Divisors

Divisors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11186		Accepted: 3324

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

Source

CTU Open 2005

思路：先将421以内的素数打表，然后我们知道一个数可以分解成素数的乘积，N=p1^k1+p2^k2+...pn^kn;

那么这个数的因子的个数就为（k1+1）（k2+1）(k3+1)....(kn+1)；

我们知道组合数的公式为（n！）/（n-m）！/（m）!；所以我们算出分子分母的各个质因数的个数，然后上下一减就可以得到组合数的质因子。

关键是要先打出各个阶乘的素数表。

还有就是数据可能给重复的，所以询问一个记下到数组，还有组合数有对称性。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
using namespace std;
typedef long long LL;
bool prime[1000]= {0};
int ans[1000];
int ak;
int cnt[500][500]= {0};
vector<int>vec[500];
queue<int>que;
int ask[500];
LL aa[500][500]= {0};
int cm[500][500];
int main(void)
{
        int i,j;
        ak=0;
        for(i=2; i<100; i++)
        {
                if(!prime[i])
                {
                        for(j=i; (i*j)<=431; j++)
                        {
                                prime[i*j]=true;
                        }
                }
        }
        for(i=2; i<=431; i++)
        {
                if(!prime[i])
                {
                        ans[ak++]=i;
                }
        }
        for(i=2; i<=431; i++)
        {
                int ss=i;
                int vv=0;
                int ff=0;
                int cn=0;
                while(ss>1)
                {
                        if(ss%ans[vv]==0&&ff==0)
                        {
                                que.push(ans[vv]);
                                cn++;
                                ff=1;
                                ss/=ans[vv];
                        }
                        else if(ss%ans[vv]==0)
                        {
                                cn++;
                                ss/=ans[vv];
                        }
                        else if(ss%ans[vv])
                        {
                                cnt[i][ans[vv]]=cn;
                                vv++;
                                cn=0;
                                ff=0;
                        }
                }
                cnt[i][ans[vv]]+=cn;
                while(!que.empty())
                {
                        int cc=que.front();
                        que.pop();
                        vec[i].push_back(cc);
                }
        }int ck;
        for(i=0;i<=431;i++)
        {
               for(j=0; j<vec[i].size(); j++)
                                {
                                         ck=vec[i][j];
                                        cm[i][ck]+=cnt[i][ck];
                                }
                                if(i>=1)
                              for(j=0;j<ak;j++)
                                {
                                    cm[i][ans[j]]+=cm[i-1][ans[j]];
                                }
        }
        int n,m;
        while(scanf("%d %d",&n,&m)!=EOF)
        {
                if(aa[n][m]!=0||aa[n][n-m]!=0)
                        printf("%lld
",aa[n][m]);
                else
                {
                        memset(ask,0,sizeof(ask));
                        for(i=0;i<ak;i++)
                        {
                            ask[ans[i]]+=cm[n][ans[i]];
                            ask[ans[i]]-=cm[m][ans[i]];
                            ask[ans[i]]-=cm[n-m][ans[i]];
                        }
                        LL sum=1;
                        for(i=0; i<ak; i++)
                        {
                                sum*=(ask[ans[i]]+1);
                        }
                        printf("%lld
",sum);
                        aa[n][m]=sum;
                        aa[n][n-m]=sum;
                }
        }
        return 0;
}