Cod sursa(job #1379365)

Utilizator corul_barbatescUNIBUC Kira96 lockmihai corul_barbatesc Data 6 martie 2015 17:34:44
Problema Suma si numarul divizorilor Scor 70
Compilator cpp Status done
Runda Arhiva educationala Marime 2.39 kb
/*
    Those without power, seek us!
    Those with power, fear us!
    We are the Order of the Black Knights!
*/

#include<fstream>
#include<cstdio>
#include<map>
#include<set>
#define FIT(a,b) for(vector<int >::iterator a=b.begin();a!=b.end();a++)
#define FITP(a,b) for(vector<pair<int,int> >::iterator a=b.begin();a!=b.end();a++)
#define RIT(a,b) for(vector<int>::reverse_iterator a=b.end();a!=b.begin();++a)
#include<stack>
#define ROF(a,b,c) for(int a=b;a>=c;--a)
#include<vector>
#include<algorithm>
#define FOR(a,b,c) for(int a=b;a<=c;++a)
#define REP(a,b) for(register int a=0;a<b;++a)
#include<cstring>
#include<ctime>
#include<bitset>
#include<cmath>
#include<iomanip>
#include<set>
#define f cin
#define g cout
#include<queue>
#define debug cerr<<"OK";
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ull unsigned int
#define mod 1000000009LL
#define SQR 350
#define db double
#define inf 1<<30
#define div fdasfasd
#define hash dsafdsfds
#define od 100003
#define mod 666013
#define DIM 60010000
#define base 256
#define bas 255
#define mod 9973
#define N 1000100
#define inf 0x3f3f3f3f
#define pdd pair<double,double>
using namespace std;
ifstream f("ssnd.in");
ofstream g("ssnd.out");
/*
    int dx[]={0,0,0,1,-1};
    int dy[]={0,1,-1,0,0};
*/
ll x,e[100];
int E[N/10],viz[N],t,n,p[100];
int main ()
{
    f>>n;
    FOR(i,2,N-5)
    if(!viz[i])
    {
        E[++t]=i;
        for(int j=i+i;j<=N-5;j+=i)
            viz[j]=1;
    }
    FOR(i,1,n)
    {
        f>>x;
        int r=0;
        FOR(j,1,t)
        {
            if(1LL*E[j]*E[j]>x)
                break;
            if(x%E[j]==0)
            {
                ++r;
                p[r]=0;
                e[r]=E[j];
                while(x%E[j]==0)
                    p[r]++,x/=E[j];
                e[r]%=mod;
            }
        }
        if(x!=1)
        {
            e[++r]=x;
            p[r]=1;
        }
        ll sol=1;
        FOR(j,1,r)
            sol=sol*(p[j]+1);
        g<<sol<<" ";
        sol=1;
        FOR(j,1,r)
        {
            ll cs=1;
            FOR(k,1,p[j]+1)
            cs*=e[j];
            cs=(cs-1)/(e[j]-1);
            cs%=mod;
            sol*=cs;
            sol%=mod;
        }
        g<<sol<<"\n";
    }
    return 0;
}