Codeforces Round #624 (Div. 3) problem C
作者：互联网
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.
题解
每次出错，重新开始写字母。
例如：abca，出错1，3
则为
a; abc; abca. (a出现四次，b出现2次，c出现2次)
#include <bits/stdc++.h>
using namespace std;
int M[200010];
void optimize_cpp_stdio() {//减少cin的时间
ios::sync_with_stdio(false);
cout.tie(NULL);
cin.tie(NULL);
}
int main() {
optimize_cpp_stdio();
int t, n, m;
cin >> t;
while(t--) {
string s;
cin >> n >> m;
cin >> s;
for (int i = 0; i < m; i++) cin >> M[i];
vector<int> a[26];//记录每个字母都在第几个
for(int i = 0;i<(s.size()); i++) {
a[s[i]-'a'].push_back(i + 1);
}
int ans[27] ={0};
for (int i = 0; i < m; i++) {
for (int j = 0; j < 26; j++) {
if (a[j].size()) {
//每次出错，编号小于等于出错的数字的字母会重复出现
//cnt记录每次出错该字母都要加多少
int cnt = upper_bound(a[j].begin(),a[j].end(),M[i]) - a[j].begin();//upper_bound返回大于M[i]的地址，减去首地址，为小于等于M[i]的数量
ans[j] += cnt;
}
}
}
for (int i = 0; i < 26; i++) {
cout << (ans[i] + a[i].size()) << " ";
}
cout << endl;
}
}
rainbowsea_1
发布了18 篇原创文章 · 获赞 17 · 访问量 322
私信
关注
标签：case,624,number,times,press,combo,test,problem,Div 来源： https://blog.csdn.net/rainbowsea_1/article/details/104562019