给定长度为n（n为偶数）的串s，每次操作：使s[i] = ‘0‘ / ‘1‘。 s为好串：s分解为若干最长连续全1或全0子串后，每一子串长度都是偶数。求使s变成好串的最少操作次数和在最少操作下最少段数

题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e9 + 5, maxm = 4e4 + 5, mod = 1e9 + 7, N = 1e6;
int a[maxn], b[maxn];
int n, m;
string s;
// int pre[maxn][2], suf[maxn];
// vector<int> G[maxn];
// int fa[maxn], dep[maxn];
// bool tag[maxn];

int qpow(int a, int b){
    int res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

int __lcm(int a, int b){
    int g = __gcd(a, b);
    return a * b / g;
}

void solve(){
    int res = 0;
    cin >> n;
    cin >> s;
    s = " " + s;
    int cnt = 1;
    char last = '#';
    for(int i = 1; i <= n; i += 2){
        if(s[i] != s[i + 1]){
            res++;
        }
        else{
            if(last != '#' && s[i] != last){
                cnt++;
            }
            last = s[i];
        }
    }
    cout << res << ' ' << cnt << '\n';
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    // fac[0] = 1;
    // for(int i = 1; i <= N; i++){
    //     fac[i] = fac[i - 1] * i % mod;
    // }
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}