// ===== REFERENCES =====
// [1] duck.ac 用户 saffah_cc_v41_agg1(本账号),提交 #104213
// <https://duck.ac/submission/104213> 用途:本文件正文的**直接父件**(tc17 = 138.447 ms)=
// #103996 + 推迟 szsub 初始化 + emit 两位查表 + **匹配回退链每级只读一次 `s_len`**。
// 本发在其之上**只叠一件**(见思路),改动 15 行。
// [2] duck.ac 用户 saffah_cc_v41_agg1(本账号),提交 #103996
// <https://duck.ac/submission/103996> 用途:间接父件(`s_nxt` 20 位打包 + 计数排序代替 len 桶链表)。
// [3] duck.ac 用户 saffah_cc_v41_260924,提交 #107908
// <https://duck.ac/submission/107908> 用途:**当前榜首 T = 137.375591 ms**,本发的对标对象
// (本发与它只差一个"回退链怎么判'有没有转移'"的表示选择)。
// [4] **本文件正文内的两条自述**:
// (a) `s_nxt` 的注释:"State ids are <= 2*MAXS+8 = 1000018 < 2^20, so 20 bits are lossless."
// ⇒ 状态号是 20 位;`sget(p,c)` 是"从 65 B 行里取出 20 位项"。
// (b) 回退环上方的注释:"The sget below is the very address the walk just tested => L1 hit."
// ⇒ 父件已经认识到"环内那次 `sget` 的地址就是停下来的那一个",本发只是**把这个洞察推到底**:
// 既然环内只需要"有没有转移"这一个布尔量,那它就不必从 65 MB 的 `s_nxt` 里取。
// [5] `problems/noi18c/notes.md` 第四轮(子代理 `n1b_`)**已经写好、本地 101 组逐字节对拍通过、
// 但【从未上过判题机定价】**的那把刀 —— 本发就是把它拿去定价。
// 许可证:以上均为 duck.ac 公开提交与公开题面,未见许可证声明;本文件按 RULES §3 逐项署名引用。
// ======================
// ===== 思路 =====
// 【口径(`python3 tools/exact.py noi18c`)】
// mine = **138.446515**(#104213)· T = **137.375591**(#107908,saffah_cc_v41_260924)
// ⇒ 宽支线 1.005*T+1us = 138.063469(现实差 **0.383046 ms**);
// ★ 我方任何新最好件都会翻严支 ⇒ 真正要跨的栏是 **0.99*T+1us = 136.002835**,
// **差 2.443680 ms(1.765%)**。本发即为此一发。
//
// 【本发 = #104213 正文 + 一件(单变量)】
// **ARM MSK —— 用"26 位转移掩码"把回退链的判据从 65 MB 的 `s_nxt` 搬到 TLB 驻留的小数组**
// * 站点:tc17 里占 **61%(85.6 ms)** 的匹配回退环
// `while(v!=1 && !sget(v,c)) v = (int)s_link[v];`
// * 现状:每一级都要**从 `s_nxt`((2*MAXS+10)×65 ≈ 65 MB,tc17 实测 ~39 MB 活跃)里取一个 20 位项**,
// 只为回答一个**布尔**问题:"状态 v 有没有字符 c 的转移?"
// * 本发:为每个状态维护一个 **26 位掩码**(`u32 s_msk[MAXST]`,tc17 上 ~2.4 MB ≈ 600 页,
// 而 `s_nxt` 是 ~9500 页)—— 位 c 置位 ⟺ `sget(state,c) != 0`(构建期 `sset(...)` 处顺手置位、
// clone 处整体复制,**转移从不删除**,所以掩码是精确的,不是近似)。
// 回退环改成 `while(v!=1 && !((s_msk[v]>>c)&1u)) v = (int)s_link[v];`
// ⇒ **环内一次 TLB-missing 的 65 MB 随机访存,换成一次 TLB 驻留的 2.4 MB 随机访存**;
// `s_nxt` 从此**只在环停下来的那个状态上被碰一次**(紧接着的 `int nx = sget(v,c);` 就是那次,
// 且是 L1 命中)。
// * 机理量级:串行回退链上每次随机访存 ~250–300 周期,其中约 100 周期是**页行走**;
// 把这一步换到 600 页的数组上,页行走基本消失。**这是"缩小被随机访问的数组"族**
// (`§2.19.779` / `§2.19.870`:先确认数组是"我方自建"—— `s_msk` 是引擎自己在构建期写的,
// 不是输入别名,符合 `§2.19.862` 的前置检查)。
// * **逐位等价**:本机把 93 组语料(覆盖 eul_ready=0/1、2/4/7/26 字母、周期串、`:full`/`:rand`、
// 含 |S|=500000 的三组大测)同时喂给父件与本发,**93/93 输出逐字节相同**。
//
// 【判题机定价】见提交结果(本发是这把刀**第一次**上判题机)。
// ⚠ 注意本席**不**用 `tools/probe.py` 给本刀定价:`problems/noi18c/notes.md` 已记录该通道的硬限制
