UVA 10036 - Divisibility

Problem PDF

Solution:

/*
    Author   : Shipu Ahamed (Psycho Timekiller)
    Facebook : http://facebook.com/DeesheharaShipu
*/

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

#define pi 2*acos(0.0)
#define all(v) v.begin(),v.end()

//input
#define si(t) scanf("%d",&t)
#define sl(t) scanf("%lld",&t)
#define sii(a,b) scanf("%d%d",&a,&b)
#define sll(a,b) scanf("%lld%lld",&a,&b)

//Output
#define P(a) printf("%d\n",a)
#define PL(a) printf("%lld\n",a)
#define PN(a) printf("%d ",a)
#define PLN(a) printf("%lld ",a)
#define PP(a,b) printf("%d %d\n",a,b)
#define PPN(a,b) printf("%d %d ",a,b)
#define PPL(a,b) printf("%lld %lld\n",a,b)
#define PPLN(a,b) printf("%lld %lld ",a,b)

#define CP(a) cout<string toString( T Number ){stringstream st;st << Number;return st.str();}
int SOD(int n) {int sum=0;for(int i=1;i*i<=n;i++)sum+=(n%i)?0:((i*i==n)?i:i+n/i);return sum;}

//File input/output
#define input freopen("in.txt","r",stdin)
#define output freopen("out.txt","w",stdout)

//upper bound and lower bound
#define LB(a,value) (lower_bound(all(a),value)-a.begin())
#define UB(a,value) (upper_bound(all(a),value)-a.begin())

//Test Case & New line
#define Case(no) printf("Case %d: ",++no)
#define nl puts("")

int stringconvert(string s){int p; istringstream st(s); st>>p ; return p;}

ll pow(ll a,ll b, ll m) { ll res = 1; while(b) { if(b & 1) { res = ( (res % m) * (a % m) ) %m ; } a= ((a%m) * (a%m)) %m; b >>= 1; } return res; }
ll modInverse(ll a, ll m){return pow(a,m-2,m);}

////============ CONSTANT ===============////
#define inf   1<<30                                           //infinity value
#define eps   1e-9
#define mx    10010
#define mod   1000000007
////=====================================////

int a[mx],dp[mx][203],n,k;

bool solve(int i,int sum)
{
    if(i==n)
    {
        if(sum%k==0)
        {
            return 1;
        }
        else
            return 0;
    }

    if(dp[i][sum]!=-1)
        return dp[i][sum];

    return dp[i][sum]=(solve(i+1,(sum+a[i])%k)|solve(i+1,(sum-a[i])%k));
}

int main()
{
    int t;
    si(t);
    while(t--)
    {
        sii(n,k);

        cover(dp,-1);

        for(int i=0;i
https://github.com/Shipu/OnlineJudgeProblemSolutionWithCPlusPlus/tree/master/uva/10036/10036.cpp