#include <bits/stdc++.h>
using namespace std;
const int maxn=3e5+1;
int n,a[maxn],q[maxn],ans[maxn],vis[maxn],maxx=0,minn=1e9+100;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;++i)
{
scanf("%d",&a[i]);
a[i+2*n]=a[i+n]=a[i];
minn=min(minn,a[i]);
maxx=max(maxx,a[i]);
}
if(maxx<=2*minn)
{
for(int i=1;i<=n;++i) printf("-1 ");
return 0;
}
int first=1,now=0;
for(int i=1;i<=3*n;++i)
{
while(now<=first&&q[first]<a[i]) first--;
q[++first]=a[i];
vis[first]=i;
while(q[now]>a[i]*2)//从前往后遍历队列内的元素
ans[vis[now]]=i-vis[now],now++;
}
for(int i=3*n;i>=1;--i) if(ans[i]==0) ans[i]=ans[i+1]+1;//队列里有些元素被弹出了,不过可以根据后面的答案来更新前面的
for(int i=1;i<=n;++i) printf("%d ",ans[i]);
}
Codeforces Global Round 5 D. Balanced Playlist(单调队列)
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转载自blog.csdn.net/qq_42479630/article/details/104435313
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