「BZOJ 4105」平方运算-线段树

现有一个长度为 $n$ 的序列 ${x_1, x_2, \cdots, x_n}$,要求支持两种操作:

  1. 0 l r 表示将 $i \in [l, r], x_i \leftarrow x_i^2 \bmod p$
  2. 1 l r 询问 $\sum\limits_{i = l} ^ r x_i$

链接

BZOJ 4105

题解

我们可以通过打表发现题目所给的模数,很快就能够使 $x_i \rightarrow x_i ^ 2 \bmod p$ 发生循环(至多 $11$ 步),同时每个环的长度都小于 $60$。

所以我们可以用线段树来维护,先预处理出循环,每次平方操作就相当于在环上走一步,对于不在环上的先暴力走,至多 $11$ 步就会进入循环。

在线段树上分解区间,如果区间里面的所有数都在环中,那么这个区间节点维护所有数的环的 $\text{lcm}$,环上的第 $i$ 个节点维护若这个区间进行 $i$ 次平方操作,和是多少。

时间复杂度为 $O(kn \log n)$,$k$ 为循环长度。

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「BZOJ 4105」平方运算 12-10-2017
* 线段树
* @author xehoth
*/
#include <bits/stdc++.h>
namespace IO {
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
return s == t ? -1 : *s++;
}
template <typename T>
inline bool read(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return false;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
iosig ? x = -x : 0;
return true;
}
inline void read(char &c) {
while (c = read(), isspace(c) && c != -1)
;
}
inline int read(char *buf) {
register int s = 0;
register char c;
while (c = read(), isspace(c) && c != -1)
;
if (c == -1) {
*buf = 0;
return -1;
}
do
buf[s++] = c;
while (c = read(), !isspace(c) && c != -1);
buf[s] = 0;
return s;
}
const int OUT_LEN = 1000000;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
oh == obuf + OUT_LEN ? (fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf) : 0;
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void print(const char *s) {
for (; *s; s++) print(*s);
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
struct InputOutputStream {
template <typename T>
inline InputOutputStream &operator>>(T &x) {
read(x);
return *this;
}
template <typename T>
inline InputOutputStream &operator<<(const T &x) {
print(x);
return *this;
}
~InputOutputStream() { flush(); }
} io;
}
namespace {
using IO::io;
const int MAXN = 100000;
const int MAX_MOD = 9977;
bool inCircle[MAX_MOD + 1];
int n, m, next[MAX_MOD + 1], mod, a[MAXN + 1];
inline int lcm(int a, int b) { return a / std::__gcd(a, b) * b; }
struct Node {
Node *lc, *rc;
int sum, len, pos, tag, circle[62];
bool in;
Node();
inline void *operator new(size_t);
inline void init(const int v) {
sum = v;
if ((in = inCircle[v])) {
len = 1, circle[pos = 0] = v;
for (register int u = next[v]; u != v; u = next[u])
circle[len++] = u;
}
}
inline void maintain() {
sum = lc->sum + rc->sum;
if ((in = lc->in & rc->in)) {
len = lcm(lc->len, rc->len);
register int tx = lc->pos, ty = rc->pos;
for (register int i = 0; i < len; i++) {
circle[i] = lc->circle[tx++] + rc->circle[ty++];
(tx == lc->len) ? tx = 0 : 0;
(ty == rc->len) ? ty = 0 : 0;
}
pos = 0;
}
}
inline void pushDown() {
if (tag) {
lc->tag += tag, rc->tag += tag, lc->pos += tag, rc->pos += tag;
(lc->pos >= lc->len) ? lc->pos %= lc->len : 0;
lc->sum = lc->circle[lc->pos];
(rc->pos >= rc->len) ? rc->pos %= rc->len : 0;
rc->sum = rc->circle[rc->pos];
tag = 0;
}
}
} pool[MAXN * 4 + 1], *null = pool, *cur = pool + 1, *root;
inline void *Node::operator new(size_t) { return cur++; }
Node::Node() : lc(null), rc(null) {}
void build(Node *&p, int l, int r) {
p = new Node();
if (l == r) {
p->init(a[l]);
return;
}
register int mid = l + r >> 1;
build(p->lc, l, mid), build(p->rc, mid + 1, r), p->maintain();
}
void modify(Node *p, int l, int r, int s, int t) {
if (l == s && t == r && p->in) {
p->tag++, p->pos++;
(p->pos == p->len) ? p->pos = 0 : 0;
p->sum = p->circle[p->pos];
return;
}
if (l == r) {
p->init(next[p->sum]);
return;
}
p->pushDown();
register int mid = l + r >> 1;
if (t <= mid)
modify(p->lc, l, mid, s, t);
else if (s > mid)
modify(p->rc, mid + 1, r, s, t);
else
modify(p->lc, l, mid, s, mid), modify(p->rc, mid + 1, r, mid + 1, t);
p->maintain();
}
int query(Node *p, int l, int r, int s, int t) {
if (l == s && t == r) return p->sum;
p->pushDown();
register int mid = l + r >> 1;
if (t <= mid)
return query(p->lc, l, mid, s, t);
else if (s > mid)
return query(p->rc, mid + 1, r, s, t);
return query(p->lc, l, mid, s, mid) + query(p->rc, mid + 1, r, mid + 1, t);
}
inline void init() {
for (register int i = 1; i <= n; i++) io >> a[i];
for (register int i = 0; i < mod; i++)
next[i] = i * i % mod, inCircle[i] = true;
static bool vis[MAX_MOD + 1];
for (register int i = 0, y; i < mod; i++) {
if (!vis[i]) {
for (y = i; !vis[y];) vis[y] = true, y = next[y];
for (register int j = i; j != y; j = next[j]) inCircle[j] = false;
}
}
}
inline void solve() {
io >> n >> m >> mod;
init();
build(root, 1, n);
for (register int cmd, l, r; m--;) {
io >> cmd >> l >> r;
switch (cmd) {
case 0: {
modify(root, 1, n, l, r);
break;
}
case 1: {
io << query(root, 1, n, l, r) << '\n';
break;
}
}
}
}
}
int main() {
#ifdef DBG
freopen("sample/1.in", "r", stdin);
#endif
solve();
return 0;
}
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