POJ 1745 Divisibility --dp

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 Northeastern Europe 1999

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All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di

给出数据任意加减问是否能整除K

dp[i][j]代表选i个数，余数是否为j，为j则值为1，否则为0。

如果dpi-1j为1，那么

dp[i][(j+s[i])%k]=1; dp[i][((j-s[i])%k+k)%k]=1;肯定也是1！

#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
#define maxn 10005
int s[maxn];
int dp[maxn][106];
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        for(int i=0;i<n;i++)
        {
            cin>>s[i];
            s[i]=(int)abs(s[i])%k;
        }
        memset(dp,0,sizeof(dp));
        dp[0][s[0]]=1;//选0个数余数为s[0]
        for(int i=1;i<n;i++)
        {
            for(int j=k-1;j>=0;j--)
            {
                if(dp[i-1][j])
                {
                    dp[i][(j+s[i])%k]=1;
                    dp[i][((j-s[i])%k+k)%k]=1;
                }
            }
        }
        if(dp[n-1][0])
        {
            cout<<"Divisible"<<endl;
        }else
        {
            cout<<"Not divisible"<<endl;
        }
    }
}