Divisibility，再练暴搜

   秉着多练练的心态，我继续ppt上面的练习题，

Divisibility

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10537		Accepted: 3755

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

Northeastern Europe 1999

 

我是用dfs做的，结果是超时的，不过还是能得出结果的

 

#include<stdio h=""> int n,k,a[10004],num; void dfs(int i,int sum) { if(i==n) { if(sum%k==0) { num=1; return ; } } else { // printf("%d\n",sum); dfs(i+1,sum+a[i]); dfs(i+1,sum-a[i]); } } int main() { // int T; // scanf("%d",&T); while(scanf("%d%d",&n,&k)!=EOF) { for(int i=0;i<n;i++) scanf("%d",&a[i]); num=0; dfs(0,0); if(num) printf("Divisible\n"); else printf("Not divisible\n"); } return 0; } </stdio>

至于如何才能不超时呢，该问题过些天再解决，嘿嘿，现在我在意暴搜