莫队算法例题，模板及小结

【莫队算法详解及例题[bzoj]2038 小z的袜子  

及知乎

以上是弱渣的学习过程，及参考的大神博客和知乎，以下是做到的题目

g - 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.

题目链接：https://vjudge.net/contest/239536#problem/g

题解：由于a^a=0;
如果a^b=k;则a^k=b;
若要a[i]^a[i+1]….^a[j] )=k;
即( a[1]^a[2]^…a[i] ) ^( a[1]^a[2]^…^a[i-1]^a[i]^a[i+1]….^a[j] )。
就可以采用前缀和思想。
让flag[ a[i] ^k]表示a[i] ^k出现的次数。
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作者：*leiju__
来源：csdn
原文：https://blog.csdn.net/*leiju__/article/details/81174608
版权声明：本文为博主原创文章，转载请附上博文链接！

ac代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define maxn 1<<20
using namespace std;

int a[maxn];
int pos[maxn];
long long ans[maxn],flag[maxn];

struct query{
int l,r,id;
}q[maxn];

bool cmp(struct query x,struct query y){
if(pos[x.l]==pos[y.l]){
return x.r<y.r;
}
return pos[x.l]<pos[y.l];
}

int n,m,k;
int l=1,r=0;
long long ans=0;

//其它基本不变，就是add和dec函数会变化 
void add(int x){
ans+=flag[a[x]^k];
flag[a[x]]++;
}

void del(int x){
flag[a[x]]--; //flag【a【x】】 是a【x】在前缀中出现的次数 
ans-=flag[a[x]^k];
}

int main(){
scanf("%d%d%d",&n,&m,&k);
int sz=sqrt(n);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
a[i]=a[i]^a[i-1]; //a[i]表示前i个值异或的结果 
pos[i]=i/sz;
}
for(int i=1;i<=m;i++){
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q+1,q+m+1,cmp);
flag[0]=1;
for(int i=1;i<=m;i++){
while(l<q[i].l){
del(l-1);
l++;
}
while(l>q[i].l){
l--;
add(l-1);
}
while(r<q[i].r){
r++;
add(r);
}
while(r>q[i].r){
del(r);
r--;
}
ans[q[i].id]=ans;
}
for(int i=1;i<=m;i++){
printf("%lld\n",ans[i]);
}
return 0;
}

 

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作者：*leiju__
来源：csdn
原文：

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