CodeForces 367B Sereja ans Anagrams （map）

题目地址：点击打开链接

题意：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。

思路：i、i+p、i+2*p肯定构成一条长链，枚举链的起点，用map维护并判断长度为m的链与b序列是否相同。

map可以直接用==判断是否相等，涨姿势了

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;
int a[maxn], b[maxn], ans[maxn];
ll n, m, p;

int main(void)
{
while(cin >> n >> m >> p)
{
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
map<int, int> M;
for(int i = 1; i <= m; i++)
scanf("%d", &b[i]), M[b[i]]++;
map<int, int> ma;
int index = 0;
for(int i = 1; i <= p; i++)
{
ma.clear();
for(ll j = 0; j <= (m-1) && i+j*p <= n; j++)
ma[a[i+j*p]]++;
if(ma == M) ans[index++] = i;
for(ll j = m; i+j*p <= n; j++)
{
int pre = a[i+(j-m)*p];
if(ma[pre] == 1) ma.erase(pre);
else ma[pre]--;
ma[a[i+j*p]]++;
if(ma == M) ans[index++] = i+(j-m+1)*p;
}
}
printf("%d\n", index);
sort(ans, ans+index);
for(int i = 0; i < index; i++)
printf("%d%c", ans[i], i==index-1 ? '\n' : ' ');
}
return 0;
}



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