codeforces 617E XOR and Favorite Number

Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pair li and ri and asks you to count the number of pairs of integers i and j, such that l ≤ i ≤ j ≤ r and the xor of the numbers ai, ai + 1, ..., aj is equal to k.

Input

The first line of the input contains integers n, m and k (1 ≤ n, m ≤ 100 000, 0 ≤ k ≤ 1 000 000) — the length of the array, the number of queries and Bob's favorite number respectively.

The second line contains n integers ai (0 ≤ ai ≤ 1 000 000) — Bob's array.

Then m lines follow. The i-th line contains integers li and ri (1 ≤ li ≤ ri ≤ n) — the parameters of the i-th query.

Output

Print m lines, answer the queries in the order they appear in the input.

Examples input 6 2 3 1 2 1 1 0 3 1 6 3 5 output 7 0 input 5 3 1 1 1 1 1 1 1 5 2 4 1 3 output 9 4 4 Note

In the first sample the suitable pairs of i and j for the first query are: (1, 2), (1, 4), (1, 5), (2, 3), (3, 6), (5, 6), (6, 6). Not a single of these pairs is suitable for the second query.

In the second sample xor equals 1 for all subarrays of an odd length.

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莫队~

原来的代码WAWAWA是因为变量k混用……改过来以后直接TLE了……直接暴力莫队果然不行啊……

T掉了的代码：

#include<cstdio> #include<iostream> #include<algorithm> using namespace std; int n,m,k,a[1000001],ans[100001],now,l,r,kkk; struct node{ int l,r,id; }q[100001]; int read() { int x=0,f=1;char ch=getchar(); while(ch<'0' || ch>'9') {if(ch=='-') f=-1;ch=getchar();} while(ch>='0' && ch<='9') {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return x*f; } bool operator < (node u,node v) { return u.l==v.l ? u.r<v.r:u.l<v.l; } int main() { n=read();m=read();kkk=read(); for(int i=1;i<=n;i++) a[i]=read(),a[i]^=a[i-1]; for(int i=1;i<=m;i++) { q[i].l=read();q[i].r=read();q[i].id=i; } sort(q+1,q+m+1);l=q[1].l;r=q[1].r; for(int i=l;i<=r;i++) for(int j=i;j<=r;j++) now+=((a[j]^a[i-1])==kkk); ans[q[1].id]=now; for(int k=2;k<=m;k++) { if(l!=q[k].l) { for(int i=l;i<q[k].l;i++) { for(int j=i;j<=r;j++) now-=((a[j]^a[i-1])==kkk); } l=q[k].l; } if(r!=q[k].r) { if(r<q[k].r) { for(int i=r+1;i<=q[k].r;i++) for(int j=l;j<=i;j++) now+=((a[i]^a[j-1])==kkk); r=q[k].r; } else { for(int i=r;i>q[k].r;i--) for(int j=l;j<=i;j++) now-=((a[i]^a[j-1])==kkk); r=q[k].r; } } ans[q[k].id]=now; } for(int i=1;i<=m;i++) printf("%d\n",ans[i]); return 0; }