XXXXX CodeForces - 1364A（思维）

Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn’t divisible by x, or determine that such subarray doesn’t exist.

An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

Input
The first line contains an integer t (1≤t≤5) — the number of test cases you need to solve. The description of the test cases follows.

The first line of each test case contains 2 integers n and x (1≤n≤105, 1≤x≤104) — the number of elements in the array a and the number that Ehab hates.

The second line contains n space-separated integers a1, a2, …, an (0≤ai≤104) — the elements of the array a.

Output
For each testcase, print the length of the longest subarray whose sum isn’t divisible by x. If there’s no such subarray, print −1.

Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn’t divisible by 3.

In the second test case, the sum of elements of the whole array is 6, which isn’t divisible by 4.

In the third test case, all subarrays have an even sum, so the answer is −1.

思路：一开始想复杂了，我们比较一下从左边删除数字和从右边删除数字，最终剩下的长度，选取最大的那个就行了。
代码如下：

#include
#define ll long long
using namespace std;

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

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&x);
		memset(sum,0,sizeof(sum));
		for(int i=1;i<=n;i++) scanf("%d",&a[i]),sum[i]=sum[i-1]+a[i];
		if(sum[n]%x) cout<<n<<endl;
		else
		{
			int ans=-1;
			for(int i=1;i<=n;i++)
			{
				if((sum[n]-sum[i])%x)
				{
					ans=max(ans,n-i);
					break;
				}
			}
			for(int i=n;i>=1;i--)
			{
				if(sum[i]%x)
				{
					ans=max(ans,i);
					break;
				}
			}
			cout<<ans<<endl;
		}
	}
	return 0;
}

努力加油a啊，(^o)/~