Codeforces Round #166 (Div. 2) DhashGood Substrings

Posted 2022-02-10 SSL_ZZL

Good Substrings

• 题目
• 题目大意
• 解题思路
• Code

Codeforces Round #166 (Div. 2) A. Beautiful Year 题解
Codeforces Round #166 (Div. 2) D.Good Substrings

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题目

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 $s=s_1{s}_2...s_{|}s|$ (where |s| is the length of string s) is string  $s_l{s}_{l+1}...s_r$.

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. $s_{[x...y]}\ne s_{[p...q]}$.

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”.

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题目大意

给出一个字符串
给出每个字母是否好坏，好是1，坏是0
一个合格的子串，坏值不应该大于k
问有多少个不同的合格子串

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解题思路

因为只有1500个字符，所以可以暴力截取字符串，和算出坏值
去重就用hash
大大大大大大大大大大大大无语，不能mod，直接ull存😶

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Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#define P 10000019
#define ll unsigned long long

using namespace std;

string s, ss;
int l, k, now, ans, che[30];
ll h[10001000], ha[2000], base[2000], num;

int find(ll d) 
	int i = 0;
	while(h[(d + i) % P] != -1 && h[(d + i) % P] != d) i ++;
	return (d + i) % P ;


int main() 
	cin >> s;
	l = s.size();
	for(int i = 0; i < 26; i ++) 
		char c = getchar();
		while(!(c >= '0' && c <= '1')) c = getchar();
		che[i] = c - '0';
		che[i] = 1 - che[i];
	
	base[0] = 1ll;
	for(int i = 1; i <= l; i ++) 
		ha[i] = (ha[i - 1] * 131ll + 1ll * (s[i - 1] - 'a' + 1)) ;
		base[i] = base[i - 1] * 131ll ;
	
	scanf("%d", &k);
	memset(h, -1, sizeof(h));
	for(int i = 1; i <= l; i ++) 
		ss = "", now = 0;
		for(int j = i; j <= l ; j ++) 
			ss = ss + s[j - 1], now += che[s[j - 1] - 'a'];
			if(now > k) break;
			num = (ha[j] - ha[i - 1] * base[j - i + 1] ) ;
			int x = find(num);
			if(h[x] != num) 
				h[x] = num;
				ans ++;
			
		
	
	printf("%d", ans);