Language:Default

Divisibility

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8396		Accepted: 2909

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

4 717 5 -21 15

Sample Output

Divisible

题意：给出n个数，在n个数中添加+，-号使得能整除k.

思路：由于状态与前一个数有关，且和第i个数+， 或-有关。dp[i][j]表示第i个数余数为j时是否为真，dp[i][j-a[i]%m]=dp[i-1][j], dp[i][j+a[i]%m]=dp[i-1][j];判断dp[m][0]是否为真即可，初始条件是dp[1][a[1]]=1;但是注意一点下标不能为负数为此+k*n是等价的。

代码如下：

#include
        
         #include
         
          #include
          
           using namespace std;int dp[10005][105], a[10005];int main(){	int i, j, k, n;	while(scanf("%d%d", &n, &k)!=EOF)	{		memset(dp, 0, sizeof(dp));		memset(a, 0, sizeof(a));		for(i=1; i<=n; i++)		{			scanf("%d", &a[i]);			a[i]%=k;		}		while(a[1]<0)			a[1]+=k; 		dp[1][a[1]%k]=1; 		for(i=2; i<=n; i++)			 for(j=0; j<=k; j++)			 	if(dp[i-1][j])				{					int aa=j-a[i], bb=j+a[i]; 					while(aa<0)						aa+=k;					while(bb<0)						bb+=k; 					dp[i][aa%k]=1, dp[i][bb%k]=1;				} 		if(dp[n][0]==1)			printf("Divisible\n");		else			printf("Not divisible\n");	}}