POJ 1745：Divisibility 枚举某一状态的DP

Divisibility

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11001		Accepted: 3933

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

题意就是给了N个数，在N-1个位置变换+ -号，问得到的结果中有没有能够整除K的，如果有，输出Divisible。没有，输出Not Divisible。

DP真是一片很深的海。

越做DP越觉得DP的花样很多，这个是我做了POJ1837觉得DP是可以做这道题的。觉得DFS也应该可以，没试。。。

POJ1837和这道题都是固定枚举其中的某个状态或者变量，这里的可以枚举的状态就是余数，给了K，所以我只需对0到K-1这些余数做枚举，然后从i的余数状态推i+1的余数状态。

就是这样：

dp[i][(j+value[i])%mod] +=dp[i-1][j];

dp[i][(j-value[i]+mod)%mod] +=dp[i-1][j];

然后这样做可能是因为数目比较大了溢出还是怎样WA了一次，于是我控制了一下数值。这样：

dp[i][(j+value[i])%mod] +=dp[i-1][j];

dp[i][(j-value[i]+mod)%mod] +=dp[i-1][j];





if(dp[i][(j+value[i])%mod]>10)

dp[i][(j+value[i])%mod]=10;

if(dp[i][(j-value[i]+mod)%mod]>10)

dp[i][(j+value[i])%mod]=10;

。。。很幼稚的方法，但还是涨姿势长见识了。。。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;

int num,mod;
int dp[10005][102];
int value[10005];

int main()
{
	int temp,i,j;
	cin>>num>>mod;

	cin>>value[1];
	value[1]=abs(value[1])%mod;
	for(i=2;i<=num;i++)
	{
		cin>>temp;
		value[i]=abs(temp);
		value[i]=value[i]%mod;
	}
	memset(dp,0,sizeof(dp));
	dp[1][value[1]]=1;

	for(i=2;i<=num;i++)
	{
		for(j=0;j<mod;j++)
		{
			dp[i][(j+value[i])%mod] +=dp[i-1][j];
			dp[i][(j-value[i]+mod)%mod] +=dp[i-1][j];

			if(dp[i][(j+value[i])%mod]>10)
				dp[i][(j+value[i])%mod]=10;
			if(dp[i][(j-value[i]+mod)%mod]>10)
				dp[i][(j+value[i])%mod]=10;
		}
	}
	if(dp[num][0])
	{
		cout<<"Divisible"<<endl;
	}
	else
	{
		cout<<"Not divisible"<<endl;
	}
	return 0;
}

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