentially a base-26 conversion with special handling for the letter 'Z'. Similar to the previous problem, storing results in an array simplifies the implementation.

Implementation:

// Store mapping, with Z at position 0
char mapping[27] = {
    'Z', 'A', 'B', 'C', 'D', 'E', 'F', 'G',
    'H', 'I', 'J', 'K', 'L', 'M', 'N',
    'O', 'P', 'Q', 'R', 'S', 'T',
    'U', 'V', 'W', 'X', 'Y'
};

// Conversion process
string result;
while(number > 0) {
    result = mapping[number % 26] + result;
    int temp = number;
    number /= 26;
    if(temp % 26 == 0) 
        number--; // Handle multiples of 26
}

**Potential Pitfalls:**1. Special handling for 'Z' character 2. Condition for multiples of 26 3. Building the result string in reverse order

Problem 3: Card Game Elimination

Score: 100

**Analysis:**This is a greedy algorithm problem. To maximize the number of players remaining, each player should take cards from the player with the fewest remaining cards. After input, we sort the array. The game continues as long as each player i has fewer cards than the number of players remaining (n-i).

Implementation:

sort(cards+1, cards+n+1);
for(int i = 1; i <= n; i++) {
    if(cards[i] >= n-i) { // If this player has enough cards that even if others take all their cards, they won't be eliminated
        printf("%d", n-i+1);
        return 0;
    }
}

**Potential Pitfalls:**The main pitfall is forgetting to sort the array before processing.

Problem 4: Salary Increase Calculation

Score: 40

**Analysis:**This problem combines greedy algorithms with fast exponentiation. To maximize total salary after m years, higher-paid employees should receive the higher multipleirs (x and y). Employees beyond x+y only receive salary in the first year if m=1.

Implementation:

// Fast exponentiation
long long power(long long base, long long exponent) {
    if(exponent == 0) return 1;
    if(exponent % 2 == 1) 
        return (base * power(base, exponent-1)) % MOD;
    else 
        return (power(base, exponent/2) * power(base, exponent/2)) % MOD;
}

// Sort in descending order
sort(salaries+1, salaries+n+1);
reverse(salaries+1, salaries+n+1);

// Calculate total salary
for(int i = 1; i <= x; i++) { // x employees get 3x multiplier
    total = (total + power(3, m) * salaries[i]) % MOD;
}
for(int i = x+1; i <= x+y; i++) { // y employees get 2x multiplier
    total = (total + power(2, m) * salaries[i]) % MOD;
}
if(m == 1)
    for(int i = x+y+1; i <= n; i++) { // Only paid in first year
        total = (total + salaries[i]) % MOD;
    }

**Potential Pitfalls:**The only pitfall is remembering that remaining employees only get paid in the first year.

Problem 5: Rich Number Sequence

Score: 0

**Analysis:**This problem requires dynamic programming with a map to handle frequency counting. Given the large possible values (up to 10^9), we need to use a map to track occurrences of each number. The state transition is: dp[i][j] = dp[i-1][j] + dp[i-1][j-1] * frequency[a[i]].

Implementation:

for(int i = 1; i <= sequence_length; i++) {
    for(int j = 1; j <= i; j++) {
        dp[i][j] = dp[i-1][j] + dp[i-1][j-1] * number_frequency[sequence[i]];
    }
}

**Potential Pitfalls:**The challenge lies in correctly counting the number of rich numbers in the sequence.