Yet Another Broken Keyboard

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一只路过的弱鸡

最新推荐文章于 2024-08-22 09:25:25 发布

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本文介绍了一道编程题目，涉及到一个长度为n的字符串s和一个包含k个可用拉丁字母的键盘。目标是计算字符串s中能用键盘上可用字母输入的子串数量。给出了输入输出示例及题目解析，解释了如何计算连续可用字符的子串总数。

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C. Yet Another Broken Keyboard

time limit per test2 seconds
memory limit per test256 megabytes
input: standard input
output: standard output

Recently, Norge found a string s=s1s2…sn consisting of n lowercase Latin letters. As an exercise to improve his typing speed, he decided to type all substrings of the string s. Yes, all n(n+1)/2 of them!

A substring of s is a non-empty string x=s[a…b]=sasa+1…sb(1≤a≤b≤n). For example, “auto” and “ton” are substrings of “automaton”.

Shortly after the start of the exercise, Norge realized that his keyboard was broken, namely, he could use only k Latin letters c1,c2,…,ck out of 26.

After that, Norge became interested in how many substrings of the string s he could still type using his broken keyboard. Help him to find this number.

Input

The first line contains two space-separated integers n and k(1≤n≤2⋅10⁵, 1≤k≤26) — the length of the string s and the number of Latin letters still available on the keyboard.

The second line contains the string s consisting of exactly n lowercase Latin letters.

The third line contains k space-separated distinct lowercase Latin letters c1,c2,…,ck— the letters still available on the keyboard.

Output

Print a single number — the number of substrings of s that can be typed using only available letters c1,c2,…,ck.

Examples

Input

7 2
abacaba
a b

Output

Input

10 3
sadfaasdda
f a d

Output

Input

7 1
aaaaaaa
b

Output

题意：

给出一个字符串，然后给出k个字符，只保留字符串里k的字符的部分，然后求有多少个子串。
某div3的c题，也是一道简单题（只要看懂题）。而我一开始没读懂题意，以为是求字符串里非k字符删去之后的字符串有多少个b不同的子串。直接丢了一个模板上去，然后本地跑的时候发现跟答案并不对。于是研究了大概五分钟后才发现，原来是求每一部分的子串和（题目连公式都给出来了！！！）
英文杀我.jpg
所以策略就是找出连续的k字符有多长然后直接丢进公式再丢进sum就好了

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,k;
string s,s1;
int a[105];
ll sum=0,num=0;

int main(){
	cin>>n>>k;
	cin>>s;
	while(k--){
		cin>>s1;
		a[s1[0]-'a']=1;
	}	
	for(int i=0;i<=n;i++){
		if(a[s[i]-'a']==1){
			num++;
		}else{
			sum = sum + num*(num+1)/2;
			num=0;
		}
	}
	cout<<sum;
	return 0;
}