Uva 10036

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

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 of the input file 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

For each case in the input file, 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

题目大意：给n个数，在这n个中放上‘+’  ‘-’使得结果能被k整除。

dp[i][j]表示 前i个数余数j 是否可能

/* ***********************************************
Author        :guanjun
Created Time  :2016/9/6 15:25:54
File Name     :uva10036.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int dp[maxn][100];
int a[maxn];
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int t,n,k;
    cin>>t;
    while(t--){
        scanf("%i %i",&n,&k);
        for(int i=1;i<=n;i++)scanf("%i",&a[i]),a[i]=abs(a[i]%k);
        cle(dp);
        dp[1][0]=1;
        for(int i=1;i<=n;i++){
            for(int j=0;j<k;j++){
                if(dp[i][j]){
                    dp[i+1][(j-a[i]+k)%k]=1;
                    dp[i+1][(j+a[i])%k]=1;
                }
            }
        }
        if(dp[n+1][0])puts("Divisible");  
        else puts("Not divisible"); 
    }
    return 0;
}

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