POJ 1745 Divisibility (线性dp)

Divisibility

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10598		Accepted: 3787

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

Source

field=source&key=Northeastern+Europe+1999">Northeastern Europe 1999



题目链接：

id=1745">http://poj.org/problem?id=1745



题目大意：给n个数，让他们通过加减运算。推断结果可不可能被k整除



题目分析：用dp[i][j]表示通过前i个数的运算得到的余数为j可不可能。先看求a % k。假设a > k，则a ＝ n * k + b。(n * k + b) % k == 0 + b % k = a % k，所以当a > k时，对求余数有影响的部分是不能被整除的部分。因此对于每一个数我们能够做a[i] = a[i] > 0 ? (a[i] % k) : -(a[i] % k)的预处理，然后就是在dp[i - 1][j]的情况下。推出下一状态。下一状态有两种可能，加和减，减的时候防止出现负数加上个k再取余，初始化dp[0][a[0]]
= true最后仅仅要推断dp[n - 1][0]及前n个数通过加减运算是否能得到被k整除的值

#include <cstdio>
#include <cstring>
int const MAX = 10005;

bool dp[MAX][105];
int a[MAX];

int main()
{
    int n, k;
    while(scanf("%d %d", &n, &k) != EOF)
    {
        memset(dp, false, sizeof(dp));
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            a[i] = a[i] > 0 ? (a[i] % k) : -(a[i] % k);
        }
        dp[0][a[0]] = true;
        for(int i = 1; i < n; i++)
        {
            for(int j = 0; j <= k; j++)
            {
                if(dp[i - 1][j])
                {
                    dp[i][(j + a[i]) % k] = true;
                    dp[i][(k + j - a[i]) % k] = true;
                }
            }
        }
        printf("%s\n", dp[n - 1][0] ? "Divisible" : "Not divisible");
    }
}

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