CodeForces

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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

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