SPOJ-ANDROUND -线段树/与操作

ANDROUND - AND Rounds

#tree

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

Submit solution!

    给出一个循环数组，定义一轮ADD操作之后a[i]=a[i-1]&a[i]&a[i+1] ,所有元素同时进行这个操作，询问k轮ADD之后数组的值。

一开始没想到ST，以为是简单的xjb。。

  我们可以发现一个规律就是a[i]进行k轮ADD后得值==(a[i-k-1]&.....&a[i-1])&a[i]&(a[i+1]&.....&a[i+k]),当k超过一个值之后，数组内的所有值都相同，就是a[1]&...&a[n],不难得出是k*2+1>=n时；

  对于另一种情况我们可以计算出k轮之后第一个数对应原数组的左右边界，然后进行&操作，如果循环计算的话会T，想到用ST将复杂度降到log级别就好了。

 #include<stdio.h>
 #include<string.h>
 #include<queue>
 #include<bits/stdc++.h>
 #include<algorithm>
 #define MAX 400005
 #define lc (id<<1)
 #define rc ((id<<1)|1)
 #define mid ((L+R)>>1)
 using namespace std;
 int res[(<<)+];
 int a[];
 void build(int id,int L,int R){
     if(L==R){
         res[id]=a[L];
         return;
     }
     build(lc,L,mid);
     build(rc,mid+,R);
     res[id]=res[lc]&res[rc];
 }
 int query(int id,int L,int R,int l,int r){
     if(R<=r&&L>=l){
         return res[id];
     }
     if(r<=mid) return query(lc,L,mid,l,r);
     else if(l>mid) return query(rc,mid+,R,l,r);
     else return (query(lc,L,mid,l,r)&query(rc,mid+,R,l,r));
 }
 int main()
 {
     int n,m,c,t,i,j,k;
     cin>>t;
     while(t--){
         cin>>n>>k;
         cin>>a[];
         int all=a[];
         for(i=;i<=n;++i) scanf("%d",a+i),all&=a[i];
         if(k*+>=n) {
             for(i=;i<=n;++i)
             printf("%d%c",all,i==n?'\n':' ');
             continue;
         }
         build(,,n);
         int l=n-k+,r=+k-;
         for(i=;i<=n;++i){
             printf("%d%c",l<=r?query(,,n,l,r):(query(,,n,,r)&query(,,n,l,n)),i==n?'\n':' ');
             r++;
             if(r==n+) r=;
             l++;
             if(l==n+) l=;
         }
     }
     return ;
 }