You will be given a sequence consists of many numbers, your task is to calculate the length of the longest consecutive subsequence , the sum of which can be divided by another given number.

Input
There are multiple test cases.
The first line of each case contains two integer N and M, (0< N <= 100000, 0 < M < 10000),then followed by a line consists of N integers a1,a2,...an,
( -100000000 <= a1,a2,...an <= 100000000 ).

Output
For each test case,Output the length of the longest consecutive subsequence , the sum of which can be divided by M.

Sample Input

2 3
1 6
3 3
2 3 6
2 5
1 3

Sample Output

1
2
0

題目重述：求長的連續子段使其子段和能被某數整除

分析：列舉所有子段和最壞情況要O（n^2）的時間 必然超時

考慮到若用一個sum求前n個數的和 sum%m的值如果在前面出現過 則說明這一段可以整除m 

因此用一個vis陣列儲存餘數為x的出現下標 當vis[x]被訪問過 則說明i-vis[x]這一段能被整除

程式碼如下

#include <cstring>
#include <cstdio>


const int maxN=100000+50;

int Sum;
int vis[maxN];
int main()
{
	int N,M,i,temp,id,ans;
	while(scanf("%d%d",&N,&M)==2)
	{
		memset(vis,-1,sizeof(vis));
		Sum=0;
		vis[0]=0;
		ans=0;
		for (i=1;i<=N;i++)
		{
			scanf("%d",&temp);
			Sum+=temp;
			id=Sum%M;
			if (id<0)
			{
				id+=M;
			}
			if (vis[id]==-1)
			{
				vis[id]=i;
			}
			else
			{
				temp=i-vis[id];
				if (ans<temp)
				{
					ans=temp;
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}