CodeForces

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define INF 0x3f3f3f3f#define rep0(i, n) for (int i = 0; i < n; i++)#define rep1(i, n) for (int i = 1; i <= n; i++)#define rep_0(i, n) for (int i = n - 1; i >= 0; i--)#define rep_1(i, n) for (int i = n; i > 0; i--)#define MAX(x, y) (((x) > (y)) ? (x) : (y))#define MIN(x, y) (((x) < (y)) ? (x) : (y))#define mem(x, y) memset(x, y, sizeof(x))#define MAXN 100000 + 10#define MAXK 10000000 + 10/**题目大意给一个数列 和数字k 多次询问[l, r]范围内 有多少各连续子序列异或值为k思路没有区间修改 用莫队需要用O(1)复杂度转移预处理 pre[x] 表示 0 ~ x 的数的异或值则 i ~ j 的异或值为 pre[i - 1] ^ pre[j] 当在右边增加时询问 区间内(包括开区间下届) pre[j] ^ k 的个数当在左边增加时询问 区间内 pre[i - 1] ^ k 的个数简单的用两个hash维护下届开区间和闭区间内的各pre个数*/using namespace std;int sqn;struct query{    int l, r;    int id;    __int64 ans;} q[MAXN];__int64 a[MAXN], pre[MAXN], book[MAXK], book1[MAXK], ans, k;bool cmp(query a, query b){    if (a.l / sqn == b.l / sqn)        return a.r < b.r;    return a.l < b.l;}bool CMP(query a, query b){    return a.id < b.id;}void add(int x, bool flag){    book[pre[x]]++;    if (x > 0)        book1[pre[x - 1]]++;    else        book1[0]++;    if (flag)    {        if (x > 0)            ans += book[pre[x - 1] ^ k];        else            ans += book[0 ^ k];    }    else    {        ans += book1[pre[x] ^ k];    }}void del(int x, bool flag){    if (flag)    {        if (x > 0)            ans -= book[pre[x - 1] ^ k];        else            ans -= book[0 ^ k];    }    else        ans -= book1[pre[x] ^ k];    book[pre[x]]--;    if (x > 0)        book1[pre[x - 1]]--;    else        book1[0]--;}int main(){    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);    #endif // ONLINE_JUDGE    int n, m;    __int64 tmp;    mem(book, 0);    mem(book1, 0);    scanf("%d %d %d", &n, &m, &k);    //cout << n << "  " << m <<endl;    sqn = sqrt(n);    for (int i = 0; i < n; i++)    {        scanf("%I64d", a + i);        if (i == 0)        {            pre[i] = a[i];            tmp = a[i];        }        else        {            pre[i] = tmp ^ a[i];            tmp = pre[i];        }    }    int l, r;    for (int i = 0; i < m; i++)    {        scanf("%d %d", &l, &r);        q[i].l = l - 1;        q[i].r = r;        q[i].id = i;    }    sort(q, q + m, cmp);    ans = 0;    l = r = 0;    for (int i = 0; i < m; i++)    {        while (r < q[i].r)        {            add(r, 0);            r++;        }        while (l > q[i].l)        {            l--;            add(l, 1);        }        while (r > q[i].r)        {            r--;            del(r, 0);        }        while (l < q[i].l)        {            del(l, 1);            l++;        }        q[i].ans = ans;    }    sort(q, q + m, CMP);    for (int i = 0; i < m; i++)        printf("%I64d\n", q[i].ans);    return 0;}