P4887 【模板】莫队二次离线（第十四分块(前体)）

Solution

简单学习了一下二次离线莫队，写了个板子题。

这题直接莫队时间复杂度为 $O(Cnn)O(Cn\sqrt n)$ ，其中 $C=(14k)C=\binom{14}{k}$ ，显然不太行。

我们考虑当前区间为 $[l, r]$ ，右端点增加 $k$ ，变为 $[l, r + k]$ 产生的贡献：

令 $f(x,[l,r])=[∑i=lrpopcount(xxorai)=k]f(x,[l,r])=[\sum_{i=l}^rpopcount(x\;xor\;a_i)=k]$

则有：
$Δans\Delta ans$

$=∑i=r+1r+kf(i,[l,i−1])=\sum_{i=r+1}^{r+k}f(i,[l,i-1])$

$=∑i=r+1r+kf(i,[1,i−1])−f(i,[1,l−1])=\sum_{i=r+1}^{r+k} f(i,[1,i-1])-f(i,[1,l-1])$

于是我们可以预处理 $f (i, [1, i - 1])$ ，并且离线求出 $∑i=r+1r+kf(i,[1,l−1])\sum_{i=r+1}^{r+k}f(i,[1,l-1])$ 。这个可以在 $O(Cn+mn)O(Cn+m\sqrt n)$ 的时间内完成。

其他端点的移动同理。

总时间复杂度为 $O(Cn+mn)O(Cn+m\sqrt n)$ 。

Code

const int MX = 1 << 14;
int b[MX], sum[MX], a[MAXN], sz, bnum = 0, Cnum = 0, n, m, K;
ll s0[MAXN], s1[MAXN], Ans[MAXN];
struct Cnode{ int l, r, x, id; ll ans; } C[MAXN << 1];
struct Qnode{ int l, r, id; } Q[MAXN];void Init() {for (int i = 0; i < MX ; ++ i) if (__builtin_popcount(i) == K) b[++ bnum] = i;sort(Q + 1, Q + m + 1, [&](Qnode a, Qnode b){ return ((a.l - 1) / sz < (b.l - 1) / sz) || ((a.l - 1) / sz == (b.l - 1) / sz && a.r < b.r); });for (int i = 1, l = 1, r = 0; i <= m ; ++ i) {if (r < Q[i].r) ++ Cnum, C[Cnum] = (Cnode){r + 1, Q[i].r, l - 1, Cnum, 0}, r = Q[i].r;if (r > Q[i].r) ++ Cnum, C[Cnum] = (Cnode){Q[i].r + 1, r, l - 1, Cnum, 0}, r = Q[i].r;if (l < Q[i].l) ++ Cnum, C[Cnum] = (Cnode){l, Q[i].l - 1, r, Cnum, 0}, l = Q[i].l;if (l > Q[i].l) ++ Cnum, C[Cnum] = (Cnode){Q[i].l, l - 1, r, Cnum, 0}, l = Q[i].l;}
}
void Work() {sort(C + 1, C + Cnum + 1, [&](Cnode a, Cnode b){ return a.x < b.x; });for (int i = 0; i < MX ; ++ i) sum[i] = 0;for (int i = 1, nw = 0; i <= Cnum ; ++ i) {while (nw < C[i].x) {++ nw;for (int j = 1; j <= bnum ; ++ j) ++ sum[a[nw] ^ b[j]];}int Sum = 0;for (int j = C[i].l; j <= C[i].r ; ++ j) Sum += sum[a[j]];C[i].ans = Sum;} for (int i = 0; i < MX ; ++ i) sum[i] = 0;for (int i = 1; i <= n ; ++ i) {s0[i] = s0[i - 1] + sum[a[i]];for (int j = 1; j <= bnum ; ++ j) ++ sum[a[i] ^ b[j]];s1[i] = s1[i - 1] + sum[a[i]];}
}
void Solve() {sort(C + 1, C + Cnum + 1, [&](Cnode a, Cnode b){ return a.id < b.id; });for (int i = 1, l = 1, r = 0, nw = 0; i <= m ; ++ i) {Ans[Q[i].id] = Ans[Q[i - 1].id];if (r < Q[i].r) ++ nw, Ans[Q[i].id] += (s0[Q[i].r] - s0[r]) - C[nw].ans, r = Q[i].r;if (r > Q[i].r) ++ nw, Ans[Q[i].id] -= (s0[r] - s0[Q[i].r]) - C[nw].ans, r = Q[i].r;if (l < Q[i].l) ++ nw, Ans[Q[i].id] += (s1[Q[i].l - 1] - s1[l - 1]) - C[nw].ans, l = Q[i].l;if (l > Q[i].l) ++ nw, Ans[Q[i].id] -= (s1[l - 1] - s1[Q[i].l - 1]) - C[nw].ans, l = Q[i].l;}
}
signed main() {
#ifndef ONLINE_JUDGEfreopen("a.in", "r", stdin);
#endifread(n), read(m), read(K), sz = (int)sqrt(n);for (int i = 1; i <= n ; ++ i) read(a[i]);for (int i = 1; i <= m ; ++ i) read(Q[i].l), read(Q[i].r), Q[i].id = i;Init();Work();Solve();for (int i = 1; i <= m ; ++ i) print(Ans[i]), putc('\n');return 0;
}