bzoj 5301 [Cqoi2018] XOR sequence Mo team

5301: [Cqoi2018] XOR sequence

Time Limit: 10 Sec Memory Limit: 512 MB
Submit: 204 Solved: 155
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Description

Given an integer sequence a[1],a[2],…,a[n] of length n, given query parameters l, r, ask how many consecutive subs are in the interval [l,r]

The sequence satisfies the XOR sum equal to k .

That is to say, for all x, y (l≤x≤y≤r), how many groups of x and y can satisfy a[x]^a[x+1]^...^a[y]=k.

Input

The first line of the input file is 3 integers n, m, k.

The second line is n integers separated by spaces, namely ai, a2, . . . an.

The next m lines, each with two integers lj, rj, represent a query.

1≤n，m≤105，O≤k，ai≤105，1≤lj≤rj≤n

Output

The output file has m lines in total, corresponding to the calculation results of each query.

Sample Input

4 5 1
1 2 3 1
1 4
1 3
2 3
2 4
4 4

Sample Output

4
2
1
2
1

Solution: Change to the prefix xor and, and then Mo team.

 1 #include<cstring>
 2 #include<cmath>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<iostream>
 6 
 7 #define N 100007
 8 #define ll long long
 9 
10 #define Wb putchar(' ')
11 #define We putchar('\n')
12 #define rg register int
13 using namespace std;
14 inline int read()
15 {
16     int x=0,f=1;char ch=getchar();
17     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
18     while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
19     return x*f;
20 }
21 inline void write(ll x)
22 {
23     if(x<0) putchar('-'),x=-x;
24     if (x==0) putchar(48);
25     int num=0;char c[20];
26     while(x) c[++num]=(x%10)+48,x/=10;
27     while(num) putchar(c[num--]);
28 }
29 
30 int n,m,K,blo;
31 int a[N],bel[N],cnt[N<<1];
32 ll ans[N];
33 struct Node
34 {
35     int l,r,id;
36     friend bool operator<(Node x,Node y)
37     {
38         if (bel[x.l]!=bel[y.l]) return bel[x.l]<bel[y.l];
39         return x.r<y.r;
40     }
41 }Q[N];
42 
43 void solve()
44 {
45     int l=1,r=0;ll res=0;
46     for (int i=1;i<=m;i++)
47     {
48         while(l<Q[i].l-1)
49             cnt[a[l]]--,res-=cnt[K^a[l]],l++;
50         while(l>Q[i].l-1)
51             l--,res+=cnt[K^a[l]],cnt[a[l]]++;
52         while(r<Q[i].r)
53             r++,res+=cnt[K^a[r]],cnt[a[r]]++;
54         while(r>Q[i].r)
55             cnt[a[r]]--,res-=cnt[K^a[r]],r--;
56         ans[Q[i].id]=res;
57     }
58 }
59 int main()
60 {
61     n=read(),m=read(),K=read(),blo=(int)sqrt(n);
62     for (rg i=1;i<=n;i++) a[i]=read()^a[i-1];
63     for (rg i=1;i<=n;i++) bel[i]=(i-1)/blo+1;
64     for (rg i=1;i<=m;i++)
65         Q[i].l=read(),Q[i].r=read(),Q[i].id=i;
66     sort(Q+1,Q+m+1);
67     solve();
68     for (rg i=1;i<=m;i++)
69         write(ans[i]),We;
70 }