BZOJ2741: 【FOTILE模拟赛】L

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 150000+5
 26 
 27 #define maxm 5000000+5
 28 #define maxk 150
 29 
 30 #define eps 1e-10
 31 
 32 #define ll long long
 33 
 34 #define pa pair<int,int>
 35 
 36 #define for0(i,n) for(int i=0;i<=(n);i++)
 37 
 38 #define for1(i,n) for(int i=1;i<=(n);i++)
 39 
 40 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 41 
 42 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 43 
 44 #define mod 1000000007
 45 
 46 using namespace std;
 47 
 48 inline int read()
 49 
 50 {
 51 
 52     int x=0,f=1;char ch=getchar();
 53 
 54     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 55 
 56     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 57 
 58     return x*f;
 59 
 60 }
 61 int n,m,q,tot,rt[maxn],id[maxm],t[maxm][2],a[maxn],b[maxk][maxn];
 62 inline void insert(int pre,int x,int k)
 63 {
 64     int now=rt[k]=++tot;id[tot]=k;
 65     for3(i,30,0)
 66     {
 67         int j=(x>>i)&1;
 68         t[now][j^1]=t[pre][j^1];//空指针指向原来的
 69         t[now][j]=++tot;id[tot]=k;
 70         now=tot;
 71         pre=t[pre][j];
 72     }
 73 }
 74 inline int ask(int l,int r,int x)
 75 {
 76     int ans=0,now=rt[r];
 77     for3(i,30,0)
 78     {
 79         int j=((x>>i)&1)^1;
 80         if(id[t[now][j]]>=l)ans|=1<<i;else j^=1;//下面的节点的id都小于l，所以要改变方向
 81         now=t[now][j];
 82     }
 83     return ans;
 84 }
 85 
 86 int main()
 87 
 88 {
 89 
 90     freopen("input.txt","r",stdin);
 91 
 92     freopen("output.txt","w",stdout);
 93 
 94     n=read();q=read();
 95     for1(i,n)a[i]=a[i-1]^read();
 96     id[0]=-1;
 97     insert(rt[0],a[0],0);//插入0
 98     for1(i,n)insert(rt[i-1],a[i],i);//挨个插入前缀异或值
 99     int len=sqrt(n);m=n/len+(n%len!=0);
100     for0(i,m-1)
101      for2(j,i*len+1,n)
102       b[i][j]=max(b[i][j-1],ask(i*len,j-1,a[j]));//分块，用a[j]去查询在i*len-j-1的最大值
103     int ans=0;
104     while(q--)
105     {
106         int x=((ll)read()+(ll)ans)%n+1,y=((ll)read()+(ll)ans)%n+1;
107         if(x>y)swap(x,y);x--;
108         int bx=x/len+(x%len!=0);
109         ans=bx*len<y?b[bx][y]:0;//大块的答案已经得到
110         for2(j,x,min(bx*len,y))
111          ans=max(ans,ask(x,y,a[j]));//用小块内的点暴力查询max
112         printf("%d\n",ans);
113     }
114 
115     return 0;
116 
117 }