BZOJ 4236: JOIOJI

记 \(cnt_{i, 0/1/2}\) 为 \(1\) 到 \(i\) 中 J O I 出现次数
当 \((l,r]\) 能成为答案时
\(cnt_{r,0}-cnt_{l,0}=cnt_{r,1}-cnt_{l,1}=cnt_{r,2}-cnt_{l,2}\)
用map记录一下二元组 \((cnt_{i,0}-cnt_{i,1},cnt_{i,1}-cnt_{i,2})\) 首次出现的位置即可

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) >> 1)
#define lowbit(i) ((i) & (-i))
#define ll long long
#define ull unsigned long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=1,head[N],to[N*2],ne[N*2];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[N*2],ne[N*2],c[N*2];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1e9 + 7;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
int gcd(int a, int b) { while (b) { a %= b; std::swap(a, b); } return a; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 2e5 + 7;
char s[N];
int cnt[N][3];
std::map< std::pii, int> mp;

int main() {
	int n;
	scanf("%d%s", &n, s + 1);
	int ans = 0;
	mp[std::pii(0, 0)] = 0;
	rep (i, 1, n + 1) {
		cnt[i][0] = cnt[i - 1][0] + (s[i] == 'J');
		cnt[i][1] = cnt[i - 1][1] + (s[i] == 'O');
		cnt[i][2] = cnt[i - 1][2] + (s[i] == 'I');
		if (mp.count(std::pii(cnt[i][0] - cnt[i][1], cnt[i][1] - cnt[i][2])))
			ans = std::max(ans, i - mp[std::pii(cnt[i][0] - cnt[i][1], cnt[i][1] - cnt[i][2])]);
		else 
			mp[std::pii(cnt[i][0] - cnt[i][1], cnt[i][1] - cnt[i][2])] = i;
	}
	printf("%d\n", ans);
	return 0;
}