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