poj3461http://poj.org/problem?id=3461
hash模板题
Description
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 Win the text T.
Sample Input
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIANSample Output
1
3
0一开始做的时候直接用pow求得差基，忘了会爆了，笑哭
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<set>
using namespace std;
#define maxn 1000010
#define base 13333
typedef unsigned long long ull;
char s1[10010];
char s2[maxn];
ull has2[maxn];
ull po[maxn];
void init()
{
po[0]=1;
for(int i=1; i<maxn; i++)
po[i]=po[i-1]*base;
}
ull gethas(int i, int l, ull has0[])
{
if(i==0) return has0[i+l-1];
return has0[i+l-1]-has0[i-1]*po[l]; //不能直接用pow，会爆，=-=忘了
}
int main()
{
int t, i;
cin>>t;
init();
while(t--)
{
scanf("%s%s", s1, s2);
ull has1 = 0;
int len1 = strlen(s1), len2 = strlen(s2);
for(i=0; i<len1; i++)
has1 = has1*base+s1[i]-'A'+1;
has2[0] = s2[0]-'A'+1;
for(i=1; i<len2; i++)
{
has2[i] = has2[i-1]*base+s2[i]-'A'+1;
}
int ans = 0;
for(i=0; i<len2-len1+1; i++)
{
if(s2[i]==s1[0])
{
if(gethas(i, len1, has2)==has1)
ans++;
}
}
cout<<ans<<endl;
}
return 0;
}