Divisors_ number of combination factors

Description

Your task in this problem are to determine the number of divisors of Cnk. Just for fun--or does need any special reason for such a useful computation?

Input

The input consists of several instances. Each instance consists of a single line containing the 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 16 310 4

Sample Output

2616

"Test instructions" to find the number of the mass factor of C (n,m).

"Theorem" sets all elements of positive integer n decomposition N=p1^a1*p2^a2*p3^a3****ps^as, then t (n) = (a1+1) * (a2+1) * (a3+1) * * * (AN+1); (Formula for number of factors)

1. Find the prime number of n

2.ei=[n/pi^1]+ [n/pi^2]+ ... + [n/pi^n] where [] is rounded. that can be int ei=0;while (N) ei+= (N/=PI);

3. Set formula calculated, m= (e1+1) * (e2+1) *......* (en+1)

#include <iostream>#include<stdio.h>#include<string.h>using namespacestd;Const intn= the;intprime[ +]={2,3,5};intk=3;Long LongN,m,cnt[n][n];voidget_prime ()//Converts the prime number within 1000 to the prime array;intFlag; intp=2;  for(intI=7; i<= +; i+=p) {Flag=0; P=6-p;//has cleverly skipped multiples of 3, improving efficiency for(intj=0;p rime[j]*prime[j]<=i;j++)        {            if(i%prime[j]==0) {flag=1;  Break; }        }        if(!flag) prime[k++]=i; }}voidinit () {memset (CNT,0,sizeof(CNT));    Get_prime (); Long LongTmp,ret;  for(intI=2; i<=431; i++)    {         for(intj=0;p rime[j]<=i;j++) {tmp=i; RET=0;  while(TMP) {tmp=tmp/Prime[j]; RET+=tmp; } Cnt[i][prime[j]]=ret;//i number of quality factors} }}intMain () {init (); Long LongRet,ans;  while(~SCANF ("%lld%lld",&n,&m)) {ans=1;  for(intI=0;p rime[i]<=n;i++) {ret=cnt[n][prime[i]]-cnt[m][prime[i]]-cnt[n-m] [Prime[i]];//c (N,m) =n!/((n-m)!m!), subtracting the number of corresponding factors, we get the result of C (n,m) decomposition ans*= (ret+1); } printf ("%lld\n", ans); }    return 0;}

Divisors_ number of combination factors