【CodeForces 1272C --- Yet Another Broken Keyboard】

题目来源：点击进入【CodeForces 1272C — Yet Another Broken Keyboard】

Description

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.

Sample Input

7 2
abacaba
a b

Sample Output

12

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
bool arr[26];
using ll = long long;

int main()
{
    SIS;
    char ch;
    int n,k;
    cin >> n >> k;
    string s;
    cin >> s;
    for(int i=0;i<k;i++)
    {
        cin >> ch;
        arr[ch-'a']=true;
    }
    ll ans=0,cnt=0;
    for(int i=0;i<n;i++)
    {
        if(arr[s[i]-'a']) cnt++;
        else cnt=0;
        ans+=cnt;
    }
    cout << ans << endl;
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
using ll = long long;

int main()
{
    SIS;
    int n,k,cnt=0;
    cin >> n >> k;
    string s;
    char ch;
    cin >> s;
    map<int,bool> m;
    for(int i=0;i<k;i++)
    {
        cin >> ch;
        m[ch-'a']=true;
    }
    vector<ll> v;
    for(int i=0;i<n;i++)
    {
        if(m[s[i]-'a']) cnt++;
        else if(cnt) v.emplace_back(cnt),cnt=0;
    }
    if(cnt) v.emplace_back(cnt);
    ll len=v.size(),ans=0;
    for(int i=0;i<len;i++)
        ans+=v[i]*(v[i]+1)/2;
    cout << ans << endl;
    return 0;
}