CodeForces - 367B Sereja ans Anagrams

题目链接

思路：用数组来存储超时，学习了用map<int,int>的用法,可以直接比较

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>

using namespace std; 

const int N = 2e5+10;
map<int,int> mchp,findp;
int cnt,n,m,p; 
int ans[N]; 
int num[N]; 

void solve( int pos ) {
	int count=0;
	queue<int> q; 
	findp.clear();
	for ( int i=pos ; i<=n; i+=p ) {
		findp[num[i]]++; 
		++count; 
		q.push(i); 
		
		if ( count==m ) {
			if ( findp==mchp ) {
				ans[cnt++] = q.front(); 
			}
			--findp[ num[ q.front()]]; 
			if ( findp[num[ q.front()]]==0 )findp.erase(num[q.front()]);//!!!!
			q.pop();  
			--count; 
		} 
	}
}
int main()
{
	int temp; 
	scanf("%d %d %d",&n,&m,&p); 
	for ( int i=1; i<=n; i++ ) scanf("%d",&num[i]); 
	for ( int i=1; i<=m; i++ ) {
		scanf("%d",&temp); 
		mchp[temp]++; 		
	}
	
	for ( int i=1; i<=p; i++ ) 
		solve( i ) ;
		
	sort(ans,ans+cnt); 
	cout<<cnt<<endl; 
	for ( int i=0; i<cnt; i++ ) 
		if ( i==cnt-1 ) printf("%d\n",ans[i]); 
		else printf("%d ",ans[i]); 
	return 0; 
}

   间隔是p，即从1到p开始枚举，若长度超过了m，则剪掉最前面的数字