题意:给一个数组,每次会删去连续重复两次的左侧部分及前面,有多个重复部分找长度最小和最靠左的部分,重复的数字最多10次
题解:根据重复数字只有10次,我们离散化后,以每两个相同数字作为起点能确定这重复的部分,一共10*n对,接下来问题就是判断这两部分是不是相同的,建sa用st表求lcp即可,然后对所有可能的对按题意sort,往后删除即可
//#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 1000000007
#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-7;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=100000+10,inf=0x3f3f3f3f;
int a[N],b[N],cnt;
vi v[N];
int 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;
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 n;
int lcp(int x,int y)
{
if(x==y)return n-x+1;
x=rk[x],y=rk[y];
if(x>y)swap(x,y);x++;
return st.query(x,y);
}
struct info{
int l,r,len;
bool operator <(const info&rhs)const{
if(len!=rhs.len)return len<rhs.len;
return l<rhs.l;
}
}p[N*20];
int main()
{
for(int i = -(Log[0]=-1); i < N; i++)
Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
scanf("%d",&n);
for(int i=0;i<n;i++)scanf("%d",&a[i]),b[cnt++]=a[i];
sort(b,b+cnt);cnt=unique(b,b+cnt)-b;
for(int i=0;i<n;i++)
{
s[i]=lower_bound(b,b+cnt,a[i])-b+1;
v[s[i]].pb(i);
}
buildSa(n+1,n+5);
st.build(n,height,-1);
int res=0;
for(int i=1;i<N;i++)
{
for(int j=0;j<v[i].size();j++)
{
for(int k=j+1;k<v[i].size();k++)
{
int x=v[i][j],y=v[i][k];
if(n-y<y-x)continue;
if(y-x<=lcp(x,y))p[++res]={x,y,y-x};
}
}
}
sort(p+1,p+1+res);
int ans=0;
for(int i=1;i<=res;i++)
{
// printf("%d %d
",p[i].l,p[i].r);
if(p[i].l>=ans)ans=p[i].r;
}
printf("%d
",n-ans);
for(int i=ans;i<n;i++)printf("%d ",a[i]);puts("");
return 0;
}
/********************
********************/