一倍经验

二倍经验

读懂题意后，第一反应应该是

这个

如果反应力不够的话，那么第二反应就应该是

这个

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好了，我们姑且认为你会了，就这些

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简述做法：考虑前缀和，分别把'N''O''I'处理为-10000000000000000（或许更大或许更小），9999999999999999（或许更大或许更小），和1。由此一来我们只需找到和为0的最长区间，至于如何找前文已经给出，

脑子都不用动一下

code:

#include<bits/stdc++.h>
#define int long long
using namespace std;
inline int reads(){
	int x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(isdigit(ch)){
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
	return x*f;
}
const int N=1e5+10;
string s,ss;
int n,now;
int ans;
int intmax=INT_MAX;
unordered_map<int,int> mp;
signed main(){
//	freopen("AKNOI最长平衡子串ex.in","r",stdin);
	n=reads();
	cin>>ss;
	s=" "+ss;
//	mp[0]=-1;
	for(int i=1;i<=n;i++){
		if(s[i]=='N')now++;
		if(s[i]=='O')now+=intmax;
		if(s[i]=='I')now-=intmax+1;
		if(mp[now]||!now){
			ans=max(ans,i-mp[now]);
//			printf("%lld时update\n",i);
		}
		else mp[now]=i;
//		printf("%lld时，ans:%lld，now:%lld\n",i,ans,now);
	}
	cout<<ans;
	return 0;
}