**【poj1745】Divisibility

Divisibility

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu

Submit Status

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n个数，和一个数k,按顺序在两个数之间添加加号或减号，使得最后结果mod k=0成立

dp[i][j]:表示前i个数经过加减运算后，对mod k=j成立，只需推出dp[n][0]=true

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[10010],dp[10010][210];//前i个数经过加减运算后，对mod k=j 是成立的int main(){int n,k;while(~scanf("%d%d",&n,&k)){for(int i=1;i<=n;i++)scanf("%d",&a[i]);memset(dp,0,sizeof(dp));dp[0][0]=1;for(int i=1;i<=n;i++){for(int j=0;j<k;j++){if(dp[i-1][j])//前i-1个数加减后 对mod k=j成立 {int t=((j+a[i])%k+k)%k;//j+a[i]和j-a[i]都有可能是负数dp[i][t]=1;t=((j-a[i])%k+k)%k;dp[i][t]=1;}}}if(dp[n][0])printf("Divisible\n");elseprintf("Not divisible\n");}return 0; }