Codeforces Round #166 (Div. 2) 271D

Codeforces Round #166 (Div. 2) 271D

Problem Description

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  slsl + 1…sr.
The substring s[l…r] is good, if among the letters  sl, sl + 1, …, sr 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.

Examples

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”.

---

题意

26个字母中，有一些字母是好的，一些字母是坏的。给定一个字符串和k，问这个字符串中有多少种子串满足坏的字符个数不大于k。

思路

hash一下，然后O(N²)直接暴力找。

坑点

双哈希会超时。
然后第8发样例非常的dog，我WA和超时一直都在第八发样例。。。

---

代码

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll MOD1=23333333333333333;
//const ll MOD2=1e9+9;
const int N=1e7+5;
const int HS1=1313131;
//const int HS2=131;

char str[1505];
int gg[30];
map<ll,int> vis1;
//unordered_map<ll,int> vis2;

int main()
{
    scanf("%s",str+1);
    int len=strlen(str+1);
    for(int i=0; i<26; i++)
    {
        scanf("%1d",&gg[i]);
    }
    int k;
    scanf("%d",&k);
    int ans=0;
    for(int i=1;i<=len;i++)
    {
        ll hs1=0;
//        int hs2=0;
        int cnt=0;
        for(int j=i;j>=1;j--)
        {
            if(gg[str[j]-'a']==0)
                cnt++;
            if(cnt>k)
                break;
            hs1=(hs1*HS1+(str[j]-'a'+1))%MOD1;
//            hs2=(hs2*HS2+(str[j]-'a'+1))%MOD2;
            if(vis1[hs1]==0/*&&vis2[hs2]==0*/)
            {
                vis1[hs1]=1;
//                vis2[hs2]=1;
                ans++;
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}