// ——输入体积 ~100 KB 封顶(|S| 只能到 ~95000),而 `s_nxt` 在 39 MB 与 4 MB 两种尺度下的
// TLB 行为**完全不同**,父件历史上"probe 好看 −5.7%~−9.3% ↔ tc17 只兑现 −1.16%"就是这条的实证。
// ⇒ **缩足迹类改动一律用正式提交定价。**
//
// 【同批还备着的下一把(本发未叠,留给下一发或下一个席位)】
// * `s_len`(4 MB) / `s_link`(4 MB) 目前都是 u32,而状态号 < 2^20(20 位)、`s_len` ≤ |S|(19 位)
// ⇒ 两者都有"宽度"余量;且 `s_link[v]` 每级都读、`s_len[v]` 只在停下的那一级读
// ⇒ 合并成一条 5 字节行可让两者同处一条 cache line(`§2.19.867`:静态声明宽度不花时间,
// **只有访问步长花** ⇒ 改动必须落在"步长/行"上,而不是只改类型)。
// * SAM(T) 阶段(13.8 ms)里 `tzero` 每状态写 104 B;arena 生命期互斥可共用地址。
// * **已证伪、不要再试**:`mmap`/`MADV_HUGEPAGE`/`MAP_POPULATE`(评测端禁用)·
// `#pragma GCC optimize("O3",...)`(评测端**反而慢 3.9%**)· `{len(link),link}` 8B 合并(probe +4~6%)。
// * **反例(别重蹈)**:把 SAM(T) 构建**融合进**匹配循环想吃 MLP ⇒ **+1.1 ms**;
// 全 `r>=|S|` 时跳过询问计数排序 ⇒ 其余 24 点都变快、**tc17 反而 +0.32 ms**。
// ===== REFERENCES =====
// [1] duck.ac 用户 saffah_codex_6s_agg2,提交 #103842 <https://duck.ac/submission/103842>
// (本题当前榜首)用途:**直接复制**了对方相对我们 #99767 的两处改动,逐字节复刻:
// (a) SAM(T) 的转移表 t_nxt 也做 3 字节打包:`u32 t_nxt[(2*MAXT+8)*26]` ->
// `u8 t_nxt[(2*MAXT+10)*80] __attribute__((aligned(16)))`;tget/tset 改成 3 字节偏移上的
// 4 字节 load + `&0xFFFFFF` / 4 字节读-改-写(保留高字节);tzero 由 3 条 32B + 1 条 16B
// store(104 B)改成 5 条对齐 16B store(80 B);clone 的 memcpy 由 104 B 变 80 B。
// (b) 区间最大值的块长 `EBS 32 -> 8`(`EBSH 5 -> 3`)。
// 本地归档:problems/noi18c/ref2/rival_103842.cpp
// [2] duck.ac 用户 saffah_cc_v41_260924,提交 #99767 <https://duck.ac/submission/99767>
// 用途:直接复制了本文件的原型(我们自己的最好件 problems/noi18c/work/nc_sn3.cpp)。
// [3] duck.ac 用户 saffah_cc_v41_260924,提交 #87737 <https://duck.ac/submission/87737>
// 用途:直接复制了本题的全部算法与代码骨架(SAM(S)/SAM(T)、后缀链接树 Euler 区间、
// 按 r 单调激活、区间最大值查询流程、输入解析与输出格式)。
// [4] duck.ac 用户 saffah_cc_v41_260924,提交 #99553 <https://duck.ac/submission/99553>
// 用途:**参考了思想**(不是复制代码)——「纯分块扫描 + 两级块和」的区间最大值结构
// (eactivate 只写 ev/ebmax/eb2 三个位置、qmax 用 32B 单条取数 + 重叠尾块取数)。
// [5] duck.ac 用户 saffah_cc_v41_260924,提交 #99482 <https://duck.ac/submission/99482>
// 用途:**参考了思想** —— Euler 区间惰性构造(只算 mxo[],区间/tin/tout 延迟到第一次
// 真的需要区间最大值时才做)。
// 以上均为公开提交,未见附带的许可证声明;此处按「作者账号 + 原始提交地址 + 所用内容」逐条注明。
// ======================
// ===== 思路 =====
// 【1】复刻 REFERENCES [1](这一步把对方相对我们 #99767 的两处差距归零)。
//
// 【2】s_nxt 由 3 字节/项(78 B/行)压到 **20 位/项(65 B/行)**,行宽 -17%。
// 状态号 <= 2*MAXS+8 = 1000018 < 2^20 ⇒ 20 位无损(严格小于 2^20 = 1048576,余量 4.6%)。
// 项 c 落在行内第 20c 位;(20c)&7 = 4*(c&1) 恒为 0 或 4,所以只要在字节偏移
// floor(20c/8) = (5c)>>1 处做一次 **4 字节** load 就一定能取到整项(20+4 = 24 <= 32):
// 取值 = (u32 load >> ((c&1)<<2)) & 0xFFFFF;
// 存值 = 同一处的 4 字节读-改-写(掩掉 20 位、其余原样写回 ⇒ 不污染邻居项/相邻行)。
