poj 1745（Divisibility）

 dp  Divisibility

                                                                         Divisibility

Time Limit: 1000MS
Memory Limit: 10000KTotal Submissions: 11474
Accepted: 4104

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

4 717 5 -21 15

Sample Output

Divisible

题意：给出几个数字，每两个数字之间的可以任意添加+或-，问该序列的和能否整除k。

题解：思路很神奇，dp[i][j] i代表前i个数，j代表前i个数取余k。注意取余成正数。存在为1，否则为0.

eg：

17%7=3，（3+5）%7=1，（1-21）%7=6，（6+15）%7=0.

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int n,k;int dp[10010][101];int a[10010];int z(int n,int x){//正数取余n%=x;while (n<0){n+=x;}return n;}int main(){int i,j;scanf ("%d %d",&n,&k);memset(dp,0,sizeof(dp));memset(a,0,sizeof(a));for (i=1;i<=n;i++){scanf ("%d",&a[i]);}dp[1][z(a[1],k)]=1;for (i=2;i<=n;i++){for (j=0;j<k;j++){if (dp[i-1][j]){dp[i][z(j+a[i],k)]=1;dp[i][z(j-a[i],k)]=1;}}}if (dp[n][0])printf ("Divisible\n");elseprintf ("Not divisible\n");return 0;}