CF1272 C Yet Another Broken Keyboard 题解+代码比对

C. Yet Another Broken Keyboard

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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.

Output

Print a single number — the number of substrings of ss that can be typed using only available letters c1,c2,…,ckc1,c2,…,ck.

Examples

input

Copy

7 2
abacaba
a b

output

Copy

12

input

Copy

10 3
sadfaasdda
f a d

output

Copy

21

input

Copy

7 1
aaaaaaa
b

output

Copy

0

Note

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].

还是先学长的
学长原贴： https://www.cnblogs.com/xyq0220/p/12036109.html

学长思路：

在字符串s中所有不能打出的字母的位置作为隔板，隔出来的每个子串(设len为子串的长度)对答案的贡献即为len×(len+1)/2。

 1 #include<bits/stdc++.h>
 2 #define fi first
 3 #define se second
 4 #define lson l,mid,p<<1
 5 #define rson mid+1,r,p<<1|1
 6 #define pb push_back
 7 #define ll long long
 8 using namespace std;
 9 const int inf=1e9;
10 const int mod=1e9+7;
11 const int maxn=2e5+10;
12 int n,k;
13 char s[maxn],t[30];
14 int main(){
15     ios::sync_with_stdio(false);
16     //freopen("in","r",stdin);
17     cin>>n>>k;
18     cin>>s+1;
19     for(int i=1;i<=k;i++){
20         char c;
21         cin>>c;
22         t[c-'a']=1;
23     }
24     ll ans=0,cnt=0;
25     for(int i=1;i<=n;i++){
26         if(!t[s[i]-'a']){
27             ans+=cnt*(cnt+1)/2;
28             cnt=0;
29         }else ++cnt;
30     }
31     ans+=cnt*(cnt+1)/2;
32     cout<<ans<<endl;
33     return 0;
34 }

我的代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     long long int n,temp=0,k,biao[26],count=0;
 5     long long int sum =0;
 6     char str[200005],tc;
 7     memset(biao,0,sizeof(biao));
 8     memset(str,'\0',sizeof(str));
 9     cin>>n>>k;
10     temp = n;
11     cin>>str;
12     while(k--){
13         cin>>tc;
14         biao[tc-'a'] = 1;
15     }
16     for(int i = 0;i<n;i++){
17         if(biao[str[i]-'a']){
18             count += 1;
19         }
20         else{
21             sum += count*(count+1)/2;
22             count = 0;
23         }
24     }
25     sum += count*(count+1)/2;
26     cout<<sum<<endl;
27     return 0;
28 } 

其实这题又和学长思路一样，把缺的字母当隔板，对每个隔间累加求总值。

可能是这题比较简单（但是比赛我就蒙了？？？），所以代码差别不大

都用了标记法，这也是这种题的常用手法。