Nk 1430 Divisors（因子数与质因数）

Time Limit: 5000 ms    Memory Limit: 10000 kB  
Total Submit : 432 (78 users)   Accepted Submit : 108 (57 users)   Page View : 3479  Font Style: Aa Aa Aa
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 4Sample Output
2
6
16

代码如下：12=2^2*3^1 因子个数就等于（2+1）*（1+2）=12

#include<iostream>
#include<cstdio>
using namespace std;
int num;
bool a[450];

struct prime
{
    int num;
    int count;
}p[200];

void init()
{
    int i,j;
    memset(a,true,sizeof(a));
    num=0;
    for(i=2;i<450;i++)
    {
        if(a[i]) p[num++].num=i;
        for(j=0;j<num&&i*p[j].num<450;j++)
        {
            a[p[j].num*i]=0;
            if(i%p[j].num==0)
                break;
        }
    }
}

__int64 Deal(int n,int m)
{
    int i,j;
    int a,b;
    __int64 sum=1;
    if(m*2<n)
        a=n,b=n-m;
    else 
        a=n,b=m;
    for(i=0;i<num;i++)
        p[i].count=0;
    for(i=b+1;i<=a;i++)
    {
        int t=i;
        for(j=0;p[j].num<=i && j<num && t!=1;j++)
        {
            while(t%p[j].num==0)
            {
                t/=p[j].num;
                p[j].count++;
            }
        }
    }
    for(i=2;i<=a-b;i++)
    {
        int t=i;
        for(j=0;p[j].num<=i && j<num && t!=1;j++)
        {
            while(t%p[j].num==0)
            {
                t/=p[j].num;
                p[j].count--;
            }
        }
    }
    for(i=0;i<num;i++)
    {
        if(p[i].count)
            sum*=(p[i].count+1);
    }
    return sum;
}

int main()
{
    int n,m;
    init();
    while(cin>>n>>m)
        printf("%I64d
",Deal(n,m));
    return 0;
}