题目链接:
E. 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.
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.
Print m lines, answer the queries in the order they appear in the input.
6 2 3
1 2 1 1 0 3
1 6
3 5
7
0
5 3 1
1 1 1 1 1
1 5
2 4
1 3
9
4
4
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.
题意:给你n个数,有m个询问,问[l,r]之间有多少对i和j满足a[i]^a[i+1]^...^a[j]=k;
思路:暴力绝对绝对绝对是不行的,所有就要用复杂度更低的算法啦,所以我就去学了莫队算法,莫队算法处理的是一种离线算法,是对询问进行分块排序来优化查询的算法;
这题还要对异或运算要了解;我要去写个位运算的小总结;
AC代码:
#include <bits/stdc++.h> using namespace std; const int N=1e6+4; struct node { friend bool operator< (node x,node y) { if(x.pos==y.pos)return x.r<y.r; return x.l<y.l; } int l,r,id; int pos; }; node qu[N]; int n,m,k; int b[N],num[2*N]; long long ans[N]; void solve() { memset(num,0,sizeof(num)); int le=1,ri=0; long long temp=0; for(int i=1;i<=m;i++) { while(ri<qu[i].r) { ri++; temp+=num[b[ri]^k];//num[]数组是当前le到ri内每个b[i](le<=i<=ri)异或一个要添加的数==k的数目;所以num[b[ri]^k]就是在已有的区间上再加一个 num[b[ri]]++;//ri前能和ri异或==k的数目;后面这些操作一样 } while(ri>qu[i].r) { num[b[ri]]--; temp-=num[b[ri]^k]; ri--; } while(le>qu[i].l-1) { le--; temp+=num[b[le]^k]; num[b[le]]++; } while(le<qu[i].l-1) { num[b[le]]--; temp-=num[b[le]^k]; le++; } ans[qu[i].id]=temp; } } int main() { b[0]=0; scanf("%d%d%d",&n,&m,&k); for(int i=1;i<=n;i++) { scanf("%d",&b[i]); b[i]^=b[i-1]; } int s=sqrt(n); for(int i=1;i<=m;i++) { scanf("%d%d",&qu[i].l,&qu[i].r); qu[i].id=i; qu[i].pos=qu[i].l/s; } sort(qu+1,qu+m+1); solve(); for(int i=1;i<=m;i++) { cout<<ans[i]<<" "; } return 0; }