UVA 11353 - A Different Kind of Sorting

Problem PDF

Solution:

/******************************************************************
***   Problem       :                                           ***
***   Author        : Shipu Ahamed (Psycho Timekiller)          ***
***   E-mail        : shipuahamed01@gmail.com                   ***
***   University    : BUBT,Dept. of CSE                         ***
***   Team          : BUBT_Psycho                               ***
***   My Blog       : http://shipuahamed.blogspot.com           ***
***   Facebook      : http://www.facebook.com/DeesheharaShipu   ***
******************************************************************/

#include 
#include 
#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 sf(t) scanf("%f",&t)
#define sd(t) scanf("%lf",&t)
#define sc(c) scanf("%c",&c)
#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 PF(a) printf("%f\n",a)
#define PDB(a) printf("%lf\n",a)
#define PN(a) printf("%d ",a)
#define PLN(a) printf("%lld ",a)
#define PFN(a) printf("%f ",a)
#define PDBN(a) printf("%lf ",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< pii;
typedef  pair piii;
typedef  pair pll;
typedef  pair psi;
typedef  vector vi;
typedef  vector vll;
typedef  vector vpii;
typedef  vector vs;
typedef  set si;
typedef  set ss;
typedef  map mii;
typedef  map mll;
typedef  map msi;

template string toString( T Number ){stringstream st;st << Number;return st.str();}
template  T SOD(T n) {__typeof(n) sum=0;for(__typeof(n) i=1;i*i<=n;i++)sum+=(n%i)?0:((i*i==n)?i:i+n/i);return sum;}


//For Define
#define forab(i,a,b) for(__typeof(b) i=(a);i<=(b);i++)
#define for0(i,n) forab(i,0,(n)-1)
#define for1(i,n) forab(i,1,n)
#define rforba(i,b,a) for(__typeof(b) i=(b);i>=(a);i--)
#define rfor0(i,n) rforba(i,(n)-1,0)
#define rfor1(i,n) rforba(i,n,1)
#define forstl(i,s) for(__typeof((s).end()) i=(s).begin(); i != (s).end(); i++)

//Debug
#define dbg(x) cout << #x << " -> " << (x) << endl;
#define dbgsarr(i,a) cout<<#a<<"["< "< "< "<>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 mx7   10000007
#define mx6   2000006
#define mx5   100005
#define inf   1<<30                                           //infinity value
#define eps   1e-9
#define mx    (1500)
#define mod   1000000007
////=====================================////

struct factcount
{
    ll value=0,cnt=0;

}A[mx6];

bool prime[mx6+10];
ll pd[mx6+10];
int k=1;

void seive()
{
    for(int i=4;i<=mx6;i+=2)
        prime[i]=1;

    prime[0]=1;
    prime[1]=1;
    pd[0]=2;

    for(ll i=3;i<=mx6-6;i+=2)
    {
        if(!prime[i])
        {
            pd[k++]=i;
            ll mul = i<<1;
            ll j=i;
            for(j*=i;j<=mx6;j+=mul)
            {
                prime[j]=1;
            }
        }
    }
//    PN(k),PL(pd[k-1]);
}

void fact(int n)
{
    int p=n;
    A[p].value=n;

    int sq = sqrt(n),i=0;
    int c=0;
    while(n!=1)
    {
        if(pd[i]>sq || i==k-1) break;

        while(n%pd[i]==0)
        {
            n/=pd[i];
            c++;
        }
        i++;
    }
    if(n!=1)
    {
        c++;
    }

    A[p].cnt=c;
}

bool cmp(factcount a, factcount b)
{
    if(a.cnt==0 || b.cnt==0) return false;
    if(a.cnt==b.cnt)
        return a.value
https://github.com/Shipu/OnlineJudgeProblemSolutionWithCPlusPlus/tree/master/uva/11353/11353.cpp