1129: Divisibility

整除问题求解

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本文介绍了一种通过动态规划解决特定整除问题的方法。对于给定的一系列整数，通过放置加减运算符来判断是否能构成一个结果被指定整数K整除的有效表达式。文章提供了完整的算法实现细节及样例输入输出。

1129: Divisibility

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	1155	290	Standard

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

There are multiple test cases, the first line is the number of test cases.
The first line of each test case 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

2
4 7
17 5 -21 15
4 5
17 5 -21 15

Sample Output

Divisible
Not divisible

This problem is used for contest: 148

#include<iostream>
#include<string.h>
using namespace std;
int dp[10002][102];//dp[i][j] 表示前i个数的结果为j 1表示存在，0表示不存在
//dp[i][j]=dp[i-1][j+ -a[j]]
int a[10002];
int main()
{
    int nn;
    cin>>nn;
    while(nn--)
    {
        int n,k,i,j;
        cin>>n>>k;
        for(i=1;i<=n;i++)
        {
            cin>>a[i];
            while(a[i]<0) a[i]+=k;
            a[i]%=k;
        }
        memset(dp,0,sizeof(dp));
        dp[1][a[1]]=1;
        for(i=2;i<=n;i++)
        {
            for(j=0;j<k;j++)
            {
                if(dp[i-1][j]==1)
                {
                    dp[i][(j+a[i])%k]=1;
                    int temp=j-a[i];
                    while(temp<0) temp+=k;
                    temp%=k;
                    dp[i][temp]=1;
                }
            }
        }
        if(dp[n][0]==1)
                cout<<"Divisible"<<endl;
        else
                cout<<"Not divisible"<<endl;
    }
    return 0;
}