POJ1745Divisibility(01背包思想)

Divisibility

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11151		Accepted: 3993

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：输入n个数，通过添加+和-能否是的结果对k取余为0
思路:智商再次背碾压

首先一个数，不用说，第一个数之前不用加符号就是本身，那么本身直接对K取余，
那么取17的时候有个余数为2
然后来了一个5，
(2 + 5)对7取余为0
(2 - 5)对7取余为4（将取余的负数变正）
那么前2个数有余数0和4
再来一个-21
（0+21）对7取余为0
（0-21）对7取余为0
（4+21）对7取余为4
（4-21）对7取余为4
再来一个-15同样是这样
（0+15）%7 = 1
（0-15）%7 = 6
（4+15）%7 = 5
（4-15）%7 = 3
同理可以找到规律，定义dp[i][j]为前i个数进来余数等于j是不是成立，1为成立，0为不成立
那么如果dp[N][0]为1那么即可以组成一个数对K取余为0
初始化dp为0

然后dp[1][a[1]%k] = 1
for i = 2 to N do
for j = 0 to K do
 if(dp[i - 1][j])
  dp[i][(j + a[i])%k] = 1;
  dp[i][(j - a[i])%k] = 1;
 if end
for end
for end

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[10000 + 10][100 + 10],a[10000 + 10];
int n,k;
int mod(int x)
{
    if(x < 0)
    {
        return x + k;
    }
    else
        return x;
}
int main()
{
    while(scanf("%d%d", &n, &k) != EOF)
    {
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        memset(dp, 0, sizeof(dp));
        dp[1][ mod(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][ mod((j + a[i]) % k)] = 1;
                    dp[i][ mod((j - a[i]) % k)] = 1;
                }
            }
        }
        if(dp[n][0])
            printf("Divisible
");
        else
            printf("Not divisible
");
    }


    return 0;
}