题意:给出\(n,m,k,a[1...n]\),对于每次询问,求\([l,r]\)中\(a[i] \ xor \ a[i+1] \ xor \ ...a[j],l<=i<=j<=r\)等于k的对数

这回看了qsc菊苣的教学,恰好是同一题,感觉理解度up

顺便把代码风格轻微改了一下,之前的维护过程太凌乱了

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<string>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#include<bitset>

#define rep(i,j,k) for(register int i=j;i<=k;i++)

#define rrep(i,j,k) for(register int i=j;i>=k;i--)

#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])

#define iin(a) scanf("%d",&a)

#define lin(a) scanf("%lld",&a)

#define din(a) scanf("%lf",&a)

#define s0(a) scanf("%s",a)

#define s1(a) scanf("%s",a+1)

#define print(a) printf("%lld",(ll)a)

#define enter putchar('\n')

#define blank putchar(' ')

#define println(a) printf("%lld\n",(ll)a)

#define IOS ios::sync_with_stdio(0)

using namespace std;

const int MAXN = 1<<20|1;

const double EPS = 1e-7;

typedef long long ll;

typedef unsigned long long ull;

const ll MOD = 1e9+7;

unsigned int SEED = 17;

const ll INF = 1ll<<60;

ll read(){

    ll x=0,f=1;register char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

struct Node{

    int l,r,id;

}Q[MAXN];

ll a[MAXN],cnt[MAXN],pre[MAXN],pos[MAXN],ans[MAXN];

ll L,R,ANS,SIZE,n,m,k;

bool cmp(Node a,Node b){

    if(pos[a.l]!=pos[b.l]) return pos[a.l]<pos[b.l];

    return a.r<b.r;

}

inline void add(int cur){

    ANS+=cnt[pre[cur]^k];//pre[cur]加入,答案加上pre[cur]^k贡献的出现次数

    cnt[pre[cur]]++;

}

inline void del(int cur){

    cnt[pre[cur]]--;//防止重复计数

    ANS-=cnt[pre[cur]^k];

}

int main(){

    while(cin>>n>>m>>k){

        memset(cnt,0,sizeof cnt);

        cnt[0]=1; SIZE=sqrt(n);

        rep(i,1,n){

            a[i]=read();

            pre[i]=pre[i-1]^a[i];

            pos[i]=i/SIZE;

        }

        rep(i,1,m){

            Q[i].l=read();

            Q[i].r=read();

            Q[i].id=i;

        }

        sort(Q+1,Q+1+m,cmp);

        L=1; R=0; ANS=0; //L-1>=0 //[1,1]也是要统计的,R=0

        rep(i,1,m){

            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;

        }

        rep(i,1,m) println(ans[i]);

    }

    return 0;

}