Sequence I
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0
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,⋯,bm is 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).
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
KMP匹配，就是把一个串分成p个串去匹配
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PI acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int inext[1000005];
ll p;
ll a[1000005];
ll b[1000005];
ll KMP_Count(ll x[],ll m,ll y[],ll n);
//void preKMP(ll x[],ll m,ll kmpNext[]);
void kmp_pre(ll x[],ll m);
int main()
{
int T;
ll n,m;
scanf("%d",&T);
for(int k=1;k<=T;k++)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(inext,0,sizeof(inext));
scanf("%lld %lld %lld",&n,&m,&p);
for(int i=0;i<n;i++)
scanf("%lld",&a[i]);
for(int i=0;i<m;i++)
scanf("%lld",&b[i]);
ll re = KMP_Count(b,m,a,n);
printf("Case #%d: %lld\n",k,re);
}
return 0;
}
void kmp_pre(ll x[],ll m)
{
ll i,j;
j=inext[0]=-1;
i=0;
while(i<m)
{
while(-1 != j&& x[i] != x[j])
j=inext[j];
inext[++i]=++j;
}
}
ll KMP_Count(ll x[],ll m,ll y[],ll n)
{
ll i,j,ci=0;
ll ans = 0;
kmp_pre(x,m);
for(int k=0;k<p;k++)//分成p个串
{
i=k;
j=0;
while(i<n)
{
while(-1!=j && y[i]!=x[j])
j=inext[j];
i+=p;
j++;
if(j>=m)
{
ans++;
j=inext[j];
}
}
}
return ans;
}