count of divisor of a factorial (also ncr)

//first we have to find the prime factorization count of that number
//we have to found out count after ncr calculation
//then add 1 with the count
//calculate if need more
#include<stdio.h>
#define ll long long
int prime[90], range=0;
int vis[500];
void inti(){
	int i,j;
	for(i=2; i*i<=431; i++){
		if(vis[i]==0){
			for(j=i*i;j<=431;j+=i){
				vis[j]=1;
			}
		}
	}
	for(i=2;i<=431;i++){
		if(vis[i]==0){
			//printf("%d\n", i);
			prime[range++]=i;
		}
	}
}

int cal(int n, int p){
	if(n<p) return 0;
	else return(n/p + cal(n/p, p));
}

int main(){
	//inti();
	int n,k,i;
	__int64 result =1,pre;
	inti();
	//printf("%d\n", range);
	while(scanf("%d%d",&n,&k)==2){
		result=1;
		if(k==0||k==n){
			printf("1\n");
			continue;
		}
		for(i=0;i<range&&prime[i]<=n;i++){
			pre=cal(n,prime[i]) - cal(n-k,prime[i]) - cal(k,prime[i]);
			result = result * (pre+1);
		}
		printf("%I64d\n", result);
	}
	return 0;
}