//给你n个数，两个数之间可以+。-两个运算 得出的所有结果只要有能被k正除，输出Divisible，否则Not divisible
# include <iostream>
# include <stdio.h>
# include <algorithm>
# include <string.h>
using namespace std;
int dp[10010][110];//第一维代表几个数相加，第二位为几个数相加%k后的余数
int b[10010];
int main()
{
    int n,k,i,j,t;
    while(~scanf("%d %d",&n,&k))
    {
        for(i=1; i<=n; i++)
            scanf("%d",&b[i]);
        for(i=0; i<=n; i++)
            for(j=0; j<=k-1; j++)
                dp[i][j]=0;
        dp[0][0]=1;
        for(i=1; i<=n; i++)
        {
            for(j=0; j<=k-1; j++)
            {
                if(dp[i-1][j]==1)
                {
                    t=((j+b[i])%k+k)%k;//可能有负数，注意取余
                    dp[i][t]=1;
                    t=((j-b[i])%k+k)%k;
                    dp[i][t]=1;
                }
            }
        }
        if(dp[n][0]==1)
            printf("Divisible\n");
        else
            printf("Not divisible\n");
    }
    return 0;
}

poj 1745 Divisibility (dp)