POJ 2992 Divisors 求组合数的约数个数

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Divisors

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9856 Accepted: 2896

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 16 310 4

Sample Output

Source

CTU Open 2005

求C(n,k)的约数的个数。

n和k都很大，直接求肯定不行。假设将一个数表示成它的质因数分解，如A=a^p1*b^p2*c^p3*...*n^pn.

那么它的约数个数就是：ans=(p1+1)*(p2+1)*(p3+1)*...*(pn+1).而C(n,k)=n!/[(k!*(n-k)!]，c[n][k]代表n的阶乘时能够分解出几个k。那么只需要求出他们的阶乘对于每一个素数的个数就可以了。公式：ai=c[n][prime[i]]-c[k][prime[i]]-c[(n-k)][prime[i]]。ans=a1*a2.*...*ak (k代表当prime[k]小于n的时候)。

//3624K610MS#include<stdio.h>#include<string.h>#define N 100007bool visit[1010];long long c[1500][1500];long long  prime[107];void init_prim()//prime存的是下标，visit存的是数。visit[5]==true。{    long long num=0;    memset(visit,true,sizeof(visit));    for(long long i=2;i<1007;++i)    {        if(visit[i]==true)        {            num++;            prime[num]=i;        }        for(long long j=1;((j<=num)&&(i*prime[j]<=10007));++j)        {            visit[i*prime[j]]=false;            if(i%prime[j]==0) break;        }    }}void init(){    for(long long i=2;i<=437;i++)        for(long long j=1;prime[j]<=i;j++)        {            long long n=i,res=0;            while(n){n/=prime[j];res+=n;}            c[i][prime[j]]=res;        }}int main(){    long long n,k;    init_prim();    init();    while(scanf("%I64d%I64d",&n,&k)!=EOF)    {        long long ans=1,a;        for(long long i=1;prime[i]<=n;i++)        {            a=c[n][prime[i]]-c[k][prime[i]]-c[n-k][prime[i]];            ans*=(1+a);        }        printf("%I64d\n",ans);    }    return 0;}

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