poj 1745

解题思路：dp表示的为dp[i][j]前i个数是否能够组成余数为j，是则为1 

#include <iostream>using namespace std;#define MAXN 10001int dp[MAXN][101];int posmod(int n,int k){//正数取余n = n % k;while(n < 0) n+=k;return n;}int main(){int n,k;int i ,j ,tmp;int a[MAXN];while(cin>>n>>k){memset(dp,0,sizeof(dp));for(i = 1;i <= n;i++) cin>>a[i];dp[1][posmod(a[1],k)] = 1;for(i = 2;i <= n;i++){for(j = 0;j < k;j++){if(dp[i - 1][j]){dp[i][posmod(j + a[i],k)] = 1;dp[i][posmod(j - a[i],k)] = 1;}}}if(dp[n][0]){cout<<"Divisible"<<endl;}else{cout<<"Not divisible"<<endl;}}return 0;}