// 行宽 65 是奇数 ⇒ 相邻行的 4KB 页内相位逐行平移 1(顺带破 4KB 同余,BRIEF §2.18.323)。
// 本方向(**只做更小更紧凑**)是本族唯一被反复验证的正面方向:
// 104B -> 78B(我们 #99767,-4.59 ms),而两次"加大足迹"(16B 结构体打包、8B 字段合并)
// 分别实测 2.1x 更差 / +4~6%。本次是同一个方向的继续。
//
// 【3】build_meta 的「按 len 分桶链表」换成 **计数排序**(去掉串行指针追逐)。
// * SAM 构造时把 `bnext[cur]=bhead[L]; bhead[L]=cur;`(一次随机写)改成 `bhead[len]++`
// (递增写,几乎顺序);**整个 bnext[] 数组(s_sz*4 B)不再存在、不再被触碰**。
// * 走一遍计数排序把状态按 len 升序摊进 ord_[](散列那趟是 2 次无链式依赖的 RMW,
// MLP 好),于是自底向上那趟不再有 `v = bnext[v]` 这条每状态一次随机访存的**串行链**。
// * ord_ 现在是 **len 升序**,所以 build_meta(自底向上)改为**逆序**扫;
// build_intervals 的两趟随之对调(正序 = 父先 = 前序编号;逆序 = 子先 = 子树大小)。
//
// 【判定机上的实测量(tools/probe.py 立即体验,真实评测机,同一驱动同一输入)】
// 三个尺度、每格都是判题机 wall time / run 内 rdtsc 相位周期:
// scale #99767 血统(3B,链) 本件(20bit+计数排序)
// sz=121437 5.010 ms / 17.94 Mc 4.725 ms / 16.91 Mc (-5.7%)
// sz=187969 (2字母) 7.506 ms / 26.92 Mc 6.808 ms / 24.41 Mc (-9.3%)
// sz= 66682 5.453 ms / 19.53 Mc 5.310 ms / 19.01 Mc (-2.6%)
// 相位:sam 14913930 -> 13826264 (-7.3%)、22131992 -> 20025232 (-9.5%);
// meta 4597358 -> 4188058 (-8.9%,2 字母大 clone 档);match/samt 基本不变。
// 本次为正式提交:直接冲击达标线。
// ================
// NOI2018 你的名字 -- JudgeDuck (MAIN style)
#include <string.h>
#include <stdint.h>
#include <immintrin.h>
#pragma GCC push_options
#pragma GCC target("avx2,popcnt,bmi,bmi2")
typedef unsigned char u8;
typedef unsigned int u32;
typedef unsigned long long u64;
typedef unsigned long ul;
struct DI {
ul abi; const char*ip; ul is; char*op; ul ol,os; char*ep; ul el,es;
const char*IBp; ul IBl; char*OBp; ul OBl; ul tsc;
} __attribute__((packed));
#define MAXS 500005
#define MAXST 1000016
#define MAXQ 100005
#define MAXT 1000008
#define MAXTT 2000016
static u32 bhead[MAXS+8], szsub[MAXST], nxt[MAXST]; // bhead = len histogram -> start offsets
static u32 tin[MAXST], tout[MAXST], mxo[MAXST];
static u32 ord_[MAXST]; static int nord_; // len-descending state order (lazy-Euler)
static int eul_ready;
// ancestor painting (see 思路): wl_last[x] = max own+1 among activated prefix-states
// whose suffix-link path passes through x == MX(x, r) exactly.
static u32 wl_last[MAXST];
static int wl_mode; static long long wl_budget;
// per-query paint-back scratch: zeroing only the states we painted keeps the
// reset O(sz) instead of O(MAXST) between queries.
static u32 plist_[MAXST];
// ======================= SAM of S (dense 26-way transitions) ==============
// transitions of state p are at s_nxt[p*26 .. p*26+26); every state index is
// used once per process, so BSS stays zero and fresh states have no transitions.
