题意比较麻烦略
题解:枚举前缀的中点,二分最远能扩展的地方,lcp来check,然后线段树维护每个点最远被覆盖的地方,然后查询线段树即可
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000009
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
using namespace std;
const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=1000000+10,maxn=1000000+10,inf=0x3f3f3f3f;
char s[N];
int sa[N], t[N], t2[N], c[N], rk[N], height[N];
void buildSa(int n, int m) {
int i, j = 0, k = 0, *x = t, *y = t2;
for(i = 0; i < m; i++) c[i] = 0;
for(i = 0; i < n; i++) c[x[i] = s[i]]++;
for(i = 1; i < m; i++) c[i] += c[i - 1];
for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
for(int k = 1; k < n; k <<= 1) {
int p = 0;
for(i = n - k; i < n; i++) y[p++] = i;
for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
for(i = 0; i < m; i++) c[i] = 0;
for(i = 0; i < n; i++) c[x[y[i]]]++;
for(i = 1; i < m; i++) c[i] += c[i - 1];
for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
swap(x, y);
p = 1; x[sa[0]] = 0;
for(int i = 1; i < n; i++) {
if(y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])
x[sa[i]] = p - 1;
else x[sa[i]] = p++;
}
if(p >= n) break;
m = p;
}
for(i = 1; i < n; i++) rk[sa[i]] = i;
for(i = 0; i < n - 1; i++) {
if(k) k--;
j = sa[rk[i] - 1];
while(s[i + k] == s[j + k]) k++;
height[rk[i]] = k;
}
}
int Log[N];
struct ST {
int dp[N][20],ty;
ST()
{
for(int i = -(Log[0]=-1); i < N; i++)
Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
}
void build(int n, int b[], int _ty) {
ty = _ty;
for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i];
for(int j = 1; j <= Log[n]; j++)
for(int i = 1; i+(1<<j)-1 <= n; i++)
dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
}
int query(int x, int y) {
int k = Log[y - x + 1];
return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);
}
}st;
int lcp(int x,int y)//from 0 begin
{
x=rk[x],y=rk[y];
if(x>y)swap(x,y);x++;
return st.query(x,y);
}
int n;
struct sgt{
int ma[N<<2],lazy[N<<2];
sgt(){memset(ma,-1,sizeof ma);memset(lazy,-1,sizeof lazy);}
void pushdown(int rt)
{
if(~lazy[rt])
{
ma[rt<<1]=max(ma[rt<<1],lazy[rt]);
ma[rt<<1|1]=max(ma[rt<<1|1],lazy[rt]);
lazy[rt<<1]=max(lazy[rt<<1],lazy[rt]);
lazy[rt<<1|1]=max(lazy[rt<<1|1],lazy[rt]);
lazy[rt]=-1;
}
}
void update(int L,int R,int v,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
ma[rt]=max(ma[rt],v);
lazy[rt]=max(lazy[rt],v);
return ;
}
pushdown(rt);
int m=(l+r)>>1;
if(L<=m)update(L,R,v,ls);
if(m<R)update(L,R,v,rs);
ma[rt]=max(ma[rt<<1],ma[rt<<1|1]);
}
int query(int pos,int l,int r,int rt)
{
if(l==r)
{
if(ma[rt]==-1)return -1;
else return (ma[rt]-l)*2+1;
}
pushdown(rt);
int m=(l+r)>>1;
if(pos<=m)return query(pos,ls);
else return query(pos,rs);
}
}sg;
int main()
{
scanf("%d%s",&n,s);
buildSa(n+1,256);
st.build(n,height,-1);
for(int i=0;i<n/2;i++)
{
if(s[i]!=s[n-i-1])continue;
int l=0,r=i+1;
while(l<r-1)
{
int m=(l+r)>>1;
if(lcp(i-m,n-i-1-m)>=2*m+1)l=m;
else r=m;
}
// if(i==6)printf("%d \n",lcp(5,11));
sg.update(i-l,i,i,0,n-1,1);
// printf("%d %d\n",i-l,i);
}
for(int i=0;i<(n+1)/2;i++)printf("%d ",sg.query(i,0,n-1,1));puts("");
return 0;
}
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