Codeforeces 617E XOR and Favorite Number（莫队+小技巧）
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.
题目链接：CF 617E
这题感觉比一般的莫队要难很多，主要问题就出在转化上，一开始看感觉没办法O（1）进行拓展，后来看了大牛的思路原来是用了一个前缀异或和的数组pre[ ]和一个维护当前区间内异或和个数的数组cnt[ ]，用pre[i-1]^pre[j]来表示ai^……aj的异或和，这样才能写出add和del函数。
假设当前已知【L，R】，如果区间增加，假设增加的那个位置的值为pre[x]，那么显然pre[x]可以和区间内所有值为 k^pre[x]的数进行两两组合，即会多出cnt[ k^pre[x] ]个组合，此时cnt就可以用到，直接加上cnt[k^pre[x]]即可，然后再更新pre[x]（顺序不能反，万一pre[x]^k和pre[x]是相同的，更新前是4，更新后是5，但是只能多出4个组合而不是5个组合）；如果区间减小，由于被减掉的那个pre[x]本来可以组合cnt[k^pre[x]]个，因此要减掉这么多个数，但是假如pre[x]=pre[x]^k的话，比如5个0去了一个0变成了4个0，组合数减少了4，因此要先自减1再更新此时的答案
代码：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1<<20;
struct info
{
int l,r;
int b,idd;
bool operator<(const info &t)const
{
if(b==t.b)
return r<t.r;
return b<t.b;
}
};
info query[N];
int pre[N];
LL cnt[N];
LL ans[N];
void init()
{
CLR(pre,0);
CLR(cnt,0);
CLR(ans,0);
}
inline void add(const int &val,LL &temp,const int &k)
{
temp+=cnt[val^k];
++cnt[val];
}
inline void del(const int &val,LL &temp,const int &k)
{
--cnt[val];
temp-=cnt[val^k];;
}
int main(void)
{
int n,m,k,i,j,val;
while (~scanf("%d%d%d",&n,&m,&k))
{
init();
for (i=1; i<=n; ++i)
{
scanf("%d",&val);
pre[i]=pre[i-1]^val;
}
int unit=sqrt(n);
for (i=0; i<m; ++i)
{
scanf("%d%d",&query[i].l,&query[i].r);
query[i].b=query[i].l/unit;
query[i].idd=i;
}
sort(query,query+m);
int L=1,R=0;
LL temp=0;
cnt[0]=1;
for (i=0; i<m; ++i)
{
while (L>query[i].l)
{
--L;
add(pre[L-1],temp,k);
}
while (L<query[i].l)
{
del(pre[L-1],temp,k);
++L;
}
while (R>query[i].r)
{
del(pre[R],temp,k);
--R;
}
while (R<query[i].r)
{
++R;
add(pre[R],temp,k);
}
ans[query[i].idd]=temp;
}
for (i=0; i<m; ++i)
printf("%I64d\n",ans[i]);
}
return 0;
}
