之前有過區域賽，簡化版問題：

給定一個小寫字符組成的字符串S，（|S|<1e5，下標從1開始）,現在有Q種操作，對於每個操作Q(Q<=1e3)，輸入opt，

如果opt==1，輸入x，c，表示把S[x]改為c，（c是小寫字母）。

如果opt==2，輸入字符串T，輸出S種有多少個字串==T（字串可以重疊）,（|T|<=10）。

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此題：

Given a string s, process q queries, each having one of the following forms:

• 1 i c — Change the i-th character in the string to c.
• 2 l r y — Consider the substring of s starting at position l and ending at position r. Output the number of times y occurs as a substring in it.

Input

The first line of the input contains the string s

The second line contains an integer q (1 ≤ q ≤ 10⁵) — the number of queries to process.

The next q lines describe the queries and may have one of the following forms:

• 1 i
   c (1 ≤ i ≤ |s|)
• 2 l r y (1 ≤ l ≤ r ≤ |s|)

c is a lowercase English letter and y is a non-empty string consisting of only lowercase English letters.

The sum of |y| over all queries of second type is at most 10⁵.

It is guaranteed that there is at least one query of second type.

All strings are 1-indexed.

|s| is the length of the string s.

Output

For each query of type 2, output the required answer in a separate line.

Example

ababababa
3
2 1 7 aba
1 5 c
2 1 7 aba

3
1

abcdcbc
5
2 1 7 bc
1 4 b
2 4 7 bc
1 2 a
2 1 4 aa

2
2
1

Note

Consider the first sample case. Initially, the string aba occurs 3 times in the range [1, 7]. Note that two occurrences may overlap.

After the update, the string becomes ababcbaba and now aba occurs only once in the range [1, 7].

思路：題意和上面的差不多，不過數據更大一點，只有Bitset或者分塊優化（後者我沒有試過）。

具體的：1，假設我們要統計S裏有多少個T，先統計S裏面字符==T[0]的是哪些，然後統計S中有T[0]的位置後面跟的字符==T[1]的有哪些，然後統計S中有T[0]的位置後面跟的字符==T[1]的而且後面跟的字符==T[2]的有哪些.....直到對比到S[len-1]。

2，最後利用位移可以得到某個區間的1的個數。

for(i=0;i<S;i++)
   ans&=(bitset[T[i]-‘a‘]>>i);

ans開始全部是1； bitset是保存的T串裏每個字符的位置；

（目測還有非常高效的方法，提交列表裏有效率為5倍以上的，還有待學習）

#include<bitset>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=100010;
bitset<maxn>s[27],ans;
char a[maxn],b[maxn],c[3];
void read(int &res)
{
    char c=getchar();
    while(c>‘9‘||c<‘0‘) c=getchar();
    for(res=0;c>=‘0‘&&c<=‘9‘;c=getchar()) res=(res<<3)+(res<<1)+c-‘0‘;
}
int main()
{
    scanf("%s",a+1);
    int T=strlen(a+1);
    int i,j,l,r,Q,opt;
    for(i=1;i<=T;i++)
        s[a[i]-‘a‘].set(i);
    read(Q);
    while(Q--){
        read(opt);
        if(opt==1){
            scanf("%d%s",&j,c);
            s[a[j]-‘a‘][j]=0;
            s[(a[j]=c[0])-‘a‘][j]=1;
        }
        else {
            read(l); read(r);
            scanf("%s",b);
            int S=strlen(b);
            if(S>r-l+1) {
                 puts("0"); continue;
            }
            ans.set();
            for(i=0;i<S;i++){
                ans&=(s[b[i]-‘a‘]>>i);
            }
            printf("%d\n",(ans>>l).count()-(ans>>(r-S+2)).count());
        }
    }
    return 0;
}

Bitset與字串