1129: Divisibility

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Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	8192K	1101	268	Standard

Consider an arbitrary sequence of integers. One can place + or - operators between integers in the sequence, thus deriving different arithmetical expressions that evaluate to different values. Let us, for example, take the sequence: 17, 5, -21, 15. There are eight possible expressions:

17 + 5 + -21 + 15 =  16
17 + 5 + -21 - 15 = -14
17 + 5 - -21 + 15 =  58
17 + 5 - -21 - 15 =  28
17 - 5 + -21 + 15 =   6
17 - 5 + -21 - 15 = -24
17 - 5 - -21 + 15 =  48
17 - 5 - -21 - 15 =  18

We call the sequence of integers divisible by K if + or - operators can be placed between integers in the sequence in such way that resulting value is divisible by K. In the above example, the sequence is divisible by 7 (17+5+-21-15=-14) but is not divisible by 5.

You are to write a program that will determine divisibility of sequence of integers.

Input

There are multiple test cases, the first line is the number of test cases.
The first line of each test case contains two integers, N and K (1 ≤ N ≤ 10000, 2 ≤ K ≤ 100) separated by a space.

The second line contains a sequence of N integers separated by spaces. Each integer is not greater than 10000 by it's absolute value.

Output

Write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.

Sample Input

2
4 7
17 5 -21 15
4 5
17 5 -21 15

Sample Output

Divisible
Not divisible

 

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This problem is used for contest: 148 

 

这道题目是比赛时候做的，总结一下

比赛时候这道题目没有AC,是因为+,-号打错了，第二天早上拿出来的时候才发现，我哭啊，否则就可以完胜大二的，拿到第一了啊，悲剧。。。。这道题目有几个经验，第一就是对负数的处理上，去摸（K+负数%K）%K，十分完美的驱魔，接着就是对各种情况的判断。最后什么情况下是成立的，也十分讲究的。

 

#include
#include
bool f[10001][101];
int a[10001];
int main()
{
 int n,i,j,k,m,t1,t2;
 freopen("in.txt","r",stdin);
 freopen("out.txt","w",stdout);
 while(scanf("%d",&m)!=EOF)
 {
  while(m--)
  {
   scanf("%d%d",&n,&k);
   for(i=0;i   memset(f,false,sizeof(f));
            f[0][0]=true;
   for(i=1;i<=n;i++)
     for(j=0;j     {
    if(j-a[i-1]<0) t1=(k+(j-a[i-1])%k)%k;
       else   t1=(j-a[i-1])%k;
    if(j+a[i-1]>0) t2=(j+a[i-1])%k;
       else   t2=(k+(j+a[i-1])%k)%k;
    if(f[i-1][t1]||f[i-1][t2])
       f[i][j]=true;  
     }
   if(f[n][0]) printf("Divisible/n");
       else    printf("Not divisible/n");
  }
 }
 return 0;
}