HDU 1745(dp)

Divisibility Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 11484 Accepted: 4110 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
本当に思い出せませんね~本当にdpをしたことがありませんね~題解:dpの思想は、1つの数がkに対して余剰を取った結果が現れるたびに、結果を保留してkに対して余剰を取って、状態方程式は:dp[i][(j+a[i])%k+k)=trueです;dp[i][((j-a[i]) % k + k) % k] = true ;

#include 
#include 
#include 
using namespace std;
#define LL long long
#define M 10010
#define INF 0x3f3f3f

bool dp[M][110];//  i      k     j     dp[i][j] = true;
int a[M];
int max(int a, int b)
{
    return a > b ? a : b;
}

int main()
{
    int n, k;
    while(scanf("%d%d", &n, &k) != EOF)
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &a[i]);
        }
        memset(dp, false, sizeof(dp));
        dp[1][(a[1] % k + k) % k] = true;
        for(int i=2; i<=n; i++)
        {
            for(int j=0; jif(dp[i-1][j])
                {
                    dp[i][((j+a[i]) % k + k) % k] = true;
                    dp[i][((j-a[i]) % k + k) % k] = true;
                }
            }
        }
        if(dp[n][0])
            printf("Divisible
");
        else
            printf("Not divisible
");
    }

    return 0;
}