POJ1745Divisibility

易水人去，明月如霜。

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

題意：給你N個數，給他們之間加入+ -問是否有一種方式使得N個數用完後得值爲K的倍數

思路：並不需要記錄前I個數的值只要記錄與K得模，再DP

代碼：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int read()
{
    char ch;int s=0,f=1;
    ch=getchar();
    while(ch>'9'||ch<'0'){ if(ch=='-') f*=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9'){ s=s*10+ch-48; ch=getchar(); }
    return s*f;
}
int dp[10005][105];
int a[10005];
int getmod(int d,int mo)
{
    int su=d%mo;
    if(su<0) su+=mo;
    return su;
}
int main()
{
int n,k;
n=read();
k=read();
for(int i=1;i<=n;i++) a[i]=read();
 dp[1][getmod(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][getmod(j+a[i],k)]=1;
             dp[i][getmod(j-a[i],k)]=1;
         }
     }
 }
 if(dp[n][0])
    printf("Divisible");
 else printf("Not divisible");
 return 0;
}