UVA 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=1617+5+-21-15=-1417+5--21+15=5817+5--21-15=2817-5+-21+15=617-5+-21-15=-2417-5--21+15=4817-5--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 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

24 717 5 -21 154 517 5 -21 15

Sample Output

DivisibleNot divisible

dp[i][j]中的i表示加上或者减去第i个数时，余数为j是否为true。最后得到dp[n-1][0]是否为真。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;int dp[11000][110];int main(){//freopen("test.in","rt",stdin);    int t,n,k,a,i,j;    cin >>t;    while (t--){        cin >>n >>k >>a;        memset(dp, 0, sizeof(dp));        dp[0][(a % k + k) % k] = 1;        for (i = 1; i < n; ++i){            cin >>a;        a = abs(a) % k;        for (j = 0; j < k; ++j)            if (dp[i - 1][j])                dp[i][(j + a) % k] = dp[i][(j - a + k) % k] = 1;        }        cout <<(dp[n - 1][0] ? "Divisible" : "Not divisible") <<endl;    }    return 0;}