CodeForces 367B. Sereja ans Anagrams

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我是靠谱客的博主霸气胡萝卜，最近开发中收集的这篇文章主要介绍CodeForces 367B. Sereja ans Anagrams，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence b consists ofm integers 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.

Sample test(s)

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

STL操作。。。。当值减小到0的时候一定要记得将这个元素删除。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
using namespace std;
const int maxn=220000;
typedef map<int,int> mII;
mII B;
int n,m,p,a[maxn],ans[maxn],cnt=0;
bool vis[maxn];
int main()
{
scanf("%d%d%d",&n,&m,&p);
B.clear();
for(int i=1;i<=n;i++) scanf("%d",a+i);
for(int i=0;i<m;i++)
{
int x;
scanf("%d",&x);
B[x]++;
}
if((m-1)*p+1>n) {printf("0n");return 0;}
for(int i=1;i<=p;i++)///P种不同的开始点
{
deque<int> q;mII tp;
tp.clear();
memset(vis,false,sizeof(vis));
for(int j=i;j<=n;j+=p)
{
q.push_back(j);
tp[a[j]]++;
if(q.size()>m)
{
int u=q.front(); q.pop_front();
tp[a[u]]--;
if(tp[a[u]]==0)
tp.erase(a[u]);
}
if(tp==B)
{
ans[cnt++]=q.front();
}
}
}
sort(ans,ans+cnt);
printf("%dn",cnt);
for(int i=0;i<cnt;i++)
{
if(i) putchar(32);
printf("%d",ans[i]);
}
putchar(10);
return 0;
}