ACM: 动态规划题 poj&nb…

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

 

题意: 给出n个数, 每两个数之间可以用+或则-操作, 结果又2^n种, 判断是否存在一种结果

     
可以整除K. 输出 Divisible, 否则输出 Not divisible

 

解题思路:

    
1. 思路很简单, 设状态dp[i][j]: 前i个数进行 + 或则 - 操作之后, 对K进行求余

       
运算是否得到结果j.

    
2. 其实从公式: (a+b)mod K = ( (a mod
K) + (b mod K) ) mod K

                   
(a-b)mod K = ( (a mod
K ) - (b mod K) ) mod K

       
只有上一个数被求余为0才可以判断下一个数进行操作后是否整除.

        if( dp[i-1][j]
== true)

            dp[i][ (j-a[i])mod
K ] = dp[i][ (j+a[i])mod K ] =
true

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 10005

#define MAXSIZE 105

int n, K;

int a[MAX];

bool dp[MAX][MAXSIZE];

inline int mod(int a)

{

 return a > 0 ? a%K : (-a)%K;

}

bool DP()

{

 memset(dp, false, sizeof(dp));

 dp[1][ mod(a[1]) ] = true;

 for(int i = 2; i <= n; ++i)

 {

  for(int j = K; j
>= 0; --j)

  {

   if( dp[i-1][
j ] )

    dp[i][
mod(j-a[i]) ] = dp[i][ mod(j+a[i]) ] = true;

  }

 }

 return dp
[0];

}

int main()

{

// freopen("input.txt", "r", stdin);

 while(scanf("%d %d", &n,
&K) != EOF)

 {

  for(int i = 1; i
<= n; ++i)

   scanf("%d",&a[i]);

  if( DP() )
printf("Divisible\n");

  else printf("Not
divisible\n");

 }

 return 0;

}