POJ1745

 

                                                                Divisibility

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8346		Accepted: 2891

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n个整数，与一个正整数k，现在需要在这n个整数直接加入n-1个加号或者减号，是否存在使得最终结果是k的倍数。
思路：
n-1个运算符，直接枚举一定超时2^(n-1)其中0<n<10000，所以不可行。
从运算结果对k取余考虑，加上负数，共有2*k种余数，即[-(k-1),(k-1)]，所以开一个2*k的数组存每一个结果即可。
感觉确实是一道不错的DP，比如输出17，那么17%7就是3，由于可以添加正负号，所以可能的结果就是-3和+3，接下来输入5,5%7还是5，那么就是-3+5或者-3-5或者+3+5或者+3-5以此类推，
因为是负数，所以就把数字都加上100即可。

#include<stdio.h>
#include<string.h>
#include <iostream>
using namespace std;
const int MAXN = 201;
///N and K (1 <= N <= 10000, 2 <= K <= 100)
int main()
{
    int i,j,n,k,num;
    bool pre[MAXN],now[MAXN];
    scanf("%d%d",&n,&k);
    memset(pre,0,sizeof(pre));
    pre[100]=1;
    for(i=1;i<=n;i++)
    {
        scanf("%d",&num);
        num%=k;
        for(j=100-k;j<100+k;j++)
        {
            now[j]=pre[j];
            pre[j]=0;
        }
        for(j=100-k;j<100+k;j++)
            if(now[j])
            {
                pre[(j-100+num)%k+100]=1;
                pre[(j-100-num)%k+100]=1;
            }
    }
    if(pre[100])  printf("Divisible\n");
    else   printf("Not divisible\n");
    return 0;
}