前缀和水题

已经断更n天了.....

最近由于忙于sql和前端的学习,已经很久没有oi了,在最近的一次oi中水的不堪入目,总结了一下,今天终于来更新一波

Background:

You want to perform the combo on your opponent in one popular fighting game. The combo is the string ss consisting of nn lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in ss. I.e. if s=s="abca" then you have to press 'a', then 'b', 'c' and 'a' again.

You know that you will spend mm wrong tries to perform the combo and during the ii-th try you will make a mistake right after pipi-th button (1≤pi<n1≤pi<n) (i.e. you will press first pipi buttons right and start performing the combo from the beginning). It is guaranteed that during the m+1m+1-th try you press all buttons right and finally perform the combo.

I.e. if s=s="abca", m=2m=2 and p=[1,3]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 tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.

Then tt test cases follow.

The first line of each test case contains two integers nn and mm (2≤n≤2⋅1052≤n≤2⋅105, 1≤m≤2⋅1051≤m≤2⋅105) — the length of ss and the number of tries correspondingly.

The second line of each test case contains the string ss consisting of nn lowercase Latin letters.

The third line of each test case contains mm integers p1,p2,…,pmp1,p2,…,pm (1≤pi<n1≤pi<n) — the number of characters pressed right during the ii-th try.

It is guaranteed that the sum of nn and the sum of mm both does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105, ∑m≤2⋅105∑m≤2⋅105).

It is guaranteed that the answer for each letter does not exceed 2⋅1092⋅109.

Output

For each test case, print the answer — 2626 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

问题分析,首先一眼看上去就有n^2版的暴力思路(其他题又何尝不是呢),但那样会tle,这道题卡常十分严格,同时间复杂度的算法会被卡常卡掉,我试了好几次接近于暴力的思路都被

卡常卡掉了,正确做法前缀和(模板),不能有任何变动,我也是服了,对于像我这样的老年选手为什么不能开放一点~~

<a herf="http://codeforces.com/problemset/problem/1311/C"></a>

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int t,n,d;
char str[200005];
int ans[30],a[200005];
int ans1[200005][30];
int main()
{
    ios::sync_with_stdio(false);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&d);
        scanf("%s",str+1);
        for(int i=1; i<=n; i++)
        {
            ans[str[i]-'a'+1]+=1;
            ans1[i][str[i]-'a'+1]+=1;
            for(int j=1; j<=26; j++)
            {
              ans1[i][j]+=ans1[i-1][j];
            }
        }
        for(int i=1; i<=d; i++)
           scanf("%d",&a[i]);
        for(int i=1; i<=d; i++)
            for(int j=1;j<=26;j++)
            {
                ans[j]+=ans1[a[i]][j];
            }
        for(int i=1; i<=26; i++)
            printf("%d ",ans[i]);
         putchar('
');
        memset(ans,0,sizeof(ans));
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=26; j++)
                ans1[i][j]=0;
        }
    }
    return 0;
}