1129: Divisibility

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

 
This problem is used for contest: 148 
 
이 제목은 시합 때 만든 것이니 총결산해 보시오
시합 때 이 문제가 AC가 없었어요. 왜냐하면 +, - 번호를 잘못 쳐서 다음날 아침에 꺼냈을 때 알았어요. 제가 울었어요. 그렇지 않으면 대학교 2학년을 완승하고 1등을 할 수 있었어요. 비극...이 문제는 몇 가지 경험이 있는데 첫 번째는 마이너스 처리에 있어 (K+마이너스%K)%K를 만져보고 완벽한 악마를 쫓는 것이다. 그 다음은 각종 상황에 대한 판단이다.마지막으로 어떤 상황이 성립되었는지도 매우 중시한다.
 
#include#includebool 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;}