Sereja ans Anagrams - CodeForces 367B 水题-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/38434775

本文详细解析了B.SerejaansAnagrams算法，包括输入输出规范、问题描述、解决思路以及AC代码实现。算法核心在于寻找特定间隔下的序列重组可能性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

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

题意：找到间隔为p的连续序列，使得可以经过调序得到第二个序列。

思路：就是一个个地找……每次向右走一个间隔，同时删除左边的那个，保证l到r的长度为m，再判断。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
map<int,int> match1,match2;
int num1[200010],ans[200010],num;
long long n,len,m;
void solve(long long l)
{ long long r=l+len*(m-1),i,j,k;
  if(r>n)
   return;
  for(i=0;i<m;i++)
  { k=num1[l+i*len];
    match1[k]++;
    if(match2[k]>=match1[k])
     num++;
  }
  if(num==m)
  { ans[0]++;
    ans[ans[0]]=l;
  }
  l+=len;r+=len;
  while(r<=n)
  { k=num1[l-len];
    match1[k]--;
    if(match2[k]>match1[k])
     num--;
    k=num1[r];
    match1[k]++;
    if(match2[k]>=match1[k])
     num++;
    if(num==m)
    { ans[0]++;
      ans[ans[0]]=l;
    }
    l+=len;r+=len;
  }
}
int main()
{ int i,j,k;
  scanf("%I64d%I64d%I64d",&n,&m,&len);
  for(i=1;i<=n;i++)
   scanf("%d",&num1[i]);
  for(i=1;i<=m;i++)
  { scanf("%d",&k);
    match2[k]++;
  }
  for(i=1;i<=len;i++)
  { match1.clear();
    num=0;
    solve(i);
  }
  k=ans[0];
  sort(ans+1,ans+1+k);
  printf("%d\n",k);
  if(k>0)
  { printf("%d",ans[1]);
    for(i=2;i<=k;i++)
     printf(" %d",ans[i]);
  }
  printf("\n");
}