D. Training Session

题目描述：

n个数，问最大能选多长的连续子区间，使得这个区间内每个数余一个大于等于2的数字后相等

思路：

做差分后用st表处理区间gcd，枚举区间的左端点，用二分去找右端点

不要用线段树来维护区间gcd + 枚举二分，会TLE

可以用线段树来维护区间gcd + 双指针

/*
Work by: Chelsea
Problem: 不知名屑题
Knowledge: 垃圾算法
Time: O(能过)
*/
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<bitset>
#include<cstdio>
#include<string>
#include<vector>
#include<sstream>
#include<cstring>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<unordered_map>
using namespace std;

//#pragma GCC optimize("Ofast")
//#pragma GCC target("fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
//#pragma GCC optimize("unroll-loops")

#define eps 1e-8
#define endl '\n'
#define inf 0x3f3f3f3f
#define NMAX 1000 + 50
#define ls p<<1
#define rs p<<1|1
//#define mid ((l + r)>>1)
#define mod 100000000
#define lowbit(x) (x & (-x))
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define sl(n) scanf("%lld",&n)
#define sll(n,m) scanf("%lld %lld",&n,&m)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d\n",n, m)
#define sddd(n,m,z) scanf("%d %d %d",&n,&m,&z)
#define slll(n,m,z) scanf("%lld %lld %lld",&n,&m,&z)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define mem(a,b) memset((a),(b),sizeof(a))
#define m_p(a,b) make_pair(a, b)
typedef  long long ll;
//typedef pair <int,int> pii;
typedef unsigned long long ull;
//不开longlong见祖宗!不改范围见祖宗!
inline int IntRead(){char ch = getchar();int s = 0, w = 1;while(ch < '0' || ch > '9'){if(ch == '-') w = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){s = s * 10 + ch - '0';ch = getchar();}return s * w;}

#define MAX 200000 + 10
int n, m, k, x, y, z, q, p;
string s;

ll ar[MAX];
ll tr[MAX];
ll st[MAX][35];
ll lg[MAX];

ll gcd(ll a, ll b){
    return b == 0 ? a : gcd(b, a % b);
}

void init(){
    lg[1] = 0;
    for(int i = 2; i <= n; ++i)lg[i] = lg[i / 2] + 1;
    for(int j = 1; j <= lg[n]; ++j){
        for(int i = 1; i + (1 << j) - 1 <= n; ++i){
            st[i][j] = gcd(st[i][j - 1], st[i + (1<<(j - 1))][j - 1]);
        }
    }
}

ll getans(ll l, ll r){
    ll d = lg[r - l + 1];
    return gcd(st[l][d], st[r - (1<<d) + 1][d]);
}

void work(){
    cin>>n;
    for(int i = 1; i <= n; ++i)cin>>ar[i];
    for(int i = 1; i < n; ++i)st[i][0] = abs(ar[i + 1] - ar[i]);
    --n;
    init();
    int ans = 0;
    for(int i = 1; i <= n; ++i){
        int l = i, r = n;
        while (l <= r) {
            int mid = (l + r) / 2;
            if(getans(i, mid) == 1)r = mid - 1;
            else l = mid + 1;
        }
        ans = max(ans, l - i);
    }
    cout<<ans + 1<<endl;
}

int main(){
    io;
    int tt;cin>>tt;
    for(int _t = 1; _t <= tt; ++_t){
//        printf("Case #%d: ", _t);
        work();
    }
//    cerr << 1.0 * clock() / CLOCKS_PER_SEC << "\n";
    return 0;
}