POJ 题目2992 Divisors（组合数因子个数）

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Divisors

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11001 Accepted: 3259

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,m)的因子个数

判断N!中有几个质因子：P(N!)=N/i+N/i^2+N/i^3+.....N/i^m(i为一个质因子,m是使N/i^m=0的最小值）

ac代码

#include<stdio.h>#include<string.h>#include<algorithm>#include<stdlib.h>#include<math.h>#define LL long longusing namespace std;LL pri[83]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431};LL c[432][432];LL slo(LL n,LL x){    LL ans=0;    LL sum=x;    while(n>=sum)    {        ans+=n/sum;        sum*=x;    }    return ans;}void fun(){    LL i,j;    for(i=1;i<=431;i++)        for(j=0;j<83;j++)            c[i][j]=slo(i,pri[j]);}int main(){    LL n,m;    fun();    while(scanf("%lld%lld",&n,&m)!=EOF)    {//        LL ans=c[n][m];        LL ans=1;        int i;        for(i=0;i<83&&pri[i]<=n;i++)        {            ans*=(c[n][i]-c[n-m][i]-c[m][i]+1);        }        printf("%lld\n",ans);    }}

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