dp写法：f[i][j]表示第i位，当前位为j，能往前找的最大的合法长度。

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'

using namespace std;

typedef pair<int, int> PII;
typedef long long ll;

const int N = 200010;

int f[N][2];

inline void solve()
{
	string s;
	cin >> s;
	int n = s.size();
	s = ' ' + s;
	
	memset(f, 0, (2 * n + 5) * sizeof (int));
	for(int i = 1; i <= n; i ++)
	{
		if(s[i] == '0')f[i][0] = f[i - 1][1] + 1;
		else if(s[i] == '1')f[i][1] = f[i - 1][0] + 1;
		else
		{
			f[i][1] = f[i - 1][0] + 1;
			f[i][0] = f[i - 1][1] + 1;
		}
	}
	ll ans = 0;
	for(int i = 1; i <= n; i ++)
	{
		if(s[i] == '0')ans += f[i][0];
		else if(s[i] == '1')ans += f[i][1];
		else ans += max(f[i][0], f[i][1]);
	}
	cout << ans << endl;
}

int main()
{
	IOS
	int _;
	cin >> _;
	while(_ --)
	{
		solve();
	}
	
	return 0;
}

 二分写法

w[i]记录第i个位置能往前找的最大合法长度，pos数组记录每个连续问号串中第一个问号的位置。

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'

using namespace std;

typedef pair<int, int> PII;
typedef long long ll;

const int N = 200010;

char s[N];
int w[N];
int l = 1;
char kep;

bool check(int u)
{
	if(u % 2 == l % 2)
	{
		if(kep == '?')
		{
			kep = s[u];
			return true;
		}
		if(kep == s[u] || s[u] == '?')return true;
		return false;
	}
	else
	{
		if(kep == '?')
		{
			if(s[u] == '1')kep = '0';
			else if(s[u] == '0')kep = '1';
			return true;
		}
		if(kep == '0' && s[u] == '1')return true;
		if(kep == '1' && s[u] == '0')return true;
		if(s[u] == '?')return true;
		return false;
	}
}

void solve()
{
	cin >> s + 1;
	int n = strlen(s + 1);
	
	vector<int> pos;
	for(int i = 1; i <= n; i ++)
	{
		if(s[i] == '?' && s[i - 1] != '?')
		{
			pos.push_back(i);
		}
	}
	pos.push_back(2e9);
	
	l = 1;
	kep = s[l];
	for(int i = 2; i <= n; i ++)
	{
		if(!check(i))
		{
			int L = 0, R = pos.size() - 1;
			int f = 0;
			if(s[i - 1] == '?')f = 1;
			
			while(L < R)
			{
				int mid = L + R + 1 >> 1;
				if(pos[mid] < i)L = mid;
				else R = mid - 1;
			}
			int tmp = i;
			if(f && pos[L] < i)tmp = pos[L];
			
			for(int j = l; j < tmp; j ++)
			{
				w[j] = i - 1;
			}
			
			l = tmp;
			kep = s[l];
			i = tmp;
		}
	}
	
	for(int i = l; i <= n; i ++)w[i] = n;
	
	ll ans = 0;
	for(int i = 1; i <= n; i ++)
	{
		ans += w[i] - i + 1;
	}
	cout << ans << endl;
}

int main()
{
	IOS
	int _;
	cin >> _;
	while(_ --)
	{
		solve();
	}
	
	return 0;
}

Problem - 1535C - Codeforces