Oulipo POJ - 3461
题目链接:https://vjudge.net/problem/POJ-3461
The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:
Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…
Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.
So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
- One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
- One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.
Sample Input
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
Sample Output
1 3 0
题目大意:输出a串在b串中出现的次数。
思路:Hash板子题 当然kmp也可以做 在这里为了存Hash板子 所以只放Hash的写法
看代码:
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #include<vector> using namespace std; typedef long long LL; typedef unsigned long long ull; const int maxn=1e6+5; const int maxm=100000+10; const ull base=2333; char a[maxn],b[maxn]; ull sum[maxn]; ull has,pw; int main() { int T;scanf("%d",&T); while(T--) { int ans=0; has=0;pw=1; scanf("%s%s",a+1,b+1); int lena=strlen(a+1);//!!!注意 这里一定是a+1而不是a int lenb=strlen(b+1); // cout<<lena<<" "<<lenb<<endl; for(int i=1;i<=lena;i++)//算出a的hash值 { pw*=base; has=has*base+a[i]; } for(int i=1;i<=lenb;i++)//算出b的前缀Hash值 { sum[i]=sum[i-1]*base+b[i]; } for(int i=lena;i<=lenb;i++) { if(sum[i]-sum[i-lena]*pw==has) ans++; } printf("%d\n",ans); } return 0; } /** */