// Transitions packed as 24-bit little-endian words: 26*3 = 78 B per state instead of
// 26*4 = 104 B. State ids are <= 2*MAXS+8 < 2^24, so 24 bits are lossless.
// BUDGET (tc17: s_sz=661320) -- MEASURED sizes, not guesses:
// touched footprint 65.6 MB -> 49.2 MB (-16.4 MB) = -4197 first-touch pages.
// At the measured 244 ns/page that alone is -1.02 ms, and every sget in the SAM
// construction and in the match loop additionally reads a 25%-smaller random
// working set (better TLB reach), which is the part that should pay off more.
// COST: sget gains one AND; sset becomes a 4-byte read-modify-write (the extra byte
// read back is the *next* entry's low byte and is written back unchanged, so
// neighbours are preserved) = ~+3..5 uops per access. Reads use a 4-byte load at
// the 3-byte offset, so the array carries 10 spare rows of padding.
/* --- s_nxt: 20-bit packed transitions (row = 26*20 bits = 65 bytes) ---------
State ids are <= 2*MAXS+8 = 1000018 < 2^20, so 20 bits are lossless.
Row stride 65 B instead of 78 B (-17% footprint on the dominant table).
Entry c of row p starts at bit 20c; (20c)&7 == 4*(c&1) is always 0 or 4, so a
single unaligned 8-byte load at byte ((5c)>>1) always contains the entry. */
static u8 s_nxt[(2*MAXS+10)*65 + 16];
/* 26-bit "has a transition by char c" mask, one u32 per state. At tc17 s_sz ~ 6e5
this is 2.4 MB (600 pages) against s_nxt's 39 MB (9500 pages): the mask stays inside
the TLB while s_nxt cannot, so a NEGATIVE lookup (the common case in the suffix-link
walk: most states have no transition by the queried character) costs a TLB-resident
load instead of a TLB-missing one inside s_nxt. The mask is exact: a bit is set iff
sget(state,c) != 0, and transitions are never deleted. */
static u32 s_msk[MAXST];
static u32 s_len[MAXST], s_link[MAXST];
static int s_sz, s_last;
static u32 s_own[MAXS];
#define S20_MASK 0xFFFFFu
static inline int sget(int p, int c){
size_t off = (size_t)p*65 + (size_t)(((unsigned)c*5u)>>1);
u32 x; __builtin_memcpy(&x, s_nxt + off, 4);
return (int)((x >> (((unsigned)c&1u)<<2)) & S20_MASK);
}
static inline void sset(int p, int c, int t){
size_t off = (size_t)p*65 + (size_t)(((unsigned)c*5u)>>1);
u32 x; __builtin_memcpy(&x, s_nxt + off, 4);
unsigned sh = ((unsigned)c&1u)<<2;
x = (x & ~(S20_MASK << sh)) | (((u32)t & S20_MASK) << sh);
__builtin_memcpy(s_nxt + off, &x, 4);
}
static void build_sam_s(const char* S, int n){
s_sz = 1; s_last = 1;
s_len[1] = 0; s_link[1] = 0;
memset(bhead, 0, (size_t)(n+2)*4);
for(int i=0;i<n;i++){
int c = S[i]-'a';
int cur = ++s_sz;
s_len[cur] = s_len[s_last]+1;
bhead[s_len[cur]]++;
mxo[cur] = (u32)(i+1);
s_own[i] = (u32)cur;
int p = s_last;
for(; p && !sget(p,c); p = (int)s_link[p]){ sset(p,c,cur); s_msk[p] |= (1u<<(unsigned)c); }
if(!p) s_link[cur] = 1;
else {
int q = sget(p,c);
if((int)s_len[p]+1 == (int)s_len[q]) s_link[cur] = (u32)q;
else {
int cl = ++s_sz;
s_len[cl] = s_len[p]+1; s_link[cl] = s_link[q];
bhead[s_len[cl]]++;
mxo[cl] = 0;
memcpy(s_nxt + (size_t)cl*65, s_nxt + (size_t)q*65, 65);
s_msk[cl] = s_msk[q];
for(; p && sget(p,c)==q; p = (int)s_link[p]) sset(p,c,cl);
s_link[q] = (u32)cl; s_link[cur] = (u32)cl;
}
}
s_last = cur;
}
}
// ======================= SAM of T, rebuilt per query (dense) ==============
static u8 t_nxt[(2*MAXT+10)*80] __attribute__((aligned(16)));
static u32 t_len[MAXTT], t_link[MAXTT];
static int t_sz, t_last;
static inline int tget(int p, int c){
u32 x; __builtin_memcpy(&x, t_nxt + (size_t)p*80 + (size_t)c*3, 4);
return (int)(x & 0xFFFFFFu);
}
static inline void tset(int p, int c, int t){
size_t off = (size_t)p*80 + (size_t)c*3;
u32 x; __builtin_memcpy(&x, t_nxt + off, 4);
x = (x & 0xFF000000u) | ((u32)t & 0xFFFFFFu);
__builtin_memcpy(t_nxt + off, &x, 4);
}
static inline void tzero(int p){
u8* row = t_nxt + (size_t)p*80;
__m128i z = _mm_setzero_si128();
_mm_store_si128((__m128i*)(row+0), z);
_mm_store_si128((__m128i*)(row+16), z);
_mm_store_si128((__m128i*)(row+32), z);
_mm_store_si128((__m128i*)(row+48), z);
_mm_store_si128((__m128i*)(row+64), z);
}
// ============ suffix-link tree: len-bucket lists + two linear passes ======
static int s_sn;
// ---- counting sort of the states by len (replaces the bucket linked list) ----
// ord_ ends up len-ASCENDING, so pass 1 (bottom-up mxo) walks it backwards and
// pass 2 (top-down preorder) walks it forwards. No linked list => no serial
// pointer chase in the sweep, and no random write per state during construction.
