POJ---1745 Divisibility[동적 계획]

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
SOURCE:클릭하여 링크 열기
제목: 모든 수를 가감 혼합 연산을 하도록 몇 개의 수를 제시한다.결과가 k로 정리되면 Divisible을 출력하고, 그렇지 않으면 Not divisible을 출력합니다
코드:

#include 
#include 
#include 
#define MAXN 10005
using namespace std;

int dp[MAXN][105],a[MAXN];//dp[i][j]   i           %k   j，    1，   0
int n,k;

int abs(int x)
{
    return x>0?x:-x;
}

int main(void)
{
    memset(dp,0,sizeof(dp));
    scanf("%d%d",&n,&k);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&a[i]);
        a[i]=abs(a[i])%k;//      ，      ，        
    }
    dp[1][a[1]%k]=1;
    for(int i=2; i<=n; i++)
        for(int j=0; j