百練1745:Divisibility題解

1745:Divisibility

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總時間限制: 

1000ms 

內存限制: 

65536kB

描述

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. 

輸入

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. 

輸出

Write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.

樣例輸入

4 7
17 5 -21 15

樣例輸出

Divisible

來源

Northeastern Europe 1999

代碼：

#include<cstdio>
#include<cstring>
int a[100],b[100];
int main(){
	int n,k,tmp;
	scanf("%d%d",&n,&k);
	scanf("%d",&tmp);
	a[(tmp%k+k)%k] = 1;
	for(int i = 2; i <= n; ++i){
		memset(b,0,sizeof(b));
		scanf("%d",&tmp);
		for(int j = 0; j < k; j++){
			if(a[j]){
				b[((j+tmp)%k+k)%k] = 1;
				b[((j-tmp)%k+k)%k] = 1;
			}
		}
		memcpy(a,b,sizeof(int)*k);
	} 
	if(a[0]) puts("Divisible");
	else puts("Not divisible");
	return 0;
}