Ultraviolet A 10036 divisibility

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 IntegersDivisibleByKIf + or-operators can be placed between integers in the sequence in such way that resulting value is divisibleK. 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

The first line of the input file contains a integer m indicating the number of cases to be analyzed. Then M couples of lines follow.
For each one of this couples, the first line contains two integers,NAndK(1 <=N& Lt; = 10000, 2 & lt; =K<= 100) separated by a space. The second line contains a sequenceNIntegers separated by spaces. Each integer is not greater than 10000 by it's absolute value.

Output

For each case in the input file, write to the output file the word "divisible" if given sequence of integers is divisibleKOr "not divisible" if it's not.

Sample Input

24 717 5 -21 154 517 5 -21 15

Sample output

DivisibleNot divisible

DP

• F [I] [J] = 1 indicates that the value of the expression formed by the first I digit is divided by K. You can add J • f [0] [0] = 1 • consider I x, suppose it is positive. • If f [I-1] [J] = 1, it indicates that the value of the expression of the previous I-1 digit can be divided by K, and then J • put "+" in front of X ", f [I] [(j + x) % K] = 1 • put "-" in front of X, F [I] [(j-x + k) % K] = 1

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5  6 int N,K; 7 bool f[2][100]; 8  9 int main()10 {11     int T ;12     scanf("%d",&T);13     while(T--) {14         scanf("%d%d" , &N ,&K) ;15         memset(f , false , sizeof(f)) ;16         f[0][0] = true ;17         int cur = 0 , x ;18         for(int i=1; i<=N; i++){19             cur ^= 1 ;20             memset(f[cur] , false , sizeof(f[cur])) ;21             scanf("%d" , &x);22             if(i>1 && x<0) x = -x;23             x%=K;24             for(int u = 0; u<K; u++) if(f[cur^1][u]) {25                 f[cur][(u+x+K)%K] = true;26                 if(i>1) f[cur][(u-x+K)%K]=true;27             }28         }29         if(f[cur][0]) puts("Divisible") ;30         else puts("Not divisible") ;31     }32     return 0;33 }