B. Reverse Binary Strings

You are given a string $s$ of even length $n$. String $s$ is binary, in other words, consists only of 0's and 1's.

String $s$ has exactly $\frac{n}{2}$ zeroes and $\frac{n}{2}$ ones ($n$ is even).

In one operation you can reverse any substring of $s$. A substring of a string is a contiguous subsequence of that string.

What is the minimum number of operations you need to make string $s$ alternating? A string is alternating if $s_i \neq s_{i + 1}$ for all $i$. There are two types of alternating strings in general: 01010101... or 10101010...

思路：

可以注意到，每次交换必定会使某对10成功匹配，因此答案为连续相邻为1的对数和连续相邻为0的对数取最大值。

通过样例也能猜出来（bushi）

#include <bits/stdc++.h>
#define fi first
#define se second
#define mm(a,b) memset(a,b,sizeof(a))
#define PI acos(-1)
#define int long long
#define no cout<<"NO\n";
#define yes cout<<"YES\n";
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e6+10;
const int mod=998244353;
int n,m,a[N],b[N];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)
    {
        string s;
        cin>>n;
        cin>>s;
        int ans=0;
        int num1=0;
        for(int i=0;i<s.size()-1;i++)
        {
            if(s[i]==s[i+1]&&s[i]=='1')
            {
                num1++;
            }
        }
        ans=max(ans,num1);
        int num2=0;
        for(int i=0;i<s.size()-1;i++)
        {
            if(s[i]==s[i+1]&&s[i]=='0')
            {
                num2++;
            }
        }
        ans=max(ans,num2);
        cout<<ans<<"\n";
    }
    return 0;
}