传送门

题意：

给你一个字符串$s$，让你求$s$中出现次数不为$1$的子串出现次数乘上该字串长度最大值。

$|s|\le 1e6$

思路：

没学明白，不多$bb$，把树建出来跑一下就行了。

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=2000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int last=1,tot=1,len;
int cnt[N];
char s[N];
struct Node {int ch[26];int len,fa;
}sam[N*2];
vector<int>v[N];void add(int c) {int p=last,np=last=++tot; cnt[tot]=1;sam[np].len=sam[p].len+1;while(p&&!sam[p].ch[c]) {sam[p].ch[c]=np;p=sam[p].fa;} if(!p) sam[np].fa=1;else {int q=sam[p].ch[c];if(sam[q].len==sam[p].len+1) sam[np].fa=q;else {int nq=++tot; sam[nq]=sam[q];sam[nq].len=sam[p].len+1;sam[q].fa=sam[np].fa=nq;while(p&&sam[p].ch[c]==q) {sam[p].ch[c]=nq;p=sam[p].fa;}}}
}LL ans=0;
void dfs(int u) {for(auto x:v[u]) dfs(x),cnt[u]+=cnt[x];if(cnt[u]!=1) ans=max(ans,1ll*cnt[u]*sam[u].len);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);scanf("%s",s+1);int len=strlen(s+1);for(int i=1;i<=len;i++) add(s[i]-'a');for(int i=2;i<=tot;i++) v[sam[i].fa].pb(i);dfs(1);cout<<ans<<endl;return 0;
}
/**/