HDU Sequence I 2016中国大学生程序设计竞赛（长春）-重现赛

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).
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

水题

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>
//#define CHECK_TIME
const int N=1e6+7;
int a[N],b[N];
bool check(int n,int m,int p,int q){
for(int i=0;i<m;++i)
if(b[i+1]!=a[q+i*p])
return false;
return true;
}
int slove(int n,int m,int p){
int ans=0;
for(int i=1;;++i){
if(i+(m-1)*p>n)
break;
ans+=check(n,m,p,i);
}
return ans;
}
int main()
{
//freopen("/home/lu/文档/r.txt","r",stdin);
//freopen("/home/lu/文档/w.txt","w",stdout);
int T,t,n,m,p;
scanf("%d",&T);
for(int t=1;t<=T;++t){
scanf("%d%d%d",&n,&m,&p);
for(int i=1;i<=n;++i)
scanf("%d",a+i);
for(int i=1;i<=m;++i)
scanf("%d",b+i);
printf("Case #%d: %dn",t,slove(n,m,p));
}
return 0;
}

最后

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