Cod sursa(job #1466400)

Utilizator om6gaLungu Adrian om6ga Data 29 iulie 2015 01:58:26
Problema Divizori Primi Scor 100
Compilator c Status done
Runda Arhiva de probleme Marime 1.74 kb
#include <stdio.h>
#include <stdlib.h>
  
#define NMAX 1000000
#define KMAX 7
  
int T, N, K, i, j, a, b, ind, mid;
int prim[NMAX+5], k_prim[KMAX+1][NMAX+5], cnt[KMAX+1];
  
void ciur()
{
    for (i = 2; i <= NMAX; i++)
        if (!prim[i])
        {
            prim[i]++;
            for (j = i<<1; j <= NMAX; j += i)
                prim[j]++;
        }
  
    for (i = 2; i <= NMAX; i++)
            k_prim[prim[i]][++cnt[prim[i]]] = i;
    k_prim[0][++cnt[prim[0]]] = 1;
}
  
  
int main()
{
    FILE *in, *out;
    in =  freopen("divprim.in", "r", stdin);
    out = freopen("divprim.out", "w", stdout);
    scanf("%d\n", &T);
      
    ciur();
    while(T)
    {
        scanf("%d %d\n", &N, &K);
        if (K == 0)
            printf("%d\n", 1);
        else
        {
            a = 1;
            b = cnt[K];
            ind = 0;
            while (b - a >= 20)
            {
                mid = (a + b)>>1;
                if (k_prim[K][mid] < N)
                    a = mid + 1;
                else if (k_prim[K][mid] > N)
                    b = mid - 1;
                else
                {
                    ind = mid;
                    break;
                }
            }
            if (!ind)
            {
                for (i = a; i <= b && k_prim[K][i] < N; i++);
                if (k_prim[K][i] == N)
                    ind = i;
                else if (k_prim[K][i] > N)
                    ind = i > 1 ? i-1 : 0;
                else
                {
                    ind = i > 2 ? i-1 : 0;
                }
            }
             
            printf("%d\n", k_prim[K][ind]);
        }
        T--;
    }

    fclose(in);
    fclose(out);
    return 0; 
}