POJ3461Oulipo 题解

题目大意：

求字符串A在字符串B中出现的次数。

思路：

KMP板题，用Hash也可水过~要学习KMP可参考http://blog.csdn.net/u011564456/article/details/20862555

代码：

KMP

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5
 6 char s1[10009],s2[1000009];
 7 int next[10009];
 8
 9 int main()
10 {
11     int n,i,l1,l2,j,ans;
12     scanf("%d",&n);
13     while (n--)
14     {
15         scanf("%s%s",s1,s2);
16         l1=strlen(s1),l2=strlen(s2);
17         for(j=next[0]=0,i=1;i<l1;i++)
18         {
19             for (;j && s1[j]^s1[i];j=next[j-1]);
20             if (s1[i]==s1[j]) j++;
21             next[i]=j;
22         }
23         for (i=j=ans=0;j<l2;j++)
24         {
25             for (;i && s1[i]^s2[j];i=next[i-1]);
26             if (s1[i]==s2[j]) i++;
27             if (i==l1) ans++,i=next[i-1];
28         }
29         printf("%d\n",ans);
30     }
31     return 0;
32 }

Hash

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 const int S=999984,mod=1000000007,N=10009,M=1000009;
 6 char s1[N],s2[M];
 7 long long Hash1[N],Hash2[M],mi[M];
 8
 9 int gethash(int l,int r)
10 {
11     return (Hash2[r]-Hash2[l-1]*mi[r-l+1]%mod+mod)%mod;
12 }
13
14 int main()
15 {
16     int n,i,j,l1,l2,ans;
17     for (mi[0]=i=1;i<=M;i++) mi[i]=mi[i-1]*S%mod;
18     scanf("%d",&n);
19     while (n--)
20     {
21         scanf("%s%s",s1+1,s2+1);
22         l1=strlen(s1+1),l2=strlen(s2+1);
23         for (i=1;i<=l1;i++) Hash1[i]=(Hash1[i-1]+s1[i])*S%mod;
24         for (i=1;i<=l2;i++) Hash2[i]=(Hash2[i-1]+s2[i])*S%mod;
25         for (ans=0,i=1;i<=l2-l1+1;i++)
26             if (gethash(i,i+l1-1)==Hash1[l1]) ans++;
27         printf("%d\n",ans);
28     }
29     return 0;
30 }