題干：

Just days before the JCPC, your internet service went down. You decided to continue your training at the ACM club at your university. Sadly, you discovered that they have changed the WiFi password. On the router, the following question was mentioned, the answer is the WiFi password padded with zeros as needed.

A subarray?[l,?r]?of an array?A?is defined as a sequence of consecutive elements?Al,?Al?+?1,?...,?Ar, the length of such subarray is?r?-?l?+?1. The bitwise OR of the subarray is defined as:?Al?OR?Al?+?1?OR ... OR?Ar, where OR is the bitwise OR operation (check the notes for details).

Given an array?A?of?n?positive integers and an integer?v, find the maximum length of a subarray such that the bitwise OR of its elements is less than or equal to?v.

Input

The first line contains an integer?T?(1?≤?T?≤?128), where?T?is the number of test cases.

The first line of each test case contains two space-separated integers?n?and?v?(1?≤?n?≤?105)?(1?≤?v?≤?3?×?105).

The second line contains?n?space-separated integers?A1,?A2,?...,?An?(1?≤?Ai?≤?2?×?105), the elements of the array.

The sum of?n?overall test cases does not exceed?106.

Output

For each test case, if no subarray meets the requirement, print?0. Otherwise, print the maximum length of a subarray that meets the requirement.

Example

Input

3 5 8 1 4 5 3 1 5 10 8 2 6 1 10 4 2 9 4 5 8

Output

5 3 0

Note

To get the value of?x?OR?y, consider both numbers in binary (padded with zeros to make their lengths equal), apply the OR operation on the corresponding bits, and return the result into decimal form. For example, the result of?10?OR?17?=?01010?OR?10001?=?11011?=?27.

題目大意：

給你n和m，n代表有n個數，然后讓你找出一個最長的區間，使得這個區間內的所有數的‘’或‘’都小于等于m。

解題報告：

? ?因為或運算對于區間長度是滿足單調不減性的，所以我們可以二分長度+check。其實尺取是最優秀的但是二分好寫一些，并且告訴你了n的總數量<=1e6，所以二分足夠了，這個總復雜度應該是nlognlogn的。

? 因為滿足區間加和的性質，所以這題也可以線段樹去做（反正不帶更新，如果帶更新就麻煩了⑧），這樣還是尺取，然后每次log去查詢區間和就好了，抽空寫一寫。

AC代碼：（592ms水過）

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 1e5 + 5; int a[MAX],n,v; int d[MAX][22]; int cnt[22]; int cal() {int res = 0;for(int j = 0; j<22; j++) {if(cnt[j]>0) res += (1<<j); }return res; } bool ok(int x) {memset(cnt,0,sizeof cnt);for(int i = 1; i<=x; i++) {for(int j = 0; j<22; j++) cnt[j] += d[i][j];}int sum = cal();if(sum <= v) return 1;for(int l = 2; l+x-1<=n; l++) {int r = l+x-1;for(int j = 0; j<22; j++) cnt[j] -= d[l-1][j];for(int j = 0; j<22; j++) cnt[j] += d[r][j];if(cal() <= v) return 1;}return 0 ; } int main() {freopen("wifi.in","r",stdin);int t;cin>>t;while(t--) {scanf("%d%d",&n,&v);memset(d,0,sizeof d);for(int i = 1; i<=n; i++) scanf("%d",a+i);for(int i = 1; i<=n; i++) {int tmp = a[i],cur = 0;while(tmp) {d[i][cur++] = tmp%2;tmp/=2;}}int l = 1,r = n,mid = (l+r)>>1,ans=0;while(l<=r) {mid=(l+r)>>1;if(ok(mid)) ans = mid,l = mid+1;else r = mid-1;}printf("%d\n",ans);}return 0 ; }

?

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總結

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