C. Watto and Mechanism
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Watto, the owner of a spare parts store, has recently got an order for the mechanism that can process strings in a certain way. Initially the memory of the mechanism is filled with n strings. Then the mechanism should be able to process queries of the following type: "Given string s, determine if the memory of the mechanism contains string t that consists of the same number of characters as s and differs from s in exactly one position".
Watto has already compiled the mechanism, all that‘s left is to write a program for it and check it on the data consisting of n initial lines and m queries. He decided to entrust this job to you.
Input
The first line contains two non-negative numbers n and m (0?≤?n?≤?3·105, 0?≤?m?≤?3·105) — the number of the initial strings and the number of queries, respectively.
Next follow n non-empty strings that are uploaded to the memory of the mechanism.
Next follow m non-empty strings that are the queries to the mechanism.
The total length of lines in the input doesn‘t exceed 6·105. Each line consists only of letters ‘a‘, ‘b‘, ‘c‘.
Output
For each query print on a single line "YES" (without the quotes), if the memory of the mechanism contains the required string, otherwise print "NO" (without the quotes).
Examples
input
Copy
2 3aaaaaacacacaaabaaccacacccaaac
output
Copy
YESNONO
暴力哈希 卡种子 CF出题人 就是强 HASH各种卡种子
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long LL;
4
5 const LL seed = 257;
6 const LL mod = 1e9 + 7;
7 const int maxn = 6e5 + 10;
8 int n,m;
9 LL p[maxn];
10 char str[maxn];
11 set<LL>st;
12 void init(){
13 p[0]=1;
14 for (int i=1 ;i<maxn ;i++)
15 p[i]=p[i-1]*seed%mod;
16 }
17 LL Hash(char s[]){
18 LL ret=0;
19 for (int i=0 ; s[i] ;i++)
20 ret=(ret*seed+s[i])%mod;
21 return ret;
22 }
23 int check(char s[]){
24 LL h=Hash(s);
25 int len=strlen(s);
26 for (int i=0 ;i<len ;i++){
27 for (int j=‘a‘ ; j<=‘c‘ ;j++) {
28 if (j==s[i]) continue;
29 LL now=((((j-s[i])*p[len-1-i]%mod)+mod)+h)%mod;
30 if (st.find(now)!=st.end()) return 1;
31 }
32 }
33 return 0;
34 }
35 int main() {
36 scanf("%d%d", &n, &m);
37 init();
38 for (int i = 0 ; i < n ; i++) {
39 scanf("%s", str);
40 st.insert(Hash(str));
41 }
42 for (int i = 0 ; i < m ; i++) {
43 scanf("%s", str);
44 if (check(str)) printf("YES\n");
45 else printf("NO\n");
46 }
47 return 0;
48 }
原文地址：https://www.cnblogs.com/qldabiaoge/p/9174539.html