AtCoder Regular Contest 108 B题11次RE的原因找到了！

AtCoder Regular Contest 108 B题11次RE的原因找到了！

这题思路没什么好说的。对于\(ffoxox\)、\(fofoxx\)、\(ffoxofoxx\)这三种情况在遍历字符串的时候特判一下就行了。前两个用欧冠递归，最后一个打的暴力，过不过都是靠运气

然而每次都是有一个点RE

我这就很不服气了啊。

怎么能对我这种21岁的老年人搞偷袭呢。

比赛结束后换成了char数组。啪一下就过了。

突然感觉可能string也是对的。错就在下面

for (ll i=0; i<s.size()-2; i++)​

真是吐了...可能std::string.size()这种操作的时间复杂度是O(N)，而并非第一感觉的O(1)。

吸取教训了。

诶...

AC代码：

/*
Seven times have I despised my soul:
——Kahlil Gibran
The first time when I saw her being meek that she might attain height.
The second time when I saw her limping before the crippled.
The third time when she was given to choose between the hard and the easy, and she chose the easy.
The fourth time when she committed a wrong, and comforted herself that others also commit wrong.
The fifth time when she forbore for weakness, and attributed her patience to strength.
The sixth time when she despised the ugliness of a face, and knew not that it was one of her own masks.
And the seventh time when she sang a song of praise, and deemed it a virtue.
*/

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
typedef long long ll;
ll n,ans,a,b,c;
string s,tt="fox";
void dg(ll i,ll j)
{
	if (i<0||j>=n) return ;
	bool pd1=0,pd2=0;
	while(i-1>=0&&s[i-1]=='0') i--,pd1=1;
	while(j+1<n&&s[j+1]=='0') j++,pd2=1;


//	cout<<i<<' '<<j<<endl;
	if (i<0||j>=n) return ;
	if (i-2>=0&&j+1<n&&s[i-2]=='f'&&s[i-1]=='o'&&s[j+1]=='x')
	{
		s[i-2]='0';
		s[i-1]='0';
		s[j+1]='0';
		ans++;
		dg(i-2,j+1);
	}
	else if (i-1>=0&&j+2<n&&s[i-1]=='f'&&s[j+1]=='o'&&s[j+2]=='x')
	{
		s[i-1]='0';
		s[j+1]='0';
		s[j+2]='0';
		ans++;
		dg(i-1,j+2);
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>s;
	for (ll i=0; i<n-2; i++)
	{
		if (s[i]=='f'&&s[i+1]=='o'&&s[i+2]=='x')
		{
			for (ll j=i; j<i+3; j++)
				s[j]='0';
			ans++;
			dg(i,i+2);
		}

	}
	ll cnt=0;
	do
	{
		cnt=0;
		for (ll i=0; i<n-2; i++)
		{
			if (s[i]=='f')
			{
				ll j=i+1,k;
				while(s[j]=='0') j++;
				if (s[j]=='o')
				{
					k=j+1;
					while(s[k]=='0') k++;
					if (s[k]=='x') {
						cnt++,ans++;
						s[i]='0';
						s[j]='0';
						s[k]='0';
					}
				}
			}
		}
	}while(cnt);
		cout<<n-3*ans;//<<endl<<s;
	return 0;

}