Divisibility

Description

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 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

4 7
17 5 -21 15

Sample Output

Divisible

题意：给你一列整数，在整数间加‘ + ’ 或 ‘ - ‘，使这个算式的值能被k整除。

用dp[ i ][ j ] 表示加上或减去第 i 个数后，所得值取模后的值能否为 j ，所以dp为bool型即可。

状态转移方程：dp[ i ][ abs( j + num[i]) % k] = true;

dp[ i ][ abs( j -  num[i]) % k] = true; (当然，必须满足dp[ i - 1 ][ j ] == true, 才能进行状态转移)

边界条件：dp[ 0 ][ 0 ] = true;

 #include"iostream"
 #include"cstdio"
 #include"cstring"
 #include"algorithm"
 #include"map"
 #include"set"
 #include"stack"
 #include"queue"
 using namespace std;
 const int ms=;
 const int mn=;
 bool dp[ms][mn];
 int a[ms];
 int N,K;
 void solve()
 {
     memset(dp,false,sizeof(dp));
     dp[][]=true;
     for(int i=;i<=N;i++)
         for(int j=;j<K;j++)
             if(dp[i-][j])
             {
                 dp[i][abs(j+a[i])%K]=true;   //涉及一点数论
                 dp[i][abs(j-a[i])%K]=true;
             }
     if(dp[N][])
         cout<<"Divisible"<<endl;
     else
         cout<<"Not divisible"<<endl;
     return ;
 }
 int main()
 {
     cin>>N>>K;
     for(int i=;i<=N;i++)
         cin>>a[i];
     solve();
     return ;
 }