static void build_ord(void){
int n = s_sn;
u32 sum = 0;
for(int L=0; L<=n; L++){ u32 c = bhead[L]; bhead[L] = sum; sum += c; }
for(u32 v=1; v<=(u32)s_sz; v++){ ord_[bhead[s_len[v]]++] = v; }
nord_ = (int)sum;
}
// Pass 1 (ALWAYS needed): mxo[v] = max own+1 over the subtree of v, by a
// len-descending sweep (children strictly before parents). The same sweep
// records the visiting order into ord_[] (sequential stores), so pass 2 never
// has to re-walk the bucket chains.
static void build_meta(void){ // bottom-up: children (larger len) first
for(int i = nord_-1; i >= 0; i--){
u32 v = ord_[i];
u32 p = s_link[v];
if(p && mxo[v] > mxo[p]) mxo[p] = mxo[v];
}
}
// Pass 2 (LAZY): subtree sizes + preorder numbers. Only the range-max over
// [tin,tout) uses these, so they are built at the first query that actually
// needs a range-max. ord_[] is len-descending, so forward = bottom-up sizes,
// backward = top-down preorder.
static void build_intervals(void){
// szsub[] is initialised here instead of in build_sam_s: it is only ever read by
// this function, so the per-state store in the construction loop (and the 2.4 MB of
// pages it dirtied) is pure overhead on tests that never build the intervals.
for(int i=0;i<nord_;i++) szsub[ord_[i]] = 1;
szsub[1] = 0; // state 1's size is never read (its link is 0), exactly as before
for(int i=nord_-1;i>=0;i--){ // len-descending = bottom-up sizes
u32 v = ord_[i]; u32 p = s_link[v];
if(p) szsub[p] += szsub[v];
}
tin[1] = 1; nxt[1] = 2;
for(int i=0;i<nord_;i++){ // len-ascending = parents first = preorder
u32 v = ord_[i];
if(v == 1) continue;
u32 p = s_link[v];
u32 t = nxt[p];
tin[v] = t; tout[v] = t + szsub[v];
nxt[p] = t + szsub[v];
nxt[v] = t + 1;
}
}
// ============ range max over the Euler array (fused block scans) ============
// Replaces the earlier "block prefix/suffix arrays + segment tree over blocks".
// MEASURED on the real judge (tools/probe.py, n=5e5 random 26-letter S, cycles):
// old eactivate 70.15M old build_meta 15.58M old build_intervals 22.38M
// old qmax ~96 cyc/call
// The old activation wrote a whole 64-entry prefix block AND a whole 64-entry
// suffix block per activation (512 B = 8 cache lines of RFO traffic), plus a
// 15-level segment-tree update -- 140 cycles per activation. With plain block
// scans the activation is 3 stores (one cold, two L1-resident) and the segment
// tree disappears.
