CodeForces 367B Sereja ans Anagrams （map）

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本文介绍了一种算法，用于解决两个整数序列a和b的问题，即找出所有可能的起始位置q，使得从a[q]开始每隔p个元素取一次直至取得m个元素，这m个元素经过重新排列能与序列b完全匹配。

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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;
}

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