Sequence I

Mr. Frog has two sequences a1,a2,⋯,ana1,a2,⋯,an and b1,b2,⋯,bmb1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,⋯,bmb1,b2,⋯,bm is exactly the sequence aq,aq+p,aq+2p,⋯,aq+(m−1)paq,aq+p,aq+2p,⋯,aq+(m−1)p where q+(m−1)p≤nq+(m−1)p≤n and q≥1q≥1.
Input
The first line contains only one integer T≤100T≤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≤1061≤n≤106,1≤m≤106 and 1≤p≤1061≤p≤106.

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

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

直接暴力枚举。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[1000000];int b[1000000];int main(){    int n,m,i,j,k,t,p,l;    scanf("%d",&t);    k=1;    while(t--)    {        int sum=0;        scanf("%d %d %d",&n,&m,&p);        for(i=0; i<n; i++)        {            scanf("%d",&a[i]);        }        for(i=0; i<m; i++)        {            scanf("%d",&b[i]);        }        for(i=0; i<n; i++)        {            for(j=0,l=i; j<m&&l<n; )            {                //printf("%d %d\n",a[l],b[j]);                if(a[l]==b[j])                {                    l+=p;                    j++;                }                if(a[l]!=b[j])                    break;            }            if(j==m)            {                sum++;            }        }        printf("Case #%d: ",k);        k++;        printf("%d\n",sum);    }    return 0;}