Cod sursa(job #799424)

Utilizator icb_mnStf Cic icb_mn Data 18 octombrie 2012 22:34:51
Problema Suma si numarul divizorilor Scor 30
Compilator cpp Status done
Runda Arhiva educationala Marime 1.65 kb
#include<fstream>
#define LL long long
#define M 9973
#define NMAX 1100000 // un numar are divizori pana la jumatatea lui => n = 10^12 =>d <= 10^6
using namespace std;

LL n, d[50], p[50], nr, sum;
int t, k,semn[NMAX + 8], ciur[NMAX + 8],lng = 0;

inline void eratosthenes()
{
    for(int i = 2; i <= NMAX; i+=2) semn[i] = 1;
    ciur[0] = 2;
    lng = 1;
    for(int i = 3; i <= NMAX; i += 2)
    {
        if(!semn[i])
        {
            ciur[lng] = i;
            lng++;
            for(int j = i + i; j + i <= NMAX; j += i)semn[j] = 1;
        }
    }
}

inline void descompune(LL n, LL &nr, LL &sum)
{
    int putere = 1;
    nr = sum = 1;
    LL numar;
    int i = 0;
    int d = 0;

    if(semn[n])
    {
        while(n > 1)
        {
        if(!(n % ciur[i]))
        {
            d = 0;
            LL s = 1;
            while(!(n % ciur[i]))
            {
                d++;
                n /= ciur[i];
            }
            putere = 1;
            for(int j = 1; j <= d; ++j)
            {
                putere = putere * ciur[i];
                s += putere;
            }
            sum = (sum * s) % M;
            nr = (nr * (1+ d)) % M;
        }
        i++;
        }
    }
    else
    {
        nr = 2;
        sum = (1 + n) % M;
    }
}
int main()
{
    ifstream f("ssnd.in");
    ofstream g("ssnd.out");

    f>>t;
    eratosthenes();

    for(int r = 1; r <= t; ++r)
    {
        f>>n;
        LL nr;
        LL sum;

        nr = 1;
        sum = 1;

        descompune(n, nr, sum);

        g<<nr<<' '<<sum <<'\n';
    }

    g.close();

    return 0;
}