Divisors pku 2992

Divisors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6789		Accepted: 1783

Description
Your task in this problem is to determine the number of divisors of Cnk. Just for fun -- or do you need any special reason for such a useful computation?
Input
The input consists of several instances. Each instance consists of a single line containing two integers n and k (0 ≤ k ≤ n ≤ 431), separated by a single space.
Output
For each instance, output a line containing exactly one integer -- the number of distinct divisors of Cnk. For the input instances, this number does not exceed 263 - 1.
Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

Source
CTU Open 2005

#include<stdio.h>
int p[83]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431};
int d[432][83]={0},m=0;
void divisors(int n)
{
    int i=n,j;
    for (j=0;n>1;j++)
    {
          while (n%p[j]==0)
          {
                d[i][j]++;
                n/=p[j];
          }
    }
    if (j>m)
          m=j;
}
int main()
{
    int i,k,l,n;
    __int64 j;
    for (i=2;i<432;i++)
    {
          for (j=0,k=i-1;j<m;j++)
                d[i][j]=d[k][j];
          divisors(i);
    }
    while (scanf("%d%d",&n,&k)!=EOF)
    {
          l=n-k;
          for (i=0,j=1;i<83;i++)
                j*=(d
[i]-d[k][i]-d[l][i]+1);
          printf("%I64d/n",j);
    }
    return 0;
}