vijos 1459 treap

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题意：中文

思路：直接一个n^2 的预处理就行了，然后有个logn，n比较小可以通过

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int maxn=1010;
ll A[maxn],ans[maxn][maxn];
inline int getint(){
    int res=0;
    char c=getchar();
    bool mi=false;
    while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
    while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
    return mi ? -res : res;
}
inline ll getll(){
    ll res=0;
    char c=getchar();
    bool mi=false;
    while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
    while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
    return mi ? -res : res;
}
struct Node{
    Node *ch[2];
    int r,s;
    ll v,sum;
    Node(ll v):v(v){ch[0]=ch[1]=NULL;r=rand();s=1;sum=0;}
    int cmp(ll x){
        if(x==v) return -1;
        return xs,sum+=ch[0]->sum;
        if(ch[1]!=NULL) s+=ch[1]->s,sum+=ch[1]->sum;
    }
};
Node *root;
void Rotate(Node* &o,int d){
    Node* k=o->ch[d^1];o->ch[d^1]=k->ch[d];k->ch[d]=o;
    o->maintain();k->maintain();o=k;
}
void Insert(Node* &o,ll x){
    if(o==NULL) o=new Node(x);
    else{
        int d=x<(o->v)? 0:1;
        Insert(o->ch[d],x);
        if(o->ch[d]->r>o->r) Rotate(o,d^1);
    }
    o->maintain();
}
ll Kth(Node* o,int k,int op,ll pp){
    int s=(o->ch[0]==NULL?0:o->ch[0]->s);
    ll lll=0,qqq=0;
    if(o->ch[1]!=NULL) lll=o->ch[1]->sum;
    if(o->ch[0]!=NULL) qqq=o->ch[0]->sum;
    if(kch[0],k,op,pp+lll+o->v);
    else if(k>s+1) return Kth(o->ch[1],k-(s+1),op,pp-qqq-o->v);
    else{
        ll tmp1=pp,tmp2=0;
        if(o->ch[1]!=NULL) tmp1+=o->ch[1]->sum;
        if(o->ch[0]!=NULL) tmp2=o->ch[0]->sum;
        if(op==1) return tmp1-(o->v)-tmp2;
        else return tmp1-tmp2;
    }
}
int main(){
    int n,m,x,y,op;
    while(scanf("%d%d",&n,&m)!=-1){
        for(int i=1;i<=n;i++) A[i]=getll();
        memset(ans,0,sizeof(ans));
        ll fans=0;
        for(int i=1;i<=n;i++){
            root=NULL;
            for(int j=i;j<=n;j++){
                Insert(root,A[j]);
                if((j-i+1)%2==0) op=1;
                else op=0;
                ans[i][j]=Kth(root,(j-i)/2+1,op,0);
            }
        }
        for(int i=0;i