poj 1745 Divisibility

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



大意：

　　输入n,k，然后有n个数，可以在n个数之间随意添加加号或减号判断最后的值能不能别k整除。

定义数组dp[i][j]表示第i个数对k取余为j，(a[1]%k+k)%k为a[1]对k取余的值，所以定义dp[i][(a[1]%k+k)%k]=true;令i从2~n，余数j从0~k-1，判断a[i-1][j]求出a[i-1]余数j，求出j+a[i]或j-a[i]对k取余的值，定义a[i][((j（+或-）a[i])%k+k)%k]=true,若dp[n][0]为真，表示a[n]加上前面的余数对k取余为0。

 

#include<cstdio>
#include<string.h>
bool dp[10010][110];
int main()
{
    int n,k,i,j,a[10010];
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        memset(dp,false,sizeof(dp));
        for(i = 1 ; i <= n ; i++)
        {
            scanf("%d",&a[i]);
        }
        dp[1][(a[1]%k+k)%k]=true;            //(a[1]%k+k)%k为a[1]对k取余的值（加k为了避免取余结果为负值） 
        for(i = 2 ; i <= n ; i++)
        {
            for(j = 0 ; j < k ; j++)
            {
                if(dp[i-1][j])                //令a[i-1]对k取余的值加上a[i]再对k取余 
                {
                    dp[i][((j+a[i])%k+k)%k]=true;
                    dp[i][((j-a[i])%k+k)%k]=true;
                }
            }
        }
        if(dp[n][0])                        //为真时说明a[n]加前面的余数对k取余为0 
        {
            printf("Divisible
");
        }
        else
        {
            printf("Not divisible
");
        }        
     } 
    return 0;
}