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
