POJ1745Divisibility（dp）

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

题意：给你n个数，每两个数之间你可以任意加减，求的每个和中是否有和可以对k整除，如果有输出Divisible，否则输出Not divisible

思路：这道题直接一个二维数组算出所有的结果在判断是不现实的。所以我们可以简化为dp[i][j],找前i个数加减后 对k整除=j是对的。层层推直到最后一个，最后判断dp[n][0]是否成立就可以了。

状态转移方程

当dp[i-1][j]非0时

1，dp[i][t] = 1其中t = ((j + a[i]) % K + K) % K. 说明一下——(j + a[i])%K后 加个K最后再取余 是为了避免值是负数。

2，dp[i][t] = 1其中t = ((j - a[i]) % K + K) % K.

代码：

#include<stdio.h>#include<iostream>using namespace std;bool dp[10001][101];long long  num[10001];int main(){int n,k;scanf("%d%d",&n,&k);for(int i=1;i<=n;i++){int a;scanf("%d",&a);num[i]=a%k;}dp[0][0]=true; for(int i=1;i<=n;i++) {  for(int j=k-1;j>=0;j--) if(dp[i-1][j]) { dp[i][((j+num[i])%k+k)%k]=1; dp[i][((k+j-num[i])%k+k)%k]=1; } } if(dp[n][0]) printf("Divisible\n"); else printf("Not divisible\n");return 0;}