Description

ZB is playing a card game where the goal is to make straights. Each card in the deck has a number between 1 and M(including 1 and M). A straight is a sequence of cards with consecutive values. Values do not wrap around, so 1 does not come after M. In addition to regular cards, the deck also contains jokers. Each joker can be used as any valid number (between 1 and M, including 1 and M).

You will be given N integers card[1] .. card[n] referring to the cards in your hand. Jokers are represented by zeros, and other cards are represented by their values. ZB wants to know the number of cards in the longest straight that can be formed using one or more cards from his hand.

Input

The first line contains an integer T, meaning the number of the cases.

For each test case:

The first line there are two integers N and M in the first line (1 <= N, M <= 100000), and the second line contains N integers card[i] (0 <= card[i] <= M).

Output

For each test case, output a single integer in a line – the longest straight ZB can get.

Sample Input

2
7 11
0 6 5 3 0 10 11
8 1000
100 100 100 101 100 99 97 103

Sample Output

5
3

題意

給出 n$n$ 張牌，其代表的數字在區間 [1,m]$[1,m]$ ，且 0$0$ 可以充當任意牌，問最長連續遞增的區間長度是多少。

思路

首先我們可以對原陣列進行處理，處理以後某兩個數牌面中間所空缺牌的數量可以由該陣列對應的差求得。

然後利用尺取法維護雙指標，保證雙指標之間代表的牌所空缺的數量小於等於 0 的數量，更新得出最大值即可。

AC 程式碼

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#define IO ios::sync_with_stdio(false);\
    cin.tie(0);\
    cout.tie(0);
using namespace std;
const int maxn = 1e6+10;
const int mod = 1e9+7;
typedef long long LL;

int n,m;
int a[maxn];
int s[maxn];

void solve()
{
    for(int i=1; i<=m; i++)
        s[i] = s[i-1] + (a[i]==0);
    int i = 0, j = 0, ans = 0;
    while(j<=m)
    {
        if(s[j]-s[i]<=a[0])
            ans = max(ans,j-i), j++;
        else
            i++;
    }
    cout<<ans<<endl;
}

void init()
{
    memset(a,0,sizeof a);
    memset(s,0,sizeof s);
}

int main()
{
    IO;
    int T;
    cin>>T;
    while(T--)
    {
        init();
        cin>>n>>m;
        for(int i=0; i<n; i++)
        {
            int x;
            cin>>x;
            a[x]++;
        }
        solve();
    }
    return 0;
}