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[500]={0},t,pp[100];
void init()
{
    int i,j;
    t=0;
    for(i=2; i<500; i++)
    {
        if(!p[i])
        {
            p[t++]=i;
            j=i*i;
            while(j<500)
            {
                p[j]=1;
                j+=i;
            }
        }
    }
}
void fun(int n,int q)
{
    int i,j,m,sum;
    for(i=0; p[i]<=n&&i<t; i++)
    {
        m=n;
        sum=0;
        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,0,sizeof(pp));
        fun(n,1);
        fun(k,0);
        fun(n-k,0);
        long long sum=1;
        for(i=0; i<t; i++)
        {
            if(pp[i])sum*=pp[i]+1;
        }
        printf("%lld
",sum);
    }
}