Problem A

A slightly trickier warm-up problem.

#include <bits/stdc++.h>
using namespace std;
int main() {
    string str;
    cin >> str;
    unordered_map<char, int> charCount;
    for (char c : str) {
        charCount[c]++;
    }
    for (int idx = 0; idx < str.size(); idx++) {
        if (charCount[str[idx]] == 1) {
            cout << idx + 1 << endl;
            return 0;
        }
    }
    return 0;
}

Problem B

Still a straightforward warm-up problem.

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;
    vector<int> pos(105);
    for (int i = 0; i < n; i++) {
        int val;
        cin >> val;
        pos[val - 1] = i;
    }
    int q_num;
    cin >> q_num;
    while (q_num--) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        cout << (pos[a] < pos[b] ? a + 1 : b + 1) << endl;
    }
    return 0;
}

Problem C

A brute-force approach will fail, so we need optimization. Use an array to track character mappings, leveraging the fact that there are only 26 lowercase letters.

#include <bits/stdc++.h>
using namespace std;
int main() {
    vector<int> originalToCurrent(26);
    for (int i = 0; i < 26; i++) {
        originalToCurrent[i] = i;
    }
    int n;
    cin >> n;
    string s;
    cin >> s;
    int q_num;
    cin >> q_num;
    while (q_num--) {
        char from, to;
        cin >> from >> to;
        int target = from - 'a';
        int replacement = to - 'a';
        for (int i = 0; i < 26; i++) {
            if (originalToCurrent[i] == target) {
                originalToCurrent[i] = replacement;
            }
        }
    }
    for (char c : s) {
        cout << (char)('a' + originalToCurrent[c - 'a']);
    }
    cout << endl;
    return 0;
}

Problem D

A number theory problem. By the Fundamental Theorem of Arithmetic, all exponents of prime factors in a perfect square are even. For each number, we repeatedly divide out squares of primes until no more divsiion is possible. We also need to handle 0 specially, since multiplying it by any number results in a perfect square.

#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200005;
int main() {
    int n;
    cin >> n;
    vector<long long> nums(MAX_N);
    unordered_map<long long, int> freq;
    long long ans = 0;
    int zeroCount = 0;
    for (int i = 1; i <= n; i++) {
        long long x;
        cin >> x;
        if (x == 0) {
            ans += i - 1 - zeroCount;
            zeroCount++;
            continue;
        }
        for (long long j = 2; j * j <= x; j++) {
            while (x % (j * j) == 0) {
                x /= (j * j);
            }
        }
        ans += freq[x];
        freq[x]++;
    }
    ans += zeroCount * (zeroCount - 1) / 2;
    cout << ans << endl;
    return 0;
}