Divisors poj2992

Divisors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9940		Accepted: 2924

Description

Your task in this problem is to determine the number of divisors of C_n^k. 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 C_n^k. For the input instances, this number does not exceed 2⁶³ - 1.

Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<time.h>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 int p[]={},t,pp[];
 void init()
 {
     int i,j;
     t=;
     for(i=; i<; i++)
     {
         if(!p[i])
         {
             p[t++]=i;
             j=i*i;
             while(j<)
             {
                 p[j]=;
                 j+=i;
             }
         }
     }
 }
 void fun(int n,int q)
 {
     int i,j,m,sum;
     for(i=; p[i]<=n&&i<t; i++)
     {
         m=n;
         sum=;
         while(m)
         {
             m/=p[i];
             sum+=m;
         }
         if(q)
         pp[i]+=sum;
         else pp[i]-=sum;
     }
 }
 int main()
 {
     init();
     int n,k,i;
     while(~scanf("%d%d",&n,&k))
     {
         memset(pp,,sizeof(pp));
         fun(n,);
         fun(k,);
         fun(n-k,);
         long long sum=;
         for(i=; i<t; i++)
         {
             if(pp[i])sum*=pp[i]+;
         }
         printf("%lld\n",sum);
     }
 }