Cow and Friend CodeForces - 1307B（思维）

Bessie has way too many friends because she is everyone’s favorite cow! Her new friend Rabbit is trying to hop over so they can play!

More specifically, he wants to get from (0,0) to (x,0) by making multiple hops. He is only willing to hop from one point to another point on the 2D plane if the Euclidean distance between the endpoints of a hop is one of its n favorite numbers: a1,a2,…,an. What is the minimum number of hops Rabbit needs to get from (0,0) to (x,0)? Rabbit may land on points with non-integer coordinates. It can be proved that Rabbit can always reach his destination.

Recall that the Euclidean distance between points (xi,yi) and (xj,yj) is (xi−xj)2+(yi−yj)2−−−−−−−−−−−−−−−−−√.

For example, if Rabbit has favorite numbers 1 and 3 he could hop from (0,0) to (4,0) in two hops as shown below. Note that there also exists other valid ways to hop to (4,0) in 2 hops (e.g. (0,0) → (2,−5–√) → (4,0)).

Here is a graphic for the first example. Both hops have distance 3, one of Rabbit’s favorite numbers.
In other words, each time Rabbit chooses some number ai and hops with distance equal to ai in any direction he wants. The same number can be used multiple times.

Input
The input consists of multiple test cases. The first line contains an integer t (1≤t≤1000) — the number of test cases. Next 2t lines contain test cases — two lines per test case.

The first line of each test case contains two integers n and x (1≤n≤105, 1≤x≤109) — the number of favorite numbers and the distance Rabbit wants to travel, respectively.

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤109) — Rabbit’s favorite numbers. It is guaranteed that the favorite numbers are distinct.

It is guaranteed that the sum of n over all the test cases will not exceed 105.

Output
For each test case, print a single integer — the minimum number of hops needed.

Example
Input
4
2 4
1 3
3 12
3 4 5
1 5
5
2 10
15 4
Output
2
3
1
2
Note
The first test case of the sample is shown in the picture above. Rabbit can hop to (2,5–√), then to (4,0) for a total of two hops. Each hop has a distance of 3, which is one of his favorite numbers.

In the second test case of the sample, one way for Rabbit to hop 3 times is: (0,0) → (4,0) → (8,0) → (12,0).

In the third test case of the sample, Rabbit can hop from (0,0) to (5,0).

In the fourth test case of the sample, Rabbit can hop: (0,0) → (5,102–√) → (10,0).
这个题目需要考虑的有点多，主要思路就是遍历每一个喜欢的数字，看看主要通过这个喜欢的数字来实现跳跃最少需要几次，然后我们再取最小值。有几种情况：
1：当前数字等于x，跳数位1。
2：当前数字大于x，跳数为2。
3：当前数字大于等于x的一半，并且x为偶数，跳数为2。
4：当前数字大于x的一半，并且x为奇数，跳数为2。
5：就是直接每次都走a[i]，直到不能走，当前走了z次，假设还剩下zz。
①zz在喜欢数字的数组中可以找到，跳数为z+1。
②zz找不到，并且不为0，跳数为z+2（这个时候，我们回退a[i]的距离，这样还剩下a[i]+zz的距离，那么这一段距离可以2次跳跃来实现）。
③zz==0，跳数为z。
代码如下：

#include
#define ll long long
#define inf 1e9+1
using namespace std;

const int maxx=1e5+100;
int a[maxx];
int n,x;

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&x);
		for(int i=1;i<=n;i++) scanf("%d",&a[i]);
		sort(a+1,a+1+n);
		int _min=inf;
		for(int i=1;i<=n;i++)
		{
			if(a[i]==x) _min=min(_min,1);
			else if(a[i]>x) _min=min(_min,2);
			else if(a[i]>=x/2&&x%2==0) _min=min(_min,2);
			else if(a[i]>x/2&&x%2==1) _min=min(_min,2);
			else
			{
				int z=x/a[i];
				int zz=x%a[i];
				if(zz&&a[lower_bound(a+1,a+1+n,zz)-a]==zz) _min=min(_min,z+1);
				else if(zz) _min=min(_min,z+1);
				else _min=min(_min,z);
			}
		}
		cout<<_min<<endl;
	}
	return 0;
}

努力加油a啊，(^o)/~