As we all know,the handsome boy, Zcat, has a fever of flat shoes. He sings it on the class, in the dormitory and the way he come back from the machinery room to his dorm.One day , he sing it on the way, just like

  "MY skating shoes are the most fashion one
  At the way back to home I couldn't stop
  rubbing rubbing
  rubbing at such smooth floor
  With the moonlight ,I was able to see my shadow was either very close or far
  I felt that my step was forced by a kind of strength
  With the skating shoes ,I will not afraid anything around the world
  One step two steps,every one was just like zhuaya
  The motion which was like a ghost was rubbing and rubbing
  I beat to beat for myself
  It's the best moment of my life
  I'd got to finish my favorite dancing
  At the terrfic street where moonlight fall
  I told myself that it really wasn't dream
  One step two steps ,every one was just like zhuaya
  The gost step was rubbing and rubbing
  rubbing at such smooth floor
  The motion which was like a ghost was rubbing and rubbing 
  One step two steps,every one was just like zhuaya
  One step two steps,every one was just like zhuaya
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯“

On he sings this,he sees a pretty girl standing on by the roadside. Zcat wants to see the pretty girl's face. Zcat knows the distance between he and her is n(1≤n≤1000000) meters. Zcat also wants to see the girl's face clearly. So he must move forward exactly n meters.To show the girl his favourite dancing,he has to follow the lyric One step two stepsof flat shoes. When the lyric is one step, he will move exactly 1 meter. When the lyric is two steps, he will move exactly 2 meters. Now, Zcat knows the whole lyric. He wants to know is there a time he can start, then move forward following the lyric, and he will see the pretty girl clearly.

Input

The first line contains a single number n(1≤n≤1000000) denotes the length of the lyric of One step and Two steps.

The second line contains n numbers.Anyone of these is 1 or 2.

The third line contains a single number q(1≤q≤3000000) denotes the time of question.

The next line contains q numbers denote then distance between Zcat and the pretty girl.

Output

To every question, if Zcat can see the girl clearly, print Yes, else print No.

Sample input and output

Sample Input Sample Output
8
2 1 1 1 1 2 2 2
6
1 3 5 7 9 11
Yes
Yes
Yes
Yes
Yes
No

解题报告

令 L = min ( s1 , s2 )

s1 为从左边开始扫,连续2的个数

s2 为从右边开始扫,连续2的个数

那么无法到达的距离为:

Len = 所有数之和 - 1

for(int i = 0 ; i < L; ++ i)
{
arrive[Len-i*2] = false;
}
这样,对于所有询问,就可以在O(1)的时间内做出回答
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = + ;
int A[maxn];
bool arrive[*+]; int main(int argc,char *argv[])
{
int n;
scanf("%d",&n);
int sum = ;
memset(arrive,true,sizeof(arrive));
for(int i = ; i < n ; ++ i)
{
scanf("%d",&A[i]);
sum += A[i];
}
int q;
int host = sum-;
int q1 = , q2 = ;
for(int i = ; i < n ; ++ i)
if (A[i] == )
break;
else
q1++;
for(int i = n- ; i >= ; --i)
if (A[i] == )
break;
else
q2++;
int thisq = min(q1,q2);
for(int i = ; i < thisq; ++ i)
{
arrive[host-i*] = false;
}
scanf("%d",&q);
while(q--)
{
int a;
scanf("%d",&a);
if (a > sum)
{
printf("No\n");
continue;
}
if(!arrive[a])
printf("No\n");
else
printf("Yes\n");
}
return ;
}
 

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