#define EBS 8
#define EBSH 3
#define GBS 128
#define GBSH 7
#define ENB ((MAXST>>EBSH)+8)
#define ESB ((ENB>>GBSH)+8)
static u32 ev[MAXST+8];
static u32 ebmax[ENB+8];
static u32 eb2[ESB+8];
static inline void eactivate(int leaf, u32 val){
ev[leaf] = val;
int b = leaf >> EBSH;
ebmax[b] = val;
eb2[b >> GBSH] = val;
}
#define AVX2T __attribute__((target("avx2")))
AVX2T static inline __m256i ldu256(const void *p){
__m256i r; __asm__("vmovdqu %1, %0" : "=x"(r) : "m"(*(const __m256i *)p)); return r; }
// accumulate max over p[0..n) into acc for ANY n>=0 (n<8 must not use the
// overlapping tail load -- it would read BEFORE p and pick up foreign values)
AVX2T static inline __m256i accblk(__m256i acc, const u32* p, int n){
if(n <= 0) return acc;
if(n < 8){
for(int i=0;i<n;i++) acc = _mm256_max_epu32(acc, _mm256_set1_epi32((int)p[i]));
return acc;
}
int i = 0;
for(; i+8 <= n; i += 8) acc = _mm256_max_epu32(acc, ldu256(p+i));
if(i < n) acc = _mm256_max_epu32(acc, ldu256(p+n-8)); // overlapping tail load
return acc;
}
AVX2T static inline u32 red256(__m256i acc){
__m128i lo = _mm256_castsi256_si128(acc), hi = _mm256_extracti128_si256(acc,1);
__m128i m = _mm_max_epu32(lo,hi);
m = _mm_max_epu32(m, _mm_shuffle_epi32(m,_MM_SHUFFLE(1,0,3,2)));
m = _mm_max_epu32(m, _mm_shuffle_epi32(m,_MM_SHUFFLE(2,3,0,1)));
return (u32)_mm_cvtsi128_si32(m);
}
AVX2T static inline u32 scanmax(const u32* p, int n){
if(n <= 0) return 0;
return red256(accblk(_mm256_setzero_si256(), p, n));
}
static inline u32 qmax(int l, int r){ // max over [l,r)
if(l>=r) return 0;
int b1=l>>EBSH, b2=(r-1)>>EBSH;
if(b1==b2) return scanmax(ev+l, r-l);
__m256i acc = _mm256_setzero_si256();
acc = accblk(acc, ev+l, ((b1+1)<<EBSH) - l);
int j = r & (EBS-1);
if(j) acc = accblk(acc, ev+(b2<<EBSH), j);
else acc = accblk(acc, ebmax+b2, 1);
int lo=b1+1, hi=b2, cnt=hi-lo; // whole blocks [lo,hi)
if(cnt > 0){
if(cnt <= GBS){
acc = accblk(acc, ebmax+lo, cnt);
} else {
int s1=(lo+GBS-1)&~(GBS-1), s2=hi&~(GBS-1);
if(s1>lo) acc = accblk(acc, ebmax+lo, s1-lo);
if(s2>s1) acc = accblk(acc, eb2+(s1>>GBSH), (s2-s1)>>GBSH);
if(hi>s2) acc = accblk(acc, ebmax+s2, hi-s2);
}
}
return red256(acc);
}
// ============================== query handling ============================
struct QE { const char* t; int tl, l, r, idx; };
static QE qs[MAXQ];
static QE qs2[MAXQ];
static long long answers[MAXQ];
static int f_[MAXS];
static int cnt_[MAXS+2];
static char* OUTP;
static const char D2[201] =
"00010203040506070809101112131415161718192021222324252627282930313233343536373839"
"40414243444546474849505152535455565758596061626364656667686970717273747576777879"
"8081828384858687888990919293949596979899";
static inline void emit(long long v){
char b[24]; int c=0;
if(v==0){ *OUTP++='0'; *OUTP++='\n'; return; }
while(v >= 100){
unsigned r = (unsigned)(v % 100); v /= 100;
b[c++] = D2[r*2+1]; b[c++] = D2[r*2];
}
if(v >= 10){ unsigned r = (unsigned)v; b[c++] = D2[r*2+1]; b[c++] = D2[r*2]; }
else if(v > 0){ b[c++] = (char)('0' + (int)v); }
while(c) *OUTP++ = b[--c];
*OUTP++='\n';
}
static void run(const u8* in, ul insz, char* out){
const char* p = (const char*)in;
const char* e = p + insz;
OUTP = out;
while(p<e && (*p<'a'||*p>'z')) p++;
const char* sstart = p;
while(p<e && *p>='a'&&*p<='z') p++;
int sn = (int)(p - sstart);
build_sam_s(sstart, sn);
while(p<e && (*p<'0'||*p>'9')) p++;
int Q=0;
while(p<e && *p>='0'&&*p<='9'){ Q = Q*10 + (*p-'0'); p++; }
if(Q>MAXQ) Q=MAXQ;
for(int i=0;i<Q;i++){
while(p<e && (*p<'a'||*p>'z')) p++;
const char* ts = p;
while(p<e && *p>='a'&&*p<='z') p++;
int tl = (int)(p-ts);
if(tl>MAXT) tl=MAXT;
while(p<e && (*p<'0'||*p>'9')) p++;
int l=0; while(p<e && *p>='0'&&*p<='9'){ l=l*10+(*p-'0'); p++; }
while(p<e && (*p<'0'||*p>'9')) p++;
int r=0; while(p<e && *p>='0'&&*p<='9'){ r=r*10+(*p-'0'); p++; }
qs[i].t = ts; qs[i].tl = tl; qs[i].l = l; qs[i].r = r; qs[i].idx = i;
}
s_sn = sn;
int sz = s_sz;
// ---- strategy choice ---------------------------------------------------
// The ancestor-painting routes are KEPT in the file but no longer selected:
// real-judge measurement showed their cost is sum_i depth(s_own[i]) steps, which
// the 256-sample estimate can under-predict (a deep tree blows it up ~4x), while
// the Euler route is now bounded and cheap: with the block-scan activation its
// fixed cost is 34 cyc/state (intervals) + 20 cyc/activation + 113 cyc/qmax.
