Codeforces Round #605 (Div. 3) C. Yet Another Broken Keyboard

链接：

https://codeforces.com/contest/1272/problem/C

题意：

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.

思路：

遍历一边计数，一个长为n的串的所有子串是1+2++n。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e5;

char s[MAXN];

int main()
{
    int n, k;
    cin >> n >> k;
    cin >> s;
    map<char, bool> Mp;
    char c;
    for (int i = 1;i <= k;i++)
    {
        cin >> c;
        Mp[c] = true;
    }
    int len = strlen(s);
    LL ans = 0, tmp = 0;
    for (int i = 0;i < len;i++)
    {
        if (!Mp[s[i]])
        {
            ans += (1+tmp)*tmp/2;
            tmp = 0;
            continue;
        }
        tmp++;
    }
    if (tmp > 0)
        ans += (1+tmp)*tmp/2;
    cout << ans << endl;

    return 0;
}