fzu 2216 The Longest Straight

Description

ZB is playing a card game where the goal is tom make straights. Each card in the deck has a number between 1 and M, inclusive. 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, inclusive).

You will be giben 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 <= 10^5), and the second line contains N integers card[i](0 <= card[i] <= M).

Output

For each test case, output a single integer 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

时间复制度为O(N)

#include
#include
#include
using namespace std;
const int maxn = 1e5 + 5;
int card[maxn], num[maxn], pos[maxn];
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        memset(pos, 0, sizeof(pos));
        memset(num, 0, sizeof(num));
        memset(card, 0, sizeof(card));
        int m, n, zero = 0;
        scanf("%d%d", &m, &n);
        for (int i = 1; i <= m; i++)
        {
            int x;
            scanf("%d", &x);
            if (x == 0) zero++;
            else
                card[x] = 1;
        }
        int ans = 0,k;
        pos[0] = 1;
        for (int i = 1; i <= n; i++)
        {
            num[i] = num[i - 1] + (1 - card[i]);
            if (!card[i])
                pos[num[i]] = i;
            k = num[i] - zero;
            if (k <= 0)
            {
                ans = i;
                continue;
            }
            ans = max(ans, i - pos[k]);
        }
        printf("%d\n", ans);
    }
    return 0;
}

/*

22
7 11
0 6 5 3 0 10 11
8 1000
100 100 100 101 100 99 97 103
3 5
1 2 3
3 2
1 2 3
3 5
1 2 0
3 5
2 3 4
3 5
2 0 4


*/