// Forcing the Euler route makes the tree phase predictable instead of data-dependent.
wl_mode = 0; wl_budget = 0;
build_ord();
if(!wl_mode) build_meta();
eul_ready = 0;
// ev / ebmax / etree are zero initialized (BSS): no clearing needed
{ // counting sort of the queries by r
int mx = sn+1;
memset(cnt_, 0, (size_t)(mx+2)*4);
for(int i=0;i<Q;i++){ int rr = qs[i].r; if(rr<0) rr=0; if(rr>mx) rr=mx; cnt_[rr]++; }
int acc=0;
for(int i=0;i<=mx;i++){ int c=cnt_[i]; cnt_[i]=acc; acc+=c; }
for(int i=0;i<Q;i++){ int rr = qs[i].r; if(rr<0) rr=0; if(rr>mx) rr=mx; qs2[cnt_[rr]++] = qs[i]; }
}
int act = 0; // positions < act already activated
long long wl_work = 0;
for(int qi=0; qi<Q; qi++){
const QE& q = qs2[qi];
int need = q.r; // activate positions 0..r-1
if(need > sn) need = sn;
if(need < 0) need = 0;
int npaint = 0;
if(wl_mode == 2){
// downward sweep, stop at the first state already painted this query:
// the first state painted gets the largest own+1, which is exactly MX(x, need)
for(int i=need-1;i>=0;i--){
int a = (int)s_own[i]; u32 val = (u32)(i+1);
while(a != 1 && !wl_last[a]){ wl_last[a] = val; plist_[npaint++] = (u32)a; a = (int)s_link[a]; }
}
} else if(wl_mode){
for(; act<need; act++){
int a = (int)s_own[act]; u32 val = (u32)(act+1);
wl_last[a] = val; wl_work++;
while(a != 1){ a = (int)s_link[a]; wl_last[a] = val; wl_work++; }
if(wl_work > wl_budget){ wl_mode = 0; break; } // blew the budget
}
if(!wl_mode){ // fell back mid-activation: replay into the Euler struct
build_meta(); build_intervals(); eul_ready = 1;
for(int i=0;i<act;i++){ int v = (int)s_own[i]; eactivate((int)tin[v], (u32)(i+1)); }
for(; act<need; act++){ int v = (int)s_own[act]; eactivate((int)tin[v], (u32)(act+1)); }
}
} else if(eul_ready){
for(; act<need; act++){
int v = (int)s_own[act];
eactivate((int)tin[v], (u32)(act+1));
}
} else if(need > act) act = need;
const char* T = q.t;
int tl = q.tl;
int l0 = q.l - 1;
int r1 = q.r - 1;
int maxlen = q.r - q.l + 1;
if(maxlen < 0) maxlen = 0;
int v = 1, len = 0, e = -1; // e = witness end position of the current match
for(int i=0;i<tl;i++){
int c = T[i]-'a';
// fast path: the match extends in place at the witness (f[i] <= f[i-1]+1, so len+1 is optimal)
if(len > 0 && e < r1 && sstart[e+1]-'a' == c){
v = sget(v,c); len++; e++;
f_[i] = len;
continue;
}
// Only the FINAL rung of this walk matters: `len` is rewritten on every step
// but read only after the loop, and once at least one step ran len == s_len[v].
// Reading s_len[] inside the loop therefore costs one extra random load per rung
// into a *separate* 2.4 MB array (one more TLB entry + one more miss) for
// nothing. The sget below is the very address the walk just tested => L1 hit.
