• AcWing269 Fotile模拟赛L(可持久化Trie)


    分块预处理,表示f[][],第i块到第j个数的答案

    这样做的时候就可以大块直接查询,小块暴力查询

    #include<bits/stdc++.h>
    #define getsz(p) (p?p->sz:0)
    using namespace std;
    typedef long long ll;
    typedef pair<ll,int> pll;
    const int N=1e5+10;
    struct node{
        ll sz;
        node * nxt[2];
    }*rt[N],pool[N*30];
    ll f[400][16000];
    ll s[N],idx,n,block,B;
    node *copynode(node *rt){
        node *p=pool+(++idx);
        pool[idx]=*rt;
        return p;
    }
    node* insert(node* rt, int x, int dep){
        node *p;
        if(rt) p = copynode(rt);
        else{
            p=pool+(++idx);
            p->sz=0;
        }
        ++p->sz;
        if(dep < 0) return p;
    
        int y = (x>>dep)&1;
        if(!y) p->nxt[0] = insert(p->nxt[0], x, dep-1);
        else p->nxt[1] = insert(p->nxt[1], x, dep-1);
    
        return p;
    }
    ll query(node* pL, node* pR, int x, int dep){
        int diff, y;
        ll ans = 0;
        for(int dep=30;dep>=0;dep--){
            y = (x>>dep)&1;
            if(!pL){
                diff = getsz(pR->nxt[!y]);
                if(diff){
                    ans += (1<<dep);
                    pR = pR->nxt[!y];
                }
                else pR = pR->nxt[y];
            }
            else{
                diff = getsz(pR->nxt[!y]) - getsz(pL->nxt[!y]);
                if(diff){
                    ans += (1<<dep);
                    pL = pL->nxt[!y];
                    pR = pR->nxt[!y];
                }
                else{
                    pL = pL->nxt[y];
                    pR = pR->nxt[y];
                }
            }
        }
        return ans;
    }
    ll dp[500][14000];
    void init(){
        B = sqrt(n);
        for(int i=0;i<=n;i+=B){
            for(int j=i+1;j<=n;j++){
                if(i==0) dp[i/B][j] = max(dp[i/B][j-1], query(0, rt[j], s[j], 30));
                else dp[i/B][j] = max(dp[i/B][j-1], query(rt[i-1], rt[j], s[j],30));
            }
        }
    }
    int main(){
        ios::sync_with_stdio(false);
        int m;
        cin>>n>>m;
        int i;
        rt[0]=insert(rt[0],0,30);
        for(i=1;i<=n;i++){
            int x;
            cin>>x;
            s[i]=s[i-1]^x;
            rt[i]=insert(rt[i-1],s[i],30);
        }
        init();
        ll last=0;
        while(m--){
            ll l,r;
            cin>>l>>r;
            l = (l+last)%n + 1;
            r = (r+last)%n + 1;
            if(l>r)
                swap(l,r);
            l--;
            int x=(l/B+1)*B;
            if(x<=r){
                last=dp[l/B+1][r];
            }
            else
                last=0;
            for(i=l;i<x&&i<=r;i++){
                last=max(last,query(rt[i],rt[r],s[i],30));
            }
            cout<<last<<endl;
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/ctyakwf/p/13492684.html
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