【线段树+Hash】Codeforces 452F Permutation

题面在这里

按顺序枚举i，则ai±k必须在同一侧（即同时出现或同时不出现）

维护权值线段树

其实就是[ai−k,ai]和[ai,ai+k]对称

线段树维护正反两个hash值就好了

示例程序：

#include<cstdio>#include<algorithm>#define mp make_pair#define Fst first#define Sec secondusing namespace std;typedef unsigned long long ull;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int red(){    int res=0,f=1;char ch=nc();    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}    while ('0'<=ch&&ch<='9') res=(res<<3)+(res<<1)+ch-48,ch=nc();    return res*f;}const int maxn=300005;int n;ull p[maxn];struct node{    node *l,*r;    int L,R;ull a,b;    node () {}    node (int _L,int _R):L(_L),R(_R),a(0),b(0) {}    void pushup(){        if (L==R) return;        a=l->a*p[r->R - r->L +1]+r->a;        b=l->b+ r->b*p[l->R - l->L +1];    }}base[maxn<<1],nil;typedef node* P_node;P_node null,Rot,len;void init(){    nil=node(0,0);null=&nil;    null->l=null->r=null;len=base;}P_node newnode(int L,int R){    *len=node(L,R);    len->l=len->r=null;    return len++;}P_node build(int L,int R){    P_node x=newnode(L,R);    if (L==R) return x;    int mid=L+R>>1;    x->l=build(L,mid);x->r=build(mid+1,R);    return x;}void insert(P_node x,int k){    if (x->L==x->R) {x->a=x->b=1;return;}    int mid=x->L+x->R>>1;    if (k<=mid) insert(x->l,k);else insert(x->r,k);    x->pushup();}pair<int,int> A,B;void queryA(P_node x,int L,int R){    if (R<x->L||x->R<L) return;    if (L<=x->L&&x->R<=R) {A.Fst=x->a*p[A.Sec]+A.Fst;A.Sec+=x->R-x->L+1;return;}    queryA(x->r,L,R);queryA(x->l,L,R);}void queryB(P_node x,int L,int R){    if (R<x->L||x->R<L) return;    if (L<=x->L&&x->R<=R) {B.Fst=x->b*p[B.Sec]+B.Fst;B.Sec+=x->R-x->L+1;return;}    queryB(x->l,L,R);queryB(x->r,L,R);}int main(){    n=red();p[0]=1;    for (int i=1;i<=n;i++) p[i]=p[i-1]*233;    init();Rot=build(1,n);    for (int i=1;i<=n;i++){        int x=red(),k=min(x-1,n-x);        A=B=mp(0,0);        queryA(Rot,x-k,x);queryB(Rot,x,x+k);        if (A.Fst!=B.Fst) return printf("YES"),0;        insert(Rot,x);    }    printf("NO");    return 0;}