Divisibility-POJ 1765-dp练习赛

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

题目大意：给出n个数字，在其各项中间添加加号或减号判断是否有一种组合使加减后的数可以被k整除。
f[i][j]表示前i个数填入符号计算后的数%K的值是否是j，如果是则为1，否则为0.
最后判断f[N-1][0]是否为1

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int f[10005][200],N,K,a[10005];
int main(){
	scanf("%d%d",&N,&K);
		memset(f,0,sizeof(f));
		for(int i=0;i<N;i++){
			scanf("%d",&a[i]);
			a[i]=abs(a[i])%K;
		}
		f[0][a[0]]=1;
		for(int i=1;i<N;i++){
			for(int j=K-1;j>=0;j--){
				if(f[i-1][j]==1){
					f[i][(j+a[i])%K]=1;
					f[i][((j-a[i])%K+K)%K]=1;
				}
			}
		}
		if(f[N-1][0]==1) printf("Divisible\n");
		else printf("Not divisible\n");
	
	return 0;
}