https://vjudge.net/contest/435353#problem/E
https://codeforces.com/problemset/problem/1234/D

You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l;r] of the string s is the string slsl+1…sr. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
1 pos c (1≤pos≤|s|, c is lowercase Latin letter): replace spos with c (set spos:=c);
2 l r (1≤l≤r≤|s|): calculate the number of distinct characters in the substring s[l;r].
Input
The first line of the input contains one string s consisting of no more than 105 lowercase Latin letters.
The second line of the input contains one integer q (1≤q≤105) — the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it — the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6

开26个树状数组/线段树，每个对应一个字符，维护1-i中出现了几次。
1操作把原来的--，新的++
2操作分别求一下所有字符的个数，非0就对答案有1个贡献，累计输出即可。

#include<bits/stdc++.h>
#define ll long long 
using namespace std;
ll a[31][100010],c[31][100010];
ll n;
char ch[100010];
ll lowbit(ll i) {return i&(-i);}
void Build()
{
    for(ll i=1;i<=n;++i)
    {
        a[(ll)(ch[i]-'a'+1)][i]=1;
    }

    for(ll o=1;o<=26;++o)
    {
        for(ll i=1;i<=n;++i)
        {
            ll j=i;
            while(j<=n) {c[o][j]+=a[o][i];j+=lowbit(j);}
        }
    }
}
void add(ll i,ll k,ll v)
{
    ll val=v-a[k][i];
    a[k][i]=v;
    while(i<=n)
        c[k][i]+=val,i+=lowbit(i);
} 
ll sum(ll k,ll i)
{
    if(i==0) return 0;
    ll ans=0;
    while(i) {ans+=c[k][i];i-=lowbit(i);}
    return ans;
}
void Replace(ll i,char c)
{
    ll k=ch[i]-'a'+1;
    add(i,k,0);
    ch[i]=c;
    k=c-'a'+1;
    add(i,k,1);
}
ll getans(ll l,ll r)
{
    ll cnt=0;
    for(ll k=1;k<=26;++k)
    {
        if(sum(k,r)-sum(k,l-1)>0)
            ++cnt;
    }
    return cnt;
}
int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);cout.tie(0);
    ll q,x,y,z;char m;
    scanf("%s",ch+1);
    n=strlen(ch+1);
    Build();
    scanf("%lld",&q);
    for(int i=1;i<=q;++i)
    {
        scanf("%lld",&x);
        if(x==1)
        {
            scanf("%lld %c",&y,&m);
            Replace(y,m);
        }
        if(x==2)
        {
            scanf("%lld%lld",&y,&z);
            printf("%lld\n",getans(y,z));
        }
    }
    return 0;
}