Problem A - Dice Magic (NiuKe Beginner Monthly Contest 98)

Check if a target number exists within a given array.

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

long long n, x;
long long arr[505];

void solve() {
    for (int i = 1; i <= n; ++i) {
        cin >> arr[i];
    }
    
    for (int i = 1; i <= n; ++i) {
        if (arr[i] == x) {
            cout << "YES" << endl;
            return;
        }
    }
    
    cout << "NO" << endl;
}

int main() {
    cin >> n >> x;
    solve();
    return 0;
}

Problem B - Minimum Remaining Count (NiuKe Beginner Monthly Contest 98)

Determine the minimal count of numbers left after pairing odd and even numbers based on their sums and products.

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

long long n;
long long nums[1000010];

void solve() {
    long long odd_count = 0, even_count = 0;
    
    for (int i = 1; i <= n; ++i) {
        if (nums[i] % 2)
            odd_count++;
        else
            even_count++;
    }
    
    if (even_count > odd_count)
        cout << even_count - odd_count << endl;
    else {
        if ((odd_count - even_count) % 2)
            cout << "1" << endl;
        else
            cout << "0" << endl;
    }
}

int main() {
    cin >> n;
    solve();
    return 0;
}

Problem C - Two Functions (NiuKe Beginner Monthly Contest 98)

Compute the result of a mathematical function involving modular arithmetic.

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

const int MOD = 998244353;
long long a, x;

void solve() {
    if (x == 1) {
        cout << (a % MOD) << endl;
        return;
    }
    
    long long term1 = (a * a) % MOD;
    long long term2 = (((x * x) - x) / 2) % MOD;
    long long result = (term1 * term2) % MOD;
    
    cout << result << endl;
}

int main() {
    int q;
    cin >> q;
    while (q--) {
        cin >> a >> x;
        solve();
    }
    return 0;
}

Problem A - Buy a Pen (AtCoder)

Determine the minimum cost among two options based on the first character of a string.

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

void solve() {
    int red, green, blue;
    cin >> red >> green >> blue;
    
    string color;
    cin >> color;
    
    char first_char = color[0];
    
    if (first_char == 'R') {
        cout << min(green, blue) << endl;
    } else if (first_char == 'B') {
        cout << min(red, green) << endl;
    } else {
        cout << min(red, blue) << endl;
    }
}

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

Problem B - Right Triangle (AtCoder)

Verify if three points form a right triangle using distance calculations.

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

void solve() {
    int x1, y1, x2, y2, x3, y3;
    cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
    
    // Check collinearity
    if ((x1 - x2) * (y2 - y3) == (x2 - x3) * (y1 - y2)) {
        cout << "No" << endl;
        return;
    }
    
    long long side1 = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
    long long side2 = (x2 - x3) * (x2 - x3) + (y2 - y3) * (y2 - y3);
    long long side3 = (x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3);
    
    if (side1 + side2 == side3 || side2 + side3 == side1 || side1 + side3 == side2) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
}

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

Problem C - Sum Equals Zero (AtCoder)

Determine if it's possible to select values from intervals such that their total sum equals zero.

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

const int MAX_N = 200010;
int left_bounds[MAX_N];
int right_bounds[MAX_N];
int result[MAX_N];

void solve() {
    int num_intervals;
    cin >> num_intervals;
    
    long long total_left = 0;
    long long total_right = 0;
    
    for (int i = 1; i <= num_intervals; ++i) {
        cin >> left_bounds[i] >> right_bounds[i];
        total_left += left_bounds[i];
        total_right += right_bounds[i];
    }
    
    if (total_left <= 0 && total_right >= 0) {
        cout << "Yes" << endl;
        long long current_sum = 0;
        
        for (int i = 1; i <= num_intervals; ++i) {
            result[i] = left_bounds[i];
            current_sum += left_bounds[i];
        }
        
        for (int i = 1; i <= num_intervals; ++i) {
            long long diff = right_bounds[i] - left_bounds[i];
            if (current_sum + diff < 0) {
                current_sum += diff;
                result[i] = right_bounds[i];
            } else {
                result[i] += (0 - current_sum);
                current_sum = 0;
                break;
            }
        }
        
        for (int i = 1; i <= num_intervals; ++i) {
            cout << result[i] << " ";
        }
        cout << endl;
    } else {
        cout << "No" << endl;
    }
}

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