参考了许多别人的代码之后成功A了这题
注意题目中的一句话
打字机会显示第 x 个打印的字符串在第 y 个打印的字符串中出现了多少次。
想到这个就是fail树上的一个性质，ac自动机中fail指针能跳到的点（b点）跟这个点（a点）的关系。
b点是a点的最长后缀
那么我们只需要沿着fail指针 一直跳 跳到另一个点为止 其实是两个点沿着fail指针有多少距离
发现一个fail树上 一个点的子树都会包含这个点 所以 就可以想到dfn序啦 之后再用树状数组维护一下前缀和就好了 另外第一个点一定是被包含在其他字符串中的所以 一开始在树状数组中的第一个dfn【0】的位置 + 1
#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define ull unsigned long long
#define PI acos(-1)
#define pb(x) push_back(x)
#define il inline
#define re register
#define IO; ios::sync_with_stdio(0);cin.tie(0);
#define ls (o<<1)
#define rs (o<<1|1)
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+5;
const int maxm = 1e5+5;
const int INF = 0x3f3f3f3f;
const ll LINF = 3e17+1;
const int mod = 1e9+7;
int n, r,h;
inline int read(){
register int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){
if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){
x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
return (f==1)?x:-x;
}
struct Tree
{
int fa;
int fail;
int vis[26];
int end;
}Aho[maxn];
struct edge
{
int v, nxt;
}E[maxn];
int c[maxn];
int head[maxn], eid;
int idx, Q[maxn], qid;
char s[maxn];
int dfn[maxn], R[maxn], tim;
vector<pii >ask[maxn];
int lowbit(int x)
{
return x&-x;
}
void update(int x, int val)
{
while (x < maxn)
{
c[x] += val;
x += lowbit(x);
}
}
int getsum(int x)
{
int res = 0;
while (x)
{
res += c[x];
x -= lowbit(x);
}
return res;
}
void add(int u, int v)
{
E[eid] = {
v,head[u]};
head[u] = eid++;
}
il void insert(char str[])
{
int len = strlen(str+1);
int now = 0;
for (int i = 1; i <= len; i++)
{
if (str[i] == 'P')
{
Q[++qid] = now;
}
else if (str[i] == 'B')
{
now = Aho[now].fa;
}
else
{
int ch = str[i] - 'a';
if (!Aho[now].vis[ch])
Aho[now].vis[ch] = ++idx;
Aho[Aho[now].vis[ch]].fa = now;
now = Aho[now].vis[ch];
}
Aho[now].end = 1;
}
}
void get_fail()
{
queue<int>q;
for (int i = 0; i < 26; i++)
{
if (Aho[0].vis[i])
{
Aho[Aho[0].vis[i]].fail = 0;
q.push(Aho[0].vis[i]);
}
}
while(!q.empty())
{
int u = q.front(); q.pop();
for (int i = 0; i < 26; i++)
{
if (Aho[u].vis[i])
{
Aho[Aho[u].vis[i]].fail = Aho[Aho[u].fail].vis[i];
q.push(Aho[u].vis[i]);
}
else
Aho[u].vis[i] = Aho[Aho[u].fail].vis[i];
}
}
}
void dfs(int u)
{
dfn[u] = ++ tim;
for (int i = head[u]; i != -1; i = E[i].nxt)
{
int v = E[i].v;
dfs(v);
}
R[u] = tim;
}
int ans[maxn];
int main()
{
scanf("%s", s+1);
insert(s);
get_fail();
scanf("%d", &n);
mem(head,-1);
for (int x,y,i = 1; i <= n; i++)
{
scanf("%d %d",&x, &y);
ask[y].push_back({
x,i});
}
for (int i = 1; i <= idx; i++)
add(Aho[i].fail, i);
dfs(0);
update(dfn[0],1);
int tt = 0;
int lens = strlen(s+1);
int now = 0;
for (int i = 1; i <= lens; i++)
{
if (s[i] == 'P')
{
tt++;
for (int j = 0; j < ask[tt].size(); j++)
{
int x = Q[ask[tt][j].first];
ans[ask[tt][j].second] = getsum(R[x]) - getsum(dfn[x] - 1);
}
}
else if (s[i] == 'B')
{
update(dfn[now], -1);
now = Aho[now].fa;
}
else
{
int ch = s[i] - 'a' ;
now = Aho[now].vis[ch];
update(dfn[now],1);
}
}
for (int i = 1; i <= n; i++)
{
printf("%d\n", ans[i]);
}
return 0;
}