description：

长度为n的字符串 m次询问 第i个位置上的字符在i之前出现几次

solution：

题意有点绕 只要用一个数组f[i] 来代表字符i在i之前出现多少次 数组标记字符出现多少次即可
也可以考虑离线操作

code:

#include <bits/stdc++.h>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define mes(x, a) memset(x, a, sizeof(x));
#define sca(a) scanf("%d", &a)
#define lowbit(x) x &(-x)
#define mk make_pair
#define pb(x) push_back(x)
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define lson v << 1
#define rson v << 1 | 1
#define pii pair<int, int>

inline void read(int &x) {
  x = 0;
  int flag_read = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag_read = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  x *= flag_read;
}

void out(int x) {
  if (x > 9) {
    out(x / 10);
  }
  putchar(x % 10 + '0');
}

const int N = 1000005;

char s[N];
int f[N],vis[35];

int main() {
  int n, m;

  scanf("%d%d", &n, &m);

  scanf("%s", s + 1);

  for(int i = 1;i <= n;i ++){
      vis[s[i] - 'a'] ++ ;
      f[i] = vis[s[i] - 'a'];
  }

  while(m --){
      int x;
      scanf("%d",&x);
      printf("%d\n",f[x]);
  }

  return 0;
}