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(chǎn)’, then ‘b’, ‘c’ and ‘a(chǎn)’ 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(chǎn)’ (here you’re making a mistake and start performing the combo from the beginning), ‘a(chǎn)’, ‘b’, ‘c’, (here you’re making a mistake and start performing the combo from the beginning), ‘a(chǎn)’ (note that at this point you will not perform the combo because of the mistake), ‘b’, ‘c’, ‘a(chǎn)’.

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(chǎn)’, 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(chǎn)’ 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：其實(shí)在草稿紙上畫畫思路就挺明白的了。這個(gè)題目，最本質(zhì)就是求每一個(gè)位置數(shù)了多少次。但是每一個(gè)位置上有字母，因此我們可以計(jì)算每一個(gè)字母計(jì)算了多少次。我的做法是樹狀數(shù)組，但是我感覺差分也可以做。我們首先將字符串反轉(zhuǎn)（這里要想明白），這樣每一個(gè)位置也要變成n-a[i]+1了。然后套用樹狀數(shù)組的板子，求出每個(gè)位置數(shù)過的次數(shù)，換算到字母上就可以了。
代碼如下：

#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;/*------樹狀數(shù)組------*/ 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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