{
int ov = v;
// The walk only needs to know WHETHER the transition exists, and the mask
// answers that from a TLB-resident array; s_nxt is then touched at most once,
// for the state where the walk stopped.
while(v!=1 && !((s_msk[v]>>(unsigned)c)&1u)) v = (int)s_link[v];
if(v != ov) len = (int)s_len[v];
}
int nx = sget(v,c);
if(nx){ v = nx; len++; } else { v = 1; len = 0; }
e = -1;
// clip the match to S[l..r]: longest suffix whose endpos hits [l+len-1, r-1]
while(len>0){
if(len > maxlen){
len = maxlen;
while(v != 1 && len <= (int)s_len[s_link[v]]) v = (int)s_link[v];
if(len == 0) break;
}
u32 m;
if(mxo[v] <= (u32)q.r) m = mxo[v];
else {
if(!eul_ready){
build_intervals();
for(int i2=0;i2<act;i2++){ int vv=(int)s_own[i2]; eactivate((int)tin[vv], (u32)(i2+1)); }
eul_ready = 1;
}
m = qmax((int)tin[v],(int)tout[v]);
}
// lengths in this state's class that can still satisfy the test
int lim = (int)m - l0;
if(lim >= len){ e = (int)m - 1; break; }
int pl = (int)s_len[s_link[v]];
if(lim > pl){ len = lim; e = (int)m - 1; break; }
len = pl;
if(len == 0) break;
v = (int)s_link[v];
}
f_[i] = len;
}
if(wl_mode == 2){ for(int k=0;k<npaint;k++) wl_last[plist_[k]] = 0; }
// SAM(T) -> g[], accumulate
t_sz=1; t_last=1;
t_len[1]=0; t_link[1]=0;
tzero(1);
long long ans=0;
for(int i=0;i<tl;i++){
int c = T[i]-'a';
int cur = ++t_sz;
t_len[cur] = t_len[t_last]+1; t_link[cur]=0;
tzero(cur);
int pp = t_last;
for(; pp && !tget(pp,c); pp = (int)t_link[pp]) tset(pp,c,cur);
int lk;
if(!pp) lk = 1;
else {
int qq = tget(pp,c);
if((int)t_len[pp]+1 == (int)t_len[qq]) lk = qq;
else {
int cl = ++t_sz;
t_len[cl] = t_len[pp]+1; t_link[cl] = t_link[qq];
memcpy(t_nxt + (size_t)cl*80, t_nxt + (size_t)qq*80, 80);
for(; pp && tget(pp,c)==qq; pp = (int)t_link[pp]) tset(pp,c,cl);
t_link[qq] = (u32)cl; lk = cl;
}
}
t_link[cur] = (u32)lk;
t_last = cur;
int g = (int)t_len[lk];
int m = g > f_[i] ? g : f_[i];
ans += (long long)(i+1) - m;
}
answers[q.idx] = ans;
}
for(int i=0;i<Q;i++) emit(answers[i]);
}
#pragma GCC pop_options
#ifndef LOCAL
int main(){ return 0; }
extern "C" void __libc_start_main(void*mm,int argc,char**argv){
ul *p=(ul*)(argv+argc+1);
while(*p) p++; p++;
DI *d=0; for(; p[0]; p+=2) if(p[0]==0x6b637564UL){ d=(DI*)p[1]; break; }
if(d){
const u8* in=(const u8*)d->IBp; ul insz=d->IBl;
if(!insz && d->is){ in=(const u8*)d->ip; insz=d->is; }
run(in,insz,d->op);
d->os=(ul)(OUTP-d->op);
}
__asm__ volatile("syscall"::"a"(60),"D"(0):"rcx","r11","memory");
for(;;);
}
#endif
| Compilation | N/A | N/A | Compile OK | Score: N/A | 显示更多 |
| Testcase #1 | 1.14 ms | 152 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #2 | 1.265 ms | 256 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #3 | 1.328 ms | 264 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #4 | 16.553 ms | 600 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #5 | 15.978 ms | 584 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #6 | 81.354 ms | 138 MB + 48 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #7 | 81.047 ms | 138 MB + 16 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #8 | 12.269 ms | 16 MB + 600 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #9 | 14.726 ms | 14 MB + 184 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #10 | 28.216 ms | 31 MB + 236 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #11 | 36.038 ms | 28 MB + 96 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #12 | 47.114 ms | 45 MB + 724 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #13 | 66.32 ms | 42 MB + 508 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #14 | 68.234 ms | 60 MB + 696 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #15 | 103.796 ms | 57 MB + 224 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #16 | 91.438 ms | 75 MB + 676 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #17 | 146.329 ms | 72 MB + 84 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #18 | 76.883 ms | 43 MB + 956 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #19 | 90.948 ms | 58 MB + 856 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #20 | 110.589 ms | 74 MB | Accepted | Score: 4 | 显示更多 |
| Testcase #21 | 121.337 ms | 89 MB + 696 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #22 | 131.678 ms | 89 MB + 568 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #23 | 129.85 ms | 89 MB + 516 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #24 | 123.909 ms | 89 MB + 664 KB | Accepted | Score: 4 | 显示更多 |
| Testcase #25 | 122.318 ms | 89 MB + 696 KB | Accepted | Score: 4 | 显示更多 |