1898A - Milica and String - CodeForces Solution

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

#define task ""
#define ll long long
#define T int
#define ii pair <T, T>
#define F first
#define S second
#define pb push_back
#define sz(x) (int) (x).size()
#define all(a) a.begin(), a.end()
#define init(f, x) memset(f, (x), sizeof(f))
#define rep(i, n) for (int i = 1; i <= (n); i++)
#define foru(i, a, b) for (int i = (a); i <= (b); i++)
#define ford(i, a, b) for (int i = (a); i >= (b); i--)

const int inf = (int) 1e9;
const ll INF = (ll) 1e18;
const int nmax = 24;
const int N = 24;
const int mod = 24;

void solve() {
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        string s;

        cin >> n >> k;
        cin >> s;

        s = '*' + s;

        int d = 0;
        for (char c : s)
            d += (c == 'B');

        if (d == k) {cout << "0\n"; continue;}

        cout << "1\n";

        if (d > k)
        {
            int cnt = 0, vt = -1;
            rep(i, n)
            {
                if (s[i] == 'B') cnt++;
                if (cnt == d - k) vt = i, i = n + 1;
            }

            cout << vt << " A\n";
        }
        else
        {
            int cnt = 0, vt = -1;
            rep(i, n)
            {
                if (s[i] == 'A') cnt++;
                if (cnt == k - d) vt = i, i = n + 1;
            }

            cout << vt << " B\n";
        }
    }

}

int main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

    if (fopen(task".inp","r"))
    {
        freopen(task".inp","r",stdin);
        freopen(task".out","w",stdout);
    }

    solve();

    return 0;
}