Codeforces 367B Sereja ans Anagrams 详细题解（map应用+维护长度）

B. Sereja ans Anagrams 
time limit per test1 second 
memory limit per test256 megabytes 
inputstandard input 
outputstandard 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 m 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.

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pre，suf来维护元素的进出跟长度，每当里面有m个元素，用map判定是否满足。

这题学到了map可以直接 == 比较的。。。唉。。菜。。。set也可以直接比较的。。另外，这题复杂度是On的。。虽然两个for但是On的，因为第一个for枚举到p，是枚举起点，没必要枚举p往后的，p往后的每次+p都是重复之前的。。第二个for从i开始每次+p，一直到最后，其实会发现每个元素都只访问了一遍。。。这题也可以用队列维护长度

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <map>using namespace std;const int maxn = 2e5 + 5;int n, m, a[maxn], p, ans, ansl[maxn];int main(){    scanf("%d%d%d", &n, &m, &p);    for(int i = 1; i <= n; i++)        scanf("%d", &a[i]);    map<int, int> b;    int num;    for(int i = 1; i <= m; i++)    {        scanf("%d", &num);        b[num]++;    }    int index = 1, pre, suf;    ans = 0;    for(int i = 1; i <= p; i++)    {        map<int, int> match;        pre = suf = i;        for(int j = i; j <= n; j += p)        {            int id = i+(pre-i)*p;            if(suf-pre < m)            {                suf++;                match[a[j]]++;            }            if(suf-pre == m) //相等的时候，先比较这一段跟b是不是相同的，然后pre往前走，suf往后走            {//                cout << pre << ' ' << suf << endl;//                cout << match[1] << ' ' << match[2] << ' ' << match[3] << endl;                if(match == b) //两个map可以直接==比较                {                    ansl[index++] = id;                    ans++;                }                match[a[id]]--;                if(match[a[id]] == 0) match.erase(a[id]); //直接比较 等于0 要把这个元素删掉                pre++;            }        }    }    sort(ansl+1, ansl+index);    printf("%d\n", ans);    for(int i = 1; i < index; i++)    {        printf("%d%c", ansl[i], i == index-1 ? '\n' : ' ');    }    return 0;}