NOIP 模拟 $27\; \rm 牛半仙的妹子序列$
题解 \(by\;zj\varphi\)
明显一道极长上升子序列的题。
直接线段树维护单调栈,最后单调栈求出可以贡献的序列,答案相加就行。
Code
#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
char buf[1<<21],*p1=buf,*p2=buf,OPUT[100];
#define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++;
template<typename T>inline void read(T &x) {
ri f=1;x=0;register char ch=gc();
while(!isdigit(ch)) {if (ch=='-') f=0;ch=gc();}
while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
x=f?x:-x;
}
template<typename T>inline void print(T x,char t) {
if (x<0) putchar('-'),x=-x;
if (!x) return putchar('0'),(void)putchar(t);
ri cnt(0);
while(x) OPUT[p(cnt)]=x%10,x/=10;
for (ri i(cnt);i;--i) putchar(OPUT[i]^48);
return (void)putchar(t);
}
}
using IO::read;using IO::print;
namespace nanfeng{
#define FI FILE *IN
#define FO FILE *OUT
template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
static const int N=2e5+7,MOD=998244353;
int pos[N],dp[N],v[N],st[N],wk[N],cnt,n,rmx,tmx,ans;
inline int MD(int x) {return x>=MOD?x-MOD:x;}
struct Seg{
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
struct segmenttree{int sum,mx,tg;}T[N<<2];
void build(int x,int l,int r) {
if (l==r) return (void)(T[x].tg=1);
int mid(l+r>>1);
build(ls(x),l,mid);
build(rs(x),mid+1,r);
}
int calc(int x,int mx) {
if (T[x].mx<=mx) return 0;
if (T[x].tg) return dp[T[x].mx];
if (T[rs(x)].mx<mx) return calc(ls(x),mx);
return MD(T[x].sum+calc(rs(x),mx));
}
inline void up(int x) {
T[x].mx=cmax(T[ls(x)].mx,T[rs(x)].mx);
T[x].sum=calc(ls(x),T[rs(x)].mx);
}
void update(int x,int k,int p,int l,int r) {
if (l==r) return (void)(T[x].mx=k);
int mid(l+r>>1);
if (p<=mid) update(ls(x),k,p,l,mid);
else update(rs(x),k,p,mid+1,r);
up(x);
}
int query(int x,int l,int r,int lt,int rt) {
if (l<=lt&&rt<=r) return tmx=rmx,rmx=cmax(rmx,T[x].mx),calc(x,tmx);
int mid(lt+rt>>1),res(0);
if (r>mid) res+=query(rs(x),l,r,mid+1,rt);
if (l<=mid) res+=query(ls(x),l,r,lt,mid);
return MD(res);
}
}T;
inline int main() {
//FI=freopen("nanfeng.in","r",stdin);
//FO=freopen("nanfeng.out","w",stdout);
read(n);
for (ri i(1);i<=n;p(i)) read(v[i]),pos[v[i]]=i;
ri tp=0;
for (ri i(1);i<=n;p(i)) {
while(tp&&v[st[tp]]<v[i]) --tp;
st[p(tp)]=i;
}
while(tp) wk[p(cnt)]=st[tp--];
T.build(1,1,n);
for (ri i(1);i<=n;p(i)) {
ri cur(pos[i]);
rmx=0;
dp[i]=T.query(1,1,cur,1,n);
if (!dp[i]) dp[i]=1;
T.update(1,i,cur,1,n);
}
for (ri i(1);i<=cnt;p(i)) ans=MD(ans+dp[v[wk[i]]]);
print(ans,'\n');
return 0;
}
}
int main() {return nanfeng::main();}