标签：decided cin pre insert asa tin frame class and
Recently, Norge found a string s=s1s2…sns=s1s2…sn consisting of nn lowercase Latin letters. As an exercise to improve his typing speed, he decided to type all substrings of the string ss. Yes, all n(n+1)2n(n+1)2 of them!
A substring of ss is a non-empty string x=s[a…b]=sasa+1…sbx=s[a…b]=sasa+1…sb (1≤a≤b≤n1≤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 kk Latin letters c1,c2,…,ckc1,c2,…,ck out of 2626.
After that, Norge became interested in how many substrings of the string ss he could still type using his broken keyboard. Help him to find this number.
The first line contains two space-separated integers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, 1≤k≤261≤k≤26) — the length of the string ss and the number of Latin letters still available on the keyboard.
The second line contains the string ss consisting of exactly nn lowercase Latin letters.
The third line contains kk space-separated distinct lowercase Latin letters c1,c2,…,ckc1,c2,…,ck — the letters still available on the keyboard.
Print a single number — the number of substrings of ss that can be typed using only available letters c1,c2,…,ckc1,c2,…,ck.
7 2 abacaba a b
12
10 3 sadfaasdda f a d
21
7 1 aaaaaaa b
0
In the first example Norge can print substrings s[1…2]s[1…2], s[2…3]s[2…3], s[1…3]s[1…3], s[1…1]s[1…1], s[2…2]s[2…2], s[3…3]s[3…3], s[5…6]s[5…6], s[6…7]s[6…7], s[5…7]s[5…7], s[5…5]s[5…5], s[6…6]s[6…6], s[7…7]s[7…7].
题倒是简单但是s=200000这样的数据用int就中招了
#include <iostream> #include <vector> #include <algorithm> #include <string> #include <set> #include <map> #include <sstream> #include <cstdio> #include <cstring> #include <numeric> #include <cmath> #define LL long long using namespace std; int main() { int n, k; cin >> n >> k; string s; cin >> s; set<char> all; for (int i = 0; i < k; i++) { char t; cin >> t; all.insert(t); } LL cnt=0,sum=0; for (int i = 0; i < s.size(); i++) { if (all.find(s[i]) != all.end()) { cnt++; } else { sum += cnt*(cnt + 1) / 2; cnt = 0; } if(i==s.size()-1 && all.find(s[i]) != all.end()) sum += cnt*(cnt + 1) / 2; } cout << sum; //system("pause"); return 0; }
标签：decided cin pre insert asa tin frame class and
原文地址：https://www.cnblogs.com/dealer/p/12269004.html
