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
1 2 3 1
1 4
1 3
2 3
2 4
4 4
Sample Output
4
2
1
2
1
2
1
2
1
I can't go on like this anymore.
Yesterday, I played the board for a day, and I feel like my brain is rusted. QAQ... The
question has been killed for a long time... Obviously the sub-sequences should be discontinuous.
At first glance, this form of the question basically didn't run,
but if the two ends join /Subtract a number, the effect of this number on the interval is close to its continuous segment,
which is obviously unmaintainable. So an XOR prefix and sum can be maintained
because a[x] xor a[x+1]...xor a[y-1] xor a[y] is equivalent to sum[y] xor sum[x-1]
then We can use Mo team to maintain the prefix sum.
Every time you add/delete the prefix sum to the Mo team, you only need to consider how many numbers are XOR equal to k with the current number, and you can open a bucket.
Because we need sum[x-1]~sum[y] when querying the interval [x,y], the left endpoint needs to be decremented by one when reading in.
Yesterday, I played the board for a day, and I feel like my brain is rusted. QAQ... The
question has been killed for a long time... Obviously the sub-sequences should be discontinuous.
At first glance, this form of the question basically didn't run,
but if the two ends join /Subtract a number, the effect of this number on the interval is close to its continuous segment,
which is obviously unmaintainable. So an XOR prefix and sum can be maintained
because a[x] xor a[x+1]...xor a[y-1] xor a[y] is equivalent to sum[y] xor sum[x-1]
then We can use Mo team to maintain the prefix sum.
Every time you add/delete the prefix sum to the Mo team, you only need to consider how many numbers are XOR equal to k with the current number, and you can open a bucket.
Because we need sum[x-1]~sum[y] when querying the interval [x,y], the left endpoint needs to be decremented by one when reading in.
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<cmath> 5 #include<algorithm> 6 #define N (500000+100) 7 using namespace std; 8 9 struct node{int x,y,num,id,ans;}ask[N]; 10 int Keg[N],sum[N],n,m,k,x,now; 11 12 bool cmp1(node a,node b){return a.id==b.id?a.y<b.y:a.id<b.id;} 13 bool cmp2(node a,node b){return a.num<b.num;} 14 15 void Ins(int pos){now+=Keg[sum[pos]^k]; Keg[sum[pos]]++;} 16 void Del(int pos){Keg[sum[pos]]--; now-=Keg[sum[pos]^k];} 17 18 int main() 19 { 20 scanf("%d%d%d",&n,&m,&k); 21 int unit=sqrt(n); 22 for (int i=1; i<=n; ++i) 23 scanf("%d",&x),sum[i]=sum[i-1]^x; 24 for (int i=1; i<=m; ++i) 25 { 26 scanf("%d%d",&ask[i].x,&ask[i].y); 27 ask[i].x--; ask[i].num=i; 28 ask[i].id=ask[i].x/unit; 29 } 30 sort(ask+1,ask+m+1,cmp1); 31 32 int l=1,r=0; 33 for (int i=1; i<=m; ++i) 34 { 35 while (l<ask[i].x) Del(l++); 36 while (l>ask[i].x) Ins(--l); 37 while (r<ask[i].y) Ins(++r); 38 while (r>ask[i].y) Del(r--); 39 ask[i].ans=now; 40 } 41 42 sort(ask+1,ask+m+1,cmp2); 43 for (int i=1; i<=m; ++i) 44 printf("%d\n",ask[i].ans); 45 }