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洛谷P4462 [CQOI2018]异或序列(莫队)

程序员文章站 2022-12-14 23:17:46
题意 "题目链接" Sol 一开始以为K每次都是给出的想了半天不会做。 然而发现读错题了维护个前缀异或和然后直接莫队搞就行,。 cpp include define Pair pair define MP(x, y) make_pair(x, y) define fi first define se ......

题意

题目链接

sol

一开始以为k每次都是给出的想了半天不会做。

然而发现读错题了维护个前缀异或和然后直接莫队搞就行,。

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, mod = 998244353, inf = 1e9 + 10;
const double eps = 1e-9;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, q, k, a[maxn], num[maxn], belong[maxn], block;
ll now, ans[maxn];
struct query {
    int l, r, id;
    bool operator < (const query &rhs) const {
        return belong[l] == belong[rhs.l] ? r < rhs.r : belong[l] < belong[rhs.l];
    }
}q[maxn];
void add(int x) {
    now += num[k ^ x];
    num[x]++;
}
void delet(int x) {
    num[x]--;
    now -= num[k ^ x];
}
void solve() {
    int l = 1, r = 0;
    for(int i = 1; i <= q; i++) {
        while(r < q[i].r) add(a[++r]);
        while(r > q[i].r) delet(a[r--]);
        while(l < q[i].l) delet(a[l++]);
        while(l > q[i].l) add(a[--l]);
        ans[q[i].id] = now;
    }
}
signed main() {
    n = read(); q = read(); k = read(); block = sqrt(n);
    for(int i = 1; i <= n; i++) a[i] = read() ^ a[i - 1], belong[i] = (i - 1) / block + 1;
    for(int i = 1; i <= q; i++) q[i].l = read() - 1, q[i].r = read(), q[i].id = i;
    sort(q + 1, q + q + 1);
    solve();
    for(int i = 1; i <= q; i++) cout << ans[i] << '\n';
    return 0;
}