ANDROUND - AND Rounds

You are given a cyclic array A having N numbers. In an AND round, each element of the array A is replaced by the bitwise AND of itself, the previous element, and the next element in the array. All operations take place simultaneously. Can you calculate A after K such AND rounds ?

Input

The first line contains the number of test cases T (T <= 50).
There follow 2T lines, 2 per test case. The first line contains two space seperated integers N (3 <= N <= 20000) and K (1 <= K <= 1000000000). The next line contains N space seperated integers Ai (0 <= Ai <= 1000000000), which are the initial values of the elements in array A.

Output

Output T lines, one per test case. For each test case, output a space seperated list of N integers, specifying the contents of array A after K AND rounds.

Example

Sample Input:
2 
3 1 
1 2 3 
5 100 
1 11 111 1111 11111 
 
Sample Output:
0 0 0 
1 1 1 1 1

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<queue>
 4 #include<bits/stdc++.h>
 5 #include<algorithm>
 6 #define MAX 400005
 7 #define lc (id<<1)
 8 #define rc ((id<<1)|1)
 9 #define mid ((L+R)>>1)
10 using namespace std;
11 int res[(20000<<2)+30];
12 int a[20010];
13 void build(int id,int L,int R){
14     if(L==R){
15         res[id]=a[L];
16         return;
17     }
18     build(lc,L,mid);
19     build(rc,mid+1,R);
20     res[id]=res[lc]&res[rc];
21 }
22 int query(int id,int L,int R,int l,int r){
23     if(R<=r&&L>=l){
24         return res[id];
25     }
26     if(r<=mid) return query(lc,L,mid,l,r);
27     else if(l>mid) return query(rc,mid+1,R,l,r);
28     else return (query(lc,L,mid,l,r)&query(rc,mid+1,R,l,r));
29 } 
30 int main()
31 {
32     int n,m,c,t,i,j,k;
33     cin>>t;
34     while(t--){
35         cin>>n>>k;
36         cin>>a[1];
37         int all=a[1];
38         for(i=2;i<=n;++i) scanf("%d",a+i),all&=a[i];
39         if(k*2+1>=n) {
40             for(i=1;i<=n;++i)
41             printf("%d%c",all,i==n?‘\n‘:‘ ‘);
42             continue;
43         }
44         build(1,1,n);
45         int l=n-k+1,r=2+k-1;
46         for(i=1;i<=n;++i){
47             printf("%d%c",l<=r?query(1,1,n,l,r):(query(1,1,n,1,r)&query(1,1,n,l,n)),i==n?‘\n‘:‘ ‘);
48             r++;
49             if(r==n+1) r=1;
50             l++;
51             if(l==n+1) l=1;
52         }
53     }
54     return 0;
55 }

SPOJ-ANDROUND -線段樹/與操作