Codeforces 271D - Good Substrings [字典树]

传送门

D. Good Substrings

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

You've got string s, consisting of small English letters. Some of the English letters are good, the rest are bad.

A substring s[l...r] (1 ≤ l ≤ r ≤ |s|) of string s  =  s₁s₂...s_|s| (where |s| is the length of string s) is string  s_ls_l + 1...s_r.

The substring s[l...r] is good, if among the letters  s_l, s_l + 1, ..., s_r there are at most k bad ones (look at the sample's explanation to understand it more clear).

Your task is to find the number of distinct good substrings of the given string s. Two substrings s[x...y] and s[p...q] are considered distinct if their content is different, i.e. s[x...y] ≠ s[p...q].

Input

The first line of the input is the non-empty string s, consisting of small English letters, the string's length is at most 1500 characters.

The second line of the input is the string of characters "0" and "1", the length is exactly 26 characters. If the i-th character of this string equals "1", then the i-th English letter is good, otherwise it's bad. That is, the first character of this string corresponds to letter "a", the second one corresponds to letter "b" and so on.

The third line of the input consists a single integer k (0 ≤ k ≤ |s|) — the maximum acceptable number of bad characters in a good substring.

Output

Print a single integer — the number of distinct good substrings of string s.

Sample test(s)

Input

ababab
01000000000000000000000000
1

Output

5

Input

acbacbacaa
00000000000000000000000000
2

Output

8

Note

In the first example there are following good substrings: "a", "ab", "b", "ba", "bab".

In the second example there are following good substrings: "a", "aa", "ac", "b", "ba", "c", "ca", "cb".

9813017

2015-02-13 06:13:52

njczy2010

271D - Good Substrings

GNU C++

Accepted

248 ms

53000 KB

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<string>

 #define N 2250005
 #define M 1505
 #define mod 10000007
 //#define p 10000007
 #define mod2 1000000000
 #define ll long long
 #define LL long long
 #define eps 1e-6
 #define inf 100000000
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int n;
 int T;
 int flag;
 int ccount;
 char ss[M];
 char f[];

 typedef struct
 {
     int v;
     vector<int>nt;
 }PP;

 PP p[N];
 int le;

 int k;
 int tot;

 void insert(char s[])
 {
     int l=strlen(s);
     int i;
     int ff;
     int st=;
     int y;
     vector<int>::iterator it;
     for(i=;i<l;i++){
         ff=;
         for(it=p[st].nt.begin();it!=p[st].nt.end();it++){
             y=*it;
             if(p[y].v==s[i]-'a'){
                 ff=;
                 st=y;
                 break;
             }
         }
         if(ff==){
             p[st].nt.push_back(tot);
             p[tot].v=s[i]-'a';
             p[tot].nt.clear();
             st=tot;
             tot++;
         }
     }
 }

 void ini()
 {
     ccount=;
     int i;
     scanf("%s",f);
     scanf("%d",&k);
     flag=;
     p[].nt.clear();
     p[].v=-;
     tot=;
     scanf("%d",&n);
     le=strlen(ss);
     for(i=;i<le;i++){
         insert(ss+i);
     }
 }

 void check(int st,int now)
 {
     //printf(" st=%d v=%d now=%d\n",st,p[st].v,now);
     int y;
     if(now>k) return;
     ccount++;
     vector<int>::iterator it;
     for(it=p[st].nt.begin();it!=p[st].nt.end();it++)
     {
         y=*it;
         //printf("  st=%d y=%d f=%c\n",st,y,f[ p[y].v ]);
         if(f[ p[y].v ]==''){
             check(y,now);
         }
         else{
             check(y,now+);
         }
     }
 }

 void solve()
 {
     //printf("solve\n");
     ccount=-;
     check(,);
 }

 void out()
 {
     printf("%d\n",ccount);
 }

 int main()
 {
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     //scanf("%d",&T);
     //for(int ccnt=1;ccnt<=T;ccnt++)
     //while(T--)
     //scanf("%d%d",&n,&m);
     while(scanf("%s",ss)!=EOF)
     {
         ini();
         solve();
         out();
     }
     return ;
 }