openjudge 1745 Divisibility（线性dp）

1745:Divisibility

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

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. 

输入

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. 

输出

Write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.

样例输入

4 7
17 5 -21 15

样例输出

Divisible

tips:N达到了1000级别；暴力枚举=暴力悲剧；

dp[i][j]代表前i个数字经过运算  取余k 的结果 能否为 j；

因为取余可能为负，我们一开始做了绝对值处理；取余过程中出现的负数我们都转化为正数进行处理；

动态转移方程：dp[i][(j+a[i])%k]=true,dp[i][((j-a[i])%k+k)%k]=true; dp[i-1][j]=true;

#include
#include
#include 
using namespace std;

int n,k;
int a[11112];
bool dp[11000][111];
//dp[i][j]表示前i个数能否对k取余得到j 
int main()
{
	memset(dp,false,sizeof(false)); 
	cin>>n>>k;
	for(int i=1;i<=n;i++)cin>>a[i],a[i]=abs(a[i])%k;
	
	dp[1][a[1]]=true; 
	for(int i=2;i<=n;i++)
	{
		for(int j=0;j