UVA 11137 - Ingenuous Cubrency

Problem PDF

Solution:

/******************************************************************
***   Author        : Shipu Ahamed (Psycho Timekiller)          ***
***   E-mail        : shipuahamed01@gmail.com                   ***
***   University    : BUBT,Dept. of CSE                         ***
***   Team          : BUBT_HIDDEN                               ***
***   My Blog       : http://shipuahamed.blogspot.com           ***
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#include 
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using namespace std;

#define sc scanf
#define pf printf
#define ll long long
#define pi 2*acos(0.0)
#define ull unsigned long long
#define all(v) v.begin(),v.end()

#define sii(t) scanf("%d",&t)
#define sll(t) scanf("%lld",&t)
#define ssii(a,b) scanf("%d%d",&a,&b)
#define ssll(a,b) scanf("%lld%lld",&a,&b)
#define Case(no) printf("Case %d: ",++no)
#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 ff first
#define se second
#define pb push_back
#define ST(v) sort(all(v))
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define maxall(v) *max_element(all(v))
#define minall(v) *min_element(all(v))
#define cover(a,d) memset(a,d,sizeof(a))
#define popcount(i) __builtin_popcount(i)                       //count one
#define input freopen("in.txt","r",stdin)
#define output freopen("out.txt","w",stdout)
#define un(v) ST(v), (v).earse(unique(all(v)),v.end())
#define common(a,b) ST(a), ST(b), a.erase(set_intersection(all(a),all(b),a.begin()),a.end())
#define uncommon(a,b) ST(a), ST(b), a.erase(set_symmetric_difference(all(a),all(b),a.begin()),a.end())

////============ CONSTANT ===============////
#define mx  (2000010)
#define inf (1<<30)                                            //infinity value
#define eps 1e-9
#define mod 1000000007
////====================================////
int main()
{
    ll coin[100],way[20000]={0},i,j,a,p;

    for(i=1,j=0;i<22;i++)
    {
        coin[j++]=i*i*i;
    }
        way[0]=1;
        for(i=0;i<22;i++)
        {
          for(j=coin[i];j<=10010;j++)
          {
                way[j]+=way[j-coin[i]];
          }
        }
        while(sii(a)==1)
        {
           PL(way[a]/2);
        }

    return 0;
}
https://github.com/Shipu/OnlineJudgeProblemSolutionWithCPlusPlus/tree/master/uva/11137/11137.cpp