HDU 6579 Operation

HDU face questions

Time limit 4000 ms

Memory limit 262144 kB

OS Windows

Source 2019 Multi-University Training Contest 1

Chinese 题意

A sequence supports two modes of operation -

• Insert a number in the back, and the sequence length is increased by one
• Asked an interval of maximum XOR

Forced online.

Problem-solving ideas

Basic is a question on Ivan and Burgers it. I have a question on the use of the solution is online. Although still not proven.

• [X] codeforces 1100F of Ivan and Burgers simple XOR maximum interrogation zone
• [X] 6579 HDU Operation plurality of operations insertion end of the data, as well as by force
• [] BZOJ 4184 Shallot this title more than the operating insertions and deletions. Actually measure what the local authority title ...... forget.
• [] UVALive 8514 XOR 2017ICPC Xi'an a question, almost all operations

Source

#include<cstdio>
#include<cstring>
#include<algorithm>

const int MAXN=1e6+5;
const int wide=31;

int T;
int n,m;

int p[MAXN][wide+2],pos[MAXN][wide+2];

void insert(int loc,int val)//location value
{
    for(int i=wide;~i;i--)
    {
        p[loc][i]=p[loc-1][i];
        pos[loc][i]=pos[loc-1][i];
    }
    int temp=loc;
    for(int i=wide;~i;i--)
    {
        if((val>>i)&1)
        {
            if(!p[loc][i])
            {
                p[loc][i]=val;
                pos[loc][i]=temp;
                return;
            }
            if(pos[loc][i]<temp)//还是不太懂
            {
                std::swap(pos[loc][i],temp);
                std::swap(p[loc][i],val);
            }
            val^=p[loc][i];
        }
    }
}


int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        memset(pos,0,sizeof(int)*(wide+2)*(n+2));
        memset(p,0,sizeof(int)*(wide+2)*(n+2));
        for(int i=1,v;i<=n;i++)
        {
            scanf("%d",&v);
            insert(i,v);
        }
        int lastans=0;
        while(m--)
        {
            int opt,l,r;
            scanf("%d",&opt);
            if(opt)
            {
                scanf("%d",&l);
                l^=lastans;
                insert(++n,l);
            }
            else
            {
                scanf("%d%d",&l,&r);
                l=(l^lastans)%n+1;
                r=(r^lastans)%n+1;
                if(l>r) std::swap(l,r);
                int ans=0;
                for(int i=wide;~i;i--)
                {
                    if((ans^p[r][i])>ans&&pos[r][i]>=l)
                        ans^=p[r][i];
                }
                printf("%d\n",lastans=ans);
            }
        }
    }
    return 0;
}