ZOJ2042 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

题意：

给定n个数，中间可以添加“+”或“-”，判断这n个数的组合是否能被k整除。

思路：

可得所有的结果数2^n个。

dp[i][j]为前i个数模k的余数为j。

因为一个数模k后只可能是0至k-1，所以写一个for循环，如果前i-1个数的余数j存在的话，再加上这次的num[i]，当然如果是减的话需要加k避免出现负数，所以最后只需要判断前n个数余数为0的状态是否存在，若存在则输出Divisible。

这里用到这个数学公式（n+m）%k=(n%k+m%k)%k。

代码如下：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[10005][100],num[10005];
int main()
{
    memset(dp,0,sizeof(dp));
    int n,k,x;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            if(x>0)
            num[i]=x%k;
            else 
            num[i]=((-x)%k);
        }
            dp[0][num[0]]=1;
            for(int i=1;i<n;i++)
            {
                for(int j=0;j<k;j++)
                {
                    if(dp[i-1][j])
                    {
                        dp[i][(j+num[i])%k]=1;
                        dp[i][(j-num[i]+k)%k]=1;
                    }
                }
            }
            if(dp[n-1][0])printf("Divisible\n");
            else printf("Not divisible\n");
    return 0;
}