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[(20000<<2)+30];
int a[20010];
void build(int id,int L,int R){
    if(L==R){
        res[id]=a[L];
        return;
    }
    build(lc,L,mid);
    build(rc,mid+1,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+1,R,l,r);
    else return (query(lc,L,mid,l,r)&query(rc,mid+1,R,l,r));
} 
int main()
{
    int n,m,c,t,i,j,k;
    cin>>t;
    while(t--){
        cin>>n>>k;
        cin>>a[1];
        int all=a[1];
        for(i=2;i<=n;++i) scanf("%d",a+i),all&=a[i];
        if(k*2+1>=n) {
            for(i=1;i<=n;++i)
            printf("%d%c",all,i==n?'
':' ');
            continue;
        }
        build(1,1,n);
        int l=n-k+1,r=2+k-1;
        for(i=1;i<=n;++i){
            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?'
':' ');
            r++;
            if(r==n+1) r=1;
            l++;
            if(l==n+1) l=1;
        }
    }
    return 0;
}