4993. FEB - AcWing题库

大佬亲笔                

将原串分成三段：

FFF|E.....B|FFF

先合并中间段，再合并两边的段

#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <queue>

#define endl '\n'
#define x first
#define y second

using namespace std;

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

int n;
string s;

void solve()
{
	cin >> n >> s;
	if (s == string(n,'F')) // 是否全为F
	{
		cout << n << endl;
		for (int i = 0;i < n;i ++)
		cout << i << endl;
	}
	
	else
	{
		int l = 0,r = n - 1;
		while(s[l] == 'F') l ++;  //分第一段
		while(s[r] == 'F') r--;    //分第三段
		
		int low = 0,high = 0;
		string str = s;
		
		for (int i = l;i <= r;i ++)  // 中间段最小值
		{
			if (str[i] == 'F')
			{
				if (str[i - 1] == 'E') str[i] = 'B';
				else str[i] = 'E';
			}
			if (i > l && str[i - 1] == str[i]) low ++;
		}
		
		str = s;
		for (int i = l;i <= r;i ++) 中间段最大值
		{
			if (str[i] == 'F')
				str[i] = str[i - 1];
			if (i > l && str[i - 1] == str[i]) high ++;
		}
		
		int d = 2;
		int len = l + n - r - 1;

		if (len) high += len,d = 1;
		
		cout << (high - low) / d + 1 << endl;
		for (int i = low;i <= high;i += d)
		cout << i << endl;
	} 
}

int main()
{
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int _ = 1;
	while(_--) solve();
	return 0;
}