POJ---1745 Divisibility【动态规划】

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:点击打开链接

题意：给出一些数，要求将所有的数都用上进行加减混合运算。如果结果能被k整除就输出Divisible，否则输出Not divisible

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#define MAXN 10005
using namespace std;

int dp[MAXN][105],a[MAXN];//dp[i][j]表示前i个数进行混合运算的结果%k是否为j，若是就为1，否则为0
int n,k;

int abs(int x)
{
    return x>0?x:-x;
}

int main(void)
{
    memset(dp,0,sizeof(dp));
    scanf("%d%d",&n,&k);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&a[i]);
        a[i]=abs(a[i])%k;//为防止数太大，进行取模处理，并且全都变为正数
    }
    dp[1][a[1]%k]=1;
    for(int i=2; i<=n; i++)
        for(int j=0; j<k; j++)
            if(dp[i-1][j])
            {
                dp[i][abs(j+a[i])%k]=1;
                dp[i][abs(j-a[i])%k]=1;
            }
    if(dp[n][0])
        cout<<"Divisible"<<endl;
    else
        cout<<"Not divisible"<<endl;
    return 0;
}