hdoj5978Sequence I【kmp】

Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 604    Accepted Submission(s): 237

Problem Description

Mr. Frog has two sequences  a1,a2,⋯,an and  b1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence  b1,b2,⋯,bmis exactly the sequence  aq,aq+p,aq+2p,⋯,aq+(m−1)p where  q+(m−1)p≤n and  q≥1.

 

Input

The first line contains only one integer  T≤100, which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers  1≤n≤106,1≤m≤106 and  1≤p≤106.

The second line contains n integers  a1,a2,⋯,an(1≤ai≤109).

the third line contains m integers  b1,b2,⋯,bm(1≤bi≤109).

 

Output

For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.

 

Sample Input

2 6 3 1 1 2 3 1 2 3 1 2 3 6 3 2 1 3 2 2 3 1 1 2 3

 

Sample Output

Case #1: 2 Case #2: 1

 

Source

2016中国大学生程序设计竞赛（长春）-重现赛

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pairpii;
const int maxn=1e6+10;
int a[maxn],b[maxn],Next[maxn];
void getNext(int n){
    int i=0,j=-1;Next[i]=j;
    while(i<=n-1){
        if(j==-1||(b[i+1]==b[j+1]&&b[i+1]==b[j+1])){
            i++;j++;
            Next[i]=j;
        }
        else j=Next[j];
    }
}
int KMP(int n,int m,int q,int x){
    int ans=0;int i=x,j=0;
    while(i<=n){
        if(j==-1||(a[i]==b[j+1]))i+=q,++j;
        else {
            j=Next[j];
        }
        if(j==m){ans++;}
    }
    return ans;
}
int main()
{
    int t,Test=1;cin>>t;
    int n,m,q;
    while(t--){
        scanf("%d%d%d",&n,&m,&q);
        memset(b,0,sizeof(b));
        for(int i=1;i<=n;++i){
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=m;++i){
            scanf("%d",&b[i]);
        }
        getNext(m);int ans=0;
        for(int i=1;i<=q&&i<=n;++i){
            ans+=KMP(n,m,q,i);
        }
        printf("Case #%d: %d\n",Test++,ans);
    }
    return 0;
}