C. Perform the Combo
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You want to perform the combo on your opponent in one popular fighting game. The combo is the string s consisting of n lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in s. I.e. if s=“abca” then you have to press ‘a’, then ‘b’, ‘c’ and ‘a’ again.

You know that you will spend m wrong tries to perform the combo and during the i-th try you will make a mistake right after pi-th button (1≤pi<n) (i.e. you will press first pi buttons right and start performing the combo from the beginning). It is guaranteed that during the m+1-th try you press all buttons right and finally perform the combo.

I.e. if s=“abca”, m=2 and p=[1,3] then the sequence of pressed buttons will be ‘a’ (here you’re making a mistake and start performing the combo from the beginning), ‘a’, ‘b’, ‘c’, (here you’re making a mistake and start performing the combo from the beginning), ‘a’ (note that at this point you will not perform the combo because of the mistake), ‘b’, ‘c’, ‘a’.

Your task is to calculate for each button (letter) the number of times you’ll press it.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤104) — the number of test cases.

Then t test cases follow.

The first line of each test case contains two integers n and m (2≤n≤2⋅105, 1≤m≤2⋅105) — the length of s and the number of tries correspondingly.

The second line of each test case contains the string s consisting of n lowercase Latin letters.

The third line of each test case contains m integers p1,p2,…,pm (1≤pi<n) — the number of characters pressed right during the i-th try.

It is guaranteed that the sum of n and the sum of m both does not exceed 2⋅105 (∑n≤2⋅105, ∑m≤2⋅105).

Output
For each test case, print the answer — 26 integers: the number of times you press the button ‘a’, the number of times you press the button ‘b’, …, the number of times you press the button ‘z’.

Example
inputCopy
3
4 2
abca
1 3
10 5
codeforces
2 8 3 2 9
26 10
qwertyuioplkjhgfdsazxcvbnm
20 10 1 2 3 5 10 5 9 4
outputCopy
4 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 9 4 5 3 0 0 0 0 0 0 0 0 9 0 0 3 1 0 0 0 0 0 0 0
2 1 1 2 9 2 2 2 5 2 2 2 1 1 5 4 11 8 2 7 5 1 10 1 5 2
Note
The first test case is described in the problem statement. Wrong tries are “a”, “abc” and the final try is “abca”. The number of times you press ‘a’ is 4, ‘b’ is 2 and ‘c’ is 2.

In the second test case, there are five wrong tries: “co”, “codeforc”, “cod”, “co”, “codeforce” and the final try is “codeforces”. The number of times you press ‘c’ is 9, ‘d’ is 4, ‘e’ is 5, ‘f’ is 3, ‘o’ is 9, ‘r’ is 3 and ‘s’ is 1.

题意：给一个长度为n的字符串 会犯m个错误 每次犯错误就要重新输入
问你所有字母共打了多少次

思路：根据会按错的位置，我们开个数组mp[i] 记录下犯错位置i的次数
然后一个数组c 来维护前缀和按错的字母的总和 然后每次碰到mp[i] 有值的话
我们去循环26个字母，把答案数组a加上当前维护的前缀的c*mp[i] 也就是当前位置犯错的次数就好了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define me(a,x) memset(a,x,sizeof a)
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define all(x) (x).begin(), (x).end()
#define pb(a) push_back(a)
#define paii pair<int,int>
#define pali pair<ll,int>
#define pail pair<int,ll>
#define pall pair<ll,ll>
#define fi first
#define se second
ll mp[200005];
int b[200005];
char s[200005];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
       ll a[130]={0};
       ll c[130]={0};
       int n,m;cin>>n>>m;
       cin>>s+1;
       for(int i=1;i<=n;i++) mp[i]=0;
       for(int i=1;i<=m;i++){
          cin>>b[i];
          mp[b[i]]++;
       }
       for(int i=1;i<=n;i++){
           a[s[i]]++;
           c[s[i]]++;
           if(mp[i]){
              for(int j='a';j<='z';j++){
                a[j]=a[j]+(mp[i]*c[j]);
              }
           }
       }
       for(int i='a';i<='z';i++) cout<<a[i]<<" ";
       puts("");

    }
    return 0;
}