Cod sursa(job #650616)

Utilizator FIIBCPBuzila Octavian Calin FIIBCP Data 18 decembrie 2011 15:49:12
Problema Suma si numarul divizorilor Scor 0
Compilator c Status done
Runda Arhiva educationala Marime 1.35 kb
#include<stdio.h>
#include<math.h>
#include<bitset>
 
const int MAX_N = 1000005;
const int MOD = 9973;

FILE *in,*out;

long long N;
int T, K, P[MAX_N];
viz[MAX_N]={0};
 
void ciur() {
    for(int i = 2; i < MAX_N; ++i) {
        if(viz[i] == 0) {
            P[++K] = i;
 
            for(int j = i+i; j < MAX_N; j += i) {
                viz[j] = 1;
            }
       }
    }
}
 
inline int putere(int x, int p) {
    int rez = 1;
    x %= MOD;
    rez=1;
    int q;
    for (q = 0; (1<<q) <= p; ++ q)
    {
        if( ((1<<q) & p) )
            rez=(rez*x)%MOD;
        x=(x*x)%MOD;
    }
    return rez;
}
 
void solve() {
    fscanf(in,"%lld",&N);
 
    int nd = 1, sd = 1;
 
    for(int i = 1; i <= K && 1LL * P[i] * P[i] <= N; ++i) {
        if(N % P[i]) continue;
        int p = 0;
 
        while(N % P[i] == 0) {
            N /= P[i];
            ++p;
       }
 
        nd *= (p+1);

        int p1 = (putere(P[i], p+1) - 1) % MOD;
       int p2 = putere(P[i]-1, MOD-2) % MOD;
 
        sd = (((sd * p1) % MOD) * p2) % MOD;
    }
 
    if(N > 1) {
       nd *= 2;
        sd = (1LL*sd*(N+1) % MOD);
    }
 
    fprintf(out,"%d %d\n",nd,sd);
}
 
int main() {
    ciur();
    in=fopen("ssnd.in","rt");
   out=fopen("ssnd.out","wt");
    fscanf(in,"%d",&T);
    for(; T; --T)
        solve();
}