解法一：

显然只需让多的在限度内最多即可

#include<iostream>
#include<algorithm>
using namespace std;
#define endl '\n'
void solve() {
	int n, k, num0 = 0, num1 = 0;
	cin >> n >> k;
	string s;
	cin >> s;
	for (int i = 0; i < s.size(); i++) {
		if (s[i] == '0') num0++;
		else if (s[i] == '1') num1++;
	}
	int mx = max(num0, num1);
	int mn = min(num0, num1);
	while (mn > 0 && k > 0) {
		mx++, mn--, k--;
	}
	cout << mx - mn << endl;
}
int main() {
	int t; cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}

解法二：

#include <iostream>
#include<algorithm>
#include<string>
using namespace std;
int main()
{
    int t, n, k;
    string str;
    cin >> t;
    for (int i = 1; i <= t; i++) {
        int tong[100] = { 0 };
        cin >> n >> k >> str;
        for (int j = 0; j < n; j++)
            tong[str[j] - '0']++;
        if (tong[1] >= tong[0]) {
            if ((tong[1] + k) > n) cout << n << endl;
            else cout << tong[1] + 2 * k - tong[0] << endl;
        }
        else {
            if ((tong[0] + k) > n) cout << n << endl;
            else cout << tong[0] - tong[1] + 2 * k << endl;
        }
    }
}