B. Sereja ans Anagrams----map维护queue

B. Sereja ans Anagrams

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Sereja has two sequences a and b and
number p. Sequence a consists
of n integers a1, a2, ..., an.
Similarly, sequence b consists of mintegers b1, b2, ..., bm.
As usual, Sereja studies the sequences he has. Today he wants to find the number of positions q(q + (m - 1)·p ≤ n; q ≥ 1),
such that sequence b can be obtained from sequence aq, aq + p, aq + 2p, ..., aq + (m - 1)p by
rearranging elements.

Sereja needs to rush to the gym, so he asked to find all the described positions of q.

Input

The first line contains three integers n, m and p (1 ≤ n, m ≤ 2·105, 1 ≤ p ≤ 2·105).
The next line contains n integers a1, a2, ..., an(1 ≤ ai ≤ 109).
The next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109).

Output

In the first line print the number of valid qs. In the second line, print the valid values in the increasing order.

Examples

input

5 3 1
1 2 3 2 1
1 2 3

output

2
1 3

input

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

output

2
1 2

题目链接：http://codeforces.com/contest/367/problem/B

题目的意思是说给定n个元素的序列a[]和m个元素的序列b[]，让你找到一个q使得a[q]、a[q+p]、…a[q+(m-1)*p]构成序列b。问你所有的q。

这个题做的和个傻子似的。。题解是看懂了，但是可能再碰到这种题还是不会。。还要加强反思。

http://blog.csdn.net/chenzhenyu123456/article/details/51038991

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#define LL long long
using namespace std;
const int mod=1e9+7;
void add(LL &x, LL y){
x+=y;
x%=mod;
}
int a[300000];
map<int, int>fp,tp;
int n,m,p;
int ans;
bool vis[300000];
void solve(int s){
tp.clear();
queue<int>Q;
for(int i=s;i<=n;i+=p){
Q.push(i);
tp[a[i]]++;
if(Q.size()==m){
if(fp==tp){
vis[Q.front()]=true;
ans++;
}
int v=a[Q.front()];
Q.pop();
if(--tp[v]==0){
tp.erase(v);
}
}
}
}
int main(){
scanf("%d%d%d",&n,&m,&p);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
vis[i]=false;
}
fp.clear();
for(int i=1;i<=m;i++){
int b;
scanf("%d",&b);
fp[b]++;
}
ans=0;
for(int i=1;i<=p;i++){
solve(i);
}
printf("%d\n",ans);
int cnt=0;
for(int i=1;i<=n;i++){
if(vis[i]){
if(cnt>0)
printf(" ");
printf("%d",i);
cnt++;
}
}
if(ans)
printf("\n");
return 0;
}