[JZOJ5185] 【NOIP2017提高组模拟6.30】tty's sequence

Description

Input

Output

Sample Input

input 1：
6 3
1 1 1 0 0 0
input 2：
6 3
1 1 0 1 0 0
input 3:
6 3
11 8 2 1 3 9

Sample Output

output 1
1 1
output 2
1 0
output 3
11 1

Data Constraint

---

或的最大值一定是全部或起来。

与的最大值一定是k个与起来， 于是用线段树维护区间与值， 然后暴力枚举。

复杂度O(NlogN);

---

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
inline int read() {
    int res=0;char ch=getchar();
    while(!isdigit(ch)) ch=getchar();
    while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
    return res;
}
#define reg register
int n, k;
int a[1000005];
long long ans, res;

int tr[1000005<<2];
#define ls(o) o << 1
#define rs(o) o << 1 | 1

void Build(int l, int r, int o)
{
    if (l == r)
    {
        tr[o] = a[l];
        return ;
    }
    int mid = l + r >> 1;
    Build(l, mid, ls(o));
    Build(mid + 1, r, rs(o));
    tr[o] = tr[ls(o)] & tr[rs(o)];
}

void query(int l, int r, int o, int ql, int qr)
{
    if (l >= ql and r <= qr)
    {
        res &= tr[o];
        return ;
    }
    int mid = l + r >> 1;
    if (mid >= ql) query(l, mid, ls(o), ql, qr);
    if (mid < qr) query(mid + 1, r, rs(o), ql, qr);
}

int main()
{
    n = read(), k = read();
    for (reg int i = 1 ; i <= n ; i ++)
        a[i] = read(), ans |= a[i];
    printf("%lld ", ans);
    Build(1, n, 1);
    ans = 0;
    for (reg int i = 1 ; i <= n - k + 1 ; i ++)
    {
        res = (1ll << 31) - 1;
        query(1, n, 1, i, i + k - 1);
        ans = max(ans, res);
    }
    printf("%lld\n", ans);
    return 0;
}