Codeforces Round #340 (Div. 2)-E-XOR and Favorite Number(莫队)
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,m次查询,每次查询一个区间中有多少子区间的所有数异或起来为k。。。。
题解:一开始想的按位建线段树,但是处理情况数的话复杂度是nlogn,总复杂度就炸了。
其实可以考虑莫队(很久之前学的了,没有想到),我们可以将查询按照莫队分块的规则排序,然后搞搞就可以了。
不懂莫队的可以看我这篇博客:http://blog.csdn.net/haut_ykc/article/details/76443756
附上惨图:(感觉现在写代码是真有毒。。。)
#include<math.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node
{
int l,r,id;
}q[100005];
int sz,a[100005],mp[1<<20];
ll ans[100005];
bool comp(node a,node b)
{
if(a.l/sz==b.l/sz)
return a.r<b.r;
return a.l<b.l;
}
void work(int m,int k)
{
ll now=0;
int l=1,r=0;
mp[a[l-1]]++;
for(int i=0;i<m;i++)
{
while(r<q[i].r)
{
r++;
now+=mp[a[r]^k];
mp[a[r]]++;
}
while(l>q[i].l)
{
l--;
now+=mp[a[l-1]^k];
mp[a[l-1]]++;
}
while(r>q[i].r)
{
mp[a[r]]--;
now-=mp[a[r]^k];
r--;
}
while(l<q[i].l)
{
mp[a[l-1]]--;
now-=mp[a[l-1]^k];
l++;
}
ans[q[i].id]=now;
}
}
int main(void)
{
int n,m,k;
scanf("%d%d%d",&n,&m,&k);
sz=(int)sqrt(n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
a[i]^=a[i-1];
for(int i=0;i<m;i++)
{
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q,q+m,comp);
work(m,k);
for(int i=0;i<m;i++)
printf("%lld\n",ans[i]);
return 0;
}