Perform the Combo CodeForces - 1311C（字符串反转+树状数组）

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).

It is guaranteed that the answer for each letter does not exceed 2⋅109.

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
Input
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
Output
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.
思路：其实在草稿纸上画画思路就挺明白的了。这个题目，最本质就是求每一个位置数了多少次。但是每一个位置上有字母，因此我们可以计算每一个字母计算了多少次。我的做法是树状数组，但是我感觉差分也可以做。我们首先将字符串反转（这里要想明白），这样每一个位置也要变成n-a[i]+1了。然后套用树状数组的板子，求出每个位置数过的次数，换算到字母上就可以了。
代码如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;const int maxx=2e5+100;
int c[maxx],a[maxx],b[maxx],d[30];
string s;
int n,m;/*------树状数组------*/
inline int lowbit(int x){return x&-x;}
inline void add(int v)
{while(v<=n+10){c[v]+=1;v+=lowbit(v);}
}
inline int query(int v)
{int ans=0;while(v){ans+=c[v];v-=lowbit(v);}return ans;
}
int main()
{int t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(int i=0;i<=n+10;i++) c[i]=0;memset(d,0,sizeof(d));cin>>s;for(int i=1;i<=m;i++) scanf("%d",&a[i]);a[m+1]=n;reverse(s.begin(),s.end());for(int i=1;i<=m+1;i++) add(n-a[i]+1);for(int i=1;i<=n;i++) b[i]=query(i);for(int i=0;i<n;i++) d[s[i]-'a']+=b[i+1];for(int i=0;i<26;i++) cout<<d[i]<<" ";cout<<endl;}return 0;
}

努力加油a啊，(^o)/~

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Perform the Combo CodeForces - 1311C（字符串反转+